Op-amp circuits worksheet with answers

Free printable Op-amp circuits worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
−+Vout​2 VRs​ = 4.7 kΩRL​10 kΩ

A sensor behaves like a 2 V source with an internal resistance Rs = 4.7 kΩ. It feeds a load RL = 10 kΩ through an ideal voltage follower (unity-gain buffer). For comparison, find the load voltage VL if the load were connected directly to the sensor without the buffer.

Vs = 2 VRs = 4.7 kΩRL = 10 kΩ
VL = V
2.Easy
−+Vout​Rf​ = 47 kΩVin​ = −1.4 VR1​ = 10 kΩ

The op-amp is ideal and runs from ±15 V supplies. Find the output voltage Vout.

Vin = −1.4 VR1 = 10 kΩRf = 47 kΩ
Vout = V
3.Medium
−+Vout​Rf​ = ?R1​ = 22 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 3.

R1 = 22 kΩAv = 3
Rf = kΩ
4.Medium
−+Vout​Rf​ = 47 kΩVin​R1​ = 2 kΩ

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. The inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.

R1 = 2 kΩRf = 47 kΩVsat = ±12 V
Vin,max = V
5.Hard
Vin​V1​Vout​stage 1: invertingR1​ = 4.7 kΩRf1​ = 10 kΩstage 2: non-invertingR2​ = 2 kΩRf2​ = 1 kΩ

Two ideal op-amp stages are connected in cascade: an inverting amplifier (R1, Rf1) followed by a non-inverting amplifier (R2 to ground, Rf2 in the feedback). Both run from ±15 V supplies. Find the overall voltage gain Av = Vout/Vin.

R1 = 4.7 kΩRf1 = 10 kΩR2 = 2 kΩRf2 = 1 kΩ
Av =
6.Hard
−+Vout​C = 0.1 µFVin​ = −1.2 VR = 22 kΩ

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±12 V. The capacitor is uncharged when a constant input Vin = −1.2 V is applied at t = 0. Find the output voltage Vout at t = 5 ms.

Vin = −1.2 VR = 22 kΩC = 0.1 µFVsat = ±12 Vt = 5 ms
Vout = V

Answer key

  1. 1. VL = 1.361 V
  2. 2. Vout = 6.58 V
  3. 3. Rf = 44 kΩ
  4. 4. Vin,max = 0.5106 V
  5. 5. Av = −3.191
  6. 6. Vout = 2.727 V

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