Homework Assignment 08 - University of Iowas-iihr64.iihr.uiowa.edu/.../Homework/HomeworkAssignment08_Solutio… · Homework Assignment 08 . ... Consider the Bode plot of a 1st order

55:041 Electronic Circuits. The University of Iowa. Fall 2013.

Homework Assignment 08

Question 1 (Short Takes) Two points each unless otherwise indicated.

1. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance large voltage gain

2. In the circuit below 𝐼𝐶 = 1 mA and all the capacitors are large enough to be considered shorts. Estimate the midband gain 𝐴𝑣 = 𝑣𝑜 𝑣𝑖⁄ (3 points)

(a) ≈ −6.8 (b) ≈ −3.4 (c) ≈ −272 (d) ≈ −136 (e) ≈ −12.1

Answer: (d)

3. In the circuit below 𝐼𝐶 = 1 mA and all the capacitors are large enough to be considered shorts. Estimate the midband gain 𝐴𝑣 = 𝑣𝑜 𝑣𝑖⁄ (3 points)

(a) ≈ −12.1 (b) ≈ −8 (c) ≈= −272 (d) ≈ −136 (e) ≈ −6.1

Answer: 𝐴𝑣 ≈ −(𝑅𝐿||𝑅𝐶) 560 = − 3.4 0.56 = −6.07⁄⁄ , so (e).

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4. In the circuit below 𝐼𝐶 = 1 mA and all the capacitors are large enough to be considered shorts. Estimate the midband gain 𝐴𝑣 = 𝑣𝑜 𝑣𝑖⁄ .

(a) ≈ −6.8 (b) ≈ −3.4 (c) −𝑔𝑚𝑅𝐶 − 0.04𝑅𝐶 = −272 (d) Need additional information

Answer: (b)

5. Sketch an npn BJT Darlington pair.

6. Sketch a two-transistor configuration using npn and pnp BJTs that is equivalent to a single pnp BJT.

7. True or false: the β of a transistor decreases with decreasing temperature.

Answer: True, it decreases with decreasing temperature.

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8. True or false: the 𝛽 of a transistor is a function of temperature, but essentially independent of collector current.

Answer: False

9. True or false: consider a BJT in the CE configuration, biased at IC = 1 mA. The small-signal input resistance πr is in the order of 500K

Answer: False. 𝑟𝜋 = 𝛽 𝑔𝑚⁄ = 𝛽 (40𝐼𝐶)⁄ . With 𝛽 = 100, 𝑟𝜋 = 2.5K

10. True or false: consider a BJT in the CE configuration, biased at IC = 1 mA. The small-signal input resistance πr is in the order of 500K .


11. Give one phrase/sentence that describes the primary advantage of an active load.

Answer: Large effective resistance large voltage gain

12. Estimate the voltage gain of the amplifier below if 𝐼𝐶𝑄 = 6.3 mA, 𝛽 = 200, and 𝐶𝐶 → ∞.

(a) −10

(b) 10

(c) 252

(d) −252

Answer: Av ≅ −RC RE = −10⁄ , so (a).

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13. Estimate the input resistance 𝑅𝑖 of the amplifier below if 𝐼𝐶𝑄 = 6.3 mA, 𝛽 = 200, 𝑟𝜋 ≅ 800Ω, and 𝐶𝐶 → ∞. (3 point)

(a) 17.35K

(b) 18.15K

(c) 37.35K

(d) 9.46K

Answer: Using BJT scaling, 𝑅𝑖 = 22𝐾‖82K‖(𝑟𝜋 + (1 + 𝛽)𝑅𝐸 ) = 22K‖82K‖(𝑟𝜋 +(1 + 𝛽)𝑅𝐸 ) = 9.46K, so (d)

14. Consider the Bode plot of a 1st order RC network. What is the attenuation of the network at 𝑓 = 60 Hz? Provide your answer in dB. (3 points)

Answer: 60 Hz is log(60 2.5⁄ ) = 1.38 decades higher than the 2.5 Hz corner frequency. The attenuation increases by 20 dB per decade, so that at 60 Hz |𝑣𝑜 𝑣𝑖⁄ | (in dB) is −3.5 − 20 × 1.38 = −31.1 dB. The attenuation is 31.1 dB. An alternate calculation is 3.5 + 20 log ��1 + (60 2.5⁄ )2� = 3.11 dB.

