Differential output amplifier formula derivation

Question

I was reading The Art of Electronics book, and I come across this circuit, which is a differential output amplifier.

The author just gave the formula without derivation (I try to derive it myself, it took me many hours, and I failed).

Can somebody prove this formula with a circuit analysis equations?

Answer 1

Apply nodal at the non inverting and inverting termial of upper Op amp

at non inveting terminal

$$\frac{V_x-V_{in}}{Rg}+\frac{V_x-V_{out2}}{R_f}=0\tag1$$

at inverting terminal

$$\frac{V_x-V_{out2}}{R_4}+\frac{V_x-V_{out1}}{R_1}=0\tag2$$

Also nodal at the lower op amp inverting terminal

$$\frac{0-V_{out2}}{R_2}+\frac{0-V_{out1}}{R_3}=0\tag3$$

As given \$R_1=R_2=R_3=R_4=R\$

So from equation \$(3)\$

$$V_{out2}=-V_{out1}$$

and from equation \$(2)\$

$$V_x=\frac{V_{out1}+V_{out2}}{2}$$ from this \$V_x=0\$.

Now from equation \$(1)\$

$$V_{out2}=-\frac{V_{in}R_f}{R_g}$$ also $$V_{out1}=\frac{V_{in}R_f}{R_g}$$

Now the desired output

$$G=\frac{V_{out1}-V_{out1}}{V_{in}}=2\frac{R_f}{R_g}$$

Answer 2

Prove at first the following useful formula:

Use this formula to express the input voltages of the opamps. If the circuit is stable, then both inputs of an opamp have the same voltage. This gives two equations to you. One (after eliminating the equal resistors in your original drawing before any edits of the question) tells that Vout2=-Vout1 and the other says the relation between Vin and Vout1. The gain is (Vout1-Vout2)/Vin ie. 2Vout/Vin

The whole calculation starting from the formula for Vz takes less than 2 minutes.