0
\$\begingroup\$

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:

enter image description here

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!

\$\endgroup\$
  • 1
    \$\begingroup\$ Write Thevinen equivalents, starting at the far right. \$\endgroup\$ – analogsystemsrf Oct 8 '17 at 23:11
  • \$\begingroup\$ So, I have to make the whole circuit into a Thevinen equivalent, making the op-amp into a dependent voltage source? The professor solved it by user the superposition principle, although I don't understand his solution. \$\endgroup\$ – Androvich Oct 8 '17 at 23:12
  • \$\begingroup\$ @Androvich Did you get your -2 from treating the inverting input as ground, shorting the output source, and then computing \$\left(R\:\mid\mid\: R\right)+R+\left(R\:\mid\mid\: R\right)= 2\: R\$?? Ah. I hadn't read enough of your writing. I see that you did. Did you also take note of what is happening to \$V_O\$ as you move from right side towards the input? \$\endgroup\$ – jonk Oct 8 '17 at 23:42
  • \$\begingroup\$ @jonk I don't think I did; I think I assumed the problem was easier than it actually is, so not really sure what the hell I'm doing. I'm trying to figure out how to solve it using the superposition principle, but yeah... no luck yet. \$\endgroup\$ – Androvich Oct 8 '17 at 23:51
  • \$\begingroup\$ @Androvich There are many methods for superposition. Look at the first node you arrive at, moving from the inverting input towards the output. This node (\$V_1\$) can spill outward in 3 ways. But all of this current has to arrive from the next node to the right (\$V_2\$.) So it must be the case that \$3\:V_1=V_2\$. Similarly, \$V_2\$ can spill current outward through 3 paths. But current arrives only from \$V_1\$ and from \$V_O\$. So it must be the case that \$3\:V_2=V_O+V_1\$. From these, what do you get for \$V_1\$ as a function of \$V_O\$? \$\endgroup\$ – jonk Oct 9 '17 at 0:09
1
\$\begingroup\$

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\$

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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:

schematic

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.

\$\endgroup\$
-1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Yeah, that formula I know, but I have no idea how to simplify a T-network.. \$\endgroup\$ – Androvich Oct 9 '17 at 0:45
  • \$\begingroup\$ It's a simple analysis with a test voltage source and finding the short circuit current, but you can also find the equation online or simulate in spice. \$\endgroup\$ – DavidG25 Oct 9 '17 at 1:12
  • \$\begingroup\$ Do you recommend any good simulators? The ones I tried didn't really work.. (that or I just suck) \$\endgroup\$ – Androvich Oct 9 '17 at 1:37
  • 1
    \$\begingroup\$ I use LTspjce and like it. \$\endgroup\$ – DavidG25 Oct 9 '17 at 1:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.