# Inverting OpAmp question

#### RFeng

Joined May 30, 2012
27
Hi,

I simulated the below inverting OpAmp: and got the below results: I don't manage to understand why is it that for input vapc0 = 1.035V we get at the output vapc = 1.967V.

It looks like that: vapc = 3V - vapc0; I don't understand why.

Thank you.

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#### nomurphy

Joined Aug 8, 2005
567
R7/R6 is a gain of 1, not 100dB, because they are both 1K ohms. To get "only" 80dB of gain, R7 would need to be a 10 Meg.

What is the point of inducing +1.5V offset into the positive pin?

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• RFeng

#### crutschow

Joined Mar 14, 2008
25,692
The gain from the non-inverting input to output is (1 + R7/R6) = +2 (when R6 is grounded or connected to a low impedance source). The gain from the inverting input to output is -R7/R6 = -1. The two signals sum at the output accordingly.

• RFeng

#### RFeng

Joined May 30, 2012
27
The gain from the non-inverting input to output is (1 + R7/R6) = +2 (when R6 is grounded or connected to a low impedance source). The gain from the inverting input to output is -R7/R6 = -1. The two signals sum at the output accordingly.
Hi crutschow, Thanks!

I think I understand you.
Is it correct to say that in general: R7/R6 is a gain of 1, not 100dB, because they are both 1K ohms. To get "only" 80dB of gain, R7 would need to be a 1meg.

What is the point of inducing +1.5V offset into the positive pin?
Hi,
Thank you too!
I do not know the purpose of the 1.5V offset, as I'm currently learning the entire system (which the circuit is just part of it).
The system is an Automatic Gain Control unit.
How did you get the 80dB? (if R7 = 1MΩ).

#### nomurphy

Joined Aug 8, 2005
567
How did you get the 80dB? (if R7 = 1MΩ).
Sorry typo, 10 Meg -- the point being that generally it is stretching the usefulness of most op amps.

• RFeng

#### crutschow

Joined Mar 14, 2008
25,692
.......................
Is it correct to say that in general: .......................
You got it. • RFeng

#### RFeng

Joined May 30, 2012
27
Thank you dear fellows #### upand_at_them

Joined May 15, 2010
555

1. R6 and R7 create a voltage divider.
2. Because you have feedback the op-amp will try to make vapc such that the inverting input is the same as the non-inverting input.
3. The non-inverting input is 1.5V.

So you have a voltage divider with 1.035V, a 1k resistor, 1.5V, a 1k resistor, and then some unknown voltage vapc. Solving for vapc gives you the result you got in simulation.

• RFeng

#### WBahn

Joined Mar 31, 2012
26,326
You got it. Huh?

If

$$Vout = V_+ \left( 1\;+\; \frac{R_7}{R_6} \right) +\;V_-\left(-\frac{R_7}{R_6} \right)$$

Then that means that

$$Vout = V_+ \;+\; (V_+\;-\;V_-) \frac{R_7}{R_6}$$

Since (V+) - (V-) is going to be a tiny voltage, this would say that Vout is always equal to V+.

The simplest way (at least to me) to get the transfer characteristic is to analyze the circuit. The current flowing from the input is:

I = (Vin-V+)/R6

The output voltage is:

Vout = (V-) - I*R7

Assuming V+ ~= V-, we have

Vout = (V+) - ((Vin-V+)/R6)*R7

Vout = (V+)(1+ R7/R6) - (Vin)(R7/R6)

NOTE: I now see how the error in the OP's equation was easily missed.

For V+ = 1.5V and R7=R6, this reduces to

Vout = 3V - Vin

• RFeng

#### crutschow

Joined Mar 14, 2008
25,692
Huh?

If

$$Vout = V_+ \left( 1\;+\; \frac{R_7}{R_6} \right) +\;V_-\left(-\frac{R_7}{R_6} \right)$$

Then that means that

$$Vout = V_+ \;+\; (V_+\;-\;V_-) \frac{R_7}{R_6}$$

Since (V+) - (V-) is going to be a tiny voltage, this would say that Vout is always equal to V+.

The simplest way (at least to me) to get the transfer characteristic is to analyze the circuit. The current flowing from the input is:

I = (Vin-V+)/R6

The output voltage is:

Vout = (V-) - I*R7

Assuming V+ ~= V-, we have

Vout = (V+) - ((Vin-V+)/R6)*R7

Vout = (V+)(1+ R7/R6) - (Vin)(R7/R6)

NOTE: I now see how the error in the OP's equation was easily missed.

For V+ = 1.5V and R7=R6, this reduces to

Vout = 3V - Vin
I don't follow the logic of your post. You start and end with the same equation while implying that there is some error.

You say "Since (V+) - (V-) is going to be a tiny voltage, this would say that Vout is always equal to V+." I don't understand why you say that. (V+) - (V-) may or may not be a tiny voltage. It depends upon their relative value.

What error in the OP's equation are you talking about? • RFeng

#### WBahn

Joined Mar 31, 2012
26,326
I am assuming that V+ is the voltage at the noninverting input of the opamp and V- is the voltage at the inverting input. This is almost always what I see these terms to mean in an opamp circuit. Using one term to mean the voltage at one of the input pins and the other to mean the voltage someplace else seems a bit ackward to me, but I guess if the intent was to mean both to be the voltage someplace else and that it just turns out that V+ also happens to be the voltage at the noninverting input, then I see what you are saying.

• RFeng

#### JMac3108

Joined Aug 16, 2010
348
RFeng,

The easiest way to analyze this kind of circuit is using superposition. The basic procedure is to consider one input at a time while grounding the other, then add the results.

In your circuit, first ground the 1.5V source and conside the input. The circuit becomes a simple inverting amp with equation, Vout1 = -Vin(Rf/Rin).

Then ground the input and consider the 1.5V source. In this case the circuit becomes a simple non-inverting amp with equation, Vout2 = Voffset (1+Rf/Rg).

Combine the results and you'll get your equation. Vout = Vout1 + Vout2.

If you do a few of these, you'll get where you can analyze them in your head. Its a really simple way to approach this kind of circuit.

• RFeng
You've given me a great help. 