homework assignment 12 - university of...
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Homework Assignment 12
Question 1 Shown the is Bode plot of the magnitude of the gain transfer function of a constant GBP amplifier. By how much will the amplifier delay a sine wave with the following frequencies? (10 points) (a) 500 Hz (b) 5 kHz (c) 10 kHz (d) 50 kHz (e) 500 kHz
Solution The transfer function is
π(π) =10
1 + π π5 Γ 103
And the phase is π = β tanβ1 π 5 Γ 103β
(a) At 500 Hz, π = 5.7Β° and the delay is (5.7 360β )(1 500β ) = 32 πs
(b) At 5 kHz Hz, π = β45Β° and the delay is (45 360β )(1 5Kβ ) = 25 πs
(c) At 10 kHz Hz, π = β63.4Β° and the delay is (63.4 360β )(1 10Kβ ) = 17.6 πs
(d) At 50 kHz Hz, π = β84.3Β° and the delay is (84.3 360β )(1 50Kβ ) = 4.7πs
(e) At 500 kHz Hz, π = β89.4Β° and the delay is (89.4 360β )(1 500Kβ ) = 0.5 πs
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 2 In the circuit, π = 10K. What should πΆ be so that the circuit delays a 5-kHz signal by 5 πs ? (6 points)
Solution
The period of a 5-kHz sine is 200 πs so that a delay of 5 πs corresponds to a phase of
360Β°5 Γ 10β6
200 Γ 10β6= 9Β°
The circuit is a low-pass filter with 3-dB frequency ππ΅ = 1 (2ππ πΆ)β . The phase is
π = β tanβ1πππ
= β tanβ1(2πππ πΆ)
Thus
9 = β tanβ1(2πππ πΆ)
π = 10K and π = 5 kHz, and solving for πΆ yields πΆ = 504 pF. In an actual circuit one would use a standard value such a πΆ = 470 pF or πΆ = 560 pF.
Below is an annotated output from a SPICE simulation, where ππΌ is a 1-V, 5-kHz sine and ππ is the output.
5 ππ
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 3 Consider an inverting amplifier with voltage gain π΄π = β100. The amplifier is driven by a sensor with an internal resistance π π = 10K. The amplifierβs input resistance is very large and may be ignored. The stray capacitance between the amplifier output and input terminals is πΆπΉ = 10 pF. A voltage step is applied to the input. What is the rise time of the output voltage? (8 points) Hint: use Miller capacitance concepts to determine an equivalent Miller capacitance and determine the amplifier bandwidth.
Solution
The gain that βworks acrossβ πΆπΉ is β100 so the Miller capacitance is πΆπ = (1 + 100)πΆπΉ =1.01 nF. This capacitor sees π π and so the time constant is π ππΆπ = 10.1 πs, so that the 3-dB bandwidth is π΅ = 1 (2ππ) =β 15.67 kHz. The rise time is then π‘π β 0.35 π΅ = 22.2 πsβ .
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 4
π 1 = 200 kΞ© π 2 = 220 kΞ© π πΆ = 2.2 kΞ© π πΈ = 1 kΞ© ππ = 100 kΞ© ππΆπΆ = 5 V π πΏ = 4.7 kΞ© π½π = 100 πΆπ = 2 pF πΆπ = 10 pF ππ΄ = β ππ΅πΈ(ππ) = 0.7 V
The coupling capacitors and bypass capacitors are large and may be treated as shorts.
