# Gain on a Differential Amplifier

For the past hour, I've been trying to calculate the gain of a differential amplifier in the following form: $\frac{V_{out}}{V_{in}} =$ some function of resistors

I went to this site and tried to figure out what they were doing but I just cannot follow. I understand the current equations, and I understand that $V_a = V_b$ for ideal op-amps but I just don't get the rest. Why do you get two different equations for $V_{out}$ when $V_b=0$ and $V_a=0$, if they are both equal. Can someone explain?

• The two solutions are because they use the superposition principle to solve the circuit. And they're not for Va = 0 and Vb = 0; they're for V1 = 0 and V2 = 0. Feb 17, 2013 at 4:48
• @ThePhoton T.T .... waste of so much time because of a typo. Okay then how are they arbitrarily saying if $R_1 = R_2$ and $R_3 = R_4$ and then getting the gain equation? Feb 17, 2013 at 5:00
• @Richard To arrive at the equation for $V_{out}$ in the yellow box. Assume $R_2 = R_1$ and $R_4 = R_3$. Substitute into the previous equation for $V_{out}$. Simplify. Feb 17, 2013 at 5:14
• @NickAlexeev yah but what I'm saying is that seems so arbitrary. Why not $R_1 = R_3$ and $R_2 = R_4$? Feb 17, 2013 at 5:15
• @Richard $R_1 = R_2$ and $R_3 = R_4$ is a special case. The resulting gain equation has a specific property that output is proportional to the difference of inputs. A circuit designer can choose just about any resistor values. Feb 17, 2013 at 5:23

The site gives you the output equation: $V_{out} = \dfrac{R_3}{R_1}(V_2-V_1)$ Difference amplifiers find the difference between two signals, and amplifies the difference. The $\dfrac{R_3}{R_1}$ factor determines the gain of the difference. The gain is relative to the difference only. If the two input signals are the same, the output will be very close to zero, depending on the common mode rejection capabilities of the op amp.