operational amplifiers mod. mcm7/ev

IPES

© COPYRIGHT BY ELETTRONICA VENETA SPA

OPERATIONAL AMPLIFIERS

mod. MCM7/EV

Volume 1/2

THEORY AND EXPERIMENTS

TEACHER / STUDENT manual

MCM07$$201E0.DOC

CONTENTS

LESSON A01: Basic characteristics of operational amplifiers 1

LESSON A02: The Inverting Amplifier 13

LESSON A03: The Non-inverting Amplifier 19

LESSON A04: The Summing Amplifier 24 LESSON A05: The Subtracting Amplifier 29 LESSON A06: The Integrator 33 LESSON A07: The Differentiator 38 LESSON A08: The Comparator 43 LESSON A09: The Logarithmic Amplifier 47 LESSON A10: The Monostable Multivibrator 52 LESSON A11: The Astable Multivibrator 57 LESSON A12: The Sine Wave Oscillator 62 LESSON A13: The Wave-form Generator 67 LESSON A14: The Voltage-to-Frequency Converter 73 LESSON A15: Low-pass Filter 79 LESSON A16: High-pass Filter 86 LESSON A17: Band-pass Filter 92

CONTENTS

CONTENTS

SAFETY RULES

Keep this handbook at hand for any further help. After the packaging has been removed, set all accessories in order so that they are not lost and check the equipment integrity. In particular, check that it shows no visible damage. Before connecting the equipment to the +/- 12V power supply, be sure that the rating corresponds to the one of the power mains. This equipment must be employed only for the use it has been conceived, i.e. as educational equipment, and must be used under the direct supervision of expert personnel. Any other use is not proper and therefore dangerous. The manufacturer cannot be held responsible for eventual damages due to inappropriate, wrong or unreasonable use.

These apparatuses and all their parts will be disposed of separately from the other wastes. At the end of their lifetime, these apparatuses will be conveyed to the proper centres of separate collection of waste. A proper separate collection ensures that the eliminated apparatuses will undergo the necessary treatment and recycling for preventing any negative effect on the environment and on men’s health.

REQUIRED ELEMENTS (Lesson with the Module alone)

• Experimental Module mod.MCM7/EV • Power Supply mod. PSLC/EV or PS1-PSU/EV • Eventual modules support mod. BOX/EV (optional)

__________ _ __________

REQUIRED ELEMENTS (Lesson with the Module and Software CBT)

• Experimental Module mod.MCM7/EV • Eventual modules support mod. BOX/EV (optional) • Power Supply mod. PSLC/EV or PS1-PSU/EV • Interfaces mod.SIS3-U/EV or mod.SIS3/EV or mod.SIS4-P/EV

with software CBT. • Personal Computer with ports USB and O.S. Windows XP or later

version.

__________ _ __________

LESSONS

In the following lessons, the questions put are of the “MULTIPLE CHOICE” type. For each question are proposed several answers, but only one is the correct.

__________ _ __________

Lesson A01: Basic characteristics of Operational Amplifiers

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OBJECTIVES

To measure the: • amplification of an operational amplifier (or “op amp”) • offset voltage of an op amp • common mode rejection ratio of an op amp • slew rate of an op amp • unity gain bandwidth of an op amp

INSTRUMENTS • digital multimeter • signal generator • dual trace oscilloscope

A01.1 BASIC THEORY

Introduction An operational amplifier, (or “op amp” as commonly known), has two inputs as shown in figure A01.1a.

fig. A01.1

The output voltage Vo is proportional to the difference in voltage between the two inputs, and is given by: Vo = GV · (Vb - Va) where GV is a large number, possibly as high as 1,000,000. Ideally, op amps should have: • infinite input impedance • zero currents flowing into the inputs • zero output impedance. However the characteristics of real op amps differ (usually only slightly) from the ideal. The key characteristics are: 1. The offset voltage - this is the voltage which when applied across

the two inputs produces a zero voltage on the output


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2. The bias and offset currents are currents flowing into the inputs

3. The input impedance (What is usually high, but not infinite), and the ouput impedance (What is low but not zero)

4. The way in which the characteristics change with frequency The operational amplifier can be considered as two identical transistor amplifiers, connected together as shown in figure A01.1b. The use of two voltage rails, +Vcc and -Vcc allows the output voltage to be either positive (if Vb > Va) or negative (if Vb < Va).

The inverting amplifier If a signal is applied to the inverting input, and the other input is connected to 0 Volts, then the output will be inverted (i.e. 1800 out of phase) as shown in fig.A01.2a.

fig. A01.2

The non-inverting amplifier

In this case the signal is applied to the non-inverting input, and the other input to 0 Volts. The output signal is then in phase with the input, i.e. 00 phase difference (fig.A01.2b).

Measuring the Voltage Gain

The voltage gain, Gv, is defined as the ratio of the output voltage and the differential input voltage: Gv = Vo / Vin (A01.1) where Vin is simply (Vb -Va). The circuit shown in figure A01.3 can be used to measure the gain. Assuming the amplifier has infinite input impedance, then the relationship between Vin and V1 can be found from the potential divider formula. This gives: Vin = V1 · R1 / (R1 + R2) (A01.2)


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In order to get the best practical measurement conditions, R2 should be much greater than R1: make the ratio R2 / R1 = 105. It is sensible to keep R1 less than100 Ω, so that R2 is not too large.

fig. A01.3

Substituting the value of Vin from A01.1 into A01.2 gives Gv = Vo·(R1 + R2) / ( V1·R1) (A01.3) This formula assumes that the input impedance is infinite. Although this is not true, it is so high that it can usually be ignored. Use this formula to find the gain of the amplifier. You will need to measure both the output voltage, Vo and the input as given by V1. The reason for measuring V1 rather than Vin (that is what we really need) is that Vin is always extremely small - typically less than a microvolt. V1 is much larger and so easier to measure.

Offset Voltage This is the voltage which must be applied across the inputs to produce a zero voltage output. Generally with no input voltage the output is not zero, as expected with a perfect amplifier. This is because of slight differences, or asymmetries in the circuit. In other words the offset voltage is the voltage needed to counterbalance the small voltage normally present at the input. The circuit used to measure the offset is shown in figure A01.4. Vo is the unwanted output voltage, caused by Vi at the input.

fig. A01.4

These two voltages depend on the resistor values of the resistive divider consisting of Ri and Rf :

Vi = Vo· Ri / (Ri+Rf)


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Since no signal is connected to the amplifier, and supposing the contribution of the bias and offset currents to be zero, then the output voltage is caused exclusively by the offset voltage. Measuring Vo allows the value of the offset voltage to be calculated.

Common Mode Rejection Ratio - CMRR With reference to figure A01.1b, we define the common mode voltage Vic as the average value (i.e. half the sum) of the voltages present across the two inputs of the operational amplifier, while we define the differential mode voltage, Vid as half the difference between the two voltages : Vic = (Via+Vib)/2 Vid = (Via-Vib)/2 The common mode rejection ratio (CMRR) is defined as: CMRR = Gd/Gc where Gc is the common mode gain i.e. the amplification obtained by connecting the same signal to both terminals, while Gd is the differential gain defined earlier: Gd = Vo/Vid Gc = Vo/Vic A possible circuit for measuring the common mode rejection ratio is shown in figure A01.5. The purpose of trimmer RV1 is to zero the effect of the offset voltage, while with trimmer RV2 you can zero the output voltage. In these conditions (zero output voltage), the common mode amplification is equal to the differential one. The ratio between the common mode voltage and the differential mode one is the common mode rejection ratio. We have: Gd·Vid + Gc·Vic = 0 from which it follows that (ignoring the minus sign): Gd/Gc = Vic/Vid Usually the common mode rejection ratio is expressed in decibel, and is then given by: 20·log10(Gd/Gc)


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fig. A01.5

Input Impedance The input impedance is defined as the impedance between the two input terminals. A suitable circuit to measure the input impedance is the one of figure A01.6. The externally applied voltage is shown as ‘V’. The actual voltage needed for the calculation is Vin. This is small, so it is measured indirectly by measuring the input at point V1, and then using the potential divider formula with resistors R1 and R2. Two measurements are needed, one with resistor R short-circuited, and the other with R in circuit. With R short-circuited, we have Vo = Gv ·Vin With R inserted: . Rin Vo = Gv · Vin · ⎯⎯⎯ Rin + R V is adjusted to make Vin the same as in the first measurement. The two measurements of output voltage then allow the input resistance to be calculated.

fig. A01.6


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Output Impedance The output of the operational amplifier is represented by an ideal voltage generator Vo, in series with a resistor, Ro. A suitable circuit for measuring Ro (output resistance) is shown in figure A01.7. Provided Rf has a much higher value than Ro, then with S open, the measured output voltage, V’o will be V’o ≈ Vo = Gv · Vin If Rc is much lower than Rf, then with S closed, Vo’ will be given by: Rc Rc Vo’= Vo · ⎯⎯⎯⎯ = Gv · Vin · ⎯⎯⎯ Rc+Ro Rc+Ro If Vin is kept constant in both measurements, Ro can be calculated : Vo - Vo’ Ro = Rc · ⎯⎯⎯⎯ Vo’

fig. A01.7


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Slew Rate The slew rate is defined as the speed with which the amplifier can change its output voltage. This parameter is measured in volts/second (or volts/μs ) and can be measured with the circuit of figure A01.8. To measure this parameter, apply a square wave signal to the amplifier input and measure the change in output voltage during a short time interval (e.g. in one μs).

fig. A01.8

Unity Gain Bandwidth The unity gain bandwidth is defined as the frequency at which the gain of the amplifier falls to a value of one. An amplifier’s gain reduces as frequency increases, due to the response speed of the device and to parasitic effects. To measure this frequency, the circuit of figure A01.8 can be used. Applying a sine wave of variable frequency to the input, the frequency is gradually increased until the output voltage becomes equal to the input. The frequency for which this occurs is the unity gain bandwidth.


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A01.2 EXERCISES

Disconnect all jumpers MCM7

Turn all switches S “OFF”

Note: some of the exercises in this lesson are quite involved, so you need to carry them out slowly and carefully!

Open loop gain measurement

• Insert jumpers J2, J8, J26, J11, J12, J21 to produce the circuit of figure A01.9.

• Connect at terminal 1 and Ground the function generator with a sine wave, 1 Hz, 10 Vpp and average value of zero.

• Connect one probe of the oscilloscope to the output of the amplifier (terminal 3).

• Using trimmer RV2, adjust the average value of the output voltage to be zero, by applying a bias voltage to the inverting input (OFFSET voltage compensation).

• Measure the peak-to-peak value of the ac output voltage. • With the oscilloscope, measure the peak-to-peak voltage present at

point A. • Using the formulae given in chapter A01.1, calculate the open-loop

gain of the operational amplifier.

