# Power Gain in a small signal Mosfet circuit

#### Kinka-Byo

Joined Apr 22, 2019
11
let's consider this circuit, which is the equivalent of a small signal Mosfet amplifier.

where RG is a gate series resistance which takes into account of the oxide leakage, and CGS and CGD are the well - known Mosfet Parasitic Capacitances.

Now, the book The Design of CMOS Radiofrequency Integrated Circuits (Lee) makes the computation of its power gain, i.e. the ratio between the power delivered to the load (by supposing matching) and the input power of the source. Regarding the last one, it is written that:

My question is: is the factor 1/2 due to the fact that in AC we consider rms values, or is it due to something related to the maximum power transfer theorem? I thought it was due to the rms, but if it is true, the formula should use the phasor of the input current (and moreover, the book does not do any assumption about a single - frequency behaviour of the input current source). Moreover, how can I compute the power delivered to the load? I do not know if I have to use the factor 1/2 due to rms or not.

#### Jony130

Joined Feb 17, 2009
5,514
The average power is equal to

P = I_rms*V_rms = I_rms * I_rms*R = I_rms^2*R

But if we use I_peak instead of I_rms we have this situation (I_rms = I_peak/√2):

$$P= \frac{I}{\sqrt{2}} * \frac{I}{\sqrt{2}} *R = \frac{I^2 R}{2}$$

#### Kinka-Byo

Joined Apr 22, 2019
11
Ok, it seems clear. But do you have an idea about why the book wrote that expression in which there are not phasors, but time domain current and voltage?

#### Jony130

Joined Feb 17, 2009
5,514
Well if this indeed is a small-signal analysis in this type of analysis we always use sinewave extortion. So I do not see a problem. But what is more interesting is what small rg represents?