the goal is to find the gain of the circuit. I don't understand the Vi/Vs part. I know they used voltage divider, but why they included the second term(in Red) in the expression? where does that come from? can anyone elaborate on this please? Thanks enter image description here

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  • \$\begingroup\$ A regular circuit diagram would help. \$\endgroup\$ – jippie Nov 8 '15 at 18:00
  • \$\begingroup\$ okay, I will add it up in a sec \$\endgroup\$ – user65652 Nov 8 '15 at 18:04
  • \$\begingroup\$ I just added up the regular circuit diagram \$\endgroup\$ – user65652 Nov 8 '15 at 18:05
  • \$\begingroup\$ OP, you got an answer (again). I see you are asking a lot of very basics questions, while I must say that you usually include everything that is needed to answer, and even your attempts, I think you should find another source of help, maybe your teacher, fellow students, or a nice book. \$\endgroup\$ – Vladimir Cravero Nov 8 '15 at 19:15

Not sure what you got so far, but in my opinion the fastest way to compute this is starting by changing your voltage generator with a current generator, with the help of Norton's theorem.

We are making the Norton equivalent of \$V_s\$ and \$R_s\$, this results in a current generator with \$I_s=\frac{V_s}{R_s}\$ in parallel with a resistor \$R_s\$.

Now \$V_i=R_{id}\cdot I_{Rid}\$, the current through the resistor can easily be found with the current divider formula:

$$ I_{Rid}=I_s\cdot\frac{R_s//2R_{ICM}}{R_s//2R_{ICM}+R_{id}+R\beta_0//2R_{ICM}} $$ now substitute \$I_{Rid}\$ and \$I_s\$ and rearrange:

$$ \frac{V_i}{V_s}=\frac{R_{id}}{R_s}\frac{R_s//2R_{ICM}}{R_s//2R_{ICM}+R_{id}+R\beta_0//2R_{ICM}}=\\ =\frac{R_{id}}{R_s//2R_{ICM}+R_{id}+R\beta_0//2R_{ICM}}\frac{2R_{ICM}}{R_s+2R_{ICM}} $$

There are a number of ways in which you can rearrange the latter result, and the one your book uses does not seem quite fit to me. The term you are unsure about seems like an easy voltage divider at the input, and indeed is there because you have a resistive divider there, but since the divider is loaded by the three rightmost resistors getting the right result is not so easy. In fact you find both \$R_s\$ and \$R_{ICM}\$ also in the first term, and not just because you have another \$R_{ICM}\$ on the right side of your circuit.

A student that is (quite properly) trying to grasp why a term is there, might see the voltage divider and say "ok that's easy I got it" and write down something like this:

$$ V_i=V_s\cdot\frac{2R_{ICM}}{R_s+2R_{ICM}}\cdot\frac{R_{id}}{R_{id}+R\beta_0//2R_{ICM}} $$

this last formula is wrong, because as I said the loading effect of the rightmost resistors on the first voltage divider can't be neglected.

My point here is that you probably saw that term, thought of the voltage divider and thought that there is some fancy way of doing things here. There is not. Sometimes math is better than intuition.

It is nice to see that things adds up though: if the leftmost resistors are quite smaller than the rightmost, i.e. you have an "heavy voltage divider"*, the right formula can be simplified and it becomes the 'wrong' one. This happens if you neglect the \$R_s//2R_{ICM}\$ term with respect to \$R_{id}+R\beta_0//2R_{ICM}\$ in the denominator of the first part of the 'right' formula.

*that's Italian slang, not sure how you natives say that, we speak of "heavy voltage divider" when the loading effects are neglectable, i.e. when you bias a bjt base for a CE amplifier we say you have to make an "heavy divider" thus sizing it so that the current flowing into it is much greater than the transistor base current.


There level of detail needed to determine the gain depends on the assumptions you make in the opamp. Clearly in this question, the opamp isn't an 'ideal' one -- not only does it not have infinite gain, there is apparently and output impedance (Ro), and two input impedances -- a differential one (between the '+' and '-' inputs), and a common-mode one (to ground at the '+' input ?).

If you examine the model for the opamp, you'll find those components, and while they make the expressions more complex, writing the answer is just a case of solving KVL or KCL at the nodes.

  • \$\begingroup\$ thanks for the input, but that does not really answer my question. \$\endgroup\$ – user65652 Nov 8 '15 at 18:24

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