Q1 What is the open loop gain ? 1 less than 20 dB 2 between 20 and 40 dB 3 between 40 and 60 dB 4 between 60 and 80 dB 5 higher than 80 Db

fig. A01.9


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Offset Voltage Measurement • Remove all jumpers, insert J11, J28, J21 to produce the circuit of

figure A0l.10. • With the multimeter (function VDC) measure the output voltage of

the operational amplifier. • Calculate the offset voltage from the formulae given in chapter

A01.1

fig. A0l.10

Q2 What is the absolute value of the offset voltage ? 1 equal to the output voltage 2 equal to the ratio R13/R4 3 between 0.1 mV and 10 mV 4 between 0.1 Volt and 10 Volt 5 zero

Measurement of the common mode rejection ratio • Remove all jumpers, insert J6, J15, J28, J17, J20 to produce the

circuit of figure A01.11. • Connect at terminal 2 and Ground the function generator with sine

wave, 10 Hz, 10 Vpp and zero average voltage value. • Adjust RV4 until the ac output voltage is zero. • With the voltmeter, measure the a.c. voltage applied to the

noninverting input of the amplifier. • Set the voltmeter for maximum sensitivity and measure the a.c.

voltage present between the two inputs of the amplifier. NOTE: If the voltmeter display shows 0 and the instrument is already on the most sensitive range, take the least significant digit as the voltage EXAMPLE : range :200 mV display : 00.0 value of the least significant digit : 100 μV value to be used : 100μV


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So in this example the display is 00.0 mV, the value to use is 0.1 mV. This is due to the difficulty in measuring voltages of this kind. Clearly if the voltmeter has a higher sensitivity (for example 10μV or 1μV, or even better) these problems are overcome

fig. A01.11 • Calculate the common mode rejection ratio in dB

Q3 What is the common mode rejection ratio in decibels ? 1 equal to the ratio Vo/Vi 2 equal to the ratio R7/R8 3 less than 40 dB 4 ranging between 70 dB and 110 dB 5 higher than 130dB

SLEW RATE MEASUREMENT • Remove J17, J20, insert J21 to produce the circuit of figure A01.12. • Apply a square-wave signal, 10 KHz, 3 Vpp and zero average

voltage value between the terminal 2 and ground. • Connect the first probe of the oscilloscope to the input of the circuit

(terminal 2) and the second to the output of the amplifier. • Measure the time taken for the output to rise from -1 V to +1V.

Q4 What is the slew rate ? 1 it is equal to the ratio of R13 to R7 2 about 0.5 V/μs 3 about 10 V/μs 4 about 100 V/μs 5 infinite


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fig. A01.12

BANDWIDTH • Keep the previous circuit as shown in fig. A01.12. • Between terminal 2 and ground, apply a sine wave, 100 Hz, 1 Vpp

and zero average voltage value. • Enter the output voltage value into table A0l.l • Increase the frequency to all the values shown on table A01.1, and

record the output voltage at each frequency. Calculate the frequency at which the output voltage drops to 0.707 of the low frequency value (corresponding to a drop of 3dB)

• Replace resistor R13 with R15 (disconnect J28 and connect J30) and repeat the measurements, completing the table. Note the frequency at which the output voltage drops to 0.707, as before

• Measure the frequency for which the amplification is equal to 1.

Q5 Replacing R13 with R15 increases the gain by a factor of 10. What else happens? 1 the unity gain bandwidth is increased 10 times 2 the unity gain bandwidth is diminished 10 times 3 the unity gain bandwidth is the same 4 the bandwidth of the amplifier is increased 10 times 5 the bandwidth of the amplifier is halved

INPUT FREQUENCY

OUTPUT VOLTAGE (R13)


100Hz

500Hz

1KHz

10KHz

20KHz

50KHz

100KHz

200KHz

500KHz

table A0l.1


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A01.3 SUMMARY QUESTIONS

Q6 Which of the following characteristics is not typical of operational amplifiers? 1 very large open loop gain 2 large common mode gain 3 ability to shift the output signal level 4 high input impedance

Q7 If with both inputs grounded, the dc output is not zero, this means 1 low common mode rejection 2 non zero offset voltage 3 high output impedance 4 high amplification

Q8 The output of an operational amplifier rises from 2V to 10V in 4 microseconds. What is the slew rate? 1 lV/μs 2 2V/μs 3 3V/μs 4 4V/μs

Q9 The bandwidth of a real operational amplifier is 1 the frequency range for which the gain is never less than 3 dB

in comparison with the maximum gain 2 the frequency range which can be applied to the input without

damaging the device 3 the frequency range in which the gain is constant 4 the frequency range which is more attenuated

Lesson A02: The inverting amplifier

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Lesson A02: The INVERTING AMPLIFIER

OBJECTIVES

• To determine the phase shift between the input and output signals • to observe the effect of negative feedback on amplifier gain • to measure the frequency response of an inverting op amp and

demonstrate the effect of negative feedback

INSTRUMENTS • function generator • dual trace oscilloscope

A02.1 BASIC THEORY The open loop gain of a real operational amplifier is so large (typical values around 100,000 or 100dB), that it is not often used. In practice, smaller and well defined gains are needed. Also very high gains can lead to unstability. To overcome these problems, feedback techniques are generally used. In this chapter, we analyse negative feedback, in which a part of the output signal is returned to the inverting input of the operational amplifier. The basic circuit for this configuration is shown in figure A02.1. We can now calculate the gain of an ideal op amp used as an inverting amplifier, and then compare it with a real device. An ideal operational amplifier has, by definition, infinite input resistance and gain. Infinite gain means, for a finite output voltage, the input differential voltage is zero. Considering the diagram of figure A02.1, this means that the inverting input has the same potential as the noninverting input, i.e. it is grounded. It is often said that the inverting input is a “virtual” ground. Using this simplification, the current through Ri will be given by I = Vin/Ri Now, exploiting the input resistance characteristic - of infinite value, we can say that all current I will pass through Rf, which then has a voltage Vo across it, given by: Vo = - Rf · I The minus sign comes from the fact that, if Vin is greater than zero, current will pass from Vin to Vo , so Vo must be at a lower voltage than the non-inverting input; since that is a virtual ground, Vo must be negative.

Lesson A02: The non inverting amplifier

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Substituting the value of I found earlier we obtain Vo = - Vin · Rf / Ri as the gain has been defined as the ratio between input and output voltages, we have: Gv = Vo / Vin = - Rf / Ri

fig. A02.1 To a good first approximation, these formulae can be used without modification. In real amplifiers, their very high gain generally means input differential voltages of some tens of microvolts, values which do not affect these formulae at all. Also the real input resistance is not generally a problem, as its value is certainly higher than Ri, with which it is in parallel.


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A02.2 EXERCISES


Turn all switches S “OFF” • Insert jumpers J6, J15, J28, J21 to produce the circuit of figure A02.2 • Connect at terminal 2 and Ground the function generator with a sine

wave, 1 KHz, l Vpp and zero average voltage. • Measure the output voltage Vo,using an oscillosope • From the measured value, calculate the voltage gain of the amplifier • Compare the theoretical result with the experimental one, using the

formulae of chapter A02.1

fig. A02.2

• Replace feedback resistor R13 of value 10KΩ with R15 of value 100kΩ. To do this, remove jumper J28 and connect jumper J30.

Q1 Keeping the amplitude of the input voltage at lVp-p, as in the last case, what is now the rms value of the output voltage Vo? 1 about 1 V rms 2 about 3.5 V rms 3 about 1 V rms 4 about l.5 V rms 5 equal to the ratio between R7 and R14

Q2 From these results, what is the maximum value of current through R7? 1 200 μA 2 50 μA 3 800 μA 4 1.2 mA 5 2 mA


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• Calculate the voltage gain using the formulae of chapter A02.1 for the new resistance value.

Q3 Connecting R15 of value 100kΩ in place of R13 of value 10kΩ what parameter changes ? 1 input impedance 2 current through R7 3 voltage at the inverting input 4 voltage at the noninverting input 5 gain of the amplifier

• Remove J30 and insert J28 to connect R13 again.

MCM7 Turn switch S23 “ON”

Q4 What is the result of this? 1 the current through R7 has changed 2 the output voltage has changed 3 the output signal frequency has changed 4 the resistor R7 has changed 5 the input impedance of the amplifier has changed

Q5 Which of the following could cause this? 1 the value of R7 has changed from 10kΩ to 500Ω 2 the open-loop gain of the amplifier has doubled 3 the value of R13 has changed from 10 kΩ to 900Ω 4 the bandwidth has increased

MCM7 Turn switch S23 “OFF”

• Connect one probe of the oscilloscope to the input signal (terminal

2) and the second probe to the output signal (terminal 3).


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Q6 What is the phase difference between the output and input signals? 1 0 2 π/4 3 π/2 4 π 5 3/2π

• Measure the output voltage and record the value in table A02.1 • Vary the frequency and record the corresponding output voltage • Determine the frequency value for which the output voltage drops to

0.707 times the maximum value • Replace resistor R13 with R15. Repeat all measurements and record

them in table A02.1

Q7 What is the effect of replacing R13 with R15? 1 the gain bandwidth product is changed 2 the unity gain bandwidth is changed 3 the bandwidth is changed 4 the output resistance is changed 5 none of the above

• From the values obtained plot a graph (see fig.A02.3), showing how

gain varies with frequency, for both cases. Indicate also the cut-off frequency, and note how it changes as a function of the feedback.

INPUT FREQUENCY



100Hz 1KHz 10KHz 20KHz 30KHz 40KHz 50KHz 100KHz 200KHz 300KHz 400KHz 500KHz 800KHz

Table A02.1


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fig.A02.3


Q8 What is the theoretical gain of the amplifier of the amplifier of figure A02.1 if Ri=2kΩ and Rf=8kΩ ? 1 2 2 4 3 6 4 8 5 equal to the ratio of R7 to R13

Q9 What is the amplitude of the input signal if the gain is 100 and the output signal is 5V ? 1 50 mV 2 500 mV 3 5 V 4 0 V

Q10 What must be done to increase the cut-off frequency of an inverting amplifier ? 1 increase the frequency of the input signal 2 lower the value of the feedback resistance 3 zero the offset voltage 4 increase the common mode rejection ratio


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Lesson A03: The NON INVERTING AMPLIFIER

OBJECTIVES

• To determine the phase shift between the output and input signals • to find the effect of negative feedback on the gain of the amplifier • to measure the cut-off frequency of an op amp in noninverting

configuration


A03. 1 BASIC THEORY The simplest circuit of a noninverting operational amplifier is shown in figure A03.1. The feedback used to stabilise the amplifier is made by returning a part of the output to the inverting input (negative feedback). If the amplifier has an ideal behaviour, the differential gain and the input resistance will be infinite. Infinite gain implies a differential voltage that is always zero. From this it follows that the inverting terminal has a voltage equal to that of the noninverting terminal. Exploiting also the characteristic of infinite input resistance, we see that all current through R2 must also pass through R1. R2 and R1 effectively form a resistive divider for the output voltage Vo.

fig. A03.1 Using these results we get : VA = Vin = Vo·Rl/(Rl+R2) from which: Vo = Vin·(Rl+R2) / R1 for which the gain will be: Gv = Vo/Vin = l + R2/Rl


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These formulae are also valid for a real op amp as its characteristic of high amplification and high input impedance produce a negligible error. From the formula for the gain of the amplifier in noninverting mode, it can be seen that the gain is never less than 1, and this value is obtained by setting R2 = 0 or R1 = ∞.

A03.2 EXERCISES



• Insert jumpers J3, J15, J29, J18 to produce the circuit of figure A03.2.

