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​4 VRs​ = 10 kΩRL​10 kΩ

A sensor behaves like a 4 V source with an internal resistance Rs = 10 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 = 4 VRs = 10 kΩRL = 10 kΩ
VL = V
2.Easy
−+Vout​Rf​ = 10 kΩVin​ = 0.8 VR1​ = 2 kΩ

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

Vin = 0.8 VR1 = 2 kΩRf = 10 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 4.

R1 = 22 kΩAv = 4
Rf = kΩ
4.Medium
−+Vout​R2​ = 2 kΩV1​ = 2.7 VR1​ = 22 kΩV2​ = −2.4 VR3​ = 22 kΩR4​ = 2 kΩ

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

V1 = 2.7 VV2 = −2.4 VR1 = R3 = 22 kΩR2 = R4 = 2 kΩ
V− = V
5.Hard
−+Vout​Rf​ = 2 kΩV1​ = 1.7 VR1​ = 4.7 kΩV2​ = −0.4 VR2​ = 100 kΩ

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the output voltage Vout.

V1 = 1.7 VV2 = −0.4 VR1 = 4.7 kΩR2 = 100 kΩRf = 2 kΩ
Vout = V
6.Hard
−+Vout​C = 0.1 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 = 0.5 ms.

Vin = 2.5 VR = 10 kΩC = 0.1 µFVsat = ±12 Vt = 0.5 ms
Vout = V

Answer key

  1. 1. VL = 2 V
  2. 2. Vout = −4 V
  3. 3. Rf = 66 kΩ
  4. 4. V− = −0.2 V
  5. 5. Vout = −0.7154 V
  6. 6. Vout = −1.25 V

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