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15. Assume the input voltage is a 1-V step function 𝑢(𝑡). What is the long-term value of 𝑣𝑜(𝑡)? That is, what is 𝑣𝑜(𝑡) for 𝑡 → ∞?

Answer: In the steady state (𝑡 → ∞), the capacitor has no effect on the circuit. The resistors form a voltage divider and 𝑣𝑜(∞) = 𝑣𝑠(𝑅𝐿 (𝑅𝐿 + 𝑅𝑆)⁄ ) = 0.5 V

16. Consider the current mirror below, and neglect base currents. What is 𝐼copy?

Answer:

𝐼copy =0.25 mA

3= 83 𝜇A

17. In the current mirrors below, neglect base currents and take 𝐼REF = 30 𝜇A, What is 𝐼copy3?

(a) 30 𝜇A (b) 30 𝜇A 3⁄ = 10 𝜇A (c) 30 𝜇A 4 = 7.5 𝜇A⁄

Answer: (a)

18. In the current mirrors below, neglect base currents and 𝐼REF = 10 𝜇A, What is 𝐼copy?

Answer: 30 𝜇A

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19. In the current mirrors below, neglect the base currents. What is 𝐼REF?

Answer: 0.25 mA

20. True or false: the β of a transistor decreases with decreasing temperature.

Answer: True, it decreases with decreasing temperature.

21. True or false: the 𝛽 of a transistor is a function of temperature, but essentially independent of collector current.

Answer: False

22. True or false: consider a BJT in the CE configuration, biased at IC = 1 mA. The small-

signal input resistance πr is in the order of 500K (2 points)


23. True or false: The input resistance of a BJT amplifier in the CB configuration, biased at 1 mA is about 25 Ω

Answer: True

24. A single-pole op-amp has an open-loop low-frequency gain of 𝐴 = 105 and an open loop, 3-dB frequency of 4 Hz. If an inverting amplifier with closed-loop low-frequency gain of �𝐴𝑓� = 50 uses this op-amp, determine the closed-loop bandwidth. (2 points) Answer: The gain-bandwidth product is 4 × 105 Hz. The bandwidth of the closed-loop amplifier is then is 4 × 105/50 = 8 kHz.

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25. A single-pole op-amp has an open-loop gain of 100 dB and a unity-gain bandwidth frequency of 2 MHz. What is the open-loop bandwidth of the op-amp? (2 points)

Answer. A gain of 100 dB corresponds to 105 and the gain-bandwidth product is 2 MHz. Thus, the open-loop bandwidth is (2 MHz) 105⁄ = 20 Hz

26. A single-pole op-amp has an open-loop gain of 100 dB and a unity-gain bandwidth frequency 5 MHz. What is the open-loop bandwidth of the amplifier? The amplifier is used as a voltage follower. What is the bandwidth of the follower?

Answer: A gain of 100 dB corresponds to 105 and the gain-bandwidth product is 5 MHz. Thus, the open-loop bandwidth is (5 MHz) 105⁄ = 50 Hz. A unity follower will have a bandwidth of 5 MHz.

27. Consider a first-order RC low-pass filter with 3-dB frequency 𝑓 = 60 Hz. By how much does it delay a 50 Hz sine wave? Express you answer in ms. (3 points) Answer: The phase shift at 60 Hz is 45° and increases at 45° / decade. 50 Hz is log(50 60) = −0.08⁄ decades higher than 60 Hz. (The negative sign implies 50 Hz is 0.08 decades before 60 Hz.) Thus, the phase shift is 45 − 0.08 × 45 = 41.4°. The period of a 50 Hz sine wave is 20 ms, so the delay is (20)(41.4 360⁄ ) = 2.3 ms An alternate and more accurate calculation for the phase is tan−1(50 60⁄ ) = 39.8° and delay of 2.2 ms.

