how to get the voltage gain of this circuit?

Assuming that the differential gain of the amplifier is Adm, and the common mode gain is Acm.

Then,could anybody help me to calculate the circuit's gain?

notice: the resistors are 1k ohms and 1M ohms.

1.) The differential part of the output signal is always zero (equal gain values for non-inv. and inv. mode of the amplifier)
2.) The common mode output is Vout,cm=Vin*1E6/(1E6+1E3)*Acm=Vin*Acm*0.999.

The above calculation assumes infinite input and zero output resistance. The differential gain Adm may be finite and small.

First, i am sorry as i late for your reply.
1.) the differential part of the output signal is not always zero,since the inv and non-inv is not equal ,just think that the output is not zero.
2.) the common mode output does not work as your description easily.

---------- Post added at 07:10 ---------- Previous post was at 07:01 ----------

what i think as follow:
Vout = Adm*Vid +Acm*Vic
Vic = Vout*R1/(R1+R2)*0.5+Vin*R2/(R1+R2)
Vid = Vout*R1/(R1+R2)

then
Vout = [Vout*R1/(R1+R2)*0.5+Vin*R2/(R1+R2)]*Acm + Vout*R1/(R1+R2)*Adm

---------- Post added at 07:15 ---------- Previous post was at 07:10 ----------

At last ,i got that
Vout/Vin is approximately equal to 1000*Acm/Adm.

Unfortunately , i was told that the answer is 2000*Acm/Adm.

Could anybody tell me why for the factor 2.

Just one comment to your calculation:

You wrote: Vid = Vout*R1/(R1+R2)

That means: Vout/Vid=1+R2/R1.

This is nothing else than the classical formula for a non-inverting gain stage with opamp.
But according to your drawing, also the inverting input is connected to the input signal. No influence?
Can you verify your result?

Remark: The shown circuit is the classical opamp-based differential amplifier with Vin=Vin1=Vin2.

LvW said:

You wrote: Vid = Vout*R1/(R1+R2)

That means: Vout/Vid=1+R2/R1.

This is nothing else than the classical formula for a non-inverting gain stage with opamp.

Click to expand...

Not quite. I'm assuming he means Vid = differential input voltage and Vic = common mode input voltage.

In that case, I agree with his reasoning up to this point:

amwygah said:

then
Vout = [Vout*R1/(R1+R2)*0.5+Vin*R2/(R1+R2)]*Acm + Vout*R1/(R1+R2)*Adm

Click to expand...

Still scratching my head about the rest. I've got as far as:

Vout/Vin = Acm / [1 + 0.001*[1 - 0.5*Acm - Adm]]
or:
Vout/Vin = 1000*Acm / [1001 - 0.5*Acm - Adm]
or:
Vout/Vin = 2000*Acm / [2002 - Acm - 2Adm]

Maybe I need to back up and try a different approach...

Remark: The shown circuit is the classical opamp-based differential amplifier with Vin=Vin1=Vin2.

Click to expand...

Respectively a CMMR test circuit.

P.S.:

Vout/Vin = 2000*Acm / [2002 - Acm - 2Adm]

Click to expand...

I think, the expression is correct, except for the Adm sign, which should be inverted. Thus the for Adm >> Acm and Adm >> 2000,

Vout/Vin = 1000*Acm/Adm

is in fact correct. A CMRR number would be used to describe the common mode behaviour.

This circuit gain is Zero.

Click to expand...

Approximately. The gain can be nevertheless calculated. Thats the given problem.

This circuit gain is Zero ( for same input as shown in #1). VCM (dc and ac) will be rejected . This circuit is used for CMRR/PSRR measurement and used to isolates two input signals.

godfreyl said:

Not quite. I'm assuming he means Vid = differential input voltage and Vic = common mode input voltage.
In that case, I agree with his reasoning ...
.

Click to expand...

Hi godfreyl, can you please explain why you can "agree with his reasoning" ?
Of course, he means Vid = differential input voltage. But how does it help to arrive at the correct result?

In words: Vid = Vout*R1/(R1+R2) is the voltage that is fed back to the inv. opamp terminal - and, thus, the inverse of the non-inverting gain (as I wrote) of a simple opamp stage with feedback.
But this is only one half of the truth because the circuit is NOT a simple non-inverter.

Hello amw
CAn u explain what r u trying to do ?
wher u got this circuit ? it is obvious that u will get zero output or very close to it due to offset in the amp.
your gain is imeg divided by ik that 1000....and 1000 is multifliyed by zero(cause the input sine wave is going to both inv and non inv inputs and the sum of them is zero) so multiply 1000 by zero and get zero....


https://www.google.co.il/#hl=iw&sit....,cf.osb&fp=1c94fc5b97bf5028&biw=1152&bih=727

https://www.electronics-tutorials.ws/opamp/opamp_8.html

CAn u explain what r u trying to do ?

Click to expand...

