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​Rf​ = 10 kΩVin​ = 0.7 VR1​ = 1 kΩ

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

Vin = 0.7 VR1 = 1 kΩRf = 10 kΩ
Vout = V
2.Easy
−+Vout​10 VRs​ = 22 kΩRL​2.2 kΩ

A sensor behaves like a 10 V source with an internal resistance Rs = 22 kΩ. It feeds a load RL = 2.2 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 = 10 VRs = 22 kΩRL = 2.2 kΩ
VL = V
3.Medium
−+Vout​R2​ = 10 kΩV1​ = −2.4 VR1​ = 4.7 kΩV2​ = 2.6 VR3​ = 4.7 kΩR4​ = 10 kΩ

The difference amplifier uses an ideal op-amp with ±15 V supplies, with R3 = R1 and R4 = R2. Find the output voltage Vout.

V1 = −2.4 VV2 = 2.6 VR1 = R3 = 4.7 kΩR2 = R4 = 10 kΩ
Vout = V
4.Medium
−+Vout​Rf​ = ?R1​ = 22 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 15.

R1 = 22 kΩAv = 15
Rf = kΩ
5.Hard
−+Vout​C = 0.22 µFVin​ = −2.9 VR = 10 kΩ

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±14 V. The capacitor is uncharged when a constant input Vin = −2.9 V is applied at t = 0. Find the output voltage Vout at t = 5 ms.

Vin = −2.9 VR = 10 kΩC = 0.22 µFVsat = ±14 Vt = 5 ms
Vout = V
6.Hard
−+Vout​C = 1 µFVin​ = −2.5 VR = 10 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 = −2.5 V is applied at t = 0. Find the output voltage Vout at t = 10 ms.

Vin = −2.5 VR = 10 kΩC = 1 µFVsat = ±12 Vt = 10 ms
Vout = V

Answer key

  1. 1. Vout = −7 V
  2. 2. VL = 0.9091 V
  3. 3. Vout = 10.64 V
  4. 4. Rf = 308 kΩ
  5. 5. Vout = 6.591 V
  6. 6. Vout = 2.5 V

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