• Connect at terminal 2 and Ground the function generator with sine wave, 1 KHz, 1 Vpp.

fig. A03.2

Q1 According to the formulae of chapter A03.1, what should the gain be with the resistance values used in this exercise? 1 1.1 2 2 3 5 4 10 5 20

• Connect the oscilloscope channel 1 at terminal 2 (Vin input signal)

and channel 2 at terminal 3 (Vo output signal ). • Measure the gain of the amplifier by comparing the two signals

displayed on the oscilloscope. • Observe the phase shift between the input and output signals.


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Q2 What is the phase shift between input and output signals for the operational amplifier in non-inverting mode? 1 0 2 π/2 3 π 4 -π/4 5 -π/2

• Compare this result with that of the inverting amplifier of lesson

A02

Q3 What is the gain if resistor R14 is increased from 1KΩ to 100KΩ ? 1 11 times 2 10 times 3 5.5 times 4 2 times 5 15 times

• Remove jumper J29 and connect J28 (R14 is replaced by R13). • Measure the output signal amplitude and calculate the gain of the

amplifier


Q4 What is the effect on the circuit? 1 the output voltage has the same amplitude as the input voltage 2 the output voltage has double the amplitude of the input 3 the output voltage is 10 times the input 4 the output voltage has average value equal to 2V 5 the non inverting input voltage is equal to 3 V

Q5 What causes this? 1 the feedback resistance is equal to 2.2kΩ 2 the feedback resistance is equal to 500 Ω 3 the feedback is simply a short-circuit 4 a continuous component, whose value is 2V at the input, has

been added to the ac component 5 the noninverting input is not connected to ground anymore, but

to a voltage of 3V

MCM7 Turn switch S24 OFF


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• Now apply a sine wave, 100 Hz, 1 Vpp and zero average voltage value

• Measure the output voltage and record the result on the table A03.1 • Repeat this procedure for the frequencies shown in the table,

checking that before each measurement, the input voltage supplied by the signal generator is always kept at 1 Vpp.

• Change the feedback resistor from 10 KΩ to 100 KΩ (disconnect J28 and connect J30).

• Apply a sine wave, 100 Hz and 1Vpp at the input. Repeat the measurements and record your results in table A03.1

• Calculate the cut-off frequency (defined as the frequency at which the output voltage drops to 0.707 of the maximum value) in the two cases.

INPUT FREQUENCY



100Hz 1KHz 10KHz 20KHz 50KHz 100KHz 200KHz

Q6 How is the ratio of the pass bands (bandwidths) B1/B2 related to the ratio of the gains G1/G2 in the two cases? 1 it is equal to the ratio of the two gains 2 it is equal to R7/R14 3 it is equal to R14/R7 4 it is equal to l+R14/R7 5 it is equal to the inverse ratio of the two gains

• Draw a graph of gain as a function of frequency (see fig A03.3), to

show the results of table 3.1

fig. A03.3

table A03.1


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Q7 In a non inverting operational amplifier what happens if the value of R2 is doubled (fig. A03.1)? 1 the gain increase is less than double 2 the gain is doubled 3 the gain remains unchanged 4 the gain is reduced

Q8 What is the value of the input voltage Vi of a non inverting op amp if the voltage gain Gv is 50 and Vo is 5 volt ? 1 10 V 2 1 V 3 100 mV 4 10 mV

Q9 What is the phase shift between the input and the output signal for a noninverting op amp? 1 it depends on the ratio R2/R1 2 180 degrees 3 90 degrees 4 zero

Q10 How can the pass band of an operational amplifier in a non-inverting configuration be increased (fig. A03.1) ? 1 by increasing the input frequency 2 by reducing the input frequency 3 by reducing R1 4 by reducing R2

Lesson A04: The summing amplifier

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Lesson A04: The SUMMING AMPLIFIER

OBJECTIVES

• To demonstrate the ability of a summing op amp to provide an output

voltage equal to the sum of the voltages present at its inputs • to demonstrate the ability of a summing op amp to provide an output

voltage equal to the average voltage at its inputs

INSTRUMENTS • digital multimeter • signal generator • dual trace oscilloscope

A04.1 BASIC THEORY Figure A04.1 shows a typical realisation of a summing circuit using an operational amplifier. Unlike previous circuits, this one has two inputs, and there can be more if needed. As can be seen in figure A04.1 the voltages V1 and V2 are applied to the inverting input of the operational amplifier via two resistors R1 and R2.

Each single input will cause an effect on the output which is independent of the other inputs. So Vo is the sum of the results of the separate inputs:

Vo VRR

VRR

= − +( • • )131

232

The minus sign indicates that the output signal is shifted by 180 degrees compared to the input signals.

fig. A04.1


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From this formula it follows that if the sum (inverted) of the input signals is to be obtained at the output, then the ratios R3/R1 and R3/R2 must equal 1. Then : Vo = -(V1+V2) If the output is to be the average of the voltages present across the input, we need a value of 0.5 for ratios R3/R1 and R3/R2 which, substituted into the general formula, gives: Vo = -( 0.5·V1 + 0.5·V2 ) = - ( Vl + V2 ) / 2 There can be any number of inputs, provided that their total amplitude does not take the operational amplifier outside its linear operation and the supplied current must remain less than the max. current indicated by the manufacturer. Finally note that the summing amplifier can operate with continuous (dc) as well as with alternating (ac) signals.


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A04.2 EXERCISES



• Insert jumpers J2, J8, J9, J28, J6, J15, J21 to produce the circuit of

figure A04.2. • From a function generator, apply the same sime wave, 1 KHz, 2Vpp

and zero average value to both terminals 1 and 2.

Fig. A04.2

Q1 According to the formulae of the last chapter, A04.1, what is the output voltage? 1 2 Vpp 2 3 Vpp 3 4 Vpp 4 6 Vpp 5 10 Vpp

• Measure the output voltage Vo • Set the resistor R13 in parallel to the resistor R11 whose value is

10 kΩ so the effective feedback resistance is 5kΩ (to do this connect jumper J26).

• Calculate the theoretical value of the output voltage

Q2 What is the measured output voltage in this case? 1 0 Vpp 2 2 Vpp 3 3 Vpp 4 4 Vpp 5 6 Vpp


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Q3 What is the peak current through R13? 1 100 μA 4 200 μA 3 400 μA 2 800 μA 5 900 μA

• Remove jumper J26 and measure the output voltage Vo.


• Measure the input voltage: if its amplitude is not 2 Vpp adjust it to

this value.

Q4 What effect do you note on the circuit? 3 the output voltage is raised from 4Vpp to 6 Vpp 2 the output voltage is halved 1 the output voltage is doubled 4 the output voltage is reduced to zero 5 the output voltage is always negative

Q5 What is the reason for this output voltage change? 1 R1, which is now 1kΩ 2 R7, which is now 5kΩ 3 R13, which is now 1kΩ 4 V1, which is now 3 Vpp 5 V2, which is now 5 Vpp



- 28 -


Q6 In a summing amplifier which of the following statements is always true? 1 the output is equal to the sum of the input voltages 2 the output is equal to the average of the input voltages 3 the output depends on the ratio between the inputs 4 the output is equal to the sum of each input voltage multiplied

by its own amplification ratio

Q7 A summing amplifier (like the one of figure A04.1) has an output voltage equal to -4V. If there are only two inputs and both have the same voltage value 1V, what is the ratio Rf/Ri, if Ri is the same for both inputs? 1 ½ 2 2 3 4 4 1/4

Q8 In the previous question, if the ratio Rf/Ri is equal to 3, what is the input voltage to the two inputs (if the output is still – 4V)? 1 2/3 V 2 3/2 V 3 ½ V 4 2 V

Q9 A summing amplifier has 4 inputs, all with the same voltage 2V, and its output is -8 V. If Ri=1kΩ for all its inputs, what is Rf ? 1 500 Ω 2 1000 Ω 3 2000 Ω 4 4000 Ω

Lesson A05: The subtracting amplifier

- 29 -

Lesson A05: The SUBTRACTING AMPLIFIER

OBJECTIVES

• To demonstrate the ability of a difference (subtracting) op amp to produce an output voltage equal to the difference between the voltages present at its inputs.

INSTRUMENTS

• signal generator • dual trace oscilloscope

A05.1 BASIC THEORY A difference amplifier provides an output voltage proportional to the difference between the two or more input voltages. The basic circuit, using an operational amplifier, is shown in figure A05.1. We can now find the formulae for the difference amplifier, supposing the op amp to be ideal. As the input impedance, which is in parallel with R4, is theoretically infinite, the input voltage of the noninverting input of the amplifier is given by: V2’ = V2·R4/(R3+R4) Because of the infinite gain of the (ideal) op amp, the voltage V1’ will be equal to the voltage V2’. With this result, the current I through the resistor R1 is given by:

1R4R3R4R2V1V

1R'1V1VI +

⋅−

=−

=

This current, as the input resistance of the amplifier is infinite, will all flow into R2. So the output voltage is given by: Vo = V1’ - I·R2 from which, using the relations found for I and V1’, we get :

1R2R

4R3R4R2V

1R2R1V

4R3R4R2VVo ⋅

+⋅

+⋅

−+⋅

=

If the ratios R2/R1 and R4/R3 are equal we obtain: Vo = (V2 - V1) · R2/R1


- 30 -

and if, in particular, R2 is equal to R1:

Vo = V2 - V1

fig. A05.1

A05.2 EXERCISES



• Insert jumpers J4, J15, J28, J17, J22 to produce the circuit of figure

A05.2.

fig. A05.2

• Connect at terminal 2 and Ground the function generator with sine

wave, 1 KHz, 2 Vpp and zero average value. • Adjust RV3 to get a dc voltage of 4V at the point V2. • Calculate the output voltage value Vo as given by the theoretical

formulae.


- 31 -

Q1 According to the formulae of chapter A05.1, what is the maximum negative output voltage under these conditions? 1 -1V 2 -2V 3 -4V 4 -5V 5 -l0V

• Measure the output voltage Vo • Compare the experimental result with the one calculated

theoretically, to test the validity of the approximations made.

Q2 What is the peak current through R10? 1 50 μA 2 100 μA 3 200 μA 4 400 mA

Q3 What is the peak current through R13? 1 100 μA 2 200 μA 3 450 μA 4 1 mA 5 2 mA


Q4 What is the effect on the circuit? 1 the output voltage is at negative saturation 2 the output voltage is halved 3 the output voltage is doubled 4 the output voltage is zero 5 the output voltage is negative

Q5 What is the reason for this output voltage change? 1 V2 is higher than V1 2 R13 is halved 3 R10 is doubled 4 the inverting input is short-circuited to ground 5 the amplifier is operating without feedback


- 32 -


• Vary the voltage V2 with RV3 and check the output voltage.


