Charles daEngineer

Minimum Stage TiA Bandwidth Calculations

Published on Sun Nov 02 2025

Introduction

This article will use the results of this post to create an n-stage tia. Then I will determine the loop gain and the bandwidth.

Bandwidth for a TiA with resistive feedback

n-stage TiA with a resistor in the feedback loop

The loop gain $L$ is equal to: $g_{m} {(\frac{- g m}{s C_{i s s}})}^{n - 1} \frac{d V_{1}}{d I_{g_{m}}}$ .

Circuit used to determine the loop gain, specifically

\frac{d V_{1}}{d I_{g_{m}}}

$\frac{d V_{1}}{d I_{g_{m}}} = - \frac{R_{l}}{s (C_{s} + C_{i s s}) (R_{f} + R_{l})}$

$∴ L = - g_{m} {(\frac{- g m}{s C_{i s s}})}^{n - 1} \frac{R_{l}}{s (C_{s} + C_{i s s}) (R_{f} + R_{l}) + 1}$

The loop gain is a low pass filter with $n - 1$ poles at the origin and one pole at $\frac{1}{(C_{s} + C_{i s s}) (R_{f} + R_{l})}$ . The bandwidth of a butterworth filter is approximated using this equation.

$ω = \sqrt[n]{g_{m} {(\frac{g m}{C_{i s s}})}^{n - 1} \frac{R_{l}}{(C_{s} + C_{i s s}) (R_{f} + R_{l})}}$ $2 π B_{ω} = (\frac{g m}{C_{i s s}}) \sqrt[n]{\frac{C_{i s s} R_{l}}{(C_{s} + C_{i s s}) (R_{f} + R_{l})}}$

Now I can simplify further by noticing that a mosfet's transit frequency is: $f_{T} = \frac{g_{m}}{2 π C_{i s s}}$ .

$B_{ω} = f_{T} \sqrt[n]{\frac{C_{i s s} R_{l}}{(C_{s} + C_{i s s}) (R_{f} + R_{l})}}$ $f_{T} = B_{ω} \sqrt[n]{(\frac{C_{s}}{C_{i s s}} + 1) (\frac{R_{f}}{R_{l}} + 1)}$

Let's consider a case where the load resistance is infinitly large, and the source capacitor $C_{s}$ is in the picofarad range with the intrinsic capacitance is in the femtofarad range; then roughly, $\frac{C_{s}}{C_{i s s}} \approx 1000$ . Using the results above, for a one stage amp the transit frequency would need to about about a thousand times bigger than the desired bandwidth; a two stage though could be 32 times bigger than the desired bandwidth and three stages 10 times bigger than the desired bandwidth.