Keith A. Hudson34Digital & Analogue Devices

M130611705 March, 2014

HNC Electrical and Electronic EngineeringYear One - 2013/14Module: Digital & Analogue Devices

OperationalAmplifiersKeith A. HudsonM130611711/03/2014

4-Way Bottle OpenerKeith A. HudsonUser Guide03 July 2012Keith A. HudsoniiPLCsM130611712 November 2013

HNC Electrical and Electronic EngineeringYear One: 2013/14

4-Way Bottle OpenerKeith A. HudsonUser Guide03 July 2012Keith A. Hudson2PLCs

M130611705 November, 2013

HNC Electrical and Electronic EngineeringYear One: 2013/14Contents1Assumptions62Adder / Subtractor Circuit (Task 1a)72.1Requirements72.2Circuit Simulation Using Proteus82.3Actual Circuit153Integration Circuit (Task 1b)173.1Requirements173.2Circuit Simulation Using Proteus183.3Actual Circuit2144-bit Analogue to Digital Convertor Circuit (Task 1c)234.1Requirements234.2Circuit Simulation Using Proteus254.3Actual Circuit345Conclusions366Bibliography37

FiguresFigure 1: Adder / Subtractor circuit8Figure 2: Testing i/p A9Figure 3: Testing i/p B10Figure 4: Testing i/p C11Figure 5: A = 5 V, B = 4 V, Output = 1.3 V12Figure 6: A = 5 V, B = 4 V, C = 2 V, Output = 0.3 V13Figure 7: A = 5 V, B = 4 V, C = 5 V, Output = -1.2 V14Figure 8: Adder / subtractor circuit15Figure 9: Multi-meters showing the actual output from each Op Amp16Figure 10: Op Amp Integrator circuit17Figure 11: The simulated Integrator Circuit18Figure 12: The input and output wave forms19Figure 13: Input / Output ranges19Figure 14: Output voltage at 0.4s20Figure 15: integrator circuit on breadboard21Figure 16: Power supply21Figure 17: Function generator producing the square wave input for the integrator21Figure 18: Setting up the PSU and initial oscilloscope output22Figure 19: Oscilloscope output22Figure 20: Input 0000, Output 025Figure 21: Input 0001, Output 126Figure 22: Input 0010, Output 226Figure 23: Input 0011, Output 327Figure 24: Input 0100, Output 427Figure 25: Input 0101, Output 528Figure 26: Input 0110, Output 628Figure 27: Input 0111, Output 729Figure 28: Input 1000, Output 829Figure 29: Input 1001, Output 930Figure 30: Input 1010, Output 1030Figure 31: Input 1011, Output 1131Figure 32: Input 1100, Output 1231Figure 33: Input 1101, Output 1332Figure 34: Input 1110, Output 1432Figure 35: Input 1111, Output 1533Figure 36: 4-bit converter34Figure 37: Duel PSUs34Figure 38: Oscilloscope showing input and output voltages35

TablesTable 1: Amplification values for each input value7Table 2: Values for A and B and the expected Output12Table 3: Values for A, B and C and the expected Output13Table 4: Same values for A and B and a different C and the expected Output14Table 5: Values for resistor, R and capacitor, C17Table 6: Input vs. output voltage18Table 7: Weighting associated with each bit23Table 8: Input values and expected outputs23Table 9: Prime factors24Table 10: Amplification values for each input bit24Table 11: Ideal vs. real Op Amp36

AssumptionsThe actual value of the resistors used will be slightly different to the manufacturers stated value. This may be as much as 10% for a resistor with a silver band. Those with a brown tolerance band are more accurate 1%. Resistors with smaller tolerances (i.e. more accurate) are available but they are more expensive. For these experiments resistors with a tolerance of 1% would be sufficiently accurate.These experiments assume the use of ideal Op Amps. These have the following characteristics:1. Voltage gain is infinite.2. Gain is independent of frequency.3. Input resistance is infinite.4. Output resistance is zero.5. Input voltage offset is zero.6. Output can swing to the positive or negative supply rails.7. Output can swing instantly to the correct value.The power supply needs to provide an accurate voltage to the Op Amp rails and as input. The model used is a HY3005D-2. The Data Sheet for this model states accuracy is: Source:constant voltage0.01%1mV Load:constant voltage0.01%5mVThese figures are accurate enough for the following circuits.