Q6 The characteristic of difference operational amplifiers is to provide an

output voltage which is : 1 equal to the difference of the signals applied to the inverting

input 2 equal to the difference of the signals applied to the

noninverting input 3 proportional to the difference of the signal applied to the

noninverting input and the one applied to the inverting input 4 equal to the sum of the signal applied to the noninverting input

and the one applied to the inverting input

Q7 Refer to figure A05.1. If the resistors R1, R2, R3, R4 are all equal to 10kΩ and the voltages V1 and V2 are respectively 4V and 6V, what is the output voltage Vo? 1 0 V 2 2 V 3 4 V 4 6 V

Q8 With reference to the last example, what is the output voltage Vo, if R2 and R4 are 5 kΩ and R1 and R3 are equal to 10 kΩ ? 1 -1 V 2 0 V 3 1 V 4 2 V

Q9 With reference to figure A05 1, if the resistances are all equal and V2 is 2 V, what must V1 be, to have an output voltage equal to Vo= -3 V ? 1 -1 V 2 +1 V 3 +3 v 4 +5 V

Lesson A06: The Integrator

- 33 -

Lesson A06: The INTEGRATOR

OBJECTIVES

• To demonstrate the ability of an integrator operational amplifier to supply the output with a signal corresponding to the integral function of the input

• to find the phase shift between the input and output signals .


A06.1 BASIC THEORY The typical circuit of an operational amplifier in integrator configuration is shown in figure A06.1 The first thing to note is the presence of a capacitor in the feedback chain. Let’s note the relations connecting the output to the input for an ideal amplifier, starting from the law governing the operation of the capacitor C. The relationship between the voltage V, and the current I for an ideal capacitor of capacitance C is: I = C · dV / dt where dV/dt indicates the derivative of the variable V with respect to time t. The inverting input can be considered as at zero voltage (virtual ground) in this configuration, too, as the infinite gain guarantees zero differential input voltage. So, the current through the resistor R is given by the ratio between the input voltage Vin and R. This current, as the input resistance is ideally infinite, will charge the capacitor which, having the left terminal (refer to figure A06.1) to ground, will present its voltage variation on the right terminal.

fig. A06.1

Then we can write: Vin ⎯⎯ = C · dV/dt R


- 34 -

from which, as Vo = -V, we obtain: 1 dVo = - ⎯⎯ ·Vin dt R·C and integrating both sides

Vo1

R • CVin dt= − ∫

This formula says that the output voltage is the integral of the input voltage divided by the constant R·C = τ. The variable τ can be defined as the time necessary for the voltage Vo to reach an amplitude equal to the one across the input, from zero starting conditions and constant input voltage. Replacing the ideal amplifier with a real one we must take account of the offset voltage, which appears as a dc voltage at the input, and when integrated appears at the output as a voltage which increases linearly. Also a part of the bias current is integrated contributing to this output voltage variation. These two causes of error will eventually take the amplifier to saturation. This is one of the limits of this circuit. The problems listed before can be partly removed by connecting a resistor between the noninverting input and ground with a value equal to R, to compensate for the effect of the bias current, and also by inserting a resistor in parallel with C to counteract the effect of the offset voltage (fig. A06.2).

fig. A06.2


- 35 -

A06.2 EXERCISES



• Insert jumpers J6, J15, J32, J21 to produce the circuit of figure A06.3 • Connect at terminal 2 and Ground the function generator with a

square-wave, 2 KHz, 2 Vpp and zero average value. • Connect a probe of the oscilloscope to the output Vo of the circuit • Connect the second probe of the oscilloscope to the input Vin • Check that the waveform of the output voltage is the integral of the

waveform of the input voltage

fig A06.3

Q1 In what way does the capacitor charge in this circuit? (ie what is the output voltage? 1 linear 2 quadratic 3 exponential 4 inverse exponential 5 logarithmic

• Measure the slope of the output curve, expressed in Volt per

microsecond. • Now apply a sine wave, 1 KHz, 1 Vpp and zero average value. • Measure the phase shift between the input and output signals.

Q2 Remembering the normal phase shift between input and output, and noting that the integral of sin(x) is -cos(x) we can say that: 1 the output is the integral of the input 2 the output is the integral of the input shifted by π/2 3 the output is the integral of the input shifted by π 4 the output is the integral of the input shifted by -π/2 5 the output is not the integral of the input


- 36 -

• Take the capacitor to lnF (remove jumper J32 and connect J31) and replace R7 with R6 (100KΩ), by removing J15 and connecting J14.

Q3 Why hasn’t the time constant changed? 1 because it depends only on the ratio R/C 2 because it depends only on the product R·C 3 because it depends only on the amplifier used 4 because it depends only on the frequency of the input signal 5 because the time constant is a parameter which does not

depend on the circuit components • Remove jumper J14 and connect J15, check the output.


Q4 What has changed in the circuit? 1 the frequency of the output signal 2 the frequency of the input signal 3 the amplitude of the input signal 4 the amplitude of the output signal 5 the formula for the capacitor charging

Q5 What is the reason for this change? 1 the insertion of a resistor in parallel with the capacitor 2 the insertion of a capacitor in series with the resistor 3 the insertion of a capacitor in parallel with the resistor R7 4 a change in the value of R7 5 a reduction of the power supply voltage



- 37 -


Q6 If the input to an integrator is a signal v(t) = sin(t) the output will be: 1 a ramp 2 a triangular function 3 the cosine function 4 the function -sin(t)

Q7 Refer to figure A06.1. If R = 1k and C = 1μF, and a step signal is applied to the input Vin with amplitude of 2V, how long will it take before the output voltage is equal to the input one? 1 100 μs 2 1 ms 3 10 ms 4 20 ms

Q8 In the last example if C were 5μF, what would then be the time taken? 1 500 μs 2 1 ms 3 5 ms 4 10 ms

Lesson A07: The Differentiator

- 38 -

Lesson A07: The DIFFERENTIATOR

OBJECTIVES

• To demonstrate the ability of a differentiator op amp to produce an

output voltage Vo equal to the derivative of the input signal Vin • to determine the gain variation as the input frequency is changed


A07.1 BASIC THEORY The circuit diagram for a differentiator made using an operational amplifier is shown in figure A07.1. A resistor is inserted in the feedback branch, while a capacitor is connected to the input. If the amplifier has an ideal behaviour (infinite bandwidth, amplification and input impedance) we can calculate the relation between the input and the output, starting from the fact that the inverting input is at zero voltage (virtual ground). So the current through R is given by: i = Vo / R As for the capacitor, the law governing its operation is: i = C · dV/dt Since the input impedance is infinite, the current into the capacitor is equal to the one through resistor R. By substitution: Vo = - R·C dVin/dt If we apply a sine wave v(t) = sin(wt) as input signal, the output voltage will be: Vo = - R·C·ω·cos(ωt) For a zero frequency, (ie dc) it is clear that the output voltage is zero, as capacitors block dc. The gain of the amplifier is then zero for the dc component. As the frequency increases, the output amplitude, as well as the gain, increase linearly, as seen from the formula: Vo is proportional to ω. Theoretically, if the frequency is infinite, the capacitor has zero impedance which means an infinite amplification from the differentiator.


- 39 -

fig. A07.1 However high amplifications make the circuit unstable. Besides, as the gain increases with frequency, interference noise (which typically has high frequencies) can be amplified so much that it swamps the original signal. To solve these problems a resistor R1 can be connected in series with the capacitor (fig.A07.2), in order to limit the gain of the differentiator to the ratio R/R1 at high frequency when the impedance of the capacitor is very low.

fig. A07.2


- 40 -

A07.2 EXERCISES



• Insert jumpers J6, J16, J29, J21 to produce the circuit of figure

A07.3 • Connect at terminal 2 and Ground the function generator with a sine

wave, 10 KHz, 150 mVpp. • Connect one probe of the oscilloscope to the input signal.

fig. A07.3

Q1 What waveform must the output signal have with this input signal?

1 5 linear 2 3 exponential 3 2 constant 4 1 sine 5 4 triangular

• Connect one probe of the oscilloscope to the output of the circuit and

check the waveform. • Measure the phase shift between the input and output.

Q2 What is the phase shift of the output compared to the input signal? 1 4 -π 2 2 π/2 3 5 0 4 3 -π/2 5 1 π

• Replace resistor R14 (lKΩ) with R13 (l0KΩ) by removing jumper

J29 and connecting J28.


- 41 -

• Measure the output signal amplitude and phase shift compared to the input for this value of feedback resistance.

Q3 What are the differences if a resistor is inserted with a value 10 times higher? 1 output amplitude and phase shift are both 10 times greater 2 same output amplitude, but phase shift is 10 times greater 3 output amplitude is 10 times greater, but phase shift is the same 4 same amplitude and same phase shift 5 same amplitude and phase shift equal to π

• Now apply at the input a triangular wave, 500 Hz and amplitude 4

Vpp.

Q4 What is the approximate rise time (defined as the time for the input to reach the output value)? 1 0.2 ms 2 l ms 3 5 ms 4 12 ms 5 25 ms

Q5 From the measurements and the analysis of the formulae given in A07.1, the rise time depends on: 1 the capacitor only 2 the product R·C 3 R/C 4 the input signal frequency 5 it is independent of the components used


Q6 What is the effect on the circuit? l the output signal frequency is changed 2 the circuit is open loop (infinite gain) 3 the output is short-circuited to the input 4 the output is short-circuited to ground 5 the amplifier has a logarithmic behaviour



- 42 -


Q7 If we apply a cos(x) type signal to the inverting input of a differentiator, the output will be a signal of type: 1 -sin(x) 2 sin(x) 3 sin(-x) 4 -cos(x)

Q8 In the last example, if we apply a signal sin(x) to the inverting input, the output will be 1 -cos ( x ) 2 cos(x) 3 cos(x+π) 4 -sin(x)

Q9 With reference to figure A07.1, if R = 1 KΩ and C = 1μF what is the approximate output amplitude if Vin is a sine wave of frequency 100 Hz and amplitude 100mVpp ? 1 45 mVpp 2 62 mVpp 3 88 mVpp 4 96 mVpp

Lesson A08: The Comparator

- 43 -

Lesson A08: The COMPARATOR

OBJECTIVES

• To examine the operation of the circuit in comparing two voltage levels

• to analyse its operation as a square wave oscillator


A08.1 BASIC THEORY

The comparator is a circuit which compares an input signal Vin with a reference signal Vref. Depending on the result, the output voltage of the comparator, Vo can only take two values: Vcc (approximately) or -Vcc (approximately). In this application, the amplifier operates in the non-linear region. Refer to figure A08.1 (where the op amp is considered as ideal). When Vin > Vref, the output of the comparator will reach the maximum positive voltage it can, while when Vin < Vref, the output goes to the maximum negative value (negative saturation). This behaviour arises from the very high gain of the amplifier, which means that only very small values of the differential voltage are needed to take the output into saturation. It can be seen that the circuit is very simple, as no external component is needed. It is mainly used as zero crossing detector and as a squarer. Let Vmin and Vmax respectively be the negative and positive saturation voltages which the output of the operational amplifier may have. As the gain is finite for a real comparator, it will not instantly go from Vmin to Vmax, when the input voltage Vin is changing, but it needs only some µV to do this. Another cause of error is the offset voltage which alters the level at which switching occurs.

fig. A08.1


- 44 -

A08.2 EXERCISES



• Insert junpers J4, J15, J18 to produce the circuit of figure A08.2. • Adjust trimmer RV3 to obtain a voltage of 0V at the inverting input

of the amplifier. • Connect at terminal 2 and Ground the function generator with a

triangular wave signal, 1 KHz, 5 Vpp and zero average value. • Connect a probe of the oscilloscope to the output signal and check

the waveform and the amplitude.

