# ECE 3410 – Homework 2 Problem 1. The non ```ECE 3410 – Homework 2
Problem 1. The non-inverting amplifier shown below uses a non-ideal op amp with a finite openloop gain.
R2
−
Vo
R1
+
vIN
−
+
(A) Calculate the ideal gain if the op amp has A → ∞, R1 = 1kΩ and R2 = 9kΩ.
Solution
R2
R1
= 10 V/ V
G? = 1 +
(B) Calculate the actual gain for the values in (A) if the gain is A = 100 V/ V.
Solution
For the non-inverting configuration:
?
G=G
A
A + G?
100
= (10 V/ V)
110
= 9.0909 V/ V
(C) Suppose the op amp has a finite open-loop gain of A = 1000 V/ V, and it is desired
to have an actual circuit gain of G = 10V/ V. What is the ratio R2 /R1 that will
achieve the desired gain?
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ECE 3410 – Homework 2
Solution
G=G
?
A
A + G?
⇒ GA + GG? = AG?
GA
⇒ G? =
A−G
Since G? = 1 + R2 /R1 , the solution is
R2
GA
=
−1
R1
A−G
= 9.10101
(D) Lastly, if the open loop gain varies from 100 to 1000 V/ V, and R1 is held constant
at 1kΩ, what is the range of adjustments needed in R2 to maintain an actual gain
of G = 10 V/ V?
Solution
Using the solution from (C), we may simply adjust the value of A to find the
range of R2 :
GA
R2 = R1
−1
A−G
= 10.1111kΩ(high end)
= 9.10101kΩ(low end)
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ECE 3410 – Homework 2
Problem 2. The transresistance amplifier shown below is to be used in a sensor interface. The
resistor Rs is an unknown resistance, which will be measured by the following method:
(i) Apply a known potential Vref to the op amp’s non-inverting input terminal. Due
to the virtual short effect, the same potential (or nearly the same) should then
appear at the inverting input terminal.
(ii) Measure the potential Vo appearing at the op amp’s output terminal.
(iii) Based on Vo , estimate the value of the current flowing through Rs . The resistance
is then
Rs ≈ Vs /Is
.
Transresistance Amplifier
Unknown Resistor
RF
Vs
Rs
−
Vo
Is
+
Vref
−
+
(A) If the op amp is ideal, obtain expressions for these items: Vs , Is , and Vo . For each
item, the left-hand side of your expression should only include Rs , Vref and/or RF .
Finally, obtain an expression for Rs .
Solution
Vs = Vref
Vref
Is =
Rs
RF
Vo = Vref 1 +
Rs
(B) Repeat (A) for the case where the op amp has finite gain equal to A V/V. This
time, the left-hand side of each expression should only include Rs , Vref , RF and/or
A.
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ECE 3410 – Homework 2
Solution
The previous equations must be modified as follows:
Vs = Vref −
Vo
A
Vs
Rs
Vo = Vs + Is RF
Is =
After some algebra, the final solutions are
&quot;
#
A
Vs = Vref
A + 1 + RRFs
&quot;
#
A
Vref
Is =
Rs A + 1 + RRF
s
&quot;
RF
A
Vo = Vref 1 +
Rs
A+1+
#
RF
Rs
(C) Suppose the op amp is ideal, Rs = 1kΩ, RF = 2kΩ, and Vref = 1V. What are the
values of Vs , Is and Vo ?
Solution
Vs = 1V
Is = 1mA
Vo = 3V
(D) Repeat (C) for an op amp with finite gain A = 100 V/V
Solution
If the op amp has finite gain, then the corresponding error coefficient is
&quot;
#
A
100
=
= 0.97087.
RF
103
A + 1 + Rs
We may obtain the solution for the finite gain case by multiplying this error
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ECE 3410 – Homework 2
Solution (cont.)
coefficient into the previous solutions from (C):
100
Vs = 1V
= 0.97087V
103
100
Is = 1mA
= 0.97087mA
103
100
= 2.91262V
Vo = 3V
103
.
(E) For the circuit described in (C) and (D), suppose Rs can vary between 500Ω and
5kΩ. What is the corresponding range of variation that may be seen in Vo ?