The purpose of the cicruit is well explained by the equations in my opinion. It's a common mode rejection test circuit, the Vout/Vin gain will be non-zero for non-zero Acm values.

https://en.wikipedia.org/wiki/Differential_amplifier
Common mode



Normal. At differential mode (the two input voltages change in opposite directions), the two voltage (emitter) followers oppose each other - while one of them tries to increase the voltage of the common emitter point, the other tries to decrease it (figuratively speaking, one of them "pulls up" the common point while the other "looses" it so that it stays immovable) and v.v. So, the common point does not change its voltage; it behaves like a virtual ground with a magnitude determined by the common-mode input voltages. The high-resistive emitter element does not play any role since it is shunted by the other low-resistive emitter follower. There is no negative feedback since the emitter voltage does not change at all when the input base voltages change. Тhe common quiescent current vigorously steers between the two transistors and the output collector voltages vigorously change. The two transistors mutually ground their emitters; so, although they are common-collector stages, they actually act as common-emitter stages with maximum gain. Bias stability and independence from variations in device parameters can be improved by negative feedback introduced via cathode/emitter resistors with relatively small resistances.


A_\text{c} is called the common-mode gain of the amplifier.
As differential amplifiers are often used to null out noise or bias-voltages that appear at both inputs, a low common-mode gain is usually desired.

The common-mode rejection ratio (CMRR), usually defined as the ratio between differential-mode gain and common-mode gain, indicates the ability of the amplifier to accurately cancel voltages that are common to both inputs. The common-mode rejection ratio is defined as:

\text{CMRR} \triangleq \frac{A_\text{d}}{A_\text{c}}

In a perfectly symmetrical differential amplifier, A_\text{c} is zero and the CMRR is infinite. Note that a differential amplifier is a more general form of amplifier than one with a single input; by grounding one input of a differential amplifier, a single-ended amplifier results.

I dont think he can calculate it is given as a part of data sheet ... he does not know how perfect his op is.. at most he can measure using actually perfect resistors..

OK, I tried working it out differently, in terms of the voltages at the inverting and non-inverting input pins (Vinvert and Vnoninv respectively), but got the same answer:

Vout/Vin = 1000 * Acm / (1001 - Acm * 0.5 - Adm)

If we assume that Adm >> Acm
and we assume that Adm >> 1000,
then we can simplify it to:

Vout/Vin = -1000 * Acm / Adm

So I agree with your answer, except for the sign.

Is this for school/college/university? If so, maybe the teacher/lecturer got it wrong. What answer did the other students get?

Sadly, even textbooks make mistakes sometimes.

edit: Wow, I missed a lot of posts while I was redoing the calculations. I'm glad FvM (almost) agrees with me.

---------- Post added at 12:00 ---------- Previous post was at 11:35 ----------

LvW said:

Hi godfreyl, can you please explain why you can "agree with his reasoning" ?

Click to expand...

I don't see any errors in his arithmetic.

LvW said:

In words: Vid = Vout*R1/(R1+R2) is the voltage that is fed back...
But this is only one half of the truth because the circuit is NOT a simple non-inverter.

Click to expand...

Yes, the other half of the truth is that the common mode input voltage, Vic = Vout*R1/(R1+R2)*0.5+Vin*R2/(R1+R2), but he took that into account as well.

I keep my opinion, that the (residual) circuit gain must be positive, according to the positive sign of Acm and Adm. The feedback provided by Adm reduces the circuit gain but can't change it's sign.

Vout/Vin = + 1000 * Acm / Adm


dselec, I don't understand what your post is targetting on. The initial poster has demonstrated in post #3, that he understands very well about differential and common mode behaviour of OPs.

Obviously, it's an exercise problem and should be treated as such. Students aren't wanted to answer it with "ideal OPs have infinite CMRR, thus we don't need to calculate the problem".

Ok FVM u tell me looking at the equation in Wikipedia how this calculation is possible even as a test.
the cmrr is a ratio given in the data sheet because it is effected mostly by the internal physics of the op.
if he knows the parameter he can simply use the formula ..i dont understand what kind of help he needs and why ?

The question is in post #3, about a given solution that is apparenly wrong.

At last ,i got that
Vout/Vin is approximately equal to 1000*Acm/Adm.

Unfortunately , i was told that the answer is 2000*Acm/Adm.

Could anybody tell me why for the factor 2.

Click to expand...

Hi godfreyl, You are right - I have done a calculation error.
More than that, I think the "sign conflict" is due to the differential part Vid. Assuming Adm>0 we must write Vid=Vi,p-Vi,n=-Vout*R1/(R1+R2).
Thanks.
LvW

FvM said:

Vout/Vin = + 1000 * Acm / Adm

Click to expand...

Yes, quite right. I found the mistake in my arithmetic at last.

dselec said:

the cmrr is a ratio given in the data sheet....
if he knows the parameter he can simply use the formula ..i dont understand what kind of help he needs and why ?

Click to expand...

His problem is basically to find the CMRR by measurement, without looking at the datasheet.

I imagine this is what manufacturers have to do to find out what value of CMRR to print on the datasheet in the first place.

p.s. Nothing wrong with your theory.

new.jpg
FVM how did u get vout ? looking at the formula what value u assumed ?

All calculations start with a set of equations like those given in post #3. (I changed the third for the usual Vid definition).


Plug in respectively and solve for Vout/Vin.

The equations in your link are assuming infinite Adm (and can ignore any finite Acm numbers in return).

May I give some explanations to the above formulas given by FvM ?

You should start with separat calculations for Vp(voltage at pos. opamp input) and Vn (neg. opamp input).
Please note, that calculation of Vn requires application of the superposition theorem for Vin and Vout.
Then, both voltages are inserted in the definitions Vic=(Vp+Vn)/2 and Vid=(Vp-Vn).

This gives the results as provided by FvM and godfreyl.