Homework Assignment 12 - University of...

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

𝑇(𝑓) =10

1 + 𝑗 𝑓5 × 103

And the phase is 𝜃 = − tan−1 𝑓 5 × 103⁄

(a) At 500 Hz, 𝜃 = 5.7° and the delay is (5.7 360⁄ )(1 500⁄ ) = 32 𝜇s

(b) At 5 kHz Hz, 𝜃 = −45° and the delay is (45 360⁄ )(1 5K⁄ ) = 25 𝜇s

(c) At 10 kHz Hz, 𝜃 = −63.4° and the delay is (63.4 360⁄ )(1 10K⁄ ) = 17.6 𝜇s

(d) At 50 kHz Hz, 𝜃 = −84.3° and the delay is (84.3 360⁄ )(1 50K⁄ ) = 4.7𝜇s

(e) At 500 kHz Hz, 𝜃 = −89.4° and the delay is (89.4 360⁄ )(1 500K⁄ ) = 0.5 𝜇s

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Question 2 In the circuit, 𝑅 = 10K. What should 𝐶 be so that the circuit delays a 5-kHz signal by 5 𝜇s ? (6 points)

Solution

The period of a 5-kHz sine is 200 𝜇s so that a delay of 5 𝜇s corresponds to a phase of

360°5 × 10−6

200 × 10−6= 9°

The circuit is a low-pass filter with 3-dB frequency 𝑓𝐵 = 1 (2𝜋𝑅𝐶)⁄ . The phase is

𝜃 = − tan−1𝑓𝑓𝑏

= − tan−1(2𝜋𝑓𝑅𝐶)

Thus

9 = − tan−1(2𝜋𝑓𝑅𝐶)

𝑅 = 10K and 𝑓 = 5 kHz, and solving for 𝐶 yields 𝐶 = 504 pF. In an actual circuit one would use a standard value such a 𝐶 = 470 pF or 𝐶 = 560 pF.

Below is an annotated output from a SPICE simulation, where 𝑉𝐼 is a 1-V, 5-kHz sine and 𝑉𝑂 is the output.

5 𝜇𝑠

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Question 3 Consider an inverting amplifier with voltage gain 𝐴𝑉 = −100. The amplifier is driven by a sensor with an internal resistance 𝑅𝑆 = 10K. The amplifier’s input resistance is very large and may be ignored. The stray capacitance between the amplifier output and input terminals is 𝐶𝐹 = 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 “works across” 𝐶𝐹 is −100 so the Miller capacitance is 𝐶𝑀 = (1 + 100)𝐶𝐹 =1.01 nF. This capacitor sees 𝑅𝑆 and so the time constant is 𝑅𝑆𝐶𝑀 = 10.1 𝜇s, so that the 3-dB bandwidth is 𝐵 = 1 (2𝜋𝜏) =⁄ 15.67 kHz. The rise time is then 𝑡𝑟 ≈ 0.35 𝐵 = 22.2 𝜇s⁄ .

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Question 4

𝑅1 = 200 kΩ 𝑅2 = 220 kΩ 𝑅𝐶 = 2.2 kΩ 𝑅𝐸 = 1 kΩ 𝑟𝑠 = 100 kΩ 𝑉𝐶𝐶 = 5 V 𝑅𝐿 = 4.7 kΩ 𝛽𝑜 = 100 𝐶𝜇 = 2 pF 𝐶𝜋 = 10 pF 𝑉𝐴 = ∞ 𝑉𝐵𝐸(𝑂𝑁) = 0.7 V

The coupling capacitors and bypass capacitors are large and may be treated as shorts.

