00:01
I'm not much of an artist, so you might want to reference this problem in the textbook.
00:04
And we are asked to, in this one, get the output voltage if vi equals 0 .8 millivolts and rd equals 40 kilooms.
00:30
Okay.
00:33
So the square law of the current.
00:38
Let's switch colors so i know what i'm doing here.
00:45
Of current id is given by id equals k times vgs minus vggs, g s, g h squared.
01:02
And that will equal 0 .4 times vgs minus 3 squared.
01:12
And the voltage vgs is vgs equals vg minus i d r s.
01:31
And this will be 10 over 50.
01:38
Did i get all my vs in there? let me go look at something.
01:41
Make sure i got.
01:44
This is, i missed one.
01:46
This one is 40 mega -ooms.
01:50
This one is 10 mega -ooms, and this one is 1 .2 kilo -homes.
01:56
This is 3 .3 kilo -ooms, and this is 1 .2 kilo -ooms.
02:04
Okay, there we go.
02:05
I forgot to put all those in, like i got everything else.
02:10
Okay.
02:11
Sorry about that.
02:13
So that's 10 over 50 times 30 minus 1 .2 r.
02:22
And this will equal 6 minus 6 minus 1 .2 r s then the line equation of the current id in millanpheres will be given by id equals negative 0 .833 v g s plus 5 and v g s plus 5 and v g s q is equal to 4 .68 volts and the transconductance gm will be equal to 2k times v g s q minus v gs t h and that'll be equal to 2 .4 times 10 to the minus 0 .3 times 4 times 4.
03:43
68 minus 3 and that will equal 1 .344 ms.
03:54
Okay.
03:55
And then we can do a graph and let me do a graph here.
04:04
I'm going to go from let's see.
04:09
I'm going to go from 10 .8.
04:20
Let me count.
04:27
I think i can make it all the way here doing 1 .8, 7, 6, 0 minus 1.
04:39
But i don't, i can't.
04:46
Let's try this.
04:55
9, 8, 7, 6, 5, 4, 3, 0.
05:12
And then from my 0, let's go over to 3.
05:25
Let's start at 3...