Adder / Subtractor Circuit(Task 1a)RequirementsA circuit is required with the following characteristics:It should have three input voltages: Input A, Input B and Input CThe value of Input A should be multiplied by 0.1, i.e. divided by 10 (we will call this o/p A)The value of Input B should be multiplied by 0.2, i.e. divided by 5 (o/p B)The value of Input C should be multiplied by 0.5, i.e. divided by 2 (o/p C)There will be one output from the circuit: Output = o/p A + o/p B o/p CFor an ideal op-amp the output voltage is calculated as follows: .Therefore, if we know what output to input ratio is required we can calculate the ratio of to (ignoring the change of sign):Table 1: Amplification values for each input valueInput AInput BInput C

The ratio is 0.1 or

The ratio is 0.2 or

The ratio is 0.2 or

If we chose an arbitrary value for, say 1k, then = 10 k

= 5 k

= 2 k

To add two numbers (i.e. voltages) together we simply connect the wires together and the op-amp will amplify them according to the resistor ratios used. The output from the op-amp will be negative (i.e. inverted), so we need to pass the result through a second inverting op-amp to invert it again, thus correcting the sign. Our circuit needs to perform the following arithmetic operation: . If we multiply both sides by -1 we get . The output from the first op-amp is . So if we add C to that output, and invert the result we are left with the required output result.

Circuit Simulation Using ProteusThe required circuit was created with using two Inverting Op-Amps, with the appropriate resistor ratios as shown in Figure 1.

Figure 1: Adder / Subtractor circuitFor each input a variable resistor was connected to a five volt source to allow an input value to be selected in the range 0-5V. Initially all the inputs were set to zero. Input A was then set to 2V to ensure the amplification was correct. If Input A is 2V, the circuit output should be 0.40V. The result can be seen in Figure 2. This was the repeated for Input B. This time the Output should be 0.20V, See Figure 3. For Input C the output should be 1.00V, see Figure 4.

Figure 2: Testing i/p A

Figure 3: Testing i/p B

Figure 4: Testing i/p C

The next step is to test that 0.1 * i/p A + 0.2 * i/p B produces the correct results.Table 2: Values for A and B and the expected Output

Figure 5: A = 5 V, B = 4 V, Output = 1.3 VAs Figure 5 shows, the actual Output matches expected Output.

Table 3: Values for A, B and C and the expected Output

Figure 6: A = 5 V, B = 4 V, C = 2 V, Output = 0.3 VAs Figure 6 shows, the actual Output matches expected Output.

Keeping A and B unchanged; a larger value of C is be used to ensure a negative result is correctly calculated.Table 4: Same values for A and B and a different C and the expected Output

Figure 7: A = 5 V, B = 4 V, C = 5 V, Output = -1.2 VAs Figure 7 shows, the actual Output matches expected Output.

Actual CircuitThe actual circuit was created using power supplies (PSU), resistors and boards each containing a 741 Op Amp. Various wires were used to connect everything together. The resultant circuit can be seen in Figure 8.

Figure 8: Adder / subtractor circuitThe equipment is as follows:Each of the white boards contains a 741 Op Amp.The PSU on the right, provide power to the rails of the 741 on the right. The PSU in the middle, provide power to the rails of the 741 on the left. The PSU on the left provides 5V for inputs A, B and C.The multi-meters show the output voltage from the respective Op Amps.The Op Amp on the left- is the adder, with inputs:

The output from the Op Amp on the left will be (minus because it is an inverting amplifier).The Op Amp on the right- is the subtractor, with inputs:

The output from the Op Amp on the right will be (minus because).

