This is not really homework, but a self-study exercise. I have to solve the following questions (the solutions are in red):

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I had no problem with part 1, so I will not ask about that. My problem is with part 2. It says the differential DC gain should be zero, so for me, since the DC part of V1 and V2 is the same, the condition is clearly that, from the transfer function $$\frac{R4}{R3}=\frac{R1}{R1+R2} \times (1 + \frac{R4}{R3})$$, and from that, simplifying, I get $$\frac{R2}{R3}=\frac{R1}{R4}$$. In the solution of the exercise, they seem to have an extra equation to make $$\frac{R2}{R3}=\frac{R1}{R4}=1$$, so that $$R2 = R3; R4=R1$$

Any idea on where this is coming from?

  • \$\begingroup\$ They want CMRR to be equal to infinity. And this can only be true if $$A_{CM} = A_{V1} + A_{V2} = \left(-\frac{R4}{R3}\right) + \left(\frac{R1}{R1+R2} \times (1 + \frac{R4}{R3}) \right) = 0$$, \$\endgroup\$ – G36 May 15 '18 at 14:52
  • \$\begingroup\$ I have looked up what CMRR is, since we have not seen it in class. Is this condition something that is commonly used, or am I supposed to infer it from the statement? \$\endgroup\$ – Bee May 15 '18 at 15:30
  • \$\begingroup\$ Your task is to get 0V at the output for V1 = V2 = 2.73V. That's all. \$\endgroup\$ – G36 May 15 '18 at 15:40
  • \$\begingroup\$ I know. But as you can see in the solution, they get two equations, while your method and my method only yields one equation. \$\endgroup\$ – Bee May 15 '18 at 15:43

The second part of the solution of the exercise says,

\$\frac{R_2}{R_3} = \frac{R_1}{R_4} = 1\$

From this, its clear that they assume the circuit to be a unity differential amplifier.

So it means, they missed out the term 'dBs' in the question. Differential gain , \$ G_d= 0 dB \implies G_d= 1\$.

It means find the relationship between \$R_1,R_2,R_3,R_4\$ for \$G_d = 0dB = 1\$

\$\implies V_{out} = V_d.G = V_d \$

\$\implies V_{out} = V_1-V_2\$

Comparing this with the Vout equation $$V_{out} = V_1.\frac{R_1}{(R_1+R_2)}.(1+\frac{R_4}{R_3})-V_2.\frac{R_4}{R_3} $$

It gives

\$\frac{R_4}{R_3} = \frac{R_1}{R_2} = 1\$


\$\frac{R_2}{R_3} = \frac{R_1}{R_4} = 1\$

  • \$\begingroup\$ On second thought, I think the second equality you mention cannot be derived from the first one right? \$\endgroup\$ – Bee May 16 '18 at 13:01
  • 1
    \$\begingroup\$ Derivable only if Gd = 1. That's what I did. \$\endgroup\$ – Mitu Raj May 16 '18 at 13:02

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