(a) Show that 𝐼𝐵 = 9.3 𝜇A. Note that you cannot assume IB = 0 (4 points) (b) Determine the numerical values for 𝑔𝑚 and 𝑟𝜋 (4 points) (c) Draw a detailed small-signal model for the amplifier showing the numerical values of the

components. Be sure to include 𝐶𝜇 and 𝐶𝜋 (6 points) (d) Determine the 3-dB frequency for the amplifier (4 points)

Solution

Part (a) The Thevenin voltage and resistance for the 𝑉𝐶𝐶 and the bias resistances are 𝑉𝑇𝐻 =5[𝑅1 (𝑅2 + 𝑅2)⁄ ] = 2.62 V, and 𝑅2‖𝑅1 = 105K respectively. Then, using BJT scaling:

−𝑉𝑇𝐻 + 𝐼𝐵 + −0.7 + (𝛽 + 1)𝐼𝐵𝑅𝐸 = 0 ⇒ 𝐼𝐵 = 9.3 𝜇A

Part (b) 𝐼𝐶 = 𝛽𝐼𝐵 = 930 𝜇A, and 𝑔𝑚 = 40𝐼𝐶 = 37 mA V⁄ , and 𝑟𝜋 = 𝛽 𝑔𝑚 = 2.69K⁄

Part (c)

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Part (d) The Miller capacitance associated with 𝐶𝜇 is 𝐶𝑀 = [1 + 𝑔𝑚(𝑅𝐿‖𝑅𝐶)]𝐶𝜇 = 113 pF. This is in parallel with 𝐶𝜋, so 𝐶𝑒𝑞 = 123 pF. The time constant is 𝜏 = 𝐶𝑒𝑞(𝑟𝑠‖𝑅1‖𝑅2‖𝑟𝜋) =315 ns. This corresponds to a 3-dB frequency of f3dB = 1 (2π × 315 × 10−9) =⁄ 505 kHz.

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Question 5 Below is the small-signal model of a BJT amplifier. Determine the so-called Miller capacitance 𝐶𝑀,and draw an equivalent small-signal circuit that incorporates 𝐶𝑀. 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 𝐶𝑀 = (1 − 𝐴𝑉)(2 pF), where 𝐴𝑣 is the voltage gain that “works across” the 2 pF capacitor. That is, the gain from the base (B) to the collector (C). For this circuit, the gain is

𝐴𝑉 = −(0.037)(2.2K||4.7K) = −55.45

Thus 𝐶𝑀 = (56.45)(2 pF) = 113 pF.

Equivalent Circuit

Circuit Time Constant

𝐶𝑀 is parallel with the 10-pF capacitor so 𝐶𝑒𝑞 = 123 pF. The time constant is 𝜏 = 𝐶𝑒𝑞(220K‖200K‖100K‖2.7K) = 315 ns. This corresponds to a 3-dB frequency of f3dB = 1 (2π × 315 × 10−9) =⁄ 505 kHz.

The amplifier has a low-pass response.

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Question 6 Below is a small-signal model of a BJT amplifier. Determine the so-called Miller capacitance 𝐶𝑀, and draw an equivalent small-signal circuit that incorporates 𝐶𝑀. 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)

𝑅𝐿 = 2 K 𝑔𝑚 = 0.04 A V⁄ 𝑟𝜋 = 5 K 𝑅𝑆 = 5 K 𝐶𝜋 = 10 pF 𝐶𝜇 = 2 pF

Solution

The gain that “works across” 𝐶𝜇 is −𝑔𝑚𝑅𝐿 = −80. The Miller capacitance is 𝐶𝑀 =(1 − 𝐴𝑉)𝐶𝜇 = (81)(2) = 162 pF. A small-signal model that incorporates 𝐶𝑀 is shown below.

The circuit time constant is 𝜏 = 𝐶𝜋||𝐶𝜇(𝑟𝜋||𝑅𝑠) = (162 pF + 10 pF)(2.5K) = 430 ns.

The 3-dB frequency is 𝑓3dB = 1 (2𝜋𝜏) = 370 kHz⁄ .

The amplifier has a low-pass response.

The midband gain is

𝐴𝑣(mid) =𝑟𝜋

𝑅𝑠 + 𝑟𝜋(−𝑔𝑚𝑅𝐿) = −40

Note: A SPICE simulation gives the 3-dB frequency as 366 kHz

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Question 7 Consider the amplifier below, which amplifies the signal from a sensor with an internal resistance of 1K. Ignore BJT’s output resistance, and assume 𝐶1 = 𝐶2 = 𝐶3 → ∞.