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Question 2 Consider the circuit below. The duty cycle and frequency of the 555 astable is 60% and 10 kHz respectively.

(a) Specify a value for 𝑅𝑙𝑖𝑚𝑖𝑡 to ensure that the average current through the IR diode does not exceed 30 mA (4 points)

(b) Explain (2 sentences maximum) the purpose of the decoupling capacitor (1 point) (c) Give a reasonable value for the decoupling capacitor (1 point) (d) If the frequency-setting capacitor is 0.1 𝜇F, what should 𝑅𝐴 and 𝑅𝐵 be? Specify E24 series

standard resistors. (4 points) (e) What are the frequency and duty cycle with the standard resistors? (2 points)

Solution Part (a) The peak current must be 𝐼𝑝𝑒𝑎𝑘 = 30 (0.6) = 50 mA⁄ . This value will give an average of 30 mA with a 60% “on” time. Assuming the 𝑉𝐵𝐸(𝑂𝑁) = 0.7 V for the BJT, then 𝑅𝑙𝑖𝑚𝑖𝑡 = 0.7 0.05 = 14 Ω⁄ . Choose the closest standard value of 15 Ω. Part (b) When the FET switches, large current spikes may appear on the supply rail, which can propagate into the IC and disturb its operation. The decoupling capacitor provides a local reservoir of energy, and ensures a clean power supply rail. Part (c) A good first try would be 0.1 𝜇F.

Part (d) Calculated (using an online applet) values are 𝑅𝐴 = 288 Ω, 𝑅𝐵 = 576 Ω with closest standard values 300 Ω and 560 Ω respectively.

Part (e) With standard values, 𝑓 = 10.1 kHz, and 𝐷 = 60.6 %.

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Question 3 The datasheet for a 5 V, three-terminal regulator indicate that the output voltage typically changes by 3 mV when the input voltage is varied from 7 V to 25 V, and by 5 mV when the load current is varied from 0.25 A to 0.75 A. Further, the ripple rejection ratio is 78 dB at 120 Hz.

(a) Estimate the typical line- and load regulation for the regulator. (4 points) (b) What is the output resistance of the regulator? (2 points) (c) Estimate the output ripple amplitude for every volt of input ripple at 120 Hz. (3 points)

Solution

Part(a)

Line Regulation =Δ𝑉𝑂Δ𝑉𝐼

× 100% =3 × 10−3

(25 − 7)× 100% = 0.017%

Load Regulation =𝑉𝑂(𝑁𝐿) − 𝑉𝑂(𝐹𝐿)

𝑉𝑂(𝑁𝐿)× 100 =

5 × 10−3

5× 100 = 0.1%

Part (b)

𝑂𝑢𝑡𝑝𝑢𝑡 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =Δ𝑉𝑂Δ𝐼𝑂

=5 × 10−3

(0.75 − 0.25)= 10 mΩ

Part (c)

Ripple Rejection (dB) = 20 log10 �𝑉𝑅𝐼𝑉𝑅𝑂

� = 78

𝑉𝑅𝑂 = (0.126 × 10−3)𝑉𝑅𝐼

Thus, a 1-V, 120-Hz ripple at the input will result in an output ripple of 0.126 mV

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Question 4 An n-channel MOSFET with 𝑉𝑇𝑁 = 1 V, 𝐾𝑛 = 0.8 mA/V2 is biased to operate in its saturation region with 𝐼𝐷 = 1 mA. Determine the transconductance 𝑔𝑚. (3 points) If 𝑣𝐺𝑆 changes with 1 mV, by how does the drain current change? (3 points)

Solution

𝑔𝑚 = 2�𝐾𝑛𝐼𝐷

= 2�(0.8 × 10−3)(1 × 10−3) = 1.78 × 10−3 A/V = 1.78 mA/V

𝑖𝑑 = 𝑔𝑚𝑣𝑔𝑠 = (1.78 × 10−3)(1 × 10−3) = 1.78 𝜇A

Question 5 For the circuit shown, 𝑉𝑃𝑆 = 5 V and = 10K . The varactor characteristics are shown in the graph. What is the bandwidth of the circuit? (3 points)

