Op-amp circuits: practice problems with step-by-step solutions

Free Op-amp circuits practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: inverting amplifier, non-inverting amplifier, summing amplifier.

Free practice problemEasy
−+Vout​Rf​ = 100 kΩVin​ = −1.6 VR1​ = 10 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 = −1.6 VR1 = 10 kΩRf = 100 kΩVsat = ±12 V
V
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What you practise

  1. Easyinverting amplifier
  2. Mediumnon-inverting amplifier
  3. Hardsumming amplifier

Key formulas

Inverting amplifier
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0

Learn Op-amp circuits from scratch

Before this: Nodal analysis

1What an op-amp is

An operational amplifier (op-amp) is a small chip that amplifies voltages. It is used in audio equipment, sensors and filters.

−+V−​V+​Vout​

It has two inputs, marked − (inverting) and + (non-inverting), and one output. It amplifies the tiny difference between the two inputs by an enormous factor, often 100 000 or more.

On its own, that gain is far too large to use. So you connect resistors from the output back to the − input. This is called negative feedback, and the resistors then set the gain.

The output can never go beyond the supply voltages, for example ±15 V.

Self-check What sets the gain of an op-amp circuit with negative feedback?
  1. The input voltage
  2. The op-amp’s own internal gain
  3. The resistors around the op-amp
  4. The supply voltages

Answer: The resistors around the op-amp. With negative feedback, the huge internal gain drops out of the result. Only the resistor values decide the gain.

2The two rules of the ideal op-amp

With negative feedback, you can analyse almost every op-amp circuit with two simple rules:

  • Rule 1: the two inputs sit at the same voltage, V− = V+. The output adjusts itself until this is true.
  • Rule 2: no current flows into either input.

If the + input is connected to ground, rule 1 says the − input is at 0 V as well. It is not wired to ground, yet it sits at 0 V. This is called a virtual ground.

Rule 2 means that a current arriving at the − input cannot go into the op-amp. It must continue through the feedback resistor Rf.

Self-check In a circuit with feedback, the + input is connected to ground. What is the voltage at the − input?
  1. +15 V
  2. 0 V
  3. The input voltage
  4. The output voltage

Answer: 0 V. Rule 1: V− = V+ = 0 V. The − input is a virtual ground.

3Inverting and non-inverting amplifiers

The inverting amplifier feeds the signal through R1 into the − input. The + input is grounded.

−+Vout​Rf​Vin​R1​
Vout = −(Rf / R1)·Vin

The minus sign means the output is upside down: a positive input gives a negative output.

The non-inverting amplifier feeds the signal straight into the + input. R1 goes from the − input to ground.

−+Vout​Rf​R1​Vin​
Vout = (1 + Rf / R1)·Vin

Its gain is always at least 1, and the output has the same sign as the input.

Self-check A non-inverting amplifier has R1 = 1 kΩ, Rf = 9 kΩ and Vin = 0.5 V. What is Vout?
  1. 0.5 V
  2. 4.5 V
  3. 5 V
  4. −4.5 V

Answer: 5 V. Gain = 1 + 9/1 = 10, so Vout = 10·0.5 = 5 V. 4.5 V forgets the “1 +”, and −4.5 V uses the inverting formula.

4The method step by step

You can use the formulas, but it is safer to know where they come from. This method works for every circuit in this topic:

  1. Find V+. Grounded: 0 V. Connected to the input: Vin.
  2. Use rule 1: V− = V+.
  3. Find the current through R1 with Ohm’s law, using the voltages at its two ends.
  4. Use rule 2: the same current flows through Rf.
  5. Walk from the − input through Rf to the output: Vout = V− − I·Rf, with I flowing towards the output.

With kΩ and V, currents come out in mA. That keeps the numbers small.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

An inverting amplifier with an ideal op-amp has R1 = 2 kΩ, Rf = 10 kΩ and Vin = 0.5 V. Find the current I1 and the output voltage Vout.

1

Look at the circuit

−+Vout​Rf​ = 10 kΩ0.5 VR1​ = 2 kΩI1​I1​0 V

The + input is grounded, so this is an inverting amplifier.

2

Virtual ground

V+ = 0 V, and the feedback forces V− = V+.

V− = 0 V
3

Current through R₁

I1 = (Vin − V−) / R1 = (0.5 − 0) / 2 = 0.25 mA
4

Output voltage

No current enters the op-amp, so the 0.25 mA continues through Rf towards the output.

Vout = 0 − 0.25·10 = −2.5 V
5

Check the answer

With the formula: Vout = −(10/2)·0.5 = −2.5 V. The gain is −5, and −2.5 V lies well inside the ±15 V supplies.

Answer

I1 = 0.25 mA and Vout = −2.5 V.

