 ### Network # How does gain apply in diff op amp?

#### 24Volts

Mar 21, 2010
164
Hello,

Intuitive deductions in reference to attachment:

1) Initial op amp assumption:
Output = 0V
Va will sense: va = v1(Ri/Rf+Ri) = 2(10K/30K) = 0.6VDC.... Therefore: 2-0.6 = aprox 1.3VDC
Vb will sense: vb = v2(R2/R1+R2) = 5(1.5K/5K) = 1.5VDC

Therefore, since the Non inverting input > inverting input, we know that Vout will be a positive voltage.

2) We also know that increasing Va (Inverting input) will cause Vout to ramp down!

3) If Vb = 1.5VDC, Then, Va should also equal to 1.5VDC

4) Iri = (v1-va)/Ri = (2v-1.5V)/10K = 50ua

5) We can see that Vri is 0.5VDC... and now we can solve for Vrf.
If Iri = 50ua, then Irf also equals 50ua, so then Vrf is:
50ua x 20K = 1VDC

6) Vout = v1 - Vri - Vrf = 2 - 0.5 - 1 = +0.5VDC

7) Op amp gain = Rf/Ri = 20K/10K = 2

8) Proof with Differential op amp formula:

Vout = V2 (R2/(R1+R2)) ((Ri+Rf)/Ri) - V1(Rf/Ri)
Vout = 5(1.5K/5K) (30K/10K) - 2(20K/10K)
Vout = 4.5 - 4 = +0.5VDC

My question is, in reference to the attachment, if the differential op amp amplifies the voltage difference between va and vb by a gain factor of 2, why is the amount divided by 2 instead of multiplied by 2. In other words, a quintillianth of a second after we power up the op amp we have a difference of 0.16666 VDC at the op amp's inputs:

1.5VDC - (2-0.6666VDC) = 0.16666VDC

Therefore, 0.16666VDC multiplied by a gain of 2:

should give 0.166666 x 2 = 0.3333VDC as vout instead of +0.5VDC ??

I am a little confused on how the gain applies here.... We know we have a gain of 2 right! So what exactly is multipied by 2??

Thanks for all the help!

Last edited:

#### duke37

Jan 9, 2011
5,364
We were here only a short time ago, look up the previous thread.

The gain is -2, offset by 1.5V.
The + and - input will be at 1.5V.
Ri drops 0.5V. The same current goes through another resistor of double the value so will drop 1V.
2V - 0.5V - 1V = 0.5V

#### 24Volts

Mar 21, 2010
164
1) Why is it -2 as opposed to +2?

Ri drops 0.5V. The same current goes through another resistor of double the value so will drop 1V.
2V - 0.5V - 1V = 0.5V

I know that, I even said it in my post didn't I... that's not the question.

My question is, what is being multiplied by a gain of 2 to get to +0.5V output ???

Or lets look at it this way, if my output is +0.5V from a gain of 2. Then obviously 0.25VDC was multiplied by 2.... but I don't see 0.25V anywhere....

I don't get what is multiplied by a gain of 2???

thanks

#### duke37

Jan 9, 2011
5,364
The gain is -2 rather than +2 since it is an inverting amplifier.
I missed point 6 because of the complicated post before and after it with many irrelevant decimal points.

The + input (1.5V) is the reference point. The - input voltage is 0.5V more than this so the output voltage will be 1V less than this.

#### 24Volts

Mar 21, 2010
164
thanks duke, .... this is exactly what I was looking for ... a simple explanation for the gain.

One more thing, if I may ask. How did you determine that this was an inverting op amp?
They say that if the + input is > than the - input, then its an non inverting op amp!!!

Isn't this our case here?

Innitially when op amp is powered, isn't +input = 1.5V and -input = 1.3VDC???

cause usually an inverting op amp has a negative output... right? We have a positive one!!!!

Last edited:

#### Laplace

Apr 4, 2010
1,252
It might help to perform a general analysis of this circuit. Write a node equation (the sum of all currents at a node must be zero) for the nodes Va and Vb. Then realize that for the op-amp to be in its linear operating region (the only region of interest here) both input terminals (+ & -) must be at the same voltage. See attachment.

Note that this circuit's inputs (V1 & V2) each have different gain to the output.

#### Attachments

• Diff-Amp-Gain.pdf
22.5 KB · Views: 101
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#### 24Volts

Mar 21, 2010
164
Hi Laplace,

I know how to calculate the output !

and yes:

-50ua + 50ua = 0
and
-1ma + 1ma = 0

So, what does this mean?

I think what I was looking for is the intuitive solution of why we have a positive voltage... and I guess I figured out my own question... its because we know va has 1.5VDC and we are dropping 1 more volt at VRF to come to +0.5VDC.

So is it safe to say that Rf/Ri equals the gain of VRi... not sure about this...

thanks

Last edited:

#### Laplace

Apr 4, 2010
1,252
What it means is that the input terminals (+ & -) of the op-amp contribute no current at nodes Va & Vb.

There are two ways to look at this circuit. The first is to see this as an asymmetric differential amplifier with ground referenced inputs where one channel has a gain of +0.9 and the other channel has a gain of -2 as shown in the derived equation for Vout.

The other way is to see this as a single input amplifier with a virtual ground reference established at +1.5VDC and a gain of -2. So V1 must then be measured relative to the virtual ground, i.e., an absolute voltage of +1.5V at V1 translates to a signal input voltage of 0 V, and a signal output voltage of -2 x 0 = 0 V relative to virtual ground (or +1.5 V absolute relative to GND). So a V1 input of +2VDC absolute is a signal of +0.5 virtual giving an output of -2 x +0.5 = -1 virtual or 0.5VDC absolute.

What could be more intuitive than that?

#### 24Volts

Mar 21, 2010
164
thanks Laplace...

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