0
\$\begingroup\$

So right now I'm studying feedback in CMOS circuits (i.e. when there is some sort of connection between the input and the output like in the picture below):

Now during the next step in the example I am looking at, they convert this circuit from a "closed-loop" circuit into an "open-loop" circuit like so:

Now what I know is that the reason we are doing is to find two coefficients A and B where A is the gain in the "open loop" circuit and B is something... I don't know what it is :. Anyway the final answer is of the form A/(1 + A*B). With that being said, I have a series of questions.

  • Why are the resistors arranged the way they are when the circuit is converted into the open loop form? In other words, how does that conversion work so that I can reproduce that process in a different but similar problem?

  • What is the meaning of B in the equation I mentioned, i.e. A/(1 + A*B)?

  • If this is not yet answered by the previous questions, why does doing this whole "closed circuit" to "open circuit" thing work? PS, here's the entire solution if that helps:

\$\endgroup\$
1
\$\begingroup\$

Brief Background
A negative feedback system looks like this (neglecting the direct forward transfer): enter image description here

The transfer function of such a unilateral system is: $$H = \frac{A}{1+AB}$$

B is known as the feedback factor which gives us the fraction of output fed back to the input.
We can also arrange the above equation like this: $$H = \frac{1}{B}\frac{L}{1+L}$$ where L = AB is the open loop gain. It is called open loop gain because if we break the loop at say the input of B and apply a test voltage (\$v_t\$) at the broken node, then the voltage that comes back at the node is \$ABv_t\$. Thus, the gain around the loop if opened is AB. If forward gain \$A -> \infty\$, \$H -> \frac{1}{B} = H_{\infty}\$.
Just as an aside, if system is bilateral we cannot define a loop gain and in that case we work with return ratios.

Why are the resistors arranged the way they are when the circuit is converted into the open loop form? In other words, how does that conversion work so that I can reproduce that process in a different but similar problem?

To find the closed loop transfer function (H) of the system (like yours) we have to find the open loop gain L as the calculations above show. That is why we are opening the loop at a certain node and we check what comes back at that node.
But in the idealized situation above neglected any loading effects. So if you break the node at \$R_F\$ and apply the test voltage, you cannot expect to calculate the correct loop gain because you changed the impedance at the output node by disconnecting it from the loop. Thus to calculate the correct loop gain you connect the loading back to the output so that the output sees the same impedance as when it was connected to the loop. Now, the impedance seen at the output node is \$R_F+R_{S1}\$.That is why you added that loading at the output.

What is the meaning of B in the equation I mentioned, i.e. A/(1 + A*B)?

As explained above B is the feedback factor. To calculate B you can calculate \$H_{\infty}\$, asymptotic transfer function, and B is just the reciprocal of it. To calculate \$H_{\infty}\$ we have to make gain infinite, so in your case you can assume \$g_{m1} -> \infty\$ and calculate the transfer function of the circuit. The overall transfer function, as shown above, will be: $$H = H_{\infty}\frac{L}{1+L}$$

\$\endgroup\$
  • \$\begingroup\$ Thanks! I had to stare at your explanation of how to open the loop but then it clicked. I'm going to go eat now and go back to this later, but I'm happy that I've finally made some actual progress in understanding this. \$\endgroup\$ – SarahK Nov 13 '17 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.