Homework Assignment 06 - University of...

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55:041 Electronic Circuits. The University of Iowa. Fall 2012. 1 Homework Assignment 06 Question 1 (Short Takes) One point each unless otherwise indicated. 1. Consider the current mirror below, and neglect base currents. What is ? Answer: 2. In the current mirrors below, neglect base currents and take , What is ? (a) (b) (c) (d) Answer: (a) 3. In the current mirrors below, neglect base currents and , What is ? Answer: 30 4. In the current mirrors below, neglect the base currents. What is ? Answer:

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Page 1: Homework Assignment 06 - University of Iowas-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55041_2012/... · 2013. 1. 15. · Homework Assignment 06 Question 1 (Short Takes) One point

55:041 Electronic Circuits. The University of Iowa. Fall 2012.

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Homework Assignment 06

Question 1 (Short Takes) One point each unless otherwise indicated.

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

Answer:

2. In the current mirrors below, neglect base currents and take , What is ?

(a)

(b) ⁄

(c) ⁄

(d)

Answer: (a)

3. In the current mirrors below, neglect base currents and , What is ?

Answer: 30

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

Answer:

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5. What is the voltage gain ⁄ of the amplifier below if and ? (3 points)

(a)

(b)

(c) Need additional information (i.e., )

(d)

(e)

Answer: ( ‖ ) ( ‖ ) so (e) is the

correct answer.

6. True of false: For both the amplifiers below, the voltage gain increases when the

drain/collector bias current is increased.

7. The small-signal output resistance of a BJT biased at = 1 mA is100K. What is

when the transistor is biased at ?

Answer: is inversely-proportional to ( ⁄ ) so that will be smaller at 5

mA, or .

8. Typically, the C-E saturation voltage for a BJT, namely ( ), is in the range of (circle

one)

(a) 0.2–1.0 V (b) 0.7 V (c) 2–3 V (d) 100 V (e) 250 V

Answer: Option (a)

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9. True or false: the Early voltage for a small-signal BJT is in the order of (circle one).

(a) 0.7 V (b) 0.2 V (c) 10 V (d) 100 V (e) 250 V

Answer: Option (d)

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

Answer: Large effective resistance large voltage gain

11. True or false: assuming that for the amplifiers below, then is larger than

. However, also more sensitive to FET parameter variation than .

Answer: True

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Question 2 Consider the amplifier shown. The transistor has an Early voltage A dc

analysis reveals that , and ⁄ . Assume that the coupling

capacitors are large enough so that they are shorts at the operating frequency. (a) Draw the

corresponding small-signal circuit. Incorporate the MOSFETs output resistance . (5 points).

(b) Determine the amplifier’s voltage gain. (10 points). (c) Determine the amplifier’s input and

output resistances. (5 points each)

Solution

Part (a) The small-signal model for the

amplifier is shown right. The numerical

values for all components are known,

except for . However, we can calculate

that from ⁄ . We can

lump and together as

‖ ‖ .

Part (b) write a KCL equation at the output node

From the small-signal model it is clear that so that the equation above becomes

Some algebraic manipulation yields

( ⁄ )

( ⁄ )

Substituting , ⁄ and yields

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Part (c) Determine the input resistance using the standard procedure: turn off independent

sources, drive the circuit with a test source and determine the current that flows. Then,

⁄ . Referring to the small signal model given above, we can simply replace with

and then

However, we already determined that so we can write so that

Determine the output resistance using the standard procedure: turn off independent sources, drive

the circuit at the output with a test source and determine the current that flows. Then,

⁄ . If we turn off the independent source, namely and add a test source, the model

below results.

Clearly the control voltage and the current source is off, so KCL at the drain gives

|| ||

Substituting the numerical values give .

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Question 3 You can assume that for all the transistors in the circuit below, is large. Show

that . Let and then determine VA, VB, VC, VD , , and Assume

( ) for all the transistors. Further, note that is small: for the purposes on this

hand-analysis, ignore its effect. That is, assume . (14 Points)

Solution

Throughout we will assume ( ) for all the transistors. Further, since is large, we

will ignore base currents and take for all the transistors.

( ) ( ) ⁄ , which is mirrored as

( ) is very small, so

Thus,

Thus, ( ) ⁄ , which is practically the same current that flows through

, and

We are ignoring the effects of , so that ( ) is the mirror of . That is ( ) ( )

.

( ) , so ( )( )

Finally,

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Question 4 The open-loop low-frequency gain of an op-amp is 100 dB. At a frequency of

the magnitude of the open-loop gain is 38 dB. The op-amp has a dominant-pole

open-loop response. Determine the frequency of the dominant pole and the unity-gain

bandwidth. (5 points)

Solution

Open-loop dominant pole response implies constant GBP, which is

The dominant-pole frequency is

The unity-gain bandwidth is the same as GBP.

Question 5 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) What is the output resistance of the regulator? (3 points)

(b) Estimate the output ripple amplitude for every volt of input ripple at 120 Hz. (4 points)

Solution

Part (a)

( )

Part (b)

( ) (

)

( )

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 6 Draw a small-signal model for the following circuit. is a bypass capacitor and

for the FET. Be sure to supply all necessary details , polarities of , labeling

the gate, source, and drain, an expression for and so on. (5 points)

Solution

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Question 7 (Midterm, 2007) Consider the amplifier

shown

(a) Draw a small-signal model for the amplifier above.

(6 points)

(b) Derive an expression for the voltage gain ⁄

(8 points)

Account for the output resistance of the transistors, the

supply voltage, and as appropriate.

is large and may be

considered a short

Solution

Part (a) R1 and Q1 is a current source that sets up a current that is mirror through Q2. We can

replace R1, Q1, and Q2 with a current source with output resistance ro2. Further CC is a coupling

capacitor and a short at ac. Thus a small-signal model is then:

Note: points are subtracted if you leave out a ground terminal, VO, the signs for Vπ, ro, etc.

Part (b)

( || || ) ( || || )

( || || )

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Question 8 (Midterm, 2009) Consider the amplifier

shown.

(c) Draw a small-signal model for the amplifier.

(6 points)

(d) Derive an expression for the voltage gain ⁄

(8 points)

Account for the output resistance of the transistors, the

supply voltage, , , and as appropriate.

Hint: √

is large and may be

considered a short

Solution

Part (a)

Note: points are subtracted if you leave out a ground terminal, VO, the signs for Vπ, ro, etc.

In the model is the output resistance of transistor .

Part (b)

( )

( )

√ ( )

√ ( )

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Question 9 The figure below is a plot of the open-loop gain function for the LF357 voltage

amplifier. An engineer will use the amplifier in a negative feedback configuration to set the mid-

frequency voltage gain at 100. Use the plot and estimate the bandwidth of the feedback

amplifier.

Solution. A gain of 100 is equivalent to a gain of ( ) . A horizontal line at

40 dB intercepts the LF357 gain curve at 100 kHz (see above). Thus, the bandwidth ~ 100 kHz.

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Question 10 Determine the output resistance of the source follower. (12 points)

Solution To find , follow the standard procedure: turn off all independent sources ( in this

case), add a test source and determine . Then, ⁄ . The small-signal model for the

amplifier is shown in (a) below. The gate current is zero, so that is zero, and the small-signal

model becomes as is shown in (b). Note that and that is in parallel with .

(a) (b)

A KCL at the source gives

||

( )

||

Rearranging gives

‖ ||