(a) Show that πΌπ΅ = 9.3 πA. Note that you cannot assume IB = 0 (4 points) (b) Determine the numerical values for ππ and ππ (4 points) (c) Draw a detailed small-signal model for the amplifier showing the numerical values of the
components. Be sure to include πΆπ and πΆπ (6 points) (d) Determine the 3-dB frequency for the amplifier (4 points)
Solution
Part (a) The Thevenin voltage and resistance for the ππΆπΆ and the bias resistances are πππ» =5[π 1 (π 2 + π 2)β ] = 2.62 V, and π 2βπ 1 = 105K respectively. Then, using BJT scaling:
βπππ» + πΌπ΅ + β0.7 + (π½ + 1)πΌπ΅π πΈ = 0 β πΌπ΅ = 9.3 πA
Part (b) πΌπΆ = π½πΌπ΅ = 930 πA, and ππ = 40πΌπΆ = 37 mA Vβ , and ππ = π½ ππ = 2.69Kβ
Part (c)
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Part (d) The Miller capacitance associated with πΆπ is πΆπ = [1 + ππ(π πΏβπ πΆ)]πΆπ = 113 pF. This is in parallel with πΆπ, so πΆππ = 123 pF. The time constant is π = πΆππ(ππ βπ 1βπ 2βππ) =315 ns. This corresponds to a 3-dB frequency of f3dB = 1 (2Ο Γ 315 Γ 10β9) =β 505 kHz.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 5 Below is the small-signal model of a BJT amplifier. Determine the so-called Miller capacitance πΆπ,and draw an equivalent small-signal circuit that incorporates πΆπ. Next, determine the circuit time constant and the 3-dB frequency. Finally, does this amplifier have a high-pass or low-pass response? (15 points)
Solution
Miller Capacitance
The Miller capacitance is πΆπ = (1 β π΄π)(2 pF), where π΄π£ is the voltage gain that βworks acrossβ the 2 pF capacitor. That is, the gain from the base (B) to the collector (C). For this circuit, the gain is
π΄π = β(0.037)(2.2K||4.7K) = β55.45
Thus πΆπ = (56.45)(2 pF) = 113 pF.
Equivalent Circuit
Circuit Time Constant
πΆπ is parallel with the 10-pF capacitor so πΆππ = 123 pF. The time constant is π = πΆππ(220Kβ200Kβ100Kβ2.7K) = 315 ns. This corresponds to a 3-dB frequency of f3dB = 1 (2Ο Γ 315 Γ 10β9) =β 505 kHz.
The amplifier has a low-pass response.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 6 Below is a small-signal model of a BJT amplifier. Determine the so-called Miller capacitance πΆπ, and draw an equivalent small-signal circuit that incorporates πΆπ. Next, determine the circuit time constant, 3-dB frequency, and the midband gain. Finally, does this amplifier have a high-pass or low-pass response? (15 points)
π πΏ = 2 K ππ = 0.04 A Vβ ππ = 5 K π π = 5 K πΆπ = 10 pF πΆπ = 2 pF
Solution
The gain that βworks acrossβ πΆπ is βπππ πΏ = β80. The Miller capacitance is πΆπ =(1 β π΄π)πΆπ = (81)(2) = 162 pF. A small-signal model that incorporates πΆπ is shown below.
The circuit time constant is π = πΆπ||πΆπ(ππ||π π ) = (162 pF + 10 pF)(2.5K) = 430 ns.
The 3-dB frequency is π3dB = 1 (2ππ) = 370 kHzβ .
The amplifier has a low-pass response.
The midband gain is
π΄π£(mid) =ππ
π π + ππ(βπππ πΏ) = β40
Note: A SPICE simulation gives the 3-dB frequency as 366 kHz
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 7 Consider the amplifier below, which amplifies the signal from a sensor with an internal resistance of 1K. Ignore BJTβs output resistance, and assume πΆ1 = πΆ2 = πΆ3 β β.
π½ = 100 πΌπΆ = 0.245 mA
(a) Determine ππ, ππ (4 points) (b) Using BJT scaling, determine π πβsee figure (4 points) (c) Using the ratio of the collector and emitter resistors, estimate the overall voltage gain
π΄π£ = π£π π£π β (4 points) (d) Calculate the voltage gain π΄π£ = π£π π£π β , but do not use the approximation that involves the
ratio of the collector and emitter resistors, but rather incorporate the π½ of the transistor (4 points)
Solution Part (a)
ππ = 40 πΌπΆ = 9.8 mS, ππ =π½ππ
= 10.2 K
Part (b)
π π = π 1|π 2|[ππ + (1 + π½)π πΈ] = 300K|160K|[10.2K + 101 Γ 3K] = 78.3K Part (c) The effective collector resistance is π πΆβ² = 22K||100K = 18K and the sensorβs internal resistance and π π form a voltage divider. Thus
π΄π£ =π£ππ£π β β
π ππ π + π π
π πΆβ²
π πΈ= β
78.31 + 78.3
18K3K
= β5.93
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Part (d) We derived the following expression (also see Neaman) for the voltage gain
π΄π£ =π£ππ£π
= βπ½π πΆ
ππ + (1 + π½)π πΈ
Taking into account the voltage division at the input and the fact that π πΏ and π πΆ are in parallel in the circuit above, the equations becomes:
π΄π£ =π£ππ£π
= βπ π
π π + π π
π½π πΆβ²
ππ + (1 + π½)π πΈ = β
78.31 + 78.3
(100)(18K)
10.2K + (101)(3K)= β5.7
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 8 Consider the CE BJT amplifier below.