Q1 What is the waveform of the output voltage? 1 saw-tooth 2 constant 3 square-wave 4 sine-wave 5 triangular

• With trimmer RV3, vary the voltage Vref applied to the inverting

input of the operational amplifier.

Q2 What is value of the input voltage Vin, at which the output voltage switches level? 1 -2V 2 -1V 3 1V 4 2V 5 at the value of the Vref voltage

fig. A08.2

• With trimmer RV3, vary the voltage Vref applied to the inverting input of the operational amplifier.


- 45 -

Q3 What effect is caused by varying the reference voltage with RV3 ? 1 the frequency of the output voltage varies 2 the output changes from square- to sine-wave 3 the output amplitude increases 4 the duty-cycle (ratio between the period in which voltage is at a

high level and the one in which it is at low level) changes 5 nothing changes

• Change the waveform of the input voltage from triangular to sine,

1 kHz, 5 Vpp and zero average value. • Repeat the measurements, varying the voltage applied to the

inverting input. • With RV3 make the voltage on the inverting input equal to 1V.


Q4 What has happened to the circuit? 1 a short-circuit has been introduced between the output and

inverting input 2 the amplitude of the input signal has been varied 3 the frequency of the input signal has been varied 4 the average value of the input voltage has been varied 5 the reference voltage Vref has been varied



- 46 -


Q5 The function of a comparator is to : 1 amplify a signal 2 invert a signal 3 compare two signals 4 differentiate a signal

Q6 If any signal of frequency F is connected to the input of a comparator, which of the following waveforms should appears at the output ? 1 a sine-wave with frequency F 2 a triangular wave with frequency F 3 a rectangular-wave with frequency F 4 a square-wave with frequency depending on the reference

voltage

Q7 An op amp used as comparator operates in which of the following regions? 1 linear 2 non linear 3 with limited gain 4 with negative feedback

Q8 If a reference voltage Vr = 1V is connected to the non-inverting input of a comparator, and its output is less than zero, the input voltage on the inverting input is: 1 more than zero 2 less than zero 3 more than 1 V 4 less than -1 V

Lesson A09: The Logarithmic Amplifier

- 47 -

Lesson A09: The LOGARITHMIC AMPLIFIER

OBJECTIVES

• To determine the relationship between the output voltage Vo and the

input Vin, for a logarithmic amplifier

INSTRUMENTS • multimeter

A09.1 BASIC THEORY The function of a logarithmic amplifier is to provide an output which is proportional to the logarithm of the input signal. One possible circuit for this amplifier is shown in figure A09.1, in which the feedback chain consists of a transistor. To explain the operation of this circuit, remember that the amplifier has a very high gain and that a very small differential voltage is sufficient to take the output into saturation. As the base of the transistor is at ground potential and the emitter is connected to the output, the output voltage is the same as the base-emitter voltage, but of opposite sign: vo = -vBE When vBE increases, so does the collector current. If the input impedance is very high (so the current entering the inverting input can be ignored), the collector current iC of the transistor then equals the current through R. This makes the differential voltage fall, and also therefore the output voltage falls. To avoid saturation, the differential voltage used must be in µV (as the minimum differential amplification is 100,000).

fig. A09.1


- 48 -

For normal operation, the transistor needs vBE values between 0.5 and 1 volt: this means a very small differential voltage (theoretically zero), and allows the inverting input to be considered as a virtual ground. The current iC which enters the collector of the transistor is then:

Rvi in

C = A09.1

The ratio between collector current iC and base current iB is given by the ratio hFE : BFEC ihi ⋅= A09.2 while the relation between the base-emitter voltage vBE and the base current is given by an exponential formula :

T

BE

Vv

B eIoi ⋅= A09.3 where: iB = base current Io = reverse leakage current of a P-N junction vBE = base-emitter voltage VT = K· T / q In the last formula, K is the Boltzmann constant, T is the absolute temperature and q is the electron charge. Using the values of these constants we obtain VT = T/11600. From the relations A09.2 and A09.3 we obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⋅=Ioh

iInVvFE

CTBE A09.4

Replacing iC with the value calculated in A09.1 we get:

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

⋅=IohR

vInVvFE

inTBE A09.5

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

⋅−=−=IohR

vInVvvoFE

inTBE A09.6

At room temperature, VT is about 26mV while Io depends very much on the actual transistor. It varies even for transistors of the same kind, and must be found experimentally. The value of hFE can be considered constant to a first approximation (a more accurate analysis takes account of its dependence on collector current).


- 49 -

AO9.2 EXERCISES



• Insert jumpers J4, J15, J9, J24, J25, J21 to produce the circuit of

figure A09.2. • Adjust trimmer RV3 until there is a voltage of about 6 Volt across

Vin and measure the output voltage • Change the resistor R7 with R6 removing jumper J15 and connecting

jumper J14 • Masure the output voltage

fig. A09.2

Q1 If both hFE and Io are constant, (independent of other parameters) what will the difference between the two output voltages be, if the collector current changes by a factor of 10? 1 it depends on the collector current 2 1V 3 ln(10) V 4 59.8 mV 5 10 mV

Q2 Comparing the value calculated in theory with the measured value, we can say that: 1 it is possible to consider hFE and Io constant, and independent

of IC 2 the measured result corresponds exactly to the theoretical one 3 replacing resistor R7 with R6 does not cause any change 4 output voltage cannot be varied by changing input voltage 5 none of the above is correct


- 50 -

• Adjust trimmer RV3 until Vin is about 0.5 Volt • Measure the output voltage and record the value in table A09.1 • Repeat this procedure for all the voltage values given in the table • Plot the graph (see figure A09.3) • Note the logarithmic behaviour of the output voltage from the curve

of figure A09.3

INPUT VOLTAGE OUTPUT VOLTAGE

0.2 0.5 1

1.5 2

2.5 3 4 5 6 7 8

table A09.1

fig. A09.3 • use RV3 to take Vin back to 6 V


- 51 -


Q3 What is the effect on the circuit? 1 the voltage has a linear behaviour with gain 10 2 the output voltage has a positive value 3 the output voltage is constantly 3V, independent of the input

voltage 4 the output voltage is almost zero (a few millivolt) for any input

voltage level 5 the amplifier is in saturation

Q4 What is the reason for this? 1 the feedback is a resistor of value is 10 KΩ 2 the feedback is simply a short-circuit 3 the amplifier has no feedback 4 the output of the amplifier is short-circuited 5 the base - emitter junction is short-circuited



Q5 The logarithmic amplifier exploits 1 the voltage-current characteristic of the p-n junction 2 positive feedback 3 the high common mode rejection 4 all of the above

Q6 For which of the following purposes can a logarithmic amplifier be used? 1 to increase the gain 2 to decrease the gain 3 to obtain a variable gain, depending on the input voltage 4 to obtain a separation between the input and output

Lesson A10: The Monostable Multivibrator

- 52 -

Lesson A10: The MONOSTABLE MULTIVIBRATOR

OBJECTIVES

• To see how an op amp, connected as monostable, can provide an output pulse of fixed duration, whenever it is “triggered” by an input pulse.

INSTRUMENTS

• function generator • dual trace oscilloscope • multimeter

A10.1 BASIC THEORY

When used as a monostable, the op amp is used in its non-linear mode, with only two output levels. These are the positive and negative voltage saturation levels. This circuit has only one stable state: with a constant input, its output never changes. If it is in the non stable state, then after a certain time it will drop back to its stable state. The classic circuit of a monostable multivibrator is shown in figure A10.1. We describe now its operation assuming that the operational amplifier has an ideal behaviour. Let Vmax and Vmin, respectively be the positive and the negative saturation voltages of the output. In stable conditions, i.e. without input signals, the output is in positive saturation (at Vmax). The voltage of the noninverting input comes from a potential divider composed of R1 and R2: Val = Vmax ·Rl/(Rl+R2) A10.1

fig. A10.1

The inverting input is at about 0.6 volt as the resistor R3 tends to charge C2 towards Vo but the diode limits the voltage across the terminals of C2 to the value of its forward bias value. As the voltage at the


- 53 -

noninverting input is greater than at the inverting input, the output will be at voltage Vmax of positive saturation. When the voltage Vin has a negative pulse, the capacitor C1 transfers it to the noninverting input and if this pulse is greater than the difference between the noninverting and inverting inputs, the output switches from Vmax to Vmin. With the output at Vmin the noninverting input in turn switches to a value:

2R1R1RminVV 2a +

⋅= A10.2

The capacitor C2 starts discharging through R3, towards the potential Vmin (the diode has no effect as it is reverse biased). When C2 reaches the voltage given by the A10.2, the differential voltage present across the inputs of the amplifier turns positive, and the output switches back to Vmax and stays there, until the next negative pulse . The output pulse duration is determined by the time for the resistor R3 to discharge the capacitor C2 from the voltage of 0.6V to the voltage given by A10.2. The capacitor discharge follows an exponential law:

2C3Rt

e)6.0minV(minVVb ⋅−

−−= A10.3

During the charging sequence, when the value given by A10.2 changes to 0.6V the capacitor voltage is:

2C3Rt

2a e)VmaxV(maxVVb ⋅−

−−= A10.4 From these formulae it can be seen that the output pulse duration T is determined by the choice of R3 and C2 and by the ratio R1/(Rl+R2). Its approximate value is: T ≈ τ · ln ( 1 + R1/R2) A10.5 with τ = R3·C2


- 54 -

A10.2 EXERCISES



• Insert jumpers J3, J5, J7, J16, J19, J28, J34 to produce the circuit of

figure A10.2.

fig.A10.2

• Measure the output voltage

Q1 From this last measurement, the output is: 1 at ground potential 2 3 Volt 3 in negative saturation 4 positive saturation 5 none of the above

• Measure the inverting input voltage

Q2 Why isn’t the capacitor C1 charged to the output voltage? 1 because, with R13 it constitutes a divider 2 because in an operational amplifier, the differential voltage is

always zero 3 because the amplifier has a finite gain 4 because the time constant R14 C1 is not zero 5 because of the presence of a diode

• Measure the non inverting input voltage


- 55 -

Q3 Why is the output voltage at saturation? 1 because the voltage of the noninverting input is higher than the

inverting input 2 because the voltage of the noninverting input is higher than

zero 3 because the voltage of the noninverting input has the same

sign as the output voltage 4 because the voltage of both inputs is higher than zero 5 because the amplifier is not linear

• Connect at terminal 2 and Ground the function generator with a

square-wave signal, 2 KHz, 5 Vpp and zero average value. • Connect the probes of the oscilloscope to the output and input signals • Determine which part of the input signal makes the output switch • Move the probe from the input signal to the inverting input • Note the voltage levels at switching, and the waveforms on the

capacitor C2


Q4 What effect can be seen on the circuit? 1 the amplitude of the output voltage has changed 2 the frequency of the output voltage has changed 3 the output remains at zero 4 the output remains constant 5 the pulse duration has changed

Q5 What is the reason for this? 1 the diode has been short-circuited 2 the resistor R13 has been replaced with one of 1 MΩ 3 the resistor R13 has been replaced with one of 1 KΩ 4 the capacitance C2 has been replaced with one of double value 5 the input frequency is halved



