Closed-Loop Voltage Gain of Circuit Problem

Question

This might be a simple problem, but I simply cannot wrap my head around it; I'm still learning about ideal op-amps and electronics in general.

The problem is:

I have tried solving it using the gain formula for an inverting op-amp configuration, which goes "Av = -(Rf/R1)".

I determined an equivalent resistance for the top 5 resistors (R||R + R||R + R = 2R) and divided that by the resistor at the inverted input, which that gave me an answer of -2, where it should be -8, according to my professor's solution.

What the hell am I doing wrong, help is greatly appreciated!

Answer 1

Write a KCL for each of the two nodes on the T-Network, namely \$V_a\$ and \$V_b\$.

Notice that we have:

\$\dfrac{V_{in}}{R}=-\dfrac{V_a}{R}\$

KCL for the two nodes:

\$\dfrac{2V_a}{R}\$ + \$\dfrac{V_a-V_b}{R}\$ = 0 ,

\$\dfrac{V_b}{R}\$ + \$\dfrac{V_b-V_a}{R}\$ + \$\dfrac{V_b-V_o}{R} \$ = 0

Thus,

\$\ 3V_a - V_b = 0 \$

\$\ 3V_b - V_a = V_o \$

Plug em into mathematica and you'll get:

\$\ V_a = \dfrac{V_o}{8}\$

And:

\$\dfrac{V_o}{V_{in}} = -8\$

Answer 2

One easy way is to take the Thevenin equivalent of the 2 resistors to the right. You have Vo/2 appearing with a source impedance of R/2. Add R to that and you have Vo/2 with a source impedance of (3/2)R.

Now you know the inverting input is a virtual ground so that source is part of a voltage divider from Vo/2 through (3/2)R to R/2. So the voltage at the resistor to the input is (Vo/2)*R/4, which is Vo/8, so for balance Vo = -Vin*8

Double check in LTSpice "operating point" with R = 10K and universal op-amp with +/-15V supplies, gives -7.9999 out.

Answer 3

For simplifying the two T-networks (replacing it by one single resistor) you have nothing to do than to apply the classical star-to-triangle transformation twice.

As a result you get three resistors - however, two of them play no role (have no influence on the gain in case of an ideal opamp): One resistor appears as a simple load at the output and the other appears between the inverting node and ground (the inverting node is at ground potential for an ideal opamp - hence, no current through this resistor).

Here is what you get after the first star-to-triangle transformation:

^{simulate this circuit – Schematic created using CircuitLab}

Therefore, for the second transformation you have to consider only the star consisting of the tree resistors R, 3R and (grounded) 3R/4. After having applied this transformation only the resulting series resistor determines the gain (and not the two grounded resistors, as explained above). And this series resistor will have the value 8R.

Answer 4

That's an inverting amplifier (G = -Rf/Rin) but Rf is composed of 2 T-networks. Simplify the t-networks into a single resistance and divide by the input resistor for the gain.