# Calculate op-amp output voltage

I am trying to calculate the output voltage of the circuit. I know that this is a voltage follower, but I am trying to understand how it works. simulate this circuit – Schematic created using CircuitLab

The first "golden rule" implies that

$$V_a = V_b = V_{in}$$

Now, the opamp has to set the output voltage to satisfy the above condition, but how do I calculate it?

I can use the second "golden rule" and assume that no current flows into the inverting input, hence there is no voltage drop across R1 and Va equals Vin. But the current does flow from input to output through R1+R2, so there will be a voltage drop across the resistors.

• If current was flowing across R1+R2 from the input to the output, there would be a voltage drop across R1. Feb 18 '15 at 2:54
• True. I was confused by the fact that op-amp inputs draw no current which lead me into thinking that the only path for the current to flow is R1+R2.
– v.m.
Feb 18 '15 at 12:56

As you said, $$V_a = V_b = V_{in}$$ As $V_{in}=V_a$, there is no current through $R_1$. Since there is no current flowing into/from the negative terminal as well, we conclude that there is no current through $R_2$ as well. So if there is no current through $R_2$, the voltage on both sides of it is equal, and it is $V_a$ which is $V_{in}$. So $V_{out}=V_{in}$ as expected. No currents through resistors.

The shown circuit is a unity gain amplifier - and one could ask if it is just an "academical" exercise or has this circuit any practical relevance?

The answer is: Yes - the circuit has some advantages if compared with the classical voltage follower. Unlike the classical follower, this circuit has a feedback factor below unity [R1/(R1+R2)] which results in a lower loop gain with an increased stability margin. This allows a step response with smaller (or even without) overshoot. Even opamps that are not unity-gain compensated can be employed. Because everything in electronics is a trade-off, we also have a kind of drawback: The bandwidth is smaller because the loop gain cross-over frequency (0 dB crossing) is reduced correspondingly.

EDIT: Step-by-step calculation is possible using superposition at the output:

• Inverting part: Vout,1=Vin(-R2/R1)

• Non-inv. Part: Vout,2=Vin(1+R2/R1)

• Together: Vout/Vin=(1+R2/R1)-R2/R1=1.

• Indeed, this is a textbook exercise. Being self-taught, it's quite hard to understand the basics sometimes :) Thank you for the explanation.
– v.m.
Feb 18 '15 at 13:00

Here is an intuitive explanation of this exotic voltage follower...

The voltage at the inverting input follows the voltage at the non-inverting input since the op-amp, obeying the H&H Golden rules, tries to keep a zero voltage difference between its inputs. To do it, the op-amp output voltage follows the voltage at the inverting input. As a result, there is no current flowing through the R1-R2 network... and all the three circuit nodes have the same voltage in regards to ground... they are equipotential... they "move" together with the same rate...

The circuit does not load the input source and has extremely high input resistance since the input source sees the virtually increased R1 (strictly speaking, R1 + R2) resistance. This phenomenon is known as "bootstrapping".

A question arises, "What is the point of this circuit, since there is a simpler circuit (without any resistors) of a voltage follower?"

We can find the answer if we disconnect the non-inverting input from the input source and connect it to the ground - then the circuit will act as an inverting amplifier with gain of -1 (an inverter). We can commutate the non-inverting input even in a simpler way by connecting a resistor between the input source and the non-inverting input and simply grounding the non-inverting input.

So the circuit can act either as follower (ungrounded non-inverting input) or inverter (grounded non-inverting input). You can see a possible application of this circuit solution in another question - Precision high speed peak detector.