Voltage Gain

Thread Starter

YMJ

Joined Oct 6, 2010
4
Hello ... This is my first post and i hope from the members here to help me ...



In the circuit above I want to calculate the gain of the amplifier ... there is two ways to do it ... one is the regular and second it to calculate the equivalent resistance for R2, R3, R4 ... Can you help me in getting the equivalent resistance?

Thanks in advance
 

Jony130

Joined Feb 17, 2009
5,599
I really don't know why buy to find equivalent resistance you need:
Fin Rin of a open-circuit ( R4 open)
Rop = R2 + R3
Then you need to find Rin of a short circuit (R4 connect to the GND):
Rsc = R2 + R3||R4
And finally equivalent resistance is equal:
Req = Rop*Rsc = (R2 + R3)* (R2 + (R3 R4)/(R3 + R4))
But I don't know why.
And I'm not sure if this is correct.
 
Last edited:

Jony130

Joined Feb 17, 2009
5,599
Well OK I solve the gain of this circuit with "classic" opamp method.
And my previous post is completely wrong.
 

The Electrician

Joined Oct 9, 2007
2,986
Hello ... This is my first post and i hope from the members here to help me ...



In the circuit above I want to calculate the gain of the amplifier ... there is two ways to do it ... one is the regular and second it to calculate the equivalent resistance for R2, R3, R4 ... Can you help me in getting the equivalent resistance?

Thanks in advance
Note that R2, R3 and R4 are in the form of a T attenuator, or T-pad:

http://en.wikipedia.org/wiki/T-pad

The T-pad has two series branches (R2 and R4) and one shunt branch (R3).

You could convert it to a PI-pad:

http://en.wikipedia.org/wiki/PI-pad

The PI-pad has two shunt branches and one series branch. The PI-pad shunt branches do nothing because the one connected to the output is just another load on the output and contributes nothing to the feedback. The shunt branch connected to the - input does nothing because that node is a virtual ground. This leaves the series branch as the only contributer to the feedback, so this is probably what your problem wants you to calculate, calling it "equivalent resistance".

You can convert the T-pad to a PI-pad using the Y-Δ transformation:

http://en.wikipedia.org/wiki/Wye-delta

Knowing ahead of time that the two shunt branches of the PI-pad do nothing in this circuit, you need not even bother calculating them. Just calculate the branch that will be the series branch of the PI-pad; that's your Req.
 

Thread Starter

YMJ

Joined Oct 6, 2010
4
Thanks alot loverly members ..
Is it possible The Electrician to show it more how do the shunt resistance have nothing to do ?
Can you prove it mathamatically if possible ...

Thanks in advance you really helped me
 

The Electrician

Joined Oct 9, 2007
2,986
Re-draw the circuit with a PI-pad arrangement of the three resistors, R2, R3 and R4.

Analyze the circuit by any standard method, and show that the shunt branches are not present in the expression for the overall voltage gain.

Can you do this?
 

Thread Starter

YMJ

Joined Oct 6, 2010
4
R a = R 2 + R 4 + ( R2 * R4 / R 3 )

do you mean that i do KVL in the below draw in the big loop and i ignore R b and R c ?


 

The Electrician

Joined Oct 9, 2007
2,986
R a = R 2 + R 4 + ( R2 * R4 / R 3 )

do you mean that i do KVL in the below draw in the big loop and i ignore R b and R c ?


Don't just ignore Rb and Rc.

Solve the network by whatever method you wish, such as KVL, including Rb and Rc in the analysis, and derive an expression for the voltage gain of the circuit.

Rb and Rc should not appear in the final gain expression, which shows that the gain is independent of Rb and Rc; in other words, Rb and Rc have no effect on the gain. This assumes that the opamp open loop gain is infinite, of course.
 
Top