## 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. 6.10

Explain what happens to Eq. (6.47) if the circuit drives a large load capacitance.

$\frac{V_X}{I_X}= \frac{1 + R_D(C_{G D} + C_{DB})s}{C_{G D}s(1 + g_m R_D + R_DC_{DB}s)}$                                                               (6.47)

## Verified Solution

Merged with $C_{DB}$, the large load capacitance reduces the numerator to $R_DC_{DB}s$ and the denominator to $C_{G D}s(R_DC_{DB}s)$, yielding $V_X /I_X ≈ 1/(C_{G D}s)$. In a manner similar to that in Example 6.7, the large load capacitance lowers the gain at high frequencies, suppressing Miller multiplication of $C_{G D}$.