[SOLVED] out put gain of this opamp arrangement

Hi my friends!
I confused by behavior of this circuit!
How can i calculate the out put gain of that?
I know all thing about op amps but this circuit is confused me! its gain is 3 . why?
Is it possible that you guide me please?

With appreciate
Goldsmith

There are several ways to calculate the gain.
Perhaps the simplest method is current-based:
The input current (through R4) equals the current through R2 - assuming that the inverting input is at virtual ground potential
and the current into the opamp is zero. It shouldn't be a problem for you to find the R2-current.

Dear LvW
Thanks for your reply. iknow that the current of r2 with assuming that the opamp is ideal , will given by :vi/R4. but i want vo/vi (gain formula) . its gain per those values is 3 but i cant understand that where 3 is from? Is it possible that again help me , please?
Thanks a lot
Goldsmith

I know that you are interested in the gain formula - but where is the problem?
I(R4) depends on vi only and I(R2) depends on vo only. Thus, by equating both currents you can solve for vo/vi, can't you?

What is your idea about R6 and R1?

I suppose, you know how a voltage divider works.
R1 and R6 act as a voltage divider, which is loaded by R2. That`s all. Don`t forget the virtual ground principle!

By the way: The feedback path as shown in your drawing (3 resistors) is used if we want to realize rather high gain values and - at the same time - avoid large resistor values.
The shown circuit (loaded voltage divider) imitates a large resistor (in your case: 3R).

Is it possible that you talk about the principles of virtual ground, please? i know that op amp wants to do V+=v- (amplifying the difference between the input terminals) so because of that the v+ is grounded , thus v- is virtual ground!
i think i cant consider that the R2 and R6 are in parallel with together ( r6 is grounded but r 2 is virtual grounded) . so what should i do?

Hello Goldsmith,
LvW tried to help you such that you would be able to find your own path to the aim. It is the best method of understanding.
However, I see that you lose the way.
Please follow the points below and think over them:
1] OpAmp is connected in negative feedback loop, thus its input difference voltage is zero, and negative input terminal behaves as a virtual ground.
2] Consider, for example, that the input voltage is 1V. Then the output voltage will be equal to the circuit gain.
3] If 1] and 2] are true, then the voltage across R4 is 1V (in left-right direction).
4] It 3] is true, then the current across R4 is 1V/1kOhm=1mA (from left to right).
5] The current from 4] cannot flow to the high-impedance OpAmp input terminal. That is why it must flow through R2.
6] If 4] and 3] are true, then voltage drop at R2 is R2*1mA=1V (from left to right).
7] Since the left terminal of R2 is connected to virtual ground (its potential is 0V), then the voltage across R6 is 1V (from bottom to upper direction!!!).
8] If 7] is true, then current flowing through R6 is 1mA (from bottom to upper terminal!!!).
9] Current flowing through R1 (from left to right) must be a sum of currents through R2 (from left to right) and R6 (from bottom to upper terminal), thus 2mA.
10] If 9] is true, then voltage across R1 (from left to right) must be 1KOhm*2mA=2V.
11] The output voltage is given by the sum of voltages across R1 (from right to left) and R6 (from top to bottom), thus -3V.
12] The gain is -3V/1V=-3.

You can repeat this procedure for general symbols of R1, R2, R4, R6, Vin in order to derive symbolic formula for the gain.
As LvW told you, this T-circuit is useful particularly for providing high gains with reasonable resistance ratios.

Dear Dalibor
Hi
Thanks very much for your valuable Guide.
the out put voltage is across the out put of op amp and bottom of R2 is it right?
Thanks again
Goldsmith

Hi,
The amplifier output is between the OpAmp output terminal and ground.
D.

goldsmith said:

........
the out put voltage is across the out put of op amp and bottom of R2 is it right?

Click to expand...

The output voltage, of course, is at the opamp output.
One additional remark to the term "virtual ground":
The feedack action forces the opamp to remain it its linear range. That means: opamp output (withn the supply limits) for a very small input differential voltage at the opamp terminals, which leads to the assumption (approximation) with V- =0.
For my opinion, the simpler approach to calculate the gain is as follows:
* The signal voltage at the common node of all 3 resistors is Vx=Vo*(R2||R6)/(R2||R6+R1). Thus, the current I2 through R2 is I2=Vx/R2.
* With I2=I4=Vin/R4 you get the wanted result.

Dear LvW
Thanks for your helpful guide.
Goldsmith