Solution

From the graph, the varactor has a capacitance of 100 pF with a 5-V reverse voltage. The time-constant of the circuit is 𝜏 = 𝑅𝐶 = (10 × 103)(100 × 10−12) = 1 𝜇s. The bandwidth is then 𝐵 = 1 (2𝜋𝜏) = 159 kHz.⁄

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Question 6

𝐾𝑛 = 93 mA V2⁄ 𝑉𝑇𝑁 = 2 V λ = 0 𝐶𝑆 = 100 𝜇F 𝐶𝐶 = 1 𝜇F

(a) Find 𝐼𝐷𝑄 for the circuit (6 points) (b) Determine 𝑔𝑚 (2 points) (c) Determine the voltage gain of the circuit (2 points)

Solution

Part (a) The gate current is zero, so

𝑉𝐺 =56

100 + 5615 = 5.39 V

Further 𝑉𝑆 = 𝐼𝐷𝑅𝑆 = 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2

Thus

𝑉𝐺𝑆 = 𝑉𝐺 − 𝑉𝑆 = 5.39 − 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2(𝑅𝑆) 𝑉𝐺𝑆 = 5.39 − 0.093(𝑉𝐺𝑆 − 2)2(100)

Solving using trial-and-error gives 2.55 V. The drain current is 𝐼𝐷 = 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2 =(0.093)(2.55 − 2)2 = 28.13 mA

Part (b) 𝑔𝑚 = 2�𝐾𝑛𝐼𝐷 = 2√0.093 × 0.0284 = 0.102 A/V.

Part (c) The small-signal model is given below where 𝑅𝐺 = 100K||56K. From the model the voltage gain is 𝐴𝑉 = −𝑔𝑚(220||2.2K) = −20.5

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Question 7 A MOEFET amplifier along with the FET and circuit parameters are shown below. 𝐶𝐶1,𝐶𝐶2 are coupling capacitors. Determine the quiescent values for 𝐼𝐷 and 𝑉𝐷𝑆 (8 points)

𝐾𝑛 = 1 mA V⁄ 2 𝑉𝑇𝑁 = 2 V 𝜆 = 0 𝑉𝐷𝐷 = 12 V 𝑅𝑆 = 2K 𝑅𝐷 = 3K 𝑅1 = 300K 𝑅2 = 200K 𝑅𝑆𝑖 = 2K 𝑅𝐿 = 3K

Solution

The gate current is zero, so

𝑉𝐺 =200

200 + 30012 = 4.8 V

Further

𝑉𝑆 = 𝐼𝐷𝑅𝑆 = 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2𝑅𝑆

Thus

𝑉𝐺𝑆 = 𝑉𝐺 − 𝑉𝑆 = 4.8 − 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2(𝑅𝑆) 𝑉𝐺𝑆 = 4.8 − 0.001(𝑉𝐺𝑆 − 2)2(2K)

Solving yields VGS = 2.96 V. Thus

𝐼𝐷 = 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2 = 0.001(2.96 − 2)2 = 0.922 mA

Further,

𝑉𝐷𝑆 = 𝑉𝐷 − 𝑉𝑆 = 12 − 𝐼𝐷𝑅𝐷 − 𝐼𝐷𝑅𝑆 = 12 − 4.61 = 7.4 V

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Question 8 For the circuit below, show that VGS = 3.55 V. State any assumptions that you make. (8 points)

𝐾𝑛 = 0.5 mA/V2, 𝑉𝑇𝑁 = 2 V, λ = 0, 𝐶𝑔𝑑 = 0.1 pF, 𝐶𝑔𝑠 = 1 pF

Solution

Assume the FET operates in saturation mode. No current flows into gate, so that

𝑉𝐺 = 10 166

166 + 234= 4.15V

𝐼𝐷 = 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2

Further,𝑉𝐺𝑆 = 𝑉𝐺 − 𝐼𝐷𝑅𝑆 = 4.15 − 𝐼𝐷𝑅𝑆, so that

𝐼𝐷 = 0.5�(4.15 − 𝐼𝐷𝑅𝑆) − 2�2

mA

If 𝑉𝐺𝑆 = 3.55 𝑉, then 𝐼𝐷 = 𝐾𝑛(𝑉𝐺𝑆 − 𝑉𝑇𝑁)2 gives 𝐼𝐷 = 1.2 mA. Substituting 𝐼𝐷 = 1.2 mA shows that the right hand side and the left hand is equal so that 𝑉𝐺𝑆 is indeed 3.55 V

