Homework Assignment 06 - University of...
Transcript of Homework Assignment 06 - University of...
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:
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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)
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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,
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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)
( || || ) ( || || )
( || || )
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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)
( )
( )
√ ( )
√ ( )
55:041 Electronic Circuits. The University of Iowa. Fall 2012.
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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
‖ ||
‖