Graded op-amp labs: an inverting gain of 10, a non-inverting sensor amplifier, a buffer, a summing amplifier, and where a 741 clips.
On the syllabus
The ideal op-amp rules: no input current, equal inputs under negative feedback
Inverting and non-inverting amplifiers
The voltage follower (buffer)
The summing amplifier
Output swing, headroom and clipping
You will be able to
Design an inverting amplifier for a given gain.
Scale a small sensor voltage with a non-inverting amplifier on one supply.
Use a follower so a load does not pull a divider down.
Add voltages with a summing amplifier.
Measure an op-amp’s output limits and explain them.
The first challenge is free. The rest, the viva and the certificate are part of Pro, and everything is here to read.
Graded challenges
Values like 4.7k, 100n or 2.2u are fine. Readings are the number only, in the unit shown. Every check measures the simulation, not a bench: ideal parts within their models, no noise.
1. Design an inverting amplifier with a gain of −10
With negative feedback an op-amp holds its two inputs at the same voltage. IN+ is grounded, so IN− sits at 0 V: a virtual ground. The input current through RIN has nowhere to go but through RF, and that sets the output.
FormulaGain = −RF / RIN
Worked exampleRIN = 4.7 kΩ and RF = 47 kΩ give a gain of −10. RIN = 2 kΩ and RF = 10 kΩ give −5, so a 1 V peak sine comes out as 5 V peak, upside down.
On the scope or meterOn the scope the output is the input flipped and scaled. Peak to peak is twice the peak.
Your task
A 741 on ±12 V is wired as an inverting amplifier: RIN = 10 kΩ, RF from the output to the inverting input. A 1 kHz sine of 0.5 V peak goes in.
Choose RF for a gain of −10, so the output swings 10 V peak to peak, within 3%.
2. Scale a sensor with a non-inverting amplifier
In a non-inverting amplifier the signal goes into IN+. Feedback makes IN− match it, so the divider RF over RG must put the input voltage on IN−. The gain is always at least 1, and the source sees an almost infinite input resistance.
FormulaGain = 1 + RF / RG
Worked exampleTo turn 0.5 V into 2.5 V you need a gain of 5, so RF / RG = 4: RG = 2.2 kΩ and RF = 8.2 kΩ gives 2.36 V, close enough for most ADCs.
On the scope or meterA steady DC output. A rail-to-rail op-amp can reach close to 5 V, so 3.3 V is no problem.
Your task
A sensor gives 0.2 V. A rail-to-rail op-amp on a single 5 V supply amplifies it for a 3.3 V ADC: RG = 1 kΩ to ground, RF from the output back.
Choose RF so the output is 3.3 V, within 5%.
3. Buffer a divider with a follower
A voltage follower is a non-inverting amplifier with RF = 0 and no RG: the output is tied to IN−, so the output copies IN+. It takes almost no current from the source and can supply mA to a load from its own rails.
FormulaVout = Vin, and I load = Vout / R load
Worked exampleA 22 kΩ / 22 kΩ divider from 9 V gives 4.5 V. Through a follower, a 470 Ω load still sees 4.5 V and draws 4.5 / 470 = 9.6 mA.
On the scope or meterThe load voltage matches the divider’s unloaded tap, and the current in the load is that voltage over its resistance.
Your task
A 10 kΩ / 10 kΩ divider makes 6 V from 12 V. Hung straight on the divider, a 1 kΩ load would pull it down to about 1 V.
Here a rail-to-rail op-amp is wired as a voltage follower between the divider and the load. Measure the voltage across the load and the current the op-amp supplies to it.
4. Add two voltages with a summing amplifier
Several inputs, each through its own resistor, feed the virtual ground at IN−. Their currents add there, and all of that current flows through RF. So the output is a weighted, inverted sum.
FormulaVout = −RF (V1 / R1 + V2 / R2)
Worked exampleInputs of 2 V and 1 V through 10 kΩ each, with RF = 5 kΩ: Vout = −5k × (0.2 mA + 0.1 mA) = −1.5 V.
On the scope or meterA steady negative DC output. Change one input and the output moves by its weight alone.
Your task
Two sources, 1 V and 0.5 V, each feed the inverting input of an op-amp through 10 kΩ. RF goes from the output back to that input.
Choose RF so the output is −3.0 V, within 3%: twice the sum, inverted.
5. Find where a 741 clips
An op-amp’s output cannot go past its supply rails. An older part like the 741 stops well short of them. Ask for more and the tops of the wave are flattened: clipping.
FormulaVout ≈ gain × Vin, until it hits V+ − headroom or V− + headroom
Worked exampleA gain of −5 with a 3 V peak input asks for 15 V peak. On ±9 V a 741 stops near ±7.5 V, so the sine comes out with flat tops and bottoms.
On the scope or meterA sine with flat tops and bottoms. The flats are the limits you are measuring.
Your task
The inverting amplifier with a gain of −10 now gets a 2 V peak sine. Ten times that, 20 V peak, is more than a ±12 V supply can give.
Put the scope on the output. Measure the highest and the lowest voltage it reaches.
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.