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Question 9 For the circuit below, determine the dc load line equation for the BJT, incorporating 𝛽. That is, do not assume that 𝐼𝐸 ≅ 𝐼𝐶. (8 points)

𝑅𝐶 = 5K 𝑅𝐸 = 5K 𝑅𝐿 = 5K 𝛽 = 100

𝑉𝐶𝐶 = +12 V 𝑉𝐷𝐷 = −5 V

Solution

The load line equation is an equation that expresses 𝐼𝐶 as a function of 𝑉𝐶𝐸. If 𝑅𝐿 were absent, one would start with, for example:

𝑉𝐶𝐶 = 𝐼𝐶𝑅𝐶 + 𝑉𝐶𝐸 + 𝐼𝐸𝑅𝐸 + 𝑉𝐷𝐷 and note that 𝐼𝐸 = ((1 + 𝛽) 𝛽⁄ )𝐼𝐶, substitute that into the equation above an express 𝐼𝐶 as a function of 𝑉𝐶𝐸. With 𝑅𝐿 present, one has to take a different approach. The simplest is to replace 𝑉𝐶𝐶 , 𝑅𝐶, and 𝑅𝐿 with a Thevenin equivalent circuit as shown below.

𝑅𝑇𝐻 = 𝑅𝐿‖𝑅𝐶 = 2.5K

𝑉𝑇𝐻 =𝑅𝐿

𝑅𝐿 + 𝑅𝐶𝑉𝐶𝐶 = 6 V

Now 𝑉𝑇𝐻 = 𝐼𝐶𝑅𝑇𝐻 + 𝑉𝐶𝐸 + 𝐼𝐸𝑅𝐸 + 𝑉𝐷𝐷

𝐼𝐸 =1 + 𝛽𝛽

𝐼𝐶 = 1.01𝐼𝐶

Combining these equations, substituting values, and rearranging, gives

𝐼𝐶 =1

7.55 × 103(11 − 𝑉𝐶𝐸)

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Question 10 For the following circuit, assume that the BJT is operating in the forward-active region.

𝑉𝐵𝐸(𝑜𝑛) = 0.7 V 𝑉𝐴 = ∞ 𝛽 = 50

(a) Determine 𝐼𝐶 (5 points) (b) Determine the power dissipated by the BJT (3 points)

Solution

Part (a) The voltage at the base is 9 − 0.7 = 8.3 V. KCL at the collector (using the convention that currents into the node is positive) gives

𝐼𝐶 −VC1K

+ 𝐼𝐵 = 0

Using 𝐼𝐵 = (8.3 − 𝑉𝐶) 18K⁄ , and 𝐼𝐶 = 50𝐼𝐵, this equation becomes

50 �8.3 − 𝑉𝐶

18K� −

VC1K

+ �8.3 − 𝑉𝐶

18K� = 0

Solving yields 𝑉𝐶 = 6.14 V, so that

𝐼𝐵 = (8.3 − 𝑉𝐶) 18K⁄ = 0.12 mA 𝐼𝐶 = 50𝐼𝐵 = 6 mA

Part (b)

𝑉𝐶 = (𝐼𝐵 + 𝐼𝐶)(1K) = 6.12 V 𝑉𝐶𝐸 = 9 − 𝑉𝐶 = 2.88 V ⇒ 𝑃 = 𝑉𝐶𝐸𝐼𝐶 = 17.28 mW

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Question 11 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) 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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Question 12 (Principles, Own) An engineer measures the bandwidth of the circuit below by driving it with a sinusoidal signal and measuring the attenuation at various frequencies. She uses a scope with an input impedance of 1 MΩ with a × 1 probe, and then a × 10 probe. Complete the following table (6 points)

True BW in Hz

Measured BW in Hz (× 1) probe

Measured BW in Hz (× 10) probe

Solution

This is a simple 1st –order system with bandwidth 𝐵 = 1 (2𝜋𝑅𝐶)⁄ . Here 𝑅 is the equivalent resistance that the capacitor 𝐶 sees.

