# Calculation of the input resistance of an op amp circuit

After I calculated that $$v_s=v_u\left(\frac{R_1}{R_1+R_2}\right)$$ I have to calculate the resistance seen by the voltage generator $v_s$.

My book, without any calculation, says it is: $+\infty$. Now, I am trying to figure out why.

I thought that the resistance seen by $v_s$ is $$\frac{v_s}{\frac{e_1-v_s}{R_s}}$$ (with the nodal analysis).

I tried to do algebraic calculations, but the results doesn't come out as $+\infty$. How should I calculate it?

simulate this circuit – Schematic created using CircuitLab

EDIT: If ideal op. amplifiers have infinite input resistance, so should be in the following circuit:

simulate this circuit

My book says in this last circuit it's $R_1$ (without any calc. too). Why?

• In the first circuit there is no current through Rs into the op-amp, hence input z is infinity. In the second circuit there is an input current, and that current flows through R1 and R2 to the op-amp output. Also, the -ve op-amp input is at virtual earth hence the magnitude of that current is Vs/R1, or in other words the input z is R1
– Chu
May 11 '15 at 23:42

When an ideal op amp is connected with negative feedback, it obeys two rules:

1. The voltages at the two input pins are equal.
2. No current flows into either pin.

In your first circuit, $V_S$ is only connected to the non-inverting input. By rule #2, no current flows into that input. This lets us calculate the equivalent input resistance:

$$I_S = 0\ \mathrm A$$ $$R_{in} = \frac{V_S}{I_S} = \frac{V_S}{0\ \mathrm A} = \infty \ \Omega$$

Your second circuit is drawn incorrectly. You've connected R2 to one of the op amp's power pins, but it should be connected to the output. In this circuit, current can flow from $V_S$ to the output, although none flows into the inverting input. Here we need rule #1, which tells us that the voltage at the inverting input is equal to the voltage at the non-inverting input -- zero volts (ground). So the circuit acts like the right side of $R_1$ is grounded, which makes $R_1$ the input resistance. We can work out the math, too:

$$I_S = \frac{V_S - 0\ \mathrm V}{R_1} = \frac{V_S}{R_1}$$

$$R_{in} = \frac{V_S}{I_S} = \frac{V_S}{\frac{V_S}{R_1}} = R_1$$

• The golden rules of Op Amps! Although, I believe a better way to say it is that the voltage of the two input pins will be equal. Your way is accurate, but might be misread at first. May 12 '15 at 5:27
• Yeah, that's a better way to phrase it. I edited the answer. May 12 '15 at 13:43

Ideal op amps have infinite input resistance, and the voltage source is connected to only the input through Rs.