# Reduce T-Network Feedback Resistance

Discussion in 'Homework Help' started by SilverKing, Jun 1, 2016.

1. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
Hi everyone,

I was asked to reduce the feedback resistance of the following T-Network:

My first though was to use Thevenin, and actually I've found some book used that method:

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
But where is your question ?

3. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
Sorry about that. ^^" My question is whether this approach OK or not. And if the feedback resistance can be reduced more.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Why do you think that this approach is wrong ?
Also you can convert the T-pad network to a PI-pad network using the Y-Δ transformation:
https://en.wikipedia.org/wiki/Y-Δ_transform
Or we can use some other network/circuit analysis techniques (KVL, KVL, Ohm's law etc.).
EDIT
For example I1 = Vin/R1 and Vx = - I1*R2 (Vx is a voltage across R4 resistor, the middle node).
I4 = Vx/R4 and I3 = (Vx - Vout)/R3. And from there we can solve for Vout/Vin

Last edited: Jun 1, 2016
5. ### crutschow Expert

Mar 14, 2008
13,501
3,375
You can make the resistance as low as you want.
There is no lower limit except for what the op amp output can drive.

6. ### MrAl Distinguished Member

Jun 17, 2014
2,561
516
Hello there,

You can check your results using actual resistor values and see if you get the right answer. For example, with R1=1k, R2=2k, R3=4k, and R4=8k, you get a gain of 7.
Also, if you end up with two resistors in series you can always lump them into one resistor.

Dont forget about Norton, which makes it simpler just like Thevenin.

Using Thevenin and Norton theorems, we can do this another way.
Turn Vout into a current source using Nortons theorem and R3, that gives you three resistors in parallel:
R2, R3, and R4,
which form a single resistor, and then use Thevenin to turn the current source and those three resistors back into a voltage source, then divide by R2 to get the current in R2, which will end up being a combination of resistors, and then equate that to 1/R5 which will be the current when there is only one resistor in the feedback circuit R5. Simply take the inverse of that equation and on the right you have "R5" and on the left you have a combination of resistors which, when the values are substituted for the variables, yields the value of the single feedback resistor.
If you are interested i'll show you the technique, which is applicable to a wide range of different circuits.
Note we dont even have to calculate the 'gain', except maybe as a means to check our final results.