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​1.5 VRs​ = 22 kΩRL​1 kΩ

A sensor behaves like a 1.5 V source with an internal resistance Rs = 22 kΩ. It feeds a load RL = 1 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 = 1.5 VRs = 22 kΩRL = 1 kΩ
VL = V
2.Easy
−+Vout​4 VRs​ = 10 kΩRL​0.47 kΩ

A sensor behaves like a 4 V source with an internal resistance Rs = 10 kΩ. It feeds a load RL = 0.47 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 = 0.47 kΩ
VL = V
3.Medium
−+Vout​Rf​ = ?R1​ = 4.7 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 31.

R1 = 4.7 kΩAv = 31
Rf = kΩ
4.Medium
−+Vout​Rf​ = 47 kΩR1​ = 47 kΩVin​ = 2.5 VRa​ = 4.7 kΩRb​ = 4.7 kΩ

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

Vin = 2.5 VRa = 4.7 kΩRb = 4.7 kΩR1 = 47 kΩRf = 47 kΩ
Vout = V
5.Hard
−+Vout​Rf​ = 2 kΩV1​ = −1.8 VR1​ = 4.7 kΩV2​ = −1.1 VR2​ = 22 kΩIf​

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the current If through the feedback resistor, flowing towards the output.

V1 = −1.8 VV2 = −1.1 VR1 = 4.7 kΩR2 = 22 kΩRf = 2 kΩ
If = mA
6.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 = 1 V. Find the output voltage Vout.

R1 = 10 kΩRf1 = 100 kΩR2 = 100 kΩRf2 = 4.7 kΩVin = 1 V
Vout = V

Answer key

  1. 1. VL = 0.06522 V
  2. 2. VL = 0.1796 V
  3. 3. Rf = 141 kΩ
  4. 4. Vout = 2.5 V
  5. 5. If = −0.433 mA
  6. 6. Vout = −10.47 V

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