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I'm trying to go through a mathematical gain analysis of a closed-loop transimpedance amplifier circuit, but I'm having issues trying to relate the block diagram to the actual circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Regarding the gain, the ideal op-amp gain function doesn't match up with the function I expected. The block diagram shown on the left can model the closed loop gain using the following equations.

$$ V_o = A_{OL} (V_i - V_f) $$ $$ V_f = \beta V_o $$ $$ A_v = \frac{V_o}{V_i} = \frac{A_{OL}}{1+A_{OL}\beta}= \Bigl(\frac{1}{\beta}\Bigl)\frac{A_{OL}\beta}{1+A_{OL}\beta}= A_{v}^{ideal}\frac{T}{1+T}$$

Based on these equations, \$A_{v}^{ideal}\$, the closed-loop gain when the op-amp is ideal, is equal to \$\frac{1}{\beta}\$. The issue I'm facing is that when I apply that equation to my transimpedance model, it doesn't match up.

Assuming an ideal op-amp, the gain can be modeled as impedance's:

$$ Z_f = R_f \vert\vert Z_{C_f} $$ $$ Z_{C_i} = \frac{1}{j\omega C_i} $$ $$ A_v^{ideal} = \frac{V_o}{V_i} = -\frac{Z_{f}}{Z_{C_i}}$$

However, when modeling the value for \$\beta\$ based on the feedback equation in the block diagram, the results are different. (Turning off the input voltage by shorting it and using voltage division):

$$ \beta = \frac{V_f}{V_o} = \frac{Z_{C_i}}{Z_{f}+Z_{C_i}} $$

Clearly, from this, \$\frac{1}{\beta}\$ does not match up with the ideal model. Am I disregarding something from my analysis that I should have, or is there something wrong with my equations?

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  • \$\begingroup\$ A transimpedance amplifier doesn't work with voltages as per your block diagram. It has zero input impedance works with currents. \$\endgroup\$ – Andy aka Nov 7 '19 at 16:31
  • \$\begingroup\$ It is a Shunt-Shunt amplifier, where the feedback network provides a voltage to current conversion through Zf. \$\endgroup\$ – sstobbe Nov 7 '19 at 16:35
  • \$\begingroup\$ @Andyaka I realize that, but in order to simplify calculations, I just did a source transformation from a current source in parallel with a capacitor to a voltage source in series with a capacitor. \$\endgroup\$ – user101402 Nov 7 '19 at 16:35
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    \$\begingroup\$ The calculations will be simpler if you stick with current as the input variable, since you just get \$v_o = Z_f i_{in}\$. (up to a point, anyway) \$\endgroup\$ – The Photon Nov 7 '19 at 16:47
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Why reinvent the wheel, when this has already been done?

Check out Transimpedance Amplifier Analysis by Erik Margan

There are further more detailed frequency models in the paper, including those that use the DC gain/open loop gain.

enter image description here
Source: http://www-f9.ijs.si/~margan/Articles/trans_z_amplifier.pdf

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  • \$\begingroup\$ Thanks for your reply and the link. One question though, for equation 2 in the link, why are the currents summed up as they are? Given the node at Vi (or V1?), how is the author envisioning the currents leaving/entering the node? Are they thinking that the current 'sources' are from Ci and (Vo-Vi)/Zf? \$\endgroup\$ – user101402 Nov 7 '19 at 20:08
  • \$\begingroup\$ They are simply summing up all the currents going in and out of the node Vi. There are only four pathways: the current source, the capacitor Ci , the resisitor Rf and the capacitor Cf. If you find the current through all those points and sum them up, you get equation 2 \$\endgroup\$ – Voltage Spike Nov 7 '19 at 20:15
  • \$\begingroup\$ Thanks again. If you don't mind me asking, when compared to how I was doing the calculations, the PDF went about it a different way, keeping the input signal as a current source whereas I used a source transformation to turn it into a voltage source. The transfer function for the closed-loop gain look really different, but why is that? Modeling the actual closed loop gain Av like shown above, I assumed that the open-loop was a logarithmic function, not like how they had it in equation 1. With that being the case, should the closed loop gain have the same shape as the open loop gain? \$\endgroup\$ – user101402 Nov 7 '19 at 22:19

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