πΌπΆ = 1 mA π½ = 185 πΆπ = 100 pF πΆπ = 14 pF ππ΄ β β πΆπΆ1 β β πΆπΆ2 β β πΆπΆ3 β β
(a) Draw a hybrid-π small signal model of the amplifier. Be sure to include πΆπ, πΆπ, and ππ. (8 points)
(b) Show that ππ = 4.5 kΞ©. (2 points) (c) Estimate the upper 3 dB bandwidth. (12 points)
Solution
(b) ππ = π½ ππ = π½ (40πΌπΆ) = 4.5 kΞ©ββ
(c) The Miller effect transforms πΆπ to a value πΆπ = (1 β π΄π)πΆπ where π΄π = βππ(π πΏβπ πΆ), the gain βworking acrossβ πΆπ. That is
πΆπ = (1 + 136)(14 pF) = 1.92 nF
The Miller capacitance is in parallel with πΆπ. A small signal model is
The time constant is π = (πΆπ + πΆπ)(π 1βπ 2βππβπ π ) = (2.02 Γ 10β9)(795) = 1.61 ππ . The upper 3 dB frequency is then
ππ» =1
2ππ= 99.1 kHz
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 9 An amplifier is designed to provide a 12 V peak-to-peak swing across a 4 Ξ© load. Assume sinusoidal signals.
(a) Assuming the amplifier has output resistance π π β 0 Ξ©, how much power will the load dissipate? (3 points)
(b) Assuming the amplifier has output resistance π π = 0.4 Ξ©, how much power will the load dissipate? (3 points)
Solution
Part (a)
π =ππππ 2
π πΏ
A 12 V peak-to-peak sinusoidal signal has an amplitude of 6 V, and an rms value of 6 β2β =4.24 V. Thus
π =(4.25)2
4= 4.5 W
Part (b)
With π π = 0.4 Ξ©, the signal amplitude across the load is
ππΏ =4
4 + 0.46 = 5.46 V
The rms value is 5.46 β2 = 3.86 Vβ , and the power is
π =(3.86)2
4= 3.72 W
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 10 An amplifier has an input resistance π π =1K, and has a voltage gain of π΄π£ = 100 when driven from a signal with internal resistance π π β 0. The amplifier is used to amplify a π£π = 1 mV signal from a sensor that has an internal resistance of π π β 20K. What is the output amplitude? (5 points)
Solution
The sensorβs internal resistance and the amplifierβs input resistance form a voltage divider so that the effective input voltage is
π£π =1
1 + 201 mV = 47.6 πV
The output voltage is π£π = π΄π£π£π = 100 Γ 47.6 Γ 10β6 = 4.76 mV
Question 11 The parameters for the transistor below are πΎπ = 0.5 mA/V2, πππ = 1.2 V, and π = 0. Determine π£π·π and π£πΊπ for πΌπ = 1 mA. (6 points)
Solution The transistor is operating the saturation region (this is always true when the gate is strapped to the drain). Thus
πΌπ· = πΎπ[π£πΊπ β πππ]2 1 Γ 10β3 = 0.5 Γ 10β3[π£πΊπ β 1.2]2
π£πΊπ = 2.61 V or π£πΊπ = β0.214 V
The correct solution is π£πΊπ = 2.61 V, because the other solution has π£πΊπ < πππ which implies the FET is off. The gate is 2.61 V higher than the source, which is at 0 V. Thus, π£πΊπ = 2.61 V, which is also π£π·π. Thus
π£π·π = 2.61 V
