# ways to compute gain

Discussion in 'Homework Help' started by screen1988, Mar 19, 2013.

1. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
Please my attached files. For common-source stage with resistive load, there are two ways to drive a gain formula of small signal model.
The first way(as in attached file):
$V_{out}=V_{DD}-R_{D} \c \dfrac {1} {2}\mu _{n} c_{ox}\dfrac {W} {L}\left( V_{GS}-V_{TH}\right) ^{2}$
Then we derivative $V_{out}$ with respect to $V_{in}$
The second way is draw small signal model and then using it to compute gain of the amplifier.
Could you explain to me why two methods are equivalent? I have computed it and they yield the same result.

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2. ### WBahn Moderator

Mar 31, 2012
20,057
5,644
Because the small signal model IS the derivative of the large signal model.

Literally.

How you develop a small signal model is you take the derivatives of the relationships between the various ports and then develop a schematic of linear components that match the value of the derivatives evaluated at the quiscent operating point.

In principle you could have nonlinear small signal models which would be developed largely the same way but which would capture the higher-order behavior of the circuit around the operating point. This would permit the small signal model to be more accurate and to be useful over a wider range of input signals. But I don't think it is done very often except in parametric simulation models.

screen1988 likes this.
3. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
As I know:
SMALL SIGNAL : The amplitude of the signal is small enough, so that the operation is over a very small portion of NON-LINEAR device (e.g. BJT or MOSFET) characteristic(graph). Such a small portion may be regarded as almost linear.
LARGE SIGNAL : Signal is so large that it causes the NON-LINEARITIES in the characteristic (graph) of the device to become evident.
I cann't imagine why small signal is the derivative of the large signal?
And why we can separate dc and ac analysis in a circuit?
Could you explain it again?

Last edited: Mar 19, 2013
4. ### #12 Expert

Nov 30, 2010
17,818
9,138
Uh...no.
As long as I can figure out the question from your words, I can help a bit, but I am not educated enough to track your math.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The derivative of the output function with respect to the input driving function at some quiescent / bias condition is the slope of the function at that condition. That slope corresponds to the apparent gain exactly at that quiescent condition given the input change is infinitesimally small. In practice, relatively small input signal excursions about the quiescent point give output signal excursions sufficiently in agreement with those expected based purely on the slope.

So one is making an assumption that the equality..

$\frac{\Delta V_{out}}{\Delta V_{in}} = \frac{dV_{out}}{dV_{in}$

is "true".

This is a basis for the small signal model gain.

6. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
Aha, today I have just known that small signal model stem from Taylor series expansion. Knowing this make it more easier for me to understand small signal model.

7. ### WBahn Moderator

Mar 31, 2012
20,057
5,644
If that turns on the light for you, great.

Of course, a Taylor series expansion is nothing more than using the derivative of a function to obtain a linear approximation to the function about a point, which is exactly what we have been saying.

screen1988 likes this.
8. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
Yes, but after knowing this I feel it is more easier for me to understand what you have said. I know why we have to derivative to obtain the small signal model.