Figure 9: Multi-meters showing the actual output from each Op AmpThe output from the adder Op Amp is 1,501 V. The output from the subtractor Op Amp is 1.017 V. The respective values are not exactly 1.5 and 1.0 for a number of reasons:The input voltage to the Op Amps was not exactly 5.0 V. (Actually 5.1 V.)The resistance of the wires may have affected the input ratios slightly.The calibration of the PSUs and multi-meters may be out.

Integration Circuit(Task 1b)RequirementsA circuit is required, such that the output voltage, , where is the input voltage.

Figure 10: Op Amp Integrator circuitThe above circuit (see Figure 10) is an example of an Op Amp Integrator circuit. The output is calculated as follows: (Bird, 2010)If we look at the required output: , it is very similar to the output from the above (Figure 10) circuit. If we ignore the (-) sign because we know this is an inverting amplifier, then we need: .Table 5: Values for resistor, R and capacitor, C

If a resistor of 5k is used then what size capacitor is needed?

Circuit Simulation Using ProteusUsing the values from Table 5, the resultant circuit can be seen in Figure 11:

Figure 11: The simulated Integrator CircuitA signal generator was used to produce a square wave (Peak-to-peak: 5V, max +2.5V, min -2.5V) input to the circuit. The input and output waves were displayed on an oscilloscope. As Figure 12 shows the output is a triangular wave. The output is the integral of the input.Figure 13 shows the input and the output values. The output starts at 0V and drops at a constant rate to -247mV. It then increases at a constant rate back to 0V. Table 6: Input vs. output voltageTime (s)00.51.0

i/p (V)+2.5+2.5-2.5-2.5

o/p (mV)0-2470+247

Table 6 shows the input and output voltages during the first oscillation of the input square wave and the corresponding triangular output wave. When the input is positive the output is negative and when the input is negative the output is positive.

Figure 12: The input and output wave forms

Figure 13: Input / Output ranges

If we take a time between 0 and 0.5 seconds, the input voltage is constantSo becomes At time, t=0.4s, the output voltage will be:

On the simulator we get , almost the same value.

Figure 14: Output voltage at 0.4s

Actual Circuit

Figure 15: integrator circuit on breadboard

Figure 16: Power supply

Figure 17: Function generator producing the square wave input for the integrator

Figure 18: Setting up the PSU and initial oscilloscope outputDue to an error in wiring the integrator circuit, the Op Amp was damaged and consequently failed to produce the desired output (see Figure 19).

Figure 19: Oscilloscope outputWiring up circuits using breadboard is extremely fiddly and mistakes are easy to make and difficult to locate. Proteus is relatively easy to use and generates errors if things are incorrectly wired. When using Proteus all the components are working. Real components are easy to damage and may not be working when even when they are new.

4-bit Analogue to Digital Convertor Circuit(Task 1c)RequirementsThis circuit requires four digital inputs. Each input will be on or off. For simplicity an input voltage of 1V will be used for on, and 0V will be used for off. In practice any set voltage could be used to represent on.Each of the four input bits will be weighted differently: Table 7: Weighting associated with each bitBitWeightBinary

110001LSB

220010

340100

481000MSB

LSB: least significant bit, MSB: most significant bit.Table 8: Input values and expected outputsInput bitsOutput ValueInput bitsOutput Value

0000010008

0001110019

00102101010

00113101111

01004110012

01015110113

01106111014

01117111115

For the input bits MSB is on the right, LSB on the leftThe adder/subtractor circuit Figure 1 can be easily adapted to fulfil the requirements of this circuit:Remove the input from the subtractor part of the circuit.Add two more inputs to the adder part of the circuit.Change the resistor ratios to match the new requirements.The second op amp (was the subtractor) will now be used to correct the sign so that the final output is positive.