𝛽 = 100 𝐼𝐶 = 0.245 mA

(a) Determine 𝑔𝑚, 𝑟𝜋 (4 points) (b) Using BJT scaling, determine 𝑅𝑖—see figure (4 points) (c) Using the ratio of the collector and emitter resistors, estimate the overall voltage gain

𝐴𝑣 = 𝑣𝑜 𝑣𝑠⁄ (4 points) (d) Calculate the voltage gain 𝐴𝑣 = 𝑣𝑜 𝑣𝑠⁄ , but do not use the approximation that involves the

ratio of the collector and emitter resistors, but rather incorporate the 𝛽 of the transistor (4 points)

Solution Part (a)

𝑔𝑚 = 40 𝐼𝐶 = 9.8 mS, 𝑟𝜋 =𝛽𝑔𝑚

= 10.2 K

Part (b)

𝑅𝑖 = 𝑅1|𝑅2|[𝑟𝜋 + (1 + 𝛽)𝑅𝐸] = 300K|160K|[10.2K + 101 × 3K] = 78.3K Part (c) The effective collector resistance is 𝑅𝐶′ = 22K||100K = 18K and the sensor’s internal resistance and 𝑅𝑖 form a voltage divider. Thus

𝐴𝑣 =𝑣𝑜𝑣𝑠≅ −

𝑅𝑖𝑅𝑆 + 𝑅𝑖

𝑅𝐶′

𝑅𝐸= −

78.31 + 78.3

18K3K

= −5.93

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Part (d) We derived the following expression (also see Neaman) for the voltage gain

𝐴𝑣 =𝑣𝑜𝑣𝑠

= −𝛽𝑅𝐶

𝑟𝜋 + (1 + 𝛽)𝑅𝐸

Taking into account the voltage division at the input and the fact that 𝑅𝐿 and 𝑅𝐶 are in parallel in the circuit above, the equations becomes:

𝐴𝑣 =𝑣𝑜𝑣𝑠

= −𝑅𝑖

𝑅𝑆 + 𝑅𝑖

𝛽𝑅𝐶′

𝑟𝜋 + (1 + 𝛽)𝑅𝐸 = −

78.31 + 78.3

(100)(18K)

10.2K + (101)(3K)= −5.7

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Question 8 Consider the CE BJT amplifier below.

𝐼𝐶 = 1 mA 𝛽 = 185 𝐶𝜋 = 100 pF 𝐶𝜇 = 14 pF 𝑉𝐴 → ∞ 𝐶𝐶1 → ∞ 𝐶𝐶2 → ∞ 𝐶𝐶3 → ∞

(a) Draw a hybrid-𝜋 small signal model of the amplifier. Be sure to include 𝐶𝜋, 𝐶𝜇, and 𝑔𝑚. (8 points)

(b) Show that 𝑟𝜋 = 4.5 kΩ. (2 points) (c) Estimate the upper 3 dB bandwidth. (12 points)

Solution

(b) 𝑟𝜋 = 𝛽 𝑔𝑚 = 𝛽 (40𝐼𝐶) = 4.5 kΩ⁄⁄

(c) The Miller effect transforms 𝐶𝜇 to a value 𝐶𝑀 = (1 − 𝐴𝑀)𝐶𝜇 where 𝐴𝑀 = −𝑔𝑚(𝑅𝐿‖𝑅𝐶), the gain “working across” 𝐶𝜇. That is

𝐶𝑀 = (1 + 136)(14 pF) = 1.92 nF

The Miller capacitance is in parallel with 𝐶𝜋. A small signal model is

The time constant is 𝜏 = (𝐶𝜋 + 𝐶𝑀)(𝑅1‖𝑅2‖𝑟𝜋‖𝑅𝑠) = (2.02 × 10−9)(795) = 1.61 𝜇𝑠. The upper 3 dB frequency is then

𝑓𝐻 =1

2𝜋𝜏= 99.1 kHz

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Question 9 An amplifier is designed to provide a 12 V peak-to-peak swing across a 4 Ω load. Assume sinusoidal signals.