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 12 Consider the following circuit, which is a simplifier schematic of an IC audio amplifier. Indicate, by circling and labeling as many of the following sub-circuits you can find: composite pnp transistor, current mirror, class AB output, Darlington pairs. (10 points)
Solution
Class AB stage
Composite pnp Current mirror
Darlington pair
Darlington pair
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 13 (N 7.4)
For (a) show that the transfer function is
π(π ) =π£π(π )π£π(π )
=π 2
π 2 + π 1(1 + π π 1πΆ1)
1 + π (π 1βπ 2)πΆ1 (π π©π¨π’π§ππ¬)
Determine the circuitβs two time constants (2 points). Sketch the Bode magnitude plot. (5 points)
For (b), determine the circuit time constant and then sketch the Bode magnitude plot. (5 points)
Solution
Part (a) The impedance of the parallel combination of π 1 and πΆ1 is π 1 (1 + π 1π πΆ1)β and using voltage division
π(π ) =π 2
π 2 + π 1 (1 + π 1π πΆ1)β =π 2(1 + π 1π πΆ1)
π 1 + π 2(1 + π 1π πΆ1) =π 2
π 2 + π 1(1 + π π 1πΆ1)
1 + π (π 1βπ 2)πΆ1
The transfer function has a zero at ππ§ = 1 (π 1πΆ1) β‘ ππ§ = 1.59 Hzβ . The transfer function has a pole at
ππ =1πΆ1π 1 + π 2π 1π 2
β‘ ππ = 2.39 Hz
At low frequencies (π β 0), |π(π )| β π 2 (π 1 + π 2) = 2 3β β‘ β3.5 dBβ . At high frequencies (π β β), |π(π )| β 1 β‘ 0 dB.
The two time constants are π1 = 1 2πππ§ = 0.1 sβ and π2 = 1 2πππ = 66.6 msβ .
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
A Bode plot is shown below.
Below is a Matlab script for plotting the Bode plot.
R1 = 10e3; R2 = 20e3; C1 = 10e-6; T = (R2/(R1+R2))*tf([R1*C1 1],[R1*R2*C1/(R1+R2) 1]) h = bodeplot(T); setoptions(h,'FreqUnits','Hz');
Part (b) The impedance of the parallel combination of π 2 and πΆ2 is π 2 (1 + π 2π πΆ2)β and using voltage division
π(π ) =π 2 (1 + π 2π πΆ2)β
π 2 + π 1=
π 2π 1 + π 2
11 + π (π 1βπ 2)πΆ2
At low frequencies (π β 0), |π(π )| β π 2 (π 1 + π 2) = 2 3β β‘ β3.5 dBβ . The transfer function has a pole at
ππ =1πΆ1π 1 + π 2π 1π 2
β‘ ππ = 2.39 Hz
A Bode plot is shown below.
The time constant is π = 1 2πππ = 66.6 msβ .
π (Hz) ππ = 2.39 ππ§ = 1.59
0
β3.5
|π(π )| (dB)
π (Hz) ππ = 2.39
β3.50
|π(π )| (dB)
Slope = β20 dB decβ
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 14
π½(πππ) = 10, π½πππ = 50
For the circuit above, make reasonable assumptions and then (a) Show that ππ1~ 156 kΞ© (4 points) (b) Use BJT impedance scaling and give a reasonable estimate for the output resistance π π
(4 points)
Solution
The three transistors are in a Darlington configuration and we can model it as a pnp transistor with π½π = π½1π½2π½3 = 25,000, and most of IQ flows through Q3. Thus, πΌπΆ3~4 mA.