- 56 -


Q6 A monostable multivibrator has 1 one stable state 2 two stable states 3 no stable state 4 none of the above

Q7 For which of the following can a monostable multivibrator be used ? 1 To obtain pulses of fixed duration 2 To obtain one pulse on the leading edge and one on the falling

edge of the input signal 3 To detect a zero crossing point 4 To detect positive and negative pulses

Lesson A11: The Astable Multivibrator

- 57 -

Lesson A11: The ASTABLE MULTIVIBRATOR

OBJECTIVES

• To see how an astable multivibrator can provide a symmetrical, square-wave output

• to produce an astable multivibrator with a non-symmetrical square-wave output

INSTRUMENTS

• function generator • dual trace oscilloscope

A11.1 BASIC THEORY With an astable multivibrator, the op amp operates only in the non-linear region. So its output has only two voltage levels, Vmin and Vmax. The astable continually switches from one state to the other, staying in each state for a fixed length of time. The circuit of an astable multivibrator is shown in figure A11.1. Note that this circuit does not need an input signal. To find out the relations governing the operation of the astable, we start with the usual hypothesis that the operational amplifier has an ideal behaviour. Suppose the output is in the state Vo = Vmax. When Vo takes this value the voltage VA1 of the non inverting input is: VA1 = Vmax · R1 / (R1 + R2) The capacitor C starts charging through resistor R towards the value Vmax. This charging continues until the voltage VB of the inverting input reaches the value VA1. At this point, as the inverting input voltage is more than the noninverting input, the output switches low, to Vmin. The voltage VA2 is now given by: VA2= Vmin · R1 / (R1 + R2) At this point, the capacitor C starts discharging through R towards the voltage Vmin until it reaches the value VA2, at which point the output switches to Vmax . The cycle then starts again. We have seen that the voltage across the capacitor C can vary from VA1 to VA2, so in the period of time when the output is low, at Vmin, the voltage on the capacitor is given by:

V (t) Vmin (VminVmax • R1

R1 R2) • eB

-t /R•C= − −+

while in the period of time when the output is at Vmax, the capacitor voltage is :


- 58 -

V (t) Vmax (VmaxVmin • R1R1 R2

) • eB-t /R•C= − −

+

The period T1 for which the output voltage is at Vmax can be found by calculating the time the capacitor voltage takes to equal VA1. So:

Vmax • R1

R1 R2(Vmin • R1R1 R2

Vmax) • e VmaxT1/R•C

+=

+− +−

from which:

T R CV

RR R

V

VR

R RV

1

11 2

11 2

=−

+

−+

• • lnmax • min

max • max

Similarly we can find the period T2 for which the output stays at Vmin :

T R C

RR R

V V

RR R

V V2

11 2

11 2

= +−

+−

• • ln• max min

• min min

Supposing that Vmin = -Vmax we obtain:

T T R C

RR R

RR R

1 21

11 2

11

1 2

= =+

+

−+

• • ln

The total period T of the square-wave is given by the sum of T1 and T2. We can see that the square-wave period and so the frequency can be varied by varying the values of R1, R2, R and C. To obtain an asymmetrical square-wave (duty cycle not 50%) you can make the capacitor charge and discharge through resistors of different values.

fig. A11.1


- 59 -

A11.2 EXERCISES



• Insert jumpers J3, J16, J22, J30, J34 to produce the circuit of figure

A11.2. • Calculate the output frequency with the formulae given in chapter

A11.l. • Connect the first probe of the oscilloscope to the output Vo of the

amplifier and the second probe to the inverting input VB • Measure the frequency with the oscilloscope, and compare it with the

theoretical result • Calculate the capacitor voltages at which output switching occurs,

according to the formulae of chapter A11.l • Measure the capacitor voltages at which output switching occurs and

compare the results with those calculated from theory

Q1 What is the maximum voltage reached across capacitor C1 ? 1 0 V 2 less than 1 V 3 less than 2 V 4 about 5.5 V 5 about 10 V

fig.A11.2 fig.A11.3

• Now connect jumper J27, to produce the circuit of figure A11.3. • Connect the first probe of the oscilloscope to the output of the

operational amplifier and the second to the inverting input VB


- 60 -

Q2 What are the approximate resistances through which the capacitor is charged and discharged ? 1 32 kΩ and 100 kΩ 2 47 kΩ and 100 kΩ 3 100 kΩ and 100 kΩ 4 10 kΩ and 47 kΩ 5 10 kΩ and 10 kΩ

• Calculate the values of T1 and T2 as given by the formulae of

chapter A11.l • measure the values of T1 and T2 with the oscilloscope.


Q3 What has happened to the circuit? 1 only the frequency of the output has changed 2 the duty-cycle of the output voltage has changed 3 the frequency and the waveform are both changed 4 the frequency and the duty-cycle of the output voltage are both

changed 5 the output voltage has changed

Q4 What is the reason for this? 1 the gain of the amplifier is zero 2 the resistor R15 has been replaced with one of 1 MΩ 3 the resistor R15 has been replaced with one of 1 KΩ 4 the resistor R10 has been replaced with one of double the

value 5 the offset has been adjusted to zero



- 61 -


Q5 What is the function of an astable multivibrator ?

1 To generate a square wave 2 To square a sine-wave 3 To generate frequency equal to a reference signal 4 All of the above

Q6 With reference to figure A11.l, if Vmin = -Vmax the frequency of oscillation of the astable is determined by: 1 The resistor R 2 The capacitor C 3 The ratio R1/R2 4 The components R, C, R1, R2

Q7 With reference to figure A11.1, what is the approximate frequency of oscillation of the astable when R = R1 = R2 = 100 k , C = 25 nF and Vmin = -Vmax ? 1 90 Hz 2 135 Hz 3 180 Hz 4 225 Hz

Q8 If in the last example you want a frequency of 500 Hz, the resistor R must be: 1 37 k 2 50 R 3 72 k 4 88 k

Lesson A12: The Sine Wave Oscillator

- 62 -

Lesson A12: The SINE WAVE OSCILLATOR

OBJECTIVES

• To determine the waveform and output frequency of a sine wave

oscillator

INSTRUMENTS • dual trace oscilloscope

A12.1 BASIC THEORY A sine oscillator can be constructed using an op amp and a feedback network. The input signal must be equal to the output one in amplitude, phase and frequency, in order to obtain an output which is a sine voltage without any input signal. This is obtained only if the feedback network contains reactive elements. The frequency at which the system oscillates is the one for which the phase change introduced by the feedback network, plus any introduced by the amplifier, is zero. Figure A12.1 shows a Wien bridge oscillator. Figure A12.2 shows the equivalent bridge circuit of figure A12.1.The input voltage will be given by the difference between the voltage present at point P1 and the voltage present at point P2.

fig. A12.1 fig. A12.2 To satisfy the required conditions, the noninverting input voltage must be in phase with the output. The voltage at point P2 is: VP2 = Vo· Z2 / (Z2 + Z1)


- 63 -

where Z1 and Z2 are given by: Z1 = R + 1 /jωC Z2 = 1 / (1/R + jωC) It is not difficult to check that Z1 and Z2 have the same phase angle (and then Vin is in phase with Vo) at a frequency: fo =1 / 2π·R·C and that for this value of frequency we have: Z1 = ( 1 - j )·R Z2 = ( 1 - j )·R / 2 From previous formulae we obtain: VP2 = Vo / 3 To obtain VP1 equal to VP2, the following must be true: VP1 = Vo· R2 / (R1 + R2) = Vo / 3 from which it follows that R1=2·R2. The potentiometer in parallel to R1 is used to correct any asymmetries in the circuit, and for component tolerances.


- 64 -

A12.2 EXERCISES



• Insert jumpers J43, J44, J49, J50, J54, J55, J56 to produce the

circuit of figure A12.3.

fig. A12.3 • Turn trimmer RV7 completely clock-wise • Connect one probe of the oscilloscope to the output of the circuit and

the second probe to point P2 of the Wien bridge • Adjust trimmer RV7 to obtain a sine wave across the output

Q1 What is the phase difference between the two signals displayed ? 1 0 2 π/4 3 π/2 4 π 5 -π/2

• Calculate the theoretical value of the frequency at which the circuit

should oscillate as given by the formulae of chapter A12.1


- 65 -

Q2 Theoretically, what is the approximate oscillation frequency of this circuit? 1 10 Hz 2 150 Hz 3 1300 Hz 4 16 kHz 5 250 kHz

• Measure the oscillation frequency with an oscilloscope • Allowing for slight measurement error and component tolerances,

check the consistency between the theoretical and measured frequency values.


Q3 What is the effect on the circuit ? 1 the output sine wave is highly distorted (trapezoidal wave) 2 the output has a triangular waveform 3 the output is constant and zero 4 only the positive half of the sine wave is present at the output 5 the output sine wave is decreased in amplitude

Q4 What is the reason for this? 1 the gain of the amplifier has been zeroed 2 the value of resistor R21 has changed 3 the ratio R26/C8 and R27 and C12 has been doubled 4 the ratio R26/C8 and R27/C12 has been set to half 5 the supply voltage of the op amp has been changed



- 66 -


Q5 Which of the following conditions must be satisfied to produce a sinusoidal output from an oscillator? 1 The input amplitude must equal the output amplitude 2 The input must be in phase with the output 3 The input and output must have the same frequency 4 All of the above

Q6 With reference to figure A12.1, what is the approximate frequency of oscillation if R= 10 kΩ and C= 100 nF ? 1 130 Hz 2 160 Hz 3 200 Hz 4 250 Hz

Q7 In the last question, if the parallel resistance of R1 with the trimmer P is 10 kΩ, what should R2 be ? 1 1 kΩ 2 5 kΩ 3 10 kΩ 4 15 kΩ

Q8 To obtain an oscillation frequency of 1000 Hz what must C be, with R = 10 kΩ ? 1 10 nF 2 13 nF 3 16 nF 4 20 nF

Lesson A13: The Waveform Generator

- 67 -

Lesson A13: The WAVEFORM GENERATOR

OBJECTIVES

• To determine the frequency and output amplitude of a triangular wave generator

• to determine the frequency and output amplitude of a ramp generator

INSTRUMENTS • dual trace oscilloscope

A13.1 BASIC THEORY

Among the waveforms that can be generated with op amps, the most common are the triangular, the ramp and the square-wave. A triangular wave can be generated with the circuit of figure A13.1, in which two operational amplifiers are used. The first operates as a comparator, while the second as an integrator. Vo is the output voltage of the integrator, Vr the output voltage of the comparator. The saturation voltages Vmax and Vmin are equal in amplitude, and so can be called +Vr and -Vr respectively. Suppose the output of the comparator is +Vr. The voltage Vo will be a negative ramp which will continue to grow until the voltage of the noninverting input of the comparator rises above zero. The minimum value of the output voltage Vo, applying the superposition principle, will be given by:

3R1RRVo

3R1R3RVr0

+⋅

++⋅

=

from which we get: Vo = - Vr ·R3 / Rl

fig. A13.1

The same principles apply for the maximum voltage the output reaches, with the only difference that the ramp is rising and the voltage Vr is negative: defining this voltage as Vo’ we have:


- 68 -

1R3RVr'Vo ⋅

=

To calculate the time T taken to rise from Vo to Vo’ remember that the capacitor C charges with a constant current given by:

2RPVrI+

=

so, from:

dtdVoCI ⋅−

=

we find that:

T)Vo'Vo(C

2RPVr −⋅

−=+

from which:

)2RP(CTVrVo'Vo

+⋅⋅

−=−

As: 1R

3RVr2Vo'Vo ⋅⋅=−

we have: 1R

C)2RP(3R2T ⋅+⋅⋅=

The time T is equal to one half period, so the output frequency F will be the inverse of twice T :

C)P2R(3R41RF

⋅+⋅⋅=

To convert the triangular wave generator to a ramp generator (also known as a saw-tooth) simply insert a diode and resistor in parallel with the potentiometer P and resistor R2. The diagram is shown in figure A13.2. Just as with the triangular wave generator, the positive ramp is generated by the current flowing through P and R2 (the diode D is reverse biased), while for the negative ramp the capacitor discharges through a smaller resistance (given by R5 in parallel with the series of P and R2) and so it is faster.


