EE 321 Analog Electronics, Fall 2013
Homework #3 solution
2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47
is given by
Rf
Rf
Rf
vO = −
vN 1 +
vN 2 + . . . +
vN n
RN 1
RN 2
RN n
RP
RP
RP
Rf
vP 1 +
vP 2 + . . . +
vP n
+ 1+
RN
RP 1
RP 2
RP n
where RN = RN 1 ||RN 2 || . . . ||RN n and RP = RP 1 ||RP 2 || . . . ||RP n ||RP 0 .
(b) Design a circuit to obtain
vO = −2vN 1 + vP 1 + 2vP 2
The smallest resistor used should be 10 kΩ.
(a) First, for the negative inputs, let’s work with the first input via superposition. The
negative input is at virtual ground, so the input current is iN 1 = RvN1
. All that current
N1
flows out of the feedback resistor because the voltage across the other input resistors
is zero. Thus, for only vN 1 on, we get
vO = −
Rf
vN 1
RN 1
Similarly, by superposition, we can see that for only vN x inputs turned on, then output
is
1
n
X
vN i
vO = −Rf
RN i
i=1
which is identical to the expression we need to show. Next, for the positive inputs, let’s
look at the first input. The voltage at the positive input is a voltage division between
RP 1 and the parallel combination of all the other resistors.
1
Pn
1
1
i=0 RP i − RP 1
v+ =
1
Pn
1
1
i=1 RP i − RP 1
1
− R1
1
RP
=
1
RP
P1
1
− R1
+ RP 1
vP 1
vP 1
P1
=
1 + RP 1
1
=
+ RP 1
RP 1
RP
1
1
RP
1+
−1
RP
=
vP 1
RP 1
−
1
RP 1
vP 1
vP 1
Next apply superposition and we get for all positive inputs on and all negative inputs
off,
n
X
vP 1
v+ = RP
RP 1
i=1
Next, this signal is amplified by the non-inverting amplifier whose gain is composed
from Rf and the parallel resistance of all negative input resistors, RN , so
n
X
vP 1
Rf
RP
vO = 1 +
RN
RP 1
i=1
which is also identical with what we had to show. Combining the output from negative
inputs and positive inputs we get the result we were asked to show.
(b) We have the equations
Rf
=2
RN 1
RP
Rf
1+
=1
RN RP 1
Note that RN = RN 1 , so that 1 +
Rf
RN
= 3, and
2
RP
Rf
1+
=2
RN RP 2
Rf
=2
RN 1
RP
1
=
RP 1
3
2
RP
=
RP 2
3
We can now select RN 1 = 10 kΩ, so that Rf = 20 kΩ. Next,
RP RP 1
2 3
RP 1
=
= × =2
RP 2
RP 2 RP
3 1
Next, we have
1
1
1
1
=
+
+
RP
RP 0 RP 1 RP 2
1=
RP
RP
RP
+
+
RP 0 RP 1 RP 2
1 2
RP
=1− − =0
RP 0
3 3
This means that RP 0 = ∞. Then we can choose RP 1 = 10 kΩ and RP 2 = 20 kΩ.
2.62. For the circuit shown in Fig. P2.62, express vO as a function of v1 and
v2 . What is the input resistance seen by v1 alone? By v2 alone? By a source
connected between the two input terminals? By a source connected to both
input terminals simultaneously?
We can use superposition to get vO . First as a function of v1 , we have vO = −v1 . Next, as a
function of v2 ,
R
R
1+
v2 = v2
vO =
R+R
R
Combining we get vO = v2 − v1 .
The input resistance on input 1 is found as
Ri1 =
dv1
di1
The relationship between the two is
v1 =
v2
+ i1 R
2
3
Plugging this into the expression above we get
Ri1 = R
On input 2 it is clear that Ri2 = 2R. For a source connected between the two inputs we have
v2 − v1 = v, and a current through it, i, and we want the differential input resistance,
dv
di
We need a relationship between v and i. Note that v = vO . To have the same current in the
two arms we must have that v − v1 = v2 . Combining that with the input output reltionship,
v = v2 − v1 , we can eliminate v1
Rid =
v − v2 = v1 = v2 − v
or
v = v2
We can also see that v2 = 2Ri. Inserting we get
v = 2Ri
and see that Ric = 2R.