Solution
Any variation in Rs affects the error coefficient. For this range of variation,
the error coefficients vary as
A
(case1) = 0.82644628
RF
A + 1 + 500Ω
A
(case2) = 0.9708737864
RF
A + 1 + 5kΩ
Then, multiplying these into the ideal results gives:
10kΩ
Vo (case1) = (1V) 1 +
(0.82644628)
500Ω
= 17.35537V
10kΩ
Vo (case2) = (1V) 1 +
(0.9708737864)
5kΩ
= 2.912621359V
(F) Estimate the value of Rs if Vo = 1.5V, RF = 10kΩ, Vref = 1V and A → ∞.
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ECE 3410 – Homework 2
Solution
Since this is the ideal case, we can obtain the solution immediately:
Rs =
Vs
Is
Vref
(Vo − Vref ) /RF
Vref
= RF
Vo − Vref
= 30kΩ.
=
Problem 3. The difference amplifier shown below uses an op amp, but suffers from mismatch in
its resistor values. By design, R2 = R4 = 10kΩ, and R1 = R3 = 1kΩ. In reality, all
resistors vary within a &plusmn;5% tolerance.
R2
−
−
vIN
R1
vOUT
+
+
R3
R4
For this circuit, recall that there are two signal paths, called the inverting path and the
non-inverting path. The goal is to balance the gains along these signal paths:
R2
inverting path : − Gi = −
R
1
R2
R4
non − inverting path : Gni = 1 +
R1
R3 + R4
Notice here that Gi is defined as a positive number. Based on these definitions, you
can show that the differential gain and the common-mode gain are:
1
(Gi + Gni )
2
= |Gi − Gni |
ACM
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ECE 3410 – Homework 2
(A) If all the resistors were perfectly matched to their specified values, what should be
the circuit’s differential gain, common-mode gain and CMRR?
Solution
With perfect matching, we have:
R2
= 10 V/ V
R1
Acm = |Gi − Gni | = 0
=∞
CMRR = 20 log10
Acm
(B) Suppose all resistors are mismatched from their specified values as follows:
R1 : 5% too large
R3 : 5% too small
R2 : 5% too small
R4 : 5% too large
In this case, calculate the resistor values, the differential gain, the common-mode
gain and the CMRR in dB.
Solution
The values are:
R1 = 1050Ω
R2 = 9500Ω
R3 = 950Ω
R4 = 10.5kΩ
With these values, the gains are
R2
= 9.0476
R
1
R2
R4
= 1+
= 9.2140
R1
R3 + R4
Gi =
Gni
Since these are mismatched, we expect to see some common-mode gain:
Acm = 0.16635
CMRR = 20 log10
Acm
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= 34.789dB
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ECE 3410 – Homework 2
Problem 4. An inverting op amp configuration is shown below. Also shown are current and voltage
sources inserted to model the op amp’s non-ideal bias current and offset voltage.
R2
vin
R1
−
v−
Ibias
vout
+
−
+
Vofs
Ibias
(A) For this problem, suppose that Vofs = 0, Ibias = 20&micro;A, and R1 = 10kΩ. The op
amp’s supply rails are at &plusmn;5V, and the op amp has infinite open-loop gain. If
the input signal is a sinusoid with zero DC offset and a peak-to-peak amplitude
10mV, what is the maximum gain (i.e. the maximum value for R2 ) that can be
allowed without saturating the op amp?
Solution
Saturation will happen when the maximum output signal voltage equals the
rail. Accounting for the effect of Ibias , we have:
R2
max vin −
+ R2 Ibias = VR
R1
The maximum value occurs where vin = −5mV, so the equation can be simplified to
R2
(5mV)
+ R2 Ibias = VR
R1
Solving for R2 gives
R2 =
5mV
R1
VR
+ Ibias
Filling in all the known values, we find that R2 can be no greater than 243.9kΩ,
corresponding to a maximum gain of 24.39 V/ V.
(B) For this problem, suppose that Ibias ≈ 0, R2 = 5kΩ, R1 = 1kΩ and the op amp
is otherwise ideal. If Vofs varies in the range of &plusmn;10mV, what is the range of DC
offset voltages that may appear at VOUT ?
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ECE 3410 – Homework 2
Solution
The offset voltage is multiplied by the op amp’s non-inverting gain, 1 +
R2 /R1 = 6. So the range of possible output offset voltages is &plusmn;60mV.
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