- 69 -

fig. A13.2 So the time T1 to go from -Vr·R3/Rl to +Vr·R3/Rl is the same as a triangular wave generator, while the time T2 of the return to -Vr·R3/R1 is given by:

1RCR3R2

2T p ⋅⋅⋅=

where: RP = (P+R2) // R5. As in module MCM7 the resistance R5 is much less than the value of P and R2, RP can be considered equal to R5. With this approximation:

1RC5R3R22T ⋅⋅⋅

=

With the values used in the module, and with the potentiometer all inserted, T2 = 4.4 μsec which can be neglected in comparison with T1 which is about few milliseconds. The frequency of the ramp generator is equal to:

C)P2R(3R21RF

⋅+⋅⋅=

To obtain a square wave generator, take the signal from the output of the first op amp, and if the charge and discharge times of the integrator are equal, the output voltage will be symmetrical. Remember though, that the specific circuit to obtain a square-wave voltage is the astable multivibrator, analysed in lesson A11.


- 70 -

A13.2 EXERCISES



Triangular waveform generator • Insert jumpers J3, J13, J33, J34, J37, J39, J44, J53, J55 to produce

the circuit of figure A13.3. • Adjust RV6 completely CCW to obtain zero resistance and measure

the output voltage value of the comparator (terminal 3) using the oscilloscope.

fig. A13.3

Q1 What is the discharge current for capacitor C11? 1 it depends on the value of the capacitor 2 it is equal to the product Rl9·C11 3 it is equal to the output voltage of IC1 divided by R19 4 1 mA 5 10 mA

• Adjust trimmer RV6 to half value. • Calculate the amplitude of the output voltage (terminal 5) using the

formulae of chapter A13.1 • Measure the amplitude of the output voltage with the oscilloscope • Calculate the output voltage frequency according to the formulae

of chapter A13.1 • Measure the output frequency with the oscilloscope • Check the presence of a square-wave at the output of the comparator

(terminal 3) • Replace capacitor C11 with C10 (remove jumper J53 and connect

J52). • Repeat the last calculations and measurements, adjusting the

resistance of trimmer RV6 as before.


- 71 -

Ramp generator • From previuos circuit, remove J52 and insert J36, J53 to produce the

circuit of figure A13.4. • Adjust RV6 completely CCW to obtain zero resistance. • Measure the amplitude of the output voltage with the oscilloscope • Calculate the output frequency according to the formulae of chapter

A13.1 • Measure the output frequency with the oscilloscope • Measure the charge and discharge times of the capacitor

fig. A13.4

Q2 What is the ratio between the rise time and fall time of the output voltage ? 1 it is approximately equal to Rl9/R18 2 it is approximately equal to the product Rl9·R18 3 it is equal to the inverse of the output frequency 4 it depends on the output voltage of the comparator 5 it depends on the output voltage of the integrator


Q3 What is the effect on the circuit ? 1 5 the output voltage becomes a square wave 2 2 the amplitude of the output voltage is changed 3 4 the output frequency is doubled 4 1 the output frequency is four times higher 5 3 the output frequency is halved

Q4 What is the reason for this?

1 3 the diode is reversed 2 5 the value of R18 is doubled 3 4 the value of Cll is halved 4 2 the offset voltage is set to zero 5 1 the value of Cll is doubled



- 72 -


Q5 To produce a triangular wave-generator the following are used : 1 a comparator with a differentiator 2 a noninverting amplifier with integrator 3 a comparator with an integrator 4 a summing with an integrator

Q6 With reference to figure A13.1, if the total resistance which discharges C is 10 kΩ, C=100 nF and R1=2⋅R3, what is the output frequency? 1 500 Hz 2 875 Hz 3 1136 Hz 4 1480 Hz

Q7 With reference to figure A13.2, what is the output amplitude if Vr varies between –15V and +15V and the values are in the previous example ? 1 6.6 Vpp 2 9.3 Vpp 3 15 Vpp 4 30 Vpp

Q8 In the circuit of fig. A13.3, what is the approximate output frequency if the maximum resistance of trimmer RV6 is inserted ? 1 about 45 Hz 2 about 135 Hz 3 about 250 Hz 4 about 300 Hz

Lesson A14: The Voltage-to-Frequency Converter

- 73 -

Lesson A14: The VOLTAGE-to-FREQUENCY CONVERTER

OBJECTIVES

• To see how a voltage-to-frequency converter can supply an output frequency proportional to the input voltage

INSTRUMENTS

• dual trace oscilloscope • multimeter

A14.1 BASIC THEORY

The voltage-to-frequency converter is a circuit which supplies an output voltage whose frequency is proportional to the input voltage. The typical circuit diagram of this device is shown in figure A14.1. It consists of two op amps, a transistor and some passive components. The purpose of the first op amp is to integrate the input signal, while the second amplifier compares the output of the first with a reference voltage. The input voltage Vi charges capacitor C through resistor R1, so the voltage at point A will be a negative ramp which increases linearly with time. The voltage VA is compared to voltage VB by the second operational amplifier. At the beginning of the cycle the voltage VA will be zero. The transistor is cut-off because the output of the comparator is low, and (ignoring the voltage drop across the diode), the voltage VB is given by (superposition principle): R3//R4 R3//R5 VB’ = -12 · ⎯⎯⎯⎯⎯ - 10 · ⎯⎯⎯⎯⎯ R3//R4 + R5 R3//R5 + R4 where: • the symbol // is the parallel combination of two resistors

(eg R3//R4 = 1/(1/R3+1/R4) • -10 V is the output voltage of IC2, in negative saturation. The ramp at point A continues to drop until it reaches the value VB’. At this value the comparator changes its output to high, taking the transistor into conduction. This means that point B goes to about -0.7 volt (VB” = -0.7V).


- 74 -

fig. A14.1 The capacitor C discharges through the transistor until the voltage at point A exceeds VB”: at this point the comparator takes its output low, so causing the transistor to cut off and point B to fall at voltage VB’. At this point the cycle is finished and can start again. Define tl as the time necessary for the capacitor to go from VB” to VB’. This time is equal to: tl = - ( VB’ - VB”)·R1·C / Vi To calculate the time t2 of discharge for the capacitor from VB‘ to VB” we need the discharge law for C, which is: t2 = -C·(VB’ + VB”) / I With the values chosen, the time t2 can in practice be ignored. From this it follows that the period of the output voltage is almost equal to tl which, (ignoring the value of VB”), can be written as : tl ·Vi = ⏐ VB’⏐· R1 · C from which: F ≈ 1/tl = Vi / (⏐VB’⏐· R1· C) which is the relation connecting the input voltage to the output frequency.


- 75 -

A14.2 EXERCISES



• Insert jumpers J4, J14, J21, J9, J23, J31, J38, J39, J44, J57, J58, to

produce the circuit of figure A14.2.

fig. A14.2 • Connect one probe of the oscilloscope to terminal 3

Q1 What is the waveform of the voltage present at the output of amplifier IC1 ? 1 square wave 2 triangular wave 3 saw-tooth wave 4 sine-wave 5 the amplifier is at saturation

• Connect the other oscilloscope probe to the output of the circuit

(transistor emitter)

Q2 When does transistor T2 conduct ? 1 for half the period of the output voltage of IC1 2 during the rise time of the output voltage of IC1 3 during the fall time of the output voltage of IC1 4 When the voltage at point A is higher than zero 5 When the output voltage of IC2 is higher than zero


- 76 -

• Note the waveform of the output signal and measure its frequency • Record this frequency value in table A14.1 (Vin = 0.5V) • Repeat this measurement for all voltage values shown in table A14.1 • Plot a graph of the voltage-frequency behaviour (figure A14.3) for

all measurements

Vin [Volts] Output Frequency [KHz]

0.5

1

1.5

2

3

4

5

6

7

8

table A14.1

fig. A14.3


- 77 -

• Take the output voltage of RV3 to 6 V


Q3 What is the effect on the circuit? 1 the output voltage of the circuit is always constant and positive 2 the output voltage of the circuit is a sine-wave 3 the output voltage of IC1 is in positive saturation 4 the output voltage of IC1 is in negative saturation 5 the output voltage of the circuit is constant and negative

Q4 What is the cause of this? 1 the base-emitter junction of transistor T2 has been short-

circuited 2 resistor R29 has been short-circuited 3 the capacitor has been short-circuited (zero gain for IC1) 4 the input voltage has been changed 5 the offset voltage has been changed



- 78 -


Q5 In the voltage-to-frequency converter the relationship between output frequency and input voltage is : 1 linear 2 quadratic 3 exponential 4 logarithmic

Q6 Refer to figure A14.1, and suppose that with the transistor OFF the point B is at a voltage of -4V. What is the approximate output frequency if R1=10kΩ, C=100 nF and the input voltage is 5V (not ignoring VB’’)? 1 1000 Hz 2 1250 Hz 3 1500 Hz 4 1750 Hz

Q7 With reference to the previous question, if R1 doubles the output frequency is: 1 one half 2 one quarter 3 double 4 four times

Q8 Again, with reference to question Q6 if the resistor R1 doubles and the capacitance C is halved the output frequency is: 1 the same 2 one half 3 double 4 less than half

Lesson A15: The Low-Pass Filter

- 79 -

Lesson A15: The LOW-PASS FILTER

OBJECTIVES

• To determine the input/output characteristic of a low-pass filter and calculate its cut-off frequency

INSTRUMENTS

• dual trace oscilloscope • function generator and frequency-meter.

A15.1 BASIC THEORY

A low-pass filter can be made with an op amp as the active element and resistors and capacitors as passive elements. The diagram of a generic filter is shown in figure A15.1, where Z1-Z5 indicate generic impedances. The characteristics of the filter are determined by the type and value of the impedances used. The input-output formula for the general diagram of figure A15.1 is given by:

Vo Z2·Z4·Z5 ⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Vin Z1·Z3·Z5 + Z1·Z2·Z5 + Z1·Z2·Z3 + Z2·Z3·Z5 - Z2·Z1·Z4

where Vo and Vin are the respective output and input voltages.

