55:041 Electronic Circuits
The University of Iowa
Fall 2010
Exam 3
Name: ___________________________
Score__________/65___
Question 1 Unless stated otherwise, each question below is 1 point.
1.
An engineer designs a class-AB amplifier to deliver 2 W (sinusoidal) signal power
to an
resistive load. Ignoring saturation in the output BJTs, what is the required
peak-to-peak voltage swing across the load? (2 points)
Answer:
2.
, so that
, so that
Consider the following circuit, which is the power output stage of an amplifier.
(a) What is the name of the shaded sub-circuit around ? (b) Write down one
sentence/phrase that describes the purpose of the sub-circuit and constant current
source. (2 points)
Answers:
(a)
multiplier
(b) Reduction of cross-over
distortion
3.
An amplifier has gain of 800. After adding negative feedback, the gain is measured
as 25. Find the loop gain. (2 points)
Answer.
yields the loop gain
1
so that
.
. Solving for
55:041 Electronic Circuits
4.
The University of Iowa
Fall 2010
An amplifier has gain of 100,000, and a 20% variation in gain over a certain
frequency range. Negative feedback is used to reduce the gain to 10. What is the
variation in gain with temperature of the feedback amplifier? (2 points)
Answer. The gain is reduced by
. The
temperature variations are reduced by the same factor, so the feedback amplifier’s
gain varies by
5.
An single-pole op-amp has an open-loop low-frequency gain of
and an
open loop, 3-dB frequency of 4 Hz. If an inverting amplifier with closed-loop lowfrequency gain of
uses this op-amp, determine the closed-loop
bandwidth. (2 points)
Answer. The gain-bandwidth product is
loop amplifier is then is
6.
. The bandwidth of the closed.
An op-amp has an open-loop gain of 120 dB and an input resistance of
. An
engineer wants to use negative feedback to obtain an amplifier with input resistance
of 5
. What is the gain (in dB) of the feedback amplifier? (2 points)
Answer. Negative feedback increases the resistance by
(or 40 dB) and reduces the gain the same factor, so the feedback amplifier’s
gain is 80 dB.
2
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 2 The multimeter in the figures below has a common-mode rejection specification of
80 dB. What possible range of output voltages can the meter indicate? (5 points)
Solution
The difference voltage is
multimeter will suppress
is thus
display anything in the range
3
, and the common-mode voltage is
A common-mode rejection specification of 80 dB means that the
by a 80 dB or by
The contribution of the common-mode error
. However, the sign is unknown. Thus, the multimeter could
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 3 In the circuit below, the FETs have
,
Determine the low frequency differential gain, namely
draw and fully label (i.e., slopes, intercepts, etc.) the Bode plot of the gain
and
. Then
. (15 points)
Solution Below is a symmetrical small-signal equivalent for the amplifier. With
we will have the same differential gain
voltages at nodes K and Z do not change, and we can treat them as virtual grounds.
The
The resulting half amplifier circuit is then
The gain from node a to x, or
4
and the half-amplifier gain is
55:041 Electronic Circuits
The University of Iowa
is twice the gain of this amplifier or
Fall 2010
.
The 6 pF capacitor between the drain and gate will experience the Miller multiplication. The
Miller gain is the gain from node a to x, or
.
Now
The Bode plot for the overall amplifier is
5
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 4
For the circuit above, make reasonable assumptions and then
(a) Show that
(4 points)
(b) Calculate a reasonable estimate for the output resistance
(4 points)
Solution
The three transistors are in a Darlington configuration and we can model it as a pnp transistor
with
, and most of IQ flows through Q3. Thus,
.
Part (a) For the composite transistor
Part (b) Use BJT impedance scaling:
6
.
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 5
The maximum transistor power is in the circuit above is
.
(a) Determine RL such that maximum power is delivered to the load. (4 points)
(b) For
, determine average power dissipated in the transistor. (6 points)
Do not calculate
and neglect the base current when calculating power.
Solution
Part (a) The transistor will dissipate the maximum power (25 W) when
From this follows that
and
Part (b) The gain of the amplifier is
, so that the amplitude of the signal output voltage is
The signal power dissipated in the resistor is
The average power dissipated in the transistor is
7
.
.
.
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 6
For the class-A amplifier shown, show that the maximum efficiency
for a sinusoidal input signal 25%. Clearly state assumptions you
make. For example, “ignore saturation…” (6 points)
Solution. Assumptions: (a) neglect bias currents, (b) saturation, and (c) for maximum power
delivered to the load, the transistor will be biased so that
.
is the given load, so
that
and the average supply power is
For a sinusoidal input voltage, the voltage across, and the current through the load is
The signal power is the rms power dissipated by the load, given by
The power conversion efficiency is then
Note: many students wrote down a number of loosely connected equations or something such
as
. This was not sufficient. Rather, students had to
show/derive the result in a logical, step-by-step manner, stating assumptions, etc.
8
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 7 Consider the bipolar difference amplifier below. A dc analysis shows that
, and
. Determine the differential-mode voltage gain
. (6 points)
You may assume that the output stage (
stage.
does not load the differential input
Solution
Transistor
,
,
form a common-emitter stage with gain
Further,
The gain of the differential stage is
The overall differential-mode gain is
9
.
55:041 Electronic Circuits
The University of Iowa
Fall 2010
Question 8 The figure below is a plot of the open-loop gain function for the LF156 voltage
amplifier. An engineer will use the amplifier in a negative feedback configuration to set the midfrequency 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 LF156 gain curve at 100 kHz (see above). Thus, the bandwidth ~ 100 kHz.
10