I'm working on generating an expression for output voltage v0 in terms of v1 for the op-amp below.

enter image description here

I was hoping you guys could walk through my logic and make sure I'm analyzing this correctly.

  1. The op-amp is providing negative feedback to the inverting input which allows me to assume that current into the op-amp is 0 and the voltage at both input nodes are equivalent.
  2. The voltage of the non-inverting input is 0V because it's connected to ground which also makes node B 0V.
  3. Therefore, the voltage drop across R1 is v1 and the current through R1 is consequently i1=v1/R1.
  4. Since current going into the op-amp is 0, the current i1 must also enter R2 making the voltage v2 across the resistor v2=i1*R2=R2/R1*v1.
  5. The voltage drop across across R2 makes the voltage at node A va=-v2=-R2/R1*v1.
  6. Using Kirchoff's current law at node A, I can generate an equation for v0 in terms of va and replace all the va terms with va = -R2/R1*v1 as found in step 5.

My final expression is $$v_0=-R_4v_1(\frac{1}{R_1}+\frac{R_2}{R_1R_3}+\frac{R_2}{R_1R_4}).$$

Even better, if there's some way for me to check/assess my answer for these types of problems, I'd like to hear it.

  • \$\begingroup\$ Note that 1/R1 is a common factor and can be pulled out. \$\endgroup\$ Mar 28, 2016 at 4:07
  • \$\begingroup\$ As an alternative analysis, you could apply the star-triangle transformation. This should simplify the calculation. More than that, this procedure allows some additional insight in the circuits properties. \$\endgroup\$
    – LvW
    Mar 28, 2016 at 8:17
  • \$\begingroup\$ Oops. I have v0 where it should read v2. \$\endgroup\$
    – user104243
    Mar 28, 2016 at 11:05
  • \$\begingroup\$ Here is the advantage of applying the star-to-triangle transformation: Two of the triangle resistors have no influence on the gain: One resistor is a pure load resistor and the second one is across the differential input of the opamp (and has also no influence). \$\endgroup\$
    – LvW
    Mar 28, 2016 at 13:58

2 Answers 2


Simple analysis allows you to say that if the voltage output were at point A then the gain at point A is -R2/R1. This has to be true because of the need to keep the voltage at point B equal to zero. Now, if the gain is -R2/R1 at point A it's a simple matter of recognizing that the gain at point "V2" is higher by a certain amount.

That certain amount is also fairly straightforward to see. R2 is virtually grounded therefore it is in parallel with R3 and together they form a potential divider with R4. If the potential divider ratio is G then V2/V1 = -G.R2/R1

I'm looking at your equation and it looks incorrect because if R4 were zero ohms, the formula doesn't reduce to -R2/R1. My formula, using my method is: -

\$\dfrac{V_2}{V_1} = \dfrac{-R_2}{R_1}\cdot \dfrac{R_2 R_3 + R_2 R_4 + R_3 R_4}{R_2 R_3}\$

  • \$\begingroup\$ If R4 were 0 ohms in my equation, I get -R2/R1*v1 still. There's a factor of R4 at the front of the expression that gets distributed. \$\endgroup\$
    – user104243
    Mar 28, 2016 at 11:37
  • \$\begingroup\$ @user50420 OK I see it now - maybe our two formulas are the same - just checking now! Not quite! \$\endgroup\$
    – Andy aka
    Mar 28, 2016 at 12:46
  • \$\begingroup\$ Yes - both are identical. \$\endgroup\$
    – LvW
    Mar 28, 2016 at 13:38

You've got the right general approach, but a few things you say are either wrong or you were being inconsistent with names.

Everything you say is correct thru point 5. The voltage at A is (-R2/R1)V1.

Point 6 doesn't make sense. You haven't defined V0, so it's unclear what it is trying to say. It doesn't help that sometimes you name voltages with single letters, and other times as VN. These kind of inconsistencies just ask for you and others to get confused.

However, solving for V2 as a function of V1 is simple, especially given what you already figured out. Basically, R3, R3, and R2 form a voltage divider from V2 to A. The overall V1 to V2 gain is then the V1 to A gain divided by the divide ratio.

One easy mistake to make is to forget that R2 is in parallel with R3 as the bottom resistor of the divider. The voltage divider output is therefore (R2//R3)/(R4 + R2//R3). The inverse of this is the gain from A to V2. You already know the gain from V1 to A is -R2/R1, so the overall system gain is:

V2   -R2 (R4 + R2//R3)  
-- = -----------------
V1    R1 (R2//R3)

I'll let you expand out the parallel resistances and simplify the result.


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