The lowest common multiple (LCM) of 1, 2, 4 and 8 is 8 (See Table 9).Table 9: Prime factors1To find the lowest common multiple, we need to think about which list has the most of each factor. In this case it is list 8, which has 2*2*2. So the lcm is 2*2*2=8. (BBC, 2014)

22

42*2

82*2*2

The feedback resistor () on the adder op amp should be a factor of 8. So an 8k resistor will be used.Table 10: Amplification values for each input bitBit 1Bit 2Bit 3Bit 4

The multiplier is 1

The multiplier is 2

The multiplier is 4

The multiplier is 8

Circuit Simulation Using ProteusThe final circuit is shown in Figure 20. Each input is connected to the 1V source via a switch. When a switch is up the input is off, i.e. 0V. When the switch is down the input is on, i.e. 1V.Figure 20 to Figure 35 show all possible input combinations and the corresponding outputs.

Figure 20: Input 0000, Output 0

Figure 21: Input 0001, Output 1

Figure 22: Input 0010, Output 2

Figure 23: Input 0011, Output 3

Figure 24: Input 0100, Output 4

Figure 25: Input 0101, Output 5

Figure 26: Input 0110, Output 6

Figure 27: Input 0111, Output 7

Figure 28: Input 1000, Output 8

Figure 29: Input 1001, Output 9

Figure 30: Input 1010, Output 10

Figure 31: Input 1011, Output 11

Figure 32: Input 1100, Output 12

Figure 33: Input 1101, Output 13

Figure 34: Input 1110, Output 14

Figure 35: Input 1111, Output 15

Actual CircuitThe circuit was created on breadboard (see Figure 36) using: Four dip-switches (top-right) Op Amp (centre-left) Resistors: feedback resistor (centre-left), various resistors (top-right) to weight each of the dip-switches as required.

Figure 36: 4-bit converter Two PSUs to power the Op Amp. One also provides power to the dip-switches.

Figure 37: Duel PSUs A selection of wires to connect everything together.

Figure 38: Oscilloscope showing input and output voltagesThe values shown in Figure 38 differ from those in the simulator primarily because different input voltages were used. Had the voltage been the same, there would probably still show some slight difference:The supply voltage to the Op Amps was not exactly 12.0 V. (Actually +12.0, -12.1 V.)The resistance of the wires may have affected the input ratios slightly.The calibration of the PSUs and oscilloscope may be out.

ConclusionsAlthough the ideal Op Amp does not exist, many of those available are close enough.Table 11: Ideal vs. real Op AmpCharacteristicIdeal Op AmpReal Op Amp

Voltage gainInfinite.Very high gain. Open-loop gain in the order of 200,000.

Gain vs. frequencyGain is independent of frequency.Gain remains constant up to about 10kHz.

Input resistanceInfinite.2M.

Output resistanceZero.Typically 75.

Input voltage offsetZero.A few mV.

Output voltageCan swing to the positive or negative supply rails.Typically 13V for an amp with rails of 15V.

Output reaction timeCan swing instantly to the correct value.Takes a finite time to reach the output value and additional time to settle (slew rate).

Effect of using a real Op Amp:1. Voltage gain was not an issue for any of the circuits because at most a gain of *8 was required.2. Frequency was not a problem, because two of the circuits were DC (i.e. a frequency of 0Hz) and the integrator circuit used a frequency of 1Hz.3. The input resistance of the Op Amp was far higher than any of the resistors used in the circuit. So for the purpose of these experiments it was effectively infinite.4. The output resistance of the Op Amp was far lower than any of the resistors used in the circuit. So for the purpose of these experiments it was as good as zero.5. The output voltage was kept well below the values of the Op Amp supply rails.6. The time taken for the output to react to the input was small enough to be of no consequence.We have shown in both simulations and actual circuits that Op Amps can be used to perform arithmetic operations on one or more input voltages and deliver the results (almost) instantly.

BibliographyBBC, 2014. BBC - GCSE Bitesize: Highest common factor and lowest common multiple. [Online] Available at: http://www.bbc.co.uk/schools/gcsebitesize/maths/number/primefactorshirev1.shtml[Accessed 06 03 2014].Bird, J., 2010. 19.8 Op amp integrator. In: Electrical and Electronic Principles and Technology. Fourth ed. Oxford: Newnes, pp. 303-304.Middlesbrough College, 2013. The Operational Amplifier. Middlesbrough: Middlesbrough College.