# Simple op amp question, finding gain and input resistance

Why is the gain -10 for the following op amp? Isn't there a voltage divider between the 100k and the leftmost 10k yielding -11 as gain? And why aren't these resistors playing any role in evaluating the input resistance? Why would Rin be merely 10k?

• Perhaps you're missing the point that op-amp "-" input remains very close to ground for any Vin? So your voltage-divider idea is incorrect. – glen_geek Jun 17 '17 at 19:40
• Just a note: the 2nd circuit in this question is the same as yours. As one answer there says, "the load resistor is "shorted" to the actual ground, so the current through that resistor does not affect the feedback section at all." Assuming an ideal op-amp (which I bet you are) the rightmost 10k resistor won't affect the transfer function relating the input/output voltages of the inverting op-amp you've shown. – Vladislav Martin Jun 17 '17 at 19:42
• How can you study all this op-amp circuit without knowing the basic stuff about inverting op-amp? – G36 Jun 17 '17 at 22:56
• @G36 I AM familiar with inverting op-amps but I can doubt my understanding at times. – peripatein Jun 17 '17 at 23:09
• Apart from completely missing the point, how do you get a gain of -11? – Chu Jun 18 '17 at 0:33

To give you a better understanding what is going on in the inverting amplifier let us at the beginning use this circuit:

We simply have an ideal voltage amplifier with different $A_O$ gains.

Now let as try to find an input resistance.

$$R_{IN} = \frac{V_{IN}}{I_{1}}$$

$$I_{1} = \frac{V_{IN} - V_{OUT}}{R_F} =\frac{V_{IN} - A_OV_{IN}}{R_F} = V_{IN} \frac{1 -A_O}{R_F}$$

$$R_{IN} = \frac{R_F}{1-A_O}$$

And for inverting input we have

$$R_{IN} = \frac{R_F}{1-( - A_O)} = \frac{R_F}{1+|A_O|}$$

For example if $A_O = 10$ (open loop gain) in input current is $I_1 = \frac{11V}{10k\Omega} = 1.1mA$ and the input resistance is :

$$R_{IN} = \frac{1V}{1.1mA} = 909.09\Omega$$

As you can see our $R_{IN}$ resistance is $(1 + |A_O|)$ smaller than $R_F$ if we have inverting amplifier. And this is what we call a Miller effect.

And now if we add a resistor between the signal source and the op-amp inverting input we created the inverting amplifier.

As you can see this time the amplifier input resistance is equal to:

$$R_{IN} = R_1 + \frac{R_F}{1+|A_O|}$$

And for the Op-amp $A_O \to \infty$ approach to infinity.

Hence we have a "virtual Ground" at the inverting input and $R_{IN} = R_1$ and the voltage gain is $A_V = -\frac{R_F}{R_1}$ due to negative feedback.

Or we can treat this circuit as voltage to current converter $R_1$ and the ideal Op-amp plus $R_F$ forms $R_1$ current to voltage converter.

The positive feedback (RF resistor between the opamp output and non-inverter input) would cause $R_{IN}$ to increases (for $A_O$ from 0 to 1, and for $A_O$ large the 1 we would create a negative resistance). But this is a different story.

Think about the Vin- pin of the opamp. For large, very large gains, even with 10 volts outpout, that Vin- will be ~~ zero volts (OK, not truly zero but within 10 or 20 microvolts of zero). The OPA211 with 130dB gain (at DC) has 3,000,000x gain, meaning with +10 volts Vout the voltage between Vin- and Vin+ is only 3.1 microVolts.

We label this behavior "virtual Ground" and visualize the currents flowing through Rin and Rfb as passing by that virtual ground, from input voltage on left side of Rin to right side of Rfb.

With Vin across Rin, and the same current flowing into Rfb to exit into the OpAmp's internal transistors (and its power pins), the Vout will be 10X the Vin.

Welcome to the useful world of "virtual ground" behavior.

• To understand WHY the inverting input is "nearly" at ground potential, you must understand how negative feedback works. – LvW Jun 18 '17 at 7:35