PROBLEMS
1)
Assume an idealized voltage amplifier
(except where indicated otherwise explicitly) and
show that the transresistance V/I is -(N+2)R; is this
an example of voltage or current feedback?
Answer The only novelty added in solving idealized opamp problems is the special properties of the
opamp. The 'rules' are straightforward. Neglect the voltage drop across the amplifier input terminals
compared to any circuit voltages, and neglect the current into the amplifier input terminals compared to
any circuit currents. The mathematical effect of this neglect is to set the amplifier input voltage and the
input currents to zero.
Any technique of circuit analysis may be applied, some more easily than others in particular
circumstances. In the present case there is no current into the inverting (-) input, and the voltage at the
noninverting (+) input is the same as that at the inverting input. Hence the voltage at the common
node of the three resistors is -IR. There is a corresponding current into the node from the R/n resistor
of IR/(NR/N) = NI, and so the current into the amplifier output is (N+1)I. And so V = -(N+1)I - IR
= -(N+2)IR. The transresistance is V/I = -(N+2)R.
2)
Show that the voltage gain Vo/Vi for the
circuit shown is 5.
The inverting input voltage is Vi, assuming the amplifier input voltage is zero. Since there is no
current into the inverting input the voltage at the middle node on the right is 2Vi. Hence the amplifier
output current is Vi/R + 2Vi/R = 3Vi/R, and Vo = 3Vi + 2Vi = 5Vi.
3)
Show that the voltage gain Vo/Vi is (1 +
(R2/R1))(-R4/R3) for the circuit shown.
Answer The voltage gain of the first stage is 1+(R2/R1), and the voltage gain of the second stage is
-R4/R3. The cascade gain is the product.
Introductory Electronics Notes
0030-OPAMP Problem Answers
1
M H Miller:University of Michigan-Dearborn
4)
Show that the current ratio I1/I2 is found
from I1R1 = I2R2 for the circuit shown.
Answer Assuming no voltage drop across the amplifier input I1R1 = I2 R2
5)
A certain OpAmp has idealized characteristics
except for a nonlinear gain Vo = 100 v2. Feedback
is added as shown in the figure. Plot Vo vs Vi for
0 ≤ Vo ≤ 600 both with and without the feedback;
use the same Vo axis and scale for both cases.
Compare the linearity of the two plots. Explain the
difference.
Answer The difference between the amplifier output voltage for a square-law characteristic and
what corresponds to a linear characteristic may be regarded as a perturbation of the output voltage. As
described in the notes the (degenerative) feedback voltage then is greater than what it would be in the
linear case, i.e., the input correction is increased.
Assuming no amplifier input current the 1KΩ and 9KΩ form a voltage divider; the inverting input
voltage then is Vo/10, and v = Vi - (Vo/10). From the (nonlinear) gain expression
Vo = 100[Vi-(Vo/10)]2, or Vi = (Vo + √Vo)/10. Differentiate to find
Note that the slope is constant within 5% for Vo > 100 volts.
Introductory Electronics Notes
2
M H Miller:University of Michigan-Dearborn
6)
The amplifier shown has idealized behavior
except that its voltage gain G is finite and varies for
one reason or another. Estimate the minimum value
of G which will insure that 9.9 ≤ Vo/Vi ≤ 10.
Answer Note that Vo = G(Vi - 0.1Vo), and therefore Vo/Vi = G/(1+0.1G). One condition is that
Vo/Vi ≤ 10; substitute to find that this requires G ≤ ∞. The other condition is that Vo/Vi ≥ 9.9.
Substitute to find G ≥ 990
7) a) Using idealized OpAmps design an amplifier with a voltage gain of -10 and an input resistance
of 1 kΩ.
b) Using idealized OpAmps design an amplifier with a voltage gain of +10 and an input resistance
of 1 kΩ.
c) Using idealized OpAmps design an amplifier with a voltage gain of -10 and an infinite input
resistance.
d) Using idealized OpAmps design an amplifier with a voltage gain of +10 and zero input
resistance.
Answer
8)
Assume idealized opamps; calculate the
voltage gain V0/Vi.
Introductory Electronics Notes
0030-OPAMP Problem Answers
3
M H Miller:University of Michigan-Dearborn
Answer Replace the two cascade amplifier
stages by an equivalent single-stage amplifier
with a voltage gain of -110 and an input
resistance of 1KΩ. Simply to provide an
estimate consider this to be an idealized
opamp. Then the feedback arrangement
within which it is imbedded provides an
‘idealized’ gain of -10. Of course the singlestage equivalent is not an idealized opamp.
