ELG3331: Assignment 1
Problem 1: Design an inverting op-amp circuit for which the gain is -5 V/V and the total
resistance used is 120 kΩ.
Solution:
100 kΩ
20 kΩ
-
vi
vo
+
vo
R
= −5 = − 2 ; R 2 = 5R1
vi
R1
R1 + R 2 = 120 kΩ
5R1 + R1 = 120 kΩ
R1 = 20 kΩ and R 2 = 100 kΩ
Problem 2. Using the following circuit and assuming an ideal op amp, design an
inverting amplifier with a gain of 26 dB having the largest possible input resistance under
the constraint of having to use resistors no larger than 10 MΩ. What is the input
resistance of your design?
R2
R1
vi
+
vo
1
20 log G = 26dB
G = −19.95 V/V
vo
R
=− 2
vi
R
R 2 = 19.95 R1
For largest possible input resistance, select R 2 = 10 MΩ and R1 = 500 kΩ
Rin = 500 kΩ
Problem 3. (a) Design an inverting amplifier with a closed-loop gain of -100 V/V and an
input resistance of 1 kΩ.
(b) If the op amp is known to have an open-loop gain of 1000 V/V, what do you expect
the closed loop gain of your circuit to be?
(c) Give the value of a resistor you can place in parallel (shunt) with R1 to restore the
closed loop gain to its nominal value.
R2
R1
v1
vi
-
vo
+
(a )
v0
R
R
= − 2 ⇒ −100 v = − 2 ⇒ R2 = 100kΩ
v
vi
R1
1kΩ
v
(b) A = 1000 v ⇒ v1 = − 0
v
A
v -v
v -v
v
Q i 0 = 1 o ⇒ o =
R1
R2
vi
R2
R1
R
(1 + 2 )
R1
1+
A
−
(c) Assume R "1 = R x || R1 when R1 = 1kΩ
=
- 100
= −90.8 v
v
101
1+
100
vo
= −100 v
v
vi
− vo
v
− o −
RR
Rx
vi - v 0 v1 - vo
vo
1 − 0 .1
100
1000
=
⇒
⇒ R '1 = R2 ×
⇒ R '1 =
= 0.899kΩ = 1 x =
− vo
1.001
R1 + R x 1 + R x
R1
R2
vi
− vo
1000
⇒ R x = 8.9kΩ ≅ 8.87 kΩ ± 1%
Problem 4. An inverting op amp circuit using an ideal op amp must be designed to have
a gain of -1000 V/V using resistors no larger than 100 kΩ.
2
(a) For the simple two resistor circuit, what input resistance would result?
(b) If a T-network is used as feedback circuit for the inverting amplifier with three
resistors of maximum value, what input resistance results? What is the value of
the smallest resistor needed?
R2
R1
= 1000 R 2 = 100kΩ ⇒ R1 = 100Ω
(a ) Rin = R1 = 100Ω
(b)
vo
R
R
R4
= − 2 (1 + 4 +
) = −1000
vi
R1
R2 R3
If R2 = R1 = R4 = 100kΩ ⇒ R3 =
100kΩ
≈ 100Ω
1000 − 2
⇒ Rin = R1 = 100kΩ
Problem 5. Consider a low pass active filter. Derive its transfer function, its DC gain and
the 3-dB frequency. Design the circuit to obtain an input resistance of 1 kΩ, a DC gain of
20 dB, and a 3-dB frequency of 4 kHz. At what frequency does the magnitude of the
transfer function reduce to unity?
Refer to fig. P2.119
R2
1
vo
R1
Z
y
R1
=− 2 =− 1 =−
=−
1
1 + sCR 2
vi
Z1
y2
+ sC
R2
R
1
which is an STC LP circuit with a dc gain of − 2 and 3 - dB frequency ω 0 =
R1
R2 C
The input resistance equal to R1 so for
Ri = 1k ⇒ R1 = 1kΩ and for dc gain of 20dB or less
⇒
R2
= 10 ⇒ R 2 = 10kΩ
R1
For 3dB frequency of 4 kHz : ω 0 = 2π × 4 × 10 3 =
1
⇒ C = 4nF
CR 2
The unity gain frequency is 0 dB is 40kHz.
3