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​3 VRs​ = 4.7 kΩRL​0.47 kΩIL​

A sensor behaves like a 3 V source with an internal resistance Rs = 4.7 kΩ. It feeds a load RL = 0.47 kΩ through an ideal voltage follower (unity-gain buffer). Find the load current IL supplied by the op-amp output.

Vs = 3 VRs = 4.7 kΩRL = 0.47 kΩ
IL = mA
2.Easy
−+Vout​Rf​ = 22 kΩVin​ = −0.6 VR1​ = 10 kΩI1​

The op-amp is ideal and runs from ±15 V supplies. Find the current I1 through R₁, flowing towards the op-amp.

Vin = −0.6 VR1 = 10 kΩRf = 22 kΩ
I1 = mA
3.Medium
−+Vout​Rf​ = 10 kΩR1​ = 2 kΩVin​ = 1.5 VRa​ = 100 kΩRb​ = 47 kΩ

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

Vin = 1.5 VRa = 100 kΩRb = 47 kΩR1 = 2 kΩRf = 10 kΩ
Vout = V
4.Medium
−+Vout​Rf​ = ?R1​ = 2 kΩVin​

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

R1 = 2 kΩAv = 21
Rf = kΩ
5.Hard
Vin​V1​Vout​stage 1: invertingR1​ = 10 kΩRf1​ = 100 kΩstage 2: non-invertingR2​ = 100 kΩRf2​ = 4.7 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 = 0.75 V. Find the output voltage Vout.

R1 = 10 kΩRf1 = 100 kΩR2 = 100 kΩRf2 = 4.7 kΩVin = 0.75 V
Vout = V
6.Hard
−+Vout​C = 4.7 µFVin​ = 0.7 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 = 0.7 V is applied at t = 0. Find the output voltage Vout at t = 5 ms.

Vin = 0.7 VR = 1 kΩC = 4.7 µFVsat = ±13 Vt = 5 ms
Vout = V

Answer key

  1. 1. IL = 6.383 mA
  2. 2. I1 = −0.06 mA
  3. 3. Vout = 2.878 V
  4. 4. Rf = 40 kΩ
  5. 5. Vout = −7.852 V
  6. 6. Vout = −0.7447 V

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