Hence the gain magnitude will be ‘somewhat
less’ than 10.
*OpAmp Illustration
VIN
R12
R23
R50
1
1
2
5
0
2
3
0
DC
0
1K
1K
1K
X1
0
3
78
4 UA741
X2
4
5
78
6 UA741
V+
7
0
DC
12v
V8
0
DC
-12V
.LIB
EVAL.LIB
.DC VIN -5 5 .1
.STEP PARAM RVAL LIST 10K 100MEG
.PROBE
.END
.PARAM RVAL = 10K
R26
R34
R56
2
3
5
6
4
6
{RVAL}
10K
10K
The PSPICE netlist for the circuit is shown above. The transfer characteristic V(6) vs V(1) is computed
for two values of the overall feedback resistor, 10KΩ and 100 MegΩ. The latter size effectively removes
the feedback allowing the voltage gain of the two ‘internal’ stages to be computed (reduced by a factor of
two because of R12. This gain is estimated above as -110/2 = -55Ω (note the unit); the computed value is
-54.96. The 10KΩ value corresponds to the actual circuit, for which the gain is estimated above to be
‘somewhat less’ than -10; the computed value is -8.4.
If the amplifier saturates at ± 12 volts (determined by power supplies) then for a gain of 110 the input
voltage range for linear operation will be approximately ±100 millevolts, while for a gain or 10 it will be
about ± 1 volt. Both cases are plotted below.
Introductory Electronics Notes
4
M H Miller:University of Michigan-Dearborn
9)
The accompanying netlist describes the
integrator circuit shown to the right. Note that
the pulse input is not zero-average. Note also
that two different values of the capacitor bypass
resistor are used; the 100 MEGΩ value is
essentially an open circuit. Compare the
expected amplifier output with the computed
output, particularly with respect to the long-term
average value..
*Integrator
VIN
3 0
+ PULSE( -.1 .12 0 1U 1U 5M 10M
RS
3 1
1K
CF
1 2
.1U IC = 0
RF
1 2
{RVAL}
RL
2 0
1K
X1
0 1
452
UA741
V+
4 0
DC12v
V5 0
DC-12V
Introductory Electronics Notes
5
.PARAM RVAL = 1
.STEP PARAM RVAL
+LIST 100MEG 220K
.LIB EVAL.LIB
.OPTIONS NOMOD
.TRAN .001M 150M 0 .01 UIC
.PROBE
.END
M H Miller:University of Michigan-Dearborn
10)
Design a cascade of noninverting 741 amplifiers for a gain of at least 40db at 100 KHz. Describe
the basis for the design choices made.
Answer This problem is intended to increase awareness of the Gain-Bandwidth constraint,
generally applicable, but here specifically in the OpAmp case. The nominal Gain-Bandwidth product
for the 741 OpAmp is 1MHz (80 db, 100 Hz). Suppose a single stage amplifier (inverting or
noninverting) is proposed. Unfortunately with a gain of 40 db the bandwidth realized is only 10 KHz!
A two stage (identical, for simplicity) cascade may be considered, with a per-stage gain of 20db; the
corresponding stage bandwidth is 100 KHz. Unfortunately the gain is down 6db at this frequency,
3db reduction for each stage. The actual 3 db frequency is 64.4 KHz.
Use three identical stages, 14db per stage, to obtain a nominal 42db overall midband gain The 3db
frequency for each stage then is 199.5 KHz. The overall gain will be
and this will be 125/√2 when f = 101.7KHz.
* 741 Frequency Response
VS 1
0
AC
1
V+ 3
0
DC
12v
V- 4
0
DC
-12V
X1 1
2
3 4 5 UA741
RPB1
2
0
.22K
RPA1
5
2
1K
X2 5
6
3 4 7 UA741
RPB2
6
0
.22K
RPA2
7
6
1K
Introductory Electronics Notes
0030-OPAMP Problem Answers
X3 7
8
3 4 9 UA741
RPB3
8
0
.22K
RPA3
9
8
1K
.AC DEC 20 10K 10MEG
.LIB
EVAL.LIB
.PROBE
.END
6
M H Miller:University of Michigan-Dearborn
Extra Comment:
The bandwidth of an amplifier also can be extended by introducing poles and zeros in the transfer
function appropriately. In this particular illustration suppose a small capacitor is added across the 220Ω
feedback resistor of each stage. Qualitatively this may be viewed as replacing the resistor by an
impedance whose magnitude decreases as the frequency increases. This decreases the feedback, and
so the gain increases (i,e,. is decreased less by the feedback). The effectiveness of this compensation
depends on the gain of the amplifier without feedback being high enough to make feedback
meaningful, and at a high enough frequency the compensation fails.