This relation assumes that the input impedance of the operational amplifier is infinite (so no current flows into its input) and that the amplification is infinite (so the inverting input can be considered a

fig. A15.1

fig. A15.2


- 80 -

virtual ground). With these assumptions, and applying the superposition principle you obtain the equation given. This filter is a low-pass filter, if Z1, Z3 and Z5 are resistors, while Z2 and Z4 are capacitors (see fig. A15.2). Changing the general impedances Zn of the formula to the impedances of the components used, we obtain the general formula for the low-pass filter:

C2•C1•R3•R21)

R31

R21

R11(•

C1j+-

C2•C1•R2•R11

VinVo

2 +++ϖ

ϖ−=

where w is the angular frequency of the input signal. The cut-off frequency Fc of the filter is determined by the value of the passive components:

FcR • R • C • C

=1

2 2 3 1 2π

The gain Go of the filter at the frequency f=0 Hz (dc) is given by: Go = - R3/R1

A15.2 EXERCISES




figure A15.3. • Connect at terminal 4 and Ground the function generator with a sine

wave signal, 10 Hz, 1 Vpp. • Connect one probe of the oscilloscope to the output of the filter and

measure the amplitude of the signal • Record this value in table A15.1 • Calculate the gain of the filter at the frequency of 10 Hz (given as the

ratio between Vo and Vin)


- 81 -

Q1 What is the approximate amplification of the filter at this frequency? 1 1 2 9 3 15 4 50 5 infinite

fig. A15.3

Q2 What is the main reason for this amplification value? 1 the value of R22 2 the value of C10 3 the value of R23 4 the value of the ratio R23/R20 5 the value of the product R22·C10

• Repeat the measurement of the output voltage (and record it in the

table) for all frequencies shown in table A15.1

INPUT FREQUENCY [Hz] OUTPUT AMPLITUDE [Vpp]

[ Vin = 1Vpp ]

10

20

40

60

80

100

150

200

500

Table A15.1


- 82 -

fig. A15.4 • Plot all the values obtained into figure A15.4 and draw the

input/output characteristic of the filter • From this graph calculate the cut-off frequency of the filter, defined

as the frequency at which the amplification drops to 0.707 times the maximum.


• Now measure the cut-off frequency of the filter again

Q3 What do you notice? 1 the cut-off frequency is increased 2 the cut-off frequency is decreased 3 the output voltage is constant 4 the output voltage is a square wave 5 the output voltage is a triangular wave

Q4 What is the cause of this? 1 the output is short-circuited 2 the inputs are short-circuited 3 the capacitor C10 has increased 4 the capacitor C10 has decreased 5 the amplifier has positive feedback



- 83 -


Q5 An ideal low-pass filter has the maximum amplification : 1 in a range between two fixed frequencies 2 below a particular frequency 3 above a particular frequency 4 for all frequencies below a frequency F’, and for all

frequencies above a frequency F” > F’

Q6 In a low-pass filter the maximum output signal is 15 Vpp. What is the approximate output voltage of this filter at the cut-off frequency? 1 4 Vpp 2 6 Vpp 3 8 Vpp 4 10 Vpp

Q7 In Question 6 the frequency at which the filter has the maximum amplification is : 1 below the cut-off frequency 2 at the cut-off frequency 3 between the cut-off frequency and double that frequency 4 more than the double the cut-off frequency

Q8 In a filter the phase shift between the input and output is : 1 dependent on frequency 2 constant and different from zero 3 constant and equal to zero 4 none of the above

Lesson A16: The High-Pass Filter

- 84 -

Lesson A16: The HIGH-PASS FILTER

OBJECTIVES

• To determine the input/output characteristic of a high-pass filter and calculate its cut-off frequency

INSTRUMENTS


A16.1 BASIC THEORY

The ideal high-pass filter is a circuit with zero amplification for all signals below a certain frequency Fc, while the amplification is constant and non-zero for all signals with frequencies higher than Fc. As with the low-pass filter, the circuit comprises an op amp, resistors and capacitors. Clearly, the real circuit for a high-pass filter cannot have ideal characteristics: it is acceptable if the transfer function has a behaviour similar to an ideal filter. To determine the behaviour of the filter, we consider the relation between input and output, from the relation written for the generic filter (fig. A16.1):


The filter of fig.A16.1 is a high-pass filter, if Z1, Z3 and Z5 are capacitors, while Z2 and Z4 are resistors (see fig.A16.2).

fig. A16.1

fig. A16.2


- 85 -

Changing the general impedances Zn of the formula to the impedances of the components used, we obtain the general formula for the low-pass filter:

C3•C2•R2•R11)

C31

C21

C3•C2C1(•

2Rj

•C3C1

VinVo

2

2

+++ω

+ω−

ω=

where w is the angular frequency of the input signal. The cut-off frequency Fc of the filter is determined by the value of the passive components:

FcR • R • C • C

=1

2 1 2 2 3π

The gain Go of the filter at the frequency f=infinite (practically, the frequencies within the pass-band region) is given by: Go = - C1/C3


- 86 -

A16.2 EXERCISES




figure A16.3.

fig. A16.3 • Connect at terminal 4 and Ground the function generator with a sine

wave signal, 5 KHz, 1 Vpp, zero average value. • Connect one probe of the oscilloscope to the output of the filter and

measure the amplitude of the signal. • Record this value in table A16.1 • Calculate the gain of the filter at the frequency of 5 KHz (given as

the ratio between Vo and Vin)

Q1 What is the gain of the filter for this frequency value? 1 0 2 less than 0.1 3 between 0.1 and 1 4 about 1.5 5 more than 10

• Repeat the measurement of the output voltage (and record it in the

table) for all the frequencies shown in table A16.1


- 87 -

INPUT FREQUENCY [KHz]

OUTPUT AMPLITUDE [Vpp]

[ Vin = 1Vpp ]

5

8

10

12

14

16

18

20

Table A16.1

fig. A16.4 • Plot all the values obtained into figure A16.4 and draw the

input/output characteristic of the filter • From this graph calculate the cut-off frequency of the filter, defined

as the frequency at which the amplification drops to 0.707 times the maximum.

Q2 What is the gain of the filter for an input frequency of l0KHz ? 1 0 2 less than 0.1 3 between 0.1 and 1 4 between 1 and 5 5 about 9


- 88 -


• Now measure the cut-off frequency of the filter again

Q3 What effect can be noted ? 1 the cut-off frequency has increased 2 the cut-off frequency has decreased 3 the output voltage is constant 4 the output frequency is a square wave 5 the output voltage is a triangular wave

Q4 What is the reason for this ? 1 the output is short-circuited 2 the inputs are short-circuited 3 the resistor R25 has increased 4 the resistor R25 has decreased 5 the amplifier has positive feedback



- 89 -


Q5 An ideal high-pass filter has maximum amplification : 1 in the range between two fixed frequencies 2 below a particular frequency 3 above a particular frequency 4 for all frequencies less than a frequency F’ and for all

frequencies above a frequency F”’ > F’

Q6 In a high-pass filter like the one of figure A16.3 which frequencies are attenuated ? 1 the low ones from the filter, and the high ones from the op amp

which has a limited band 2 only the low ones 3 only the high ones 4 none of the above

Q7 With reference to figure A16.2 what must be done to vary the cut-off frequency of the high-pass filter ? 1 Change the value of the resistors R1 and R2 2 Change the value of the capacitors C2 and C3 3 Change the value of at least one of the passive components 4 Any of the above

Q8 In a high-pass filter the phase shift between the input and output signal is due to: l the resistive component of the amplifier network 2 to the reactance of the amplifier network 3 to the very high input impedance of the amplifier 4 none of the above

Lesson A17: The Band-Pass Filter

- 90 -

Lesson A17: The BAND-PASS FILTER

OBJECTIVES

• To determine the input/output characteristic of the band-pass filter and calculate its center frequency

INSTRUMENTS


A17.1 BASIC THEORY

The ideal band-pass filter is a circuit which has zero amplification for all signals whose frequency is below a certain frequency f’, and also for those whose frequency is above a frequency f”. Between these two frequencies the amplification is constant. As before, the circuit used to make the filter consists of an op amp, resistors and capacitors.

The real circuit for the band-pass filter cannot have ideal characteristics: it is acceptable if the transfer function has a behaviour similar to the ideal filter. Again, we start from the relation between the input and output. The formula for the generic filter of fig.A17.1 is :


This filter will be a band-pass if Z1, Z2 and Z4 are resistors and Z3 and Z5 are capacitors (fig. A17.2). Substituting into the general relation, we obtain:

fig. A17.1

fig. A17.2


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)2R

11R

1(•2C•1C•3R

1)2C

11C

1(•3R

11C•1R

j

VinVo

2 ++++ω−

ω

−=

where w is the angular frequency of the input signal. The center frequency Fo of the filter is determined by the value of the passive components:

Fo

R C C R R

=+

1

21

3 1 211

12

π• •

• ( )

The gain Go of the filter at the center frequency is given by:

GoRR C

C

= −+

31

1

112

•


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A17.2 EXERCISES




figure A17.3.

fig. A17.3 • Connect at terminal 4 and Ground the function generator with a sine

wave signal, 500 Hz, 1 Vpp, zero average value. • Connect one probe of the oscilloscope to the output of the filter and

measure the amplitude of the signal • Record this value in table A17.1 • Repeat the measurement of the output voltage (and record it in the

table) for all the frequencies shown in table A17.1 • Calculate the gain of the filter at the frequency of 1 KHz (given as

the ratio between Vo and Vin)

Q1 What is the gain of the filter at 1000 Hz? 1 0 2 less than 0.1 3 about 3.5 4 about 10 5 more than 15

Q2 What is the gain of the filter in the band centre? 1 0 2 less than 0.1 3 about 1.5 4 about 7 5 more than 15


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INPUT FREQUENCY [Hz] OUTPUT AMPLITUDE [Vpp] [ Vin = 1Vpp ]

500

750

1000

1250

1500

1750

2000

2250

2500

2750

3000

3500

4000

Table A17.1

Q3 What is the filter gain for an input frequency of l0KHz ? 1 0 2 less than 0.1 3 between 0.5 and 1.5 4 between 5 and 10 5 more than 10

• Plot all the values obtained into figure A17.4 and draw the

input/output characteristic of the filter

fig. A17.4


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• From this graph calculate the cut-off frequency of the filter, defined as the frequency at which the amplification drops to 0.707 times the maximum.

17.3 SUMMARY QUESTIONS

Q4 An ideal high-pass filter has maximum amplification: 1 in the range between two fixed frequencies 2 below a particular frequency 3 above a particular frequency 4 for all frequencies below a frequency F’ and for all

frequencies above a frequency F’’ > F’

Q5 In a band-pass filter like the one of figure A17.3 which frequencies are attenuated? 1 the low ones from the filter and the high ones from the

operational amplifier which has a limited band 2 only the low ones 3 only the high ones 4 those below a frequency F’ and those above a frequency

F”>F’

Q6 With reference to figure A17.2 what must be done to vary the center frequency of the band-pass filter? 1 Change the value of the capacitors C1 and C2 2 Change the value of the resistors R1, R2 and R3 3 Change the value of at least one of the passive components 4 Any of the above

Q7 In a band-pass filter the phase shift between the input and output signal is due to : 1 the resistive components of the amplifier network 2 the reactance of the amplifier network 3 the very high input impedance of the amplifier 4 none of the above

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