(a) Assuming the amplifier has output resistance 𝑅𝑜 ≈ 0 Ω, how much power will the load dissipate? (3 points)

(b) Assuming the amplifier has output resistance 𝑅𝑜 = 0.4 Ω, how much power will the load dissipate? (3 points)

Solution

Part (a)

𝑃 =𝑉𝑟𝑚𝑠2

𝑅𝐿

A 12 V peak-to-peak sinusoidal signal has an amplitude of 6 V, and an rms value of 6 √2⁄ =4.24 V. Thus

𝑃 =(4.25)2

4= 4.5 W

Part (b)

With 𝑅𝑜 = 0.4 Ω, the signal amplitude across the load is

𝑉𝐿 =4

4 + 0.46 = 5.46 V

The rms value is 5.46 √2 = 3.86 V⁄ , and the power is

𝑃 =(3.86)2

4= 3.72 W

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Question 10 An amplifier has an input resistance 𝑅𝑖 =1K, and has a voltage gain of 𝐴𝑣 = 100 when driven from a signal with internal resistance 𝑅𝑠 ≈ 0. The amplifier is used to amplify a 𝑣𝑠 = 1 mV signal from a sensor that has an internal resistance of 𝑅𝑠 ≈ 20K. What is the output amplitude? (5 points)

Solution

The sensor’s internal resistance and the amplifier’s input resistance form a voltage divider so that the effective input voltage is

𝑣𝑖 =1

1 + 201 mV = 47.6 𝜇V

The output voltage is 𝑣𝑜 = 𝐴𝑣𝑣𝑖 = 100 × 47.6 × 10−6 = 4.76 mV

Question 11 The parameters for the transistor below are 𝐾𝑛 = 0.5 mA/V2, 𝑉𝑇𝑁 = 1.2 V, and 𝜆 = 0. Determine 𝑣𝐷𝑆 and 𝑣𝐺𝑆 for 𝐼𝑄 = 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

𝐼𝐷 = 𝐾𝑛[𝑣𝐺𝑆 − 𝑉𝑇𝑁]2 1 × 10−3 = 0.5 × 10−3[𝑣𝐺𝑆 − 1.2]2

𝑣𝐺𝑆 = 2.61 V or 𝑣𝐺𝑆 = −0.214 V

The correct solution is 𝑣𝐺𝑆 = 2.61 V, because the other solution has 𝑣𝐺𝑆 < 𝑉𝑇𝑁 which implies the FET is off. The gate is 2.61 V higher than the source, which is at 0 V. Thus, 𝑣𝐺𝑆 = 2.61 V, which is also 𝑣𝐷𝑆. Thus

𝑣𝐷𝑆 = 2.61 V

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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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Question 13 (N 7.4)

For (a) show that the transfer function is

𝑇(𝑠) =𝑣𝑜(𝑠)𝑣𝑖(𝑠)

=𝑅2

𝑅2 + 𝑅1(1 + 𝑠𝑅1𝐶1)

1 + 𝑠(𝑅1‖𝑅2)𝐶1 (𝟔 𝐩𝐨𝐢𝐧𝐭𝐬)

Determine the circuit’s 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 𝑅1 and 𝐶1 is 𝑅1 (1 + 𝑅1𝑠𝐶1)⁄ and using voltage division

𝑇(𝑠) =𝑅2

𝑅2 + 𝑅1 (1 + 𝑅1𝑠𝐶1)⁄ =𝑅2(1 + 𝑅1𝑠𝐶1)

𝑅1 + 𝑅2(1 + 𝑅1𝑠𝐶1) =𝑅2

𝑅2 + 𝑅1(1 + 𝑠𝑅1𝐶1)

1 + 𝑠(𝑅1‖𝑅2)𝐶1

The transfer function has a zero at 𝜔𝑧 = 1 (𝑅1𝐶1) ≡ 𝑓𝑧 = 1.59 Hz⁄ . The transfer function has a pole at

𝜔𝑝 =1𝐶1𝑅1 + 𝑅2𝑅1𝑅2

≡ 𝑓𝑝 = 2.39 Hz

At low frequencies (𝑠 → 0), |𝑇(𝑠)| → 𝑅2 (𝑅1 + 𝑅2) = 2 3⁄ ≡ −3.5 dB⁄ . At high frequencies (𝑠 → ∞), |𝑇(𝑠)| → 1 ≡ 0 dB.