The singularity is set (ad hoc) at about the 200KHz pole of the inherent amplifier. The response for
several values of capacitance is shown below. (For a 0.1 nF capacitance the response is essentially
that absent the capacitor.
11)
Assume an idealized opamp; calculate the transfer function.
Answer The inverting input of the opamp
provides a virtual ground. Hence the diode will
be OFF
When the diode is OFF Vo = 0. When the diode
conducts the diode current is (Vi/R) - (VB/Ro)
and
Introductory Electronics Notes
0030-OPAMP Problem Answers
7
M H Miller:University of Michigan-Dearborn
Answer The use of the word 'phase' implies V i is sinusoidal . Calculate the voltage at the
noninverting input as
This also is the voltage at the inverting input. Hence the input current into R is
and so
Answer The opamp is used to fix the voltage across C2 to Vo; the current through C2 then is
jωC2Vo, and the voltage at the R-R common node is (1+jωRC2)Vo. Apply KCL at this node to
obtain
Substitute the condition C1 = 2C2 and solve for
Introductory Electronics Notes
0030-OPAMP Problem Answers
8
M H Miller:University of Michigan-Dearborn
14)
A weighted summing amplifier is drawn
below. The amplifier output voltage is
where the relative weights (resistor ratios) form a
binary sequence. Examination of the pulse periods
specified indicate that the output voltage cyclically
decreases from -14 to 0 (relative to the common
pulse amplitude) in seven incremental steps of 2
each. For the PSPICE analysis following the pulse
amplitude is set to 0.5 volt, so that the output ranges
from -7 to 0 volts in steps of 1 volt. (Resistor values
are chosen for illustrative simplicity rather than as
standard values.) Compute the summer output and
evaluate the computation.
Answer
*Summer Illustration
V0
V2
V4
1
2
3
0
0
0
R0
R1
R2
RF
1
2
3
4
4
4
4
5
PULSE(0 .5 0 1U 1U 8M 16M)
PULSE(0 .5 0 1U 1U 4M 8M)
PULSE(0 .5 0 1U 1U 2M 4M)
1K
2K
4K
8K
Introductory Electronics Notes
0030-OPAMP Problem Answers
9
X1 0 4 6 7
V+ 6 0 DC
V7 0 DC
.LIB EVAL.LIB
.OP
.TRAN .1m 20m
.PROBE
.END
5
UA741
12v
-12V
M H Miller:University of Michigan-Dearborn
15)
The unity gain summing amplifier drawn to
the right has a 5KHz unit-amplitude sinusoidal
voltage applied to node 1. A positive 5 millisecond
pulse train (amplitude 3v, period 5 millisecond) is
applied to node 2. As a result the opamp output is
regularly shifted down by -3v. The diode network
recognizes (at node 6) only voltages more negative
than (about) -1.7v. Review the expected
performance of the circuit, and then compute the
performance using a 741 opamp and a 1N4148
switching diode.
Answer
*'Window' Circuit
VSIN
1
0
SIN(0, 1, 5K)
VPULSE2 2
0
+ PULSE(0, 3, 1M, .01M, .01M, .5M, 2M)
X741
0
3
7 8 4 UA741
V+ 7
0
DC
10v
V- 8
0
DC
-10V
RS 1
3
1K
RDC
2
3
1K
RF 3
4
1K
D1 5
4
D1N4148
VDC
6
5
DC
1
RL 6
0
1K
.TRAN 1U
4M 0 5U
.LIB
EVAL.LIB
.PROBE
.END
This circuit is used here to sample that portion of the sinusoid occurring in a specific 'window' defined
by the pulse. In general the window provides a time interval within which the occurrence (or nonoccurrence) of a signal can be determined.
Introductory Electronics Notes
0030-OPAMP Problem Answers
10
M H Miller:University of Michigan-Dearborn