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The Ideal Operational Amplifier and Its Basic Configurations

8 min read · bitGateLab Team

The operational amplifier, or op-amp, is the most versatile building block in analog electronics. With the same chip and two or three resistors you can amplify, invert, add or subtract signals, or isolate one stage from another. Best of all, if you know two simple rules, you can analyze almost any op-amp circuit in just a few lines. In this topic you'll see the ideal model, those golden rules and the five configurations you will run into again and again.

The ideal op-amp

An op-amp has two inputs, the inverting (−) and the non-inverting (+), and one output. It amplifies the difference between its inputs:

Vout = A · (V+ − V−)

where A is the open-loop gain. The ideal model assumes that:

What you always do have to respect is the power supply: the output cannot go beyond the voltages of the supplies that power the op-amp. If the calculation gives a value outside that range, the output saturates near the corresponding rail.

The two golden rules

With such a large gain, the op-amp is only useful as a linear amplifier if part of the output is fed back to the inverting input: that is called negative feedback. When it is present, the op-amp adjusts its output so that the difference between its inputs is practically zero. The golden rules follow from that:

  1. Voltage rule: with negative feedback, V+ = V−.
  2. Current rule: no current flows into either of the two inputs.
The voltage rule holds only with negative feedback and without saturation. An op-amp with no feedback works as a comparator: its output jumps to one rail or the other, and V+ is not equal to V−.

Inverting amplifier

The signal enters through Rin to the inverting input, Rf connects the output to that same input, and the non-inverting input goes to ground.

Vin Rin Rf −+ Vout
Inverting amplifier. The + input is grounded, so the voltage rule puts the − input at 0 V (virtual ground).

Analysis with the rules: since V+ = 0, V− = 0 as well (virtual ground). The current through Rin is Vin/Rin, and since it does not enter the op-amp, all of it flows through Rf, which leaves the output at −Rf·I:

Vout = −(Rf / Rin) · Vin

The minus sign indicates a 180° inversion. The input impedance seen by the source is simply Rin.

Example 1. Rin = 10 kΩ, Rf = 47 kΩ, ±12 V supply.

  1. Gain: −47/10 = −4.7.
  2. With Vin = 0.5 V: Vout = −4.7 · 0.5 = −2.35 V.
  3. With Vin = 3 V the calculation gives −14.1 V, which is outside the supply range. The output saturates near −12 V (a little less in practice) and the circuit is no longer linear.

Non-inverting amplifier and follower

Now the signal goes straight into the + terminal. Rf runs from the output to the − input, and Rg from the − input to ground. By the voltage rule, V− = Vin, and Rf and Rg form a divider that must produce that value:

Vout = (1 + Rf / Rg) · Vin

The output is in phase with the input, the gain is never less than 1, and the input impedance is very high because the signal goes directly to the + terminal.

If Rf = 0 (output tied to the − input) and you remove Rg, the gain is exactly 1: this is the voltage follower, or buffer. It does not amplify, but it copies the voltage without drawing current from the source and can deliver current to the load.

Example 2. Non-inverting amplifier with Rf = 20 kΩ and Rg = 10 kΩ: gain = 1 + 20/10 = 3. With Vin = 1.2 V, Vout = 3.6 V.

Why the follower is useful. A sensor with an internal resistance of 10 kΩ generates 1 V and has to drive a 1 kΩ load. Connected directly, the divider leaves 1 · 1/(10 + 1) = 0.091 V across the load. With a follower in between, the sensor delivers no current, the op-amp copies the 1 V and the load receives 1 V.

Inverting summing amplifier

If you connect several inputs to the inverting input, each through its own resistor, the currents add at the virtual ground and flow together through Rf:

Vout = −Rf · (V1/R1 + V2/R2 + …)

Because the − input is at 0 V, the sources do not affect one another. With all the resistors equal to Rf, the output is the negated sum of the inputs; that is how an audio mixer works.

Example 3. R1 = 10 kΩ, R2 = 20 kΩ, Rf = 20 kΩ, with V1 = 0.5 V and V2 = −0.2 V.

  1. Weight of each input: Rf/R1 = 2 and Rf/R2 = 1.
  2. Weighted sum: 2 · 0.5 + 1 · (−0.2) = 0.8 V.
  3. Vout = −0.8 V.

Difference amplifier

This circuit amplifies the difference between two signals. V1 enters through R1 to the − terminal, with R2 from the output to that same terminal. V2 enters through R3 to the + terminal, with R4 from there to ground. If you choose R2/R1 = R4/R3:

Vout = (R2 / R1) · (V2 − V1)

Any voltage common to both inputs cancels out. That cancellation depends on the resistor ratios matching; with tight-tolerance resistors the rejection is better.

Example 4. R1 = R3 = 10 kΩ and R2 = R4 = 100 kΩ (gain of 10). A sensor bridge delivers V2 = 1.25 V and V1 = 1.20 V.

Vout = 10 · (1.25 − 1.20) = 0.5 V. The 1.2 V that both inputs have in common does not appear at the output.

Common mistakes

Try it in the simulator

Build the configurations and check the gains with the DC analysis and with the virtual oscilloscope.

  1. Place an op-amp and power it from two DC sources of +12 V and −12 V with respect to ground.
  2. Wire up the inverting amplifier with Rin = 10 kΩ and Rf = 47 kΩ and a 0.5 V DC input.
  3. Run the DC analysis and measure the output and the voltage at the − input, which should be very close to 0 V.
  4. Change the input to 3 V and observe the saturation.
  5. Replace the input with a 1 V peak sine wave and look at the inverted 4.7 V peak output on the oscilloscope.
  6. Rearrange the resistors to form the non-inverting, summing and difference amplifiers, and compare with your calculations.

Exercises

  1. An inverting amplifier has Rin = 2.2 kΩ and Rf = 22 kΩ. What is the output for Vin = 0.3 V?

    Show answer

    Gain = −22/2.2 = −10. Vout = −10 · 0.3 = −3 V.

  2. You need a non-inverting amplifier with a gain of 11 and Rg = 1 kΩ. What Rf do you use?

    Show answer

    1 + Rf/Rg = 11, so Rf/Rg = 10 and Rf = 10 kΩ.

  3. A summing amplifier has three inputs with 30 kΩ resistors and Rf = 10 kΩ. If the inputs are 1 V, 2 V and 3 V, what is the output?

    Show answer

    Each weight is 10/30 = 1/3. Vout = −(1 + 2 + 3)/3 = −2 V: the circuit delivers the negated average.

  4. A difference amplifier with a gain of 5 receives V2 = 2.0 V and V1 = 2.3 V. What is the output?

    Show answer

    Vout = 5 · (2.0 − 2.3) = −1.5 V.