Quantcast
Channel: Active questions tagged transfer-function - Electrical Engineering Stack Exchange
Viewing all articles
Browse latest Browse all 237

If I have an operational amplifier with a finite open loop gain, can I calculate the transfer function neglecting the open loop gain?

$
0
0

opamp

I'm talking about this op-amp. In the solution it finds an expression for the transfer function using a virtual short circuit without considering the open loop gain \$ A_{vol} \$ at all. I wanted to approach it using the following expression which takes in consideration the open loop gain:$$ V_{u} = A_{vol}(V^+ - V^-)$$values for \$V^+\$ and \$V^-\$ are:$$V^+ = \frac{R_1C_1s}{R_1C_1s+1}V_s$$$$V^-=\frac{R_3C_2s+R_3}{R_3C_2s+1}V_u $$then I wanted to proceed plugging them into the first expression.

The solution however completely neglects \$ A_{vol} \$ and makes use of a virtual short circuit, so \$V^+ = V^-\$ and it proceeds from there to find the transfer function. I can't understand why it's being neglected, there are many other circuits where it's neglected and I don't know if it's just negligible for these circuits in my book, or if it doesn't matter when finding the transfer function.

I also have another question, how do I calculate gain? I'm assuming that once I have the transfer function I can calculate gain at low frequencies and as for higher frequencies I'd have to plot the Bode diagram since gain would be dependent on gain?

Thanks for the help.


Viewing all articles
Browse latest Browse all 237

Trending Articles