The two time constants are 𝜏1 = 1 2𝜋𝑓𝑧 = 0.1 s⁄ and 𝜏2 = 1 2𝜋𝑓𝑝 = 66.6 ms⁄ .

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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 𝑅2 and 𝐶2 is 𝑅2 (1 + 𝑅2𝑠𝐶2)⁄ and using voltage division

𝑇(𝑠) =𝑅2 (1 + 𝑅2𝑠𝐶2)⁄

𝑅2 + 𝑅1=

𝑅2𝑅1 + 𝑅2

11 + 𝑠(𝑅1‖𝑅2)𝐶2

At low frequencies (𝑠 → 0), |𝑇(𝑠)| → 𝑅2 (𝑅1 + 𝑅2) = 2 3⁄ ≡ −3.5 dB⁄ . The transfer function has a pole at

𝜔𝑝 =1𝐶1𝑅1 + 𝑅2𝑅1𝑅2

≡ 𝑓𝑝 = 2.39 Hz

A Bode plot is shown below.

The time constant is 𝜏 = 1 2𝜋𝑓𝑝 = 66.6 ms⁄ .

𝑓 (Hz) 𝑓𝑝 = 2.39 𝑓𝑧 = 1.59

0

−3.5

|𝑇(𝑠)| (dB)

𝑓 (Hz) 𝑓𝑝 = 2.39

−3.50

|𝑇(𝑠)| (dB)

Slope = −20 dB dec⁄

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Question 14

𝛽(𝑝𝑛𝑝) = 10, 𝛽𝑛𝑝𝑛 = 50

For the circuit above, make reasonable assumptions and then (a) Show that 𝑟𝜋1~ 156 kΩ (4 points) (b) Use BJT impedance scaling and give a reasonable estimate for the output resistance 𝑅𝑂

(4 points)

Solution

The three transistors are in a Darlington configuration and we can model it as a pnp transistor with 𝛽𝑐 = 𝛽1𝛽2𝛽3 = 25,000, and most of IQ flows through Q3. Thus, 𝐼𝐶3~4 mA.

Part (a). To determine 𝑟𝜋1, we need to find IC1 and then use 𝑟𝜋1 = 𝛽1/𝑔𝑚1

𝐼𝐶1 ≈𝐼𝐸2𝛽𝑛𝑝𝑛

=𝐼𝐶3

𝛽𝑛𝑝𝑛2

𝑔𝑚1 ≈ 40𝐼𝐶1 =40𝐼𝐶3𝛽𝑛𝑝𝑛

2

𝑟𝜋1 = 𝛽1/𝑔𝑚1 ≈𝛽𝑝𝑛𝑝 𝛽𝑛𝑝𝑛

2

40𝐼𝐶3= 156 kΩ

Alternatively, the transistors is a composite transistor, and 𝑟𝜋1 = 𝑟𝜋𝑐 = 𝛽𝑐/𝑔𝑚𝑐 ≈ 𝛽𝑐/40𝐼𝐶3

Part (b). Use BJT impedance scaling:

𝑅𝑜 =Resistance in base network

1 + 𝛽≈

156K + 25K

𝛽𝑝𝑛𝑝 𝛽𝑛𝑝𝑛2 = 7.24 Ω

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Question 15 (Final Exam, 2006) In the following circuit, the three transistors are matched and in the same thermal environment. Determine the values for 𝑅𝑅 and 𝑅𝑀 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 2𝑉𝐵𝐸 and we ignore base currents, so the voltage drop across the base-emitter of the output transistor is 𝑉𝐵𝐸, and the voltage drop across 𝑅𝑀 is 𝑉𝐵𝐸. Assume VBE = 0.7 V, so that

𝑅𝑀 = 0.7 (0.4 × 10−3)⁄ = 1.75 kΩ

and

2𝑉𝐵𝐸 = 8 − 𝐼𝐶𝑅𝑅 𝑅𝑅 = 16.5 kΩ

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Question 16 Consider the MOSFET amplifier below. Draw the small-signal model, incorporating 𝑟𝑜. (5 points) Determine the voltage gain 𝑣𝑜 𝑣𝑖⁄ (8 points) and the output resistance 𝑅𝑂 (8 points).

