# Transimpedance Amplifier Gain Analysis

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. 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?

• A transimpedance amplifier doesn't work with voltages as per your block diagram. It has zero input impedance works with currents. – Andy aka Nov 7 '19 at 16:31
• It is a Shunt-Shunt amplifier, where the feedback network provides a voltage to current conversion through Zf. – sstobbe Nov 7 '19 at 16:35
• @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. – BestQualityVacuum Nov 7 '19 at 16:35
• 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) – The Photon Nov 7 '19 at 16:47

## 1 Answer

Why reinvent the wheel, when this has already been done?

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

• 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? – BestQualityVacuum Nov 7 '19 at 20:08
• 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 – Voltage Spike Nov 7 '19 at 20:15
• 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? – BestQualityVacuum Nov 7 '19 at 22:19