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​6 VRs​ = 22 kΩRL​4.7 kΩIL​

A sensor behaves like a 6 V source with an internal resistance Rs = 22 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 = 6 VRs = 22 kΩRL = 4.7 kΩ
IL = mA
2.Easy
−+Vout​Rf​ = 10 kΩVin​ = 1.3 VR1​ = 2 kΩ

The circuit is an inverting amplifier. The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. Find the output voltage Vout.

Vin = 1.3 VR1 = 2 kΩRf = 10 kΩVsat = ±12 V
Vout = V
3.Medium
−+Vout​Rf​ = ?R1​ = 2 kΩVin​

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

R1 = 2 kΩAv = 3
Rf = kΩ
4.Medium
−+Vout​R2​ = 100 kΩV1​ = 2.1 VR1​ = 47 kΩV2​ = 2.8 VR3​ = 47 kΩR4​ = 100 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 = 2.1 VV2 = 2.8 VR1 = R3 = 47 kΩR2 = R4 = 100 kΩ
V− = V
5.Hard
−+Vout​C = 2.2 µFVin​ = −1.6 VR = 1 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 = −1.6 V is applied at t = 0. Find the output voltage Vout at t = 10 ms.

Vin = −1.6 VR = 1 kΩC = 2.2 µFVsat = ±12 Vt = 10 ms
Vout = V
6.Hard
−+Vout​Rf​ = 47 kΩV1​ = 0.2 VR1​ = 22 kΩV2​ = −0.2 VR2​ = 2 kΩ

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

V1 = 0.2 VV2 = −0.2 VR1 = 22 kΩR2 = 2 kΩRf = 47 kΩ
Vout = V

Answer key

  1. 1. IL = 1.277 mA
  2. 2. Vout = −6.5 V
  3. 3. Rf = 4 kΩ
  4. 4. V− = 1.905 V
  5. 5. Vout = 7.273 V
  6. 6. Vout = 4.273 V

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