6Common mistakes

  • Forgetting the minus sign of the inverting amplifier.
  • Using Rf/R1 for the non-inverting gain. It is 1 + Rf/R1.
  • Mixing up which input the signal goes to. Look at the drawing: signal into − through R1 means inverting.
  • In a summing amplifier, using only one input. The currents of all inputs add up in Rf.
Self-check An inverting amplifier has R1 = 1 kΩ, Rf = 4.7 kΩ and Vin = −0.2 V. What is Vout?
  1. 0.94 V
  2. 1.14 V
  3. −0.94 V
  4. −1.14 V

Answer: 0.94 V. Vout = −(4.7/1)·(−0.2) = +0.94 V: two minus signs give a plus. −0.94 V drops one sign, and ±1.14 V uses the non-inverting gain 5.7.

7How exam questions look

  • Easy: an inverting amplifier. You find Vout or the current I1 through R1. The input can be negative.
  • Medium: a non-inverting amplifier. You find Vout, or choose Rf for a given gain: Rf = (gain − 1)·R1.
  • Hard: a summing amplifier with two inputs, each through its own resistor into the − input.

In a summing amplifier the − input is a virtual ground, so each input sends its own current V/R into the node. The currents add up and all flow through Rf:

If = V1/R1 + V2/R2 Vout = −Rf·If

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
−+Vout​Rf​ = 47 kΩVin​ = 1.5 VR1​ = 10 kΩ

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

Vin = 1.5 VR1 = 10 kΩRf = 47 kΩ

Find Vout (V)

Show hints
  1. The + input is grounded, so the − input is a virtual ground: V₋ = 0.
  2. No current enters the op-amp, so the current through R₁ continues through Rf.
  3. I1 = 0.15 mA.
Show the answer and the step-by-step solution

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Virtual ground

V₊ = 0 and the feedback forces V₋ = V₊.

V₋ = V₊ = 0 V
I1 = (Vin − 0)/R1 = 1.5/10 = 0.15 mA

Output voltage

The same current flows through Rf, from the − input towards the output.

Vout = 0 − I1·Rf = −0.15·47 = −7.05 V
(gain = −Rf/R1 = −4.7)

Answer

Vout = −7.05 V
Example 2Medium
−+Vout​Rf​ = 47 kΩR1​ = 1 kΩVin​

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. The non-inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.

R1 = 1 kΩRf = 47 kΩVsat = ±12 V

Find Vin,max (V)

Show hints
  1. For a non-inverting amplifier the gain is 1 + Rf/R₁.
  2. The output peak |gain|·Vin must stay below Vsat.
  3. |Av| = 48.
Show the answer and the step-by-step solution

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Gain

Av = 1 + Rf/R1 = 1 + 47/1 = 48

Largest input before clipping

Vin,max = Vsat/|Av| = 12/48 = 0.25 V

Answer

Vin,max = 0.25 V
Example 3Hard
−+Vout​C = 4.7 µFVin​ = −2.6 VR = 22 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 = −2.6 V is applied at t = 0. How long after t = 0 does the output reach saturation? Find t.

Vin = −2.6 VR = 22 kΩC = 4.7 µFVsat = ±12 V

Find t (ms)

Show hints
  1. The − input is a virtual ground, so a constant current Vin/R flows through R and on into the capacitor.
  2. A constant current charges the capacitor linearly: Vout(t) = −Vin·t/(RC).
  3. RC = 103.4 ms.
Show the answer and the step-by-step solution

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Input current and ramp rate

kΩ times µF gives milliseconds.

RC = 22 kΩ · 4.7 µF = 103.4 ms
ramp rate = −Vin/(RC) = −(−2.6)/103.4 = 0.02515 V/ms

Time to saturation

|Vin|·t/(RC) = Vsat → t = Vsat·RC/|Vin| = 12·103.4/2.6 = 477.2 ms

Answer

t = 477.2 ms

Questions and answers

What is Op-amp circuits?

An operational amplifier (op-amp) is a small chip that amplifies voltages. It is used in audio equipment, sensors and filters.

It has two inputs, marked − (inverting) and + (non-inverting), and one output. It amplifies the tiny difference between the two inputs by an enormous factor, often 100 000 or more.

How do I solve Op-amp circuits problems step by step?

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

What are common mistakes in Op-amp circuits?
  • That is the non-inverting gain. This is an inverting amplifier: Vout = −(Rf/R₁)·Vin.
  • The output is the input times the gain, so the input limit is Vsat divided by the gain.
  • One time constant RC is not the answer here: the output ramps linearly at Vin/(RC) volts per millisecond until it hits Vsat.
What do the Op-amp circuits exercises cover?
  • Easy: inverting amplifier
  • Medium: non-inverting amplifier
  • Hard: summing amplifier

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

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Nodal analysis

Calculators

Op-amp gain calculator

Courses with this topic

Circuit Analysis 1Electronics 1Measurement and InstrumentationFE Electrical and Computer exam practice

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Printable worksheet

Op-amp circuits worksheet with answers
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