2.111. An op-amp intended for operation with a closed-loop gain of −100 V/V
uses feedback resistors of 10 kΩ and 1 MΩ with a bias-current-compensation
resistor R3 . What should the value of R3 be? With input grounded, the output
offset voltage is found to be +0.21 V. Estimate the input offset current assuming
zero input offset voltage. If the input offset voltage can be as large as 1 mV of
unknown polarity, what range of offset current is possible? What current injected
into, or extracted from, the nongrounded end of R3 would reduce the op-amp
output voltage to zero? For available ±15 V supplies, what resistor and supply
voltage would you use?
The bias-current compensating resistor R3 = R1 ||R2 = 9.9 kΩ. With the bias compensated
and zero offset voltage the output is equal to the offset current flowing through the feedback
resistor, R2 ,
vO
0.21
=
= 210 nA
R2
106
If the input offset voltage is up to ±1 V, the output offset voltage can be as large as ±0.1 V.
In that case, the output offset due to bias current can range between 0.21 − 0.1 = 0.11 V
to 0.21 + 0.1 = 0.31 V. That results in a minimum offset current of IOS,min = 110 nA and a
maximum offset current of IOS,max = 310 nA.
In order to reduce the output to zero we need to add a voltage on the positive input
which cancels the output of 0.21 V. That means we need to add a negative voltage on the
positive input which is that output divided by the non-inverting gain, 101. The voltage we
IOS =
4
need to add is Vcompensation = −0.21 V/101 = −2.08 mV. That voltage is produced by a
current through R3 . That current should be
−2.08 mV
Vcompensation
=
= 210 nA
R3
9.9 kΩ
To get that current from the negative supply we need a resistor,
Icompensation = −
Rcompensation =
Vsupply
Icompensation
=
15
= 71.4 MΩ
210 × 10−9
3.2. For the circuits shown in Fig. P3.2 using ideal diodes, find the values of the
voltages and currents indicated.
(a) In this case the diode is conducting, such that the voltage at the top end of the diode
V
equals that at the bottom end, so V = −3 V. Then, I = 106 kΩ
= 0.6 mA.
(b) The diode is not conducting, so I = 0, and V = 3 V.
(c) The diode is conducting so I = 0.6 mA as before, but now V = 3 V.
(d) The diode is not conducting, so I = 0 and V = −3 V.
3.4. In each of the ideal-diode circuits shown in Fig. P3.4, vI is a 1 kHz, 10 V
peak sine wave. Sketch the waveform resulting at vO . What are its positive and
negative peak values?
5
(My plotting program is temporarily out of commission so you get a description in words and
equations instead. Note that this procedure is probably the best way to go about plotting
the waveform anyway; getting an expression and then plottinng that expression.)
(a) Piecewise expression for vO
(
0
vO =
vI
vI ≤ 0
vI > 0
Minimum value: 0 V. Maximum value: 10 V
(b) Piecewise expression for vO
(
vI
vO =
0
vI ≤ 0
vI > 0
Minimum value: −10 V. Maximum value: 0 V.
(c) Current never flows, so vO = 0 always.
(d) This case is identical to (a)
(e) Current always flows so vO = vI . Minimum values: −10 V. Maximum value: 10 V.
(f) This case is identical to (a)
(g) Piecwise expression for vO
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(
vI
vO =
0
vI ≤ 0
vI > 0
Minimum value: −10 V. Maximum value: 0 V.
(h) The output is always connected to ground, vO = 0 always.
(i) When vI < 0 this acts as a voltage divider. Otherwise they two are equal. Piecewise
expression is
vO =
(
vI
2
vI
vI ≤ 0
vI > 0
Minimum value: −5 V. Maximum value: 10 V.
(j) When vI > 0 the output is shorted to the input. When vI < 0 the output is obtained
from the voltage division. The expression for vO is identical to (i).
(k) The current source causes 1 V voltage drop across the resistor, such that vO is always
one volt higher than the voltage at the bottom of the resistor. When vI > 0 the voltage
at the bottom of the resistor is ground. When vI < 0 the voltage at the bottom of the
resistor is vI . The piecewise expression for vO is
(
vI + 1 V vI ≤ 0
vO =
1V
vI > 0
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