00:01
So we have to sketch the ac equivalent circuits.
00:07
So the small signal ac equivalent circuits, given the amplifier, i'll just draw a rough sketch.
00:33
For the first one, we have a supply of our alternating kinds.
00:52
And let's call it b sub s.
00:59
And that's connected to the ground.
01:13
We have another resistor connected in parallel to another resistor, all the way to the ground and we have a capacity c1 and just to label it, just call this rg sub 1, and this called this c sub 1.
02:33
Our second image, we have the currents connected to the ground is to resistors of connecting their parallel to r -s -l.
04:17
So v -0, let's label this gm, g -sup m -v -g -s -v -g -s.
04:34
This will be r knots and r sub l and c knots.
04:50
So now we have our diagram, our ac equivalent circuits.
04:58
Now for c sub 1 and c knots, the input capacity, the input capacitance c sub 1 is given by the formula.
05:48
Now using the input similar capacities, c sub m is equal to 1 minus a sub v times the capacity gd.
06:17
So av is negative 2 .47.
06:21
So we have 1 minus negative 2 .47 times gd which is 8 times 10 to the negative 12.
06:55
And that is equal to 27.
07:01
27 .76.
07:10
Tika fluoride.
07:12
So if we use equation 1 here, the input capacitance, c sub 1 or c sub i, c sub 1 is equal to 4 plus 12 plus what we have here, 27 .26.
07:54
And we get 43 .76.
08:05
Now we could find c .nox.
08:10
C not.
08:10
C not to be.
08:11
The equivalent output capacities, which is equal to, that's the form of calculated c knots.
08:36
So the equivalent capacitance, according to the method's law, this would be 1 minus 1 over a sub v times b, d...