I am trying to understand how to calculate the gain when the input is AC voltage. I am familiar with ideal Op Amp calculations, figuring out the voltage will be the same at both input! terminals. But how does this change when the voltage is AC?!enter image description here

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    \$\begingroup\$ Is this homework? \$\endgroup\$ – Warren Young Jan 23 '13 at 5:42
  • \$\begingroup\$ Looks like a trick question to me. \$\endgroup\$ – Brian Drummond Jan 23 '13 at 9:02

A simple procedure when calculating AC versus DC gain is to consider all the capacitors open for the DC case and shorts for the AC case. Then you can look more closely and see what resistance are accross or in series with the capacitors and use the formula:

  F = 1 / (2πRC)

to see at what frequency the high pass or low pass rolloff is at. When R is in Ohms, C in Farads, then F is in Hertz.

From the above, it should be pretty obvious that the DC gain is 10, and the AC gain at high frequencies is 1. Now, can you tell us what the rolloff frequency of this low pass filter is?

  • \$\begingroup\$ Thanks alot, boiling it down to reactance then I can think of it in more in terms of DC, which I understand. I was given a frequency of 60Hz, so I'll just do nodal analysis to find the gain. Do you know where I can read more on this? I find it hard to find things online that both answer my questions and I can understand! \$\endgroup\$ – natharra Jan 24 '13 at 3:17

While that equation is true for any single frequency, the AC case doesn't yield a single answer.

\$ A_o = 1+ \dfrac{Z_f}{Z_i} \$

\$ Z_i = R_i \$

\$ \begin{align} Z_f & = ((R_f)^{-1} + (Z_c)^{-1})^{-1} \\ & = \left(\dfrac{1}{R_f} + j \omega C_f \right)^{-1} \\ & = \left (\dfrac{1 + j \omega R_f C_f}{R_f} \right)^{-1} \end{align} \$ \$ Z_f = \dfrac {R_f}{1 + j \omega R_f C_f} \$

\$ \dfrac{Z_f}{Z_i} = \left(\dfrac{\dfrac{R_f}{1 + j \omega R_f C_f}}{R_i}\right) \left(\dfrac{\dfrac{1}{R_i}}{\dfrac{1}{R_i}}\right)\$

\$ A_o = 1 + \dfrac{R_f}{R_i(1 + j \omega R_f C_f)} \$

  • \$\begingroup\$ Please proofread my MathML before deleting the picture! \$\endgroup\$ – Adam Lawrence Jan 23 '13 at 18:27

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