Part (a). To determine ππ1, we need to find IC1 and then use ππ1 = π½1/ππ1
πΌπΆ1 βπΌπΈ2π½πππ
=πΌπΆ3
π½πππ2
ππ1 β 40πΌπΆ1 =40πΌπΆ3π½πππ
2
ππ1 = π½1/ππ1 βπ½πππ π½πππ
2
40πΌπΆ3= 156 kΞ©
Alternatively, the transistors is a composite transistor, and ππ1 = πππ = π½π/πππ β π½π/40πΌπΆ3
Part (b). Use BJT impedance scaling:
π π =Resistance in base network
1 + π½β
156K + 25K
π½πππ π½πππ2 = 7.24 Ξ©
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 15 (Final Exam, 2006) In the following circuit, the three transistors are matched and in the same thermal environment. Determine the values for π π and π π to produce an output current of 0.4 mA. You may ignore base currents and make reasonable assumptions about VBE. (5 points)
Solution
The voltage across the diode-strapped transistors is 2ππ΅πΈ and we ignore base currents, so the voltage drop across the base-emitter of the output transistor is ππ΅πΈ, and the voltage drop across π π is ππ΅πΈ. Assume VBE = 0.7 V, so that
π π = 0.7 (0.4 Γ 10β3)β = 1.75 kΞ©
and
2ππ΅πΈ = 8 β πΌπΆπ π π π = 16.5 kΞ©
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 16 Consider the MOSFET amplifier below. Draw the small-signal model, incorporating ππ. (5 points) Determine the voltage gain π£π π£πβ (8 points) and the output resistance π π (8 points).
πΎπ = 1 mA Vβ πππ = 1.2 V π = 0.01 Vβ1 πΌπ·π = 1 mA
Solution
The small-signal model for the amplifier is shown in (a) below. In (b) the model is redrawn to ease analysis.
(a) (b) In both models, ππ = 2πΎππΌπ·π = 2 mA/V, and ππ = 1 (ππΌπ·π)β = 100K. From (b)
π£π = πππ£ππ (ππ||π πΏ) π£ππ = π£π β π£π
β π£π = ππ(π£π β π£π)(ππ||π πΏ) Rewriting yields
π΄π =ππ(ππ||π πΏ)
1 + ππ(ππ||π πΏ)=
(2 Γ 10β3)(100K||4K)1 + (2 Γ 10β3)(100K||4K)
= 0.885
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
To determine the output resistance, turn of independent sources, add a test voltage ππ, and determine the current πΌπ that flows. The resulting small-signal model is shown below. Note that π πΏ is absent from the diagram, to be consistent with the definition of π π in the problem statement.
KCL at the source gives
πππ£ππ + πΌπ =ππππ
From the diagram follows that π£ππ = βππ, so that
βππππ + πΌπ =ππππ
πΌπ = ππ 1ππ
+ ππ
Then
π π =πππΌπ
= 1 1ππ
+ ππ = 0.498K
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 17 For the amplifier below, π πΏ = 500 Ξ©. Determine, π ππ ,π π and the small-signal voltage gain π΄π£ = π£π π£πβ . (15 points)
π π = 10K π+ = 3 V πβ = β3 V πΌπ = 2 mA π½ = 100 ππ΄ = 100 V CC β β
Solution
πΌπΆ =π½
1 + π½πΌπΈ = 1.98 mA
ππ β 40πΌπΆ = 79.2 mA Vβ , ππ = π½ ππβ = 1.26K, ππ =ππ΄πΌπ
= 50.5K
Use BJT impedance scaling to determine output resistance:
π π =π π + ππ1 + π½
=10πΎ + 1.26πΎ
101= 112 Ξ©
Use BJT impedance scaling to determine input resistance looking into the base
π ππ = ππ + (1 + π½)(ππ||π πΏ) = 51.26K
The amplifier is an emitter follower with gain from base to emitter of β 1. The overall gain is therefore
π΄π£ β (1)π ππ
π π + π ππ= (1)
51.26K10K + 51.26K
= 0.837
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 18 (N EX 7.14)
π½ = 125, πΆπ = 3 pF, πΆπ = 24 pF, ππ΄ = 200, ππ΅πΈ(ππ) = 0.7 V
A dc analysis shows that ICQ = 0.84 mA.
(a) Draw a detailed small-signal model of the amplifier showing the numerical values of the components. (6 points)
(b) Calculate the Miller capacitance. (3 points) (c) Determine the upper 3-dB frequency. (3 points) (d) Determine the small-signal mid-band voltage gain. (4 points)
Part (a)
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