𝐾𝑛 = 1 mA V⁄ 𝑉𝑇𝑁 = 1.2 V 𝜆 = 0.01 V−1 𝐼𝐷𝑄 = 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, 𝑔𝑚 = 2𝐾𝑛𝐼𝐷𝑄 = 2 mA/V, and 𝑟𝑜 = 1 (𝜆𝐼𝐷𝑄)⁄ = 100K. From (b)

𝑣𝑜 = 𝑔𝑚𝑣𝑔𝑠(𝑟𝑜||𝑅𝐿) 𝑣𝑔𝑠 = 𝑣𝑖 − 𝑣𝑜

⇒ 𝑣𝑜 = 𝑔𝑚(𝑣𝑖 − 𝑣𝑜)(𝑟𝑜||𝑅𝐿) Rewriting yields

𝐴𝑉 =𝑔𝑚(𝑟𝑜||𝑅𝐿)

1 + 𝑔𝑚(𝑟𝑜||𝑅𝐿)=

(2 × 10−3)(100K||4K)1 + (2 × 10−3)(100K||4K)

= 0.885

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To determine the output resistance, turn of independent sources, add a test voltage 𝑉𝑋, and determine the current 𝐼𝑋 that flows. The resulting small-signal model is shown below. Note that 𝑅𝐿 is absent from the diagram, to be consistent with the definition of 𝑅𝑜 in the problem statement.

KCL at the source gives

𝑔𝑚𝑣𝑔𝑠 + 𝐼𝑋 =𝑉𝑋𝑟𝑜

From the diagram follows that 𝑣𝑔𝑠 = −𝑉𝑋, so that

−𝑔𝑚𝑉𝑋 + 𝐼𝑋 =𝑉𝑋𝑟𝑜

𝐼𝑋 = 𝑉𝑋 1𝑟𝑜

+ 𝑔𝑚

Then

𝑅𝑜 =𝑉𝑋𝐼𝑋

= 1 1𝑟𝑜

+ 𝑔𝑚 = 0.498K

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Question 17 For the amplifier below, 𝑅𝐿 = 500 Ω. Determine, 𝑅𝑖𝑏 ,𝑅𝑜 and the small-signal voltage gain 𝐴𝑣 = 𝑣𝑜 𝑣𝑖⁄ . (15 points)

𝑅𝑆 = 10K 𝑉+ = 3 V 𝑉− = −3 V 𝐼𝑄 = 2 mA 𝛽 = 100 𝑉𝐴 = 100 V CC → ∞

Solution

𝐼𝐶 =𝛽

1 + 𝛽𝐼𝐸 = 1.98 mA

𝑔𝑚 ≅ 40𝐼𝐶 = 79.2 mA V⁄ , 𝑟𝜋 = 𝛽 𝑔𝑚⁄ = 1.26K, 𝑟𝑜 =𝑉𝐴𝐼𝑐

= 50.5K

Use BJT impedance scaling to determine output resistance:

𝑅𝑜 =𝑅𝑆 + 𝑟𝜋1 + 𝛽

=10𝐾 + 1.26𝐾

101= 112 Ω

Use BJT impedance scaling to determine input resistance looking into the base

𝑅𝑖𝑏 = 𝑟𝜋 + (1 + 𝛽)(𝑟𝑜||𝑅𝐿) = 51.26K

The amplifier is an emitter follower with gain from base to emitter of ≅ 1. The overall gain is therefore

𝐴𝑣 ≅ (1)𝑅𝑖𝑏

𝑅𝑠 + 𝑅𝑖𝑏= (1)

51.26K10K + 51.26K

= 0.837

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Question 18 (N EX 7.14)

𝛽 = 125, 𝐶𝜇 = 3 pF, 𝐶𝜋 = 24 pF, 𝑉𝐴 = 200, 𝑉𝐵𝐸(𝑂𝑁) = 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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Part (b)

Part (c)

Part (d)

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Homework Assignment 12 - University of...

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