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​10 VRs​ = 10 kΩRL​4.7 kΩIL​

A sensor behaves like a 10 V source with an internal resistance Rs = 10 kΩ. It feeds a load RL = 4.7 kΩ through an ideal voltage follower (unity-gain buffer). Find the load current IL supplied by the op-amp output.

Vs = 10 VRs = 10 kΩRL = 4.7 kΩ
IL = mA
2.Easy
−+Vout​Rf​ = 47 kΩVin​ = −1.1 VR1​ = 4.7 kΩ

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

Vin = −1.1 VR1 = 4.7 kΩRf = 47 kΩ
Vout = V
3.Medium
−+Vout​Rf​ = ?R1​ = 1 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 21.

R1 = 1 kΩAv = 21
Rf = kΩ
4.Medium
−+Vout​R2​ = 1 kΩV1​ = −2.2 VR1​ = 4.7 kΩV2​ = 0.8 VR3​ = 4.7 kΩR4​ = 1 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 = −2.2 VV2 = 0.8 VR1 = R3 = 4.7 kΩR2 = R4 = 1 kΩ
Vout = V
5.Hard
−+Vout​C = 2.2 µFVin​ = −2.6 VR = 1 kΩ

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±13 V. The capacitor is uncharged when a constant input Vin = −2.6 V is applied at t = 0. Find the output voltage Vout at t = 5 ms.

Vin = −2.6 VR = 1 kΩC = 2.2 µFVsat = ±13 Vt = 5 ms
Vout = V
6.Hard
Vin​V1​Vout​stage 1: invertingR1​ = 1 kΩRf1​ = 4.7 kΩstage 2: non-invertingR2​ = 10 kΩRf2​ = 2 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. The input is Vin = −0.8 V. Find the output voltage Vout.

R1 = 1 kΩRf1 = 4.7 kΩR2 = 10 kΩRf2 = 2 kΩVin = −0.8 V
Vout = V

Answer key

  1. 1. IL = 2.128 mA
  2. 2. Vout = 11 V
  3. 3. Rf = 20 kΩ
  4. 4. Vout = 0.6383 V
  5. 5. Vout = 5.909 V
  6. 6. Vout = 4.512 V

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