## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 6.6

A student considers only $C_{G D}$ in Fig. 6.13(a) so as to obtain a one-pole response, reasons that the voltage gain drops by 3 dB (by a factor of $= \sqrt{2})$ at the pole frequency, and concludes that a better approximation of the Miller effect should multiply $C_{G D} by 1 + g_m R_D \sqrt{2}$. Explain the flaw in this reasoning.

## Verified Solution

Setting $C_{GS} and C_{DB}$ to zero, we obtain

$\frac{V_{out}}{V_{in}}(s) =\frac{(C_{G D}s − g_m)R_D}{\frac{s}{ω_0}+ 1}$                  (6.31)

where $ω_0 = R_S(1+ g_m R_D)C_{G D} + R_DC_{G D}$. We note that $C_{G D}$ is multiplied by $1+ g_m R_D$ in this exact analysis. So where is the flaw in the student’s argument? It is true that the voltage gain in Fig. 6.13(a) falls by $\sqrt{2}$ at $ω_0$, but this gain would be from $V_{in} to V_{out}$ and not the gain seen by $C_{G D}$. The reader can readily express the transfer function from node X to $V_{out}$ as

$\frac{V_{out}}{V_x}(s) =\frac{(C_{G D}s − g_m)R_D}{R_DC_{G D} + 1}$                                     (6.32)

observing that this gain begins to roll off at a higher frequency, namely, at $1/(R_DC_{G D})$. Thus, the multiplication of $C_{G D} by 1 + g_m R_D$ is still justified.