True bandwidth calculation: the Thevenin equivalent resistance that 𝐶 sees is simply 𝑅1, so

𝐵𝑇𝑟𝑢𝑒 =1

2𝜋𝑅𝐶=

12𝜋(910 × 103)(1 × 10−9) = 175 Hz

× 𝟏 Probe bandwidth calculation: the probe+scope has a 1M resistance that is effectively in parallel with 𝑅1, so that the Thevenin equivalent resistance that 𝐶 sees is 1M‖𝑅1, so that

𝐵×1 𝑝𝑟𝑜𝑏𝑒 =1

2𝜋(𝑅�𝑅𝑠𝑐𝑜𝑝𝑒)𝐶=

12𝜋(910 × 103)‖(1 × 106)(1 × 10−9) = 334 Hz

× 𝟏𝟎 Probe bandwidth calculation: × 10 probes increase the 1 MΩ probe+scope resistance to 10 MΩ, so that the Thevenin equivalent resistance that 𝐶 sees is 10M‖𝑅1, so that

𝐵×1 𝑝𝑟𝑜𝑏𝑒 =1

2𝜋(𝑅�𝑅𝑠𝑐𝑜𝑝𝑒)𝐶=

12𝜋(910 × 103)‖(10 × 106)(1 × 10−9) = 191 Hz

True BW in Hz 175 Hz

Measured BW in Hz (× 1) probe 334 Hz

Measured B in Hz (× 10) probe 1901 Hz

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Question 13 For the circuit below 𝛽 = 120,𝑉𝐵𝐸(𝑂𝑁) = 0.7 V, and 𝑉𝐴 = ∞

𝑅1 = 110 kΩ 𝑅2 = 82 kΩ 𝛽 = 120 𝑉𝐴 = ∞

𝑉𝐵𝐸(𝑂𝑁 ) = 0.7 V

(a) Determine 𝐼𝐶𝑄 , 𝑔𝑚, and 𝑟𝜋 (8 points) (b) Determine the output resistance 𝑅𝑜 using BJT scaling (4 points)

Solution

Part (a) The Thevenin equivalent circuit for the base-emitter bias circuit is

𝑅𝑇𝐻 = 𝑅1||𝑅2 = 47K

𝑉𝑇𝐻 = 12𝑅2

𝑅1 + 𝑅2= 5.1 V

𝑉𝐵𝐸 = 0.7 V 𝛽 = 120

𝐼𝐵 =𝑉𝑇𝐻 − 0.7

𝑅𝑇𝐻 + (1 + 𝛽)𝑅𝐸= 8.3 𝜇A

𝐼𝐶 = 𝛽𝐼𝐵 = 1 mA Part (b) 𝑔𝑚 = 40 𝐼𝐶 = 40 mS, and 𝑟𝜋 = 𝛽 𝑔𝑚 = 3K⁄

Part (c) Using BJT impedance scaling

𝑅𝑜 = 𝑅𝐸 ��𝑟𝜋

1 + 𝛽� = 4K ��

3K1 + 120

� ≅ 25 Ω

18 Version 2


The table below compares the design values with a SPICE simulation that started with the 2N2222 BJT with the Early voltage and 𝛽 replaced with 𝐕𝐀𝐅 = 1G and 𝐁𝐅 = 120.

Parameter Calculated/Design SPICE 𝐼𝐶𝑄 1 mA 1.02 mA 𝑉𝐵𝐸 0.7 V 0.642 V 𝐼𝐵𝑄 8.33 𝜇A 12.5 µA 𝑉𝑅𝐸 4 V 3.89 V

19 Version 2

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Homework Assignment 08 - University of Iowas-iihr64.iihr.uiowa.edu/.../Homework/HomeworkAssignment08_Solutio… · Homework Assignment 08 . ... Consider the Bode plot of a 1st order