[SOLVED] Another NMOS-transistor problem


by Molecular
Tags: nmostransistor, solved
Molecular
Molecular is offline
#1
Nov17-07, 11:16 AM
P: 29
I'm sorry for asking again so soon, but these transistors really give my head a spin.

1. The problem statement, all variables and given/known data
The problem is pretty much summed up by the following photograph:


3. The attempt at a solution
To be quite honest I'm really stumped here. I can go over what I know (Or at least what I think I know):

Since everything displayed here is in series, I'm presuming the current over each element is the same. The voltage over R1 should be 10 - v1, basically meaning if I could find v1 I should be able to find the current R1 and thus the current in every other element connected aswell, which would make the problem easy.

Vgs for the first transistor is 5 volts, and vgs=vds for the second transistor. R1 and R2 should have the same voltage drop across them.
All in all, 15 volts dissipate over this circuits as vdd = 10 volts and vss = -5.

After this, it completely stops. I've got the correct answer, which is
v1 = 6 v
and
v2 = 2 v

So by my logic, the current over the first resistance is (10-6)/1000 = 4 mA.
This is where I get confused, because if v2 = 2v, doesn't that mean the voltage drop across the first transistor is 4v aswell?
But if it is, this indicates that 16 mA runs through it, which isn't really possible if only 4 mA runs through the first resistance.
I tried checking what the voltage vds over the first transistor had to have been in order to allow for 4mA to pass through, and as far as I can remember the answer I got was sqrt(2)+1. Which is a fairly ugly number so I'm presuming this is wrong.

Anyone got a hint that can push me in the correct direction? It would be greatly appreciated. I'm dying to understand these blasted transistors.
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dlgoff
dlgoff is offline
#2
Nov18-07, 02:04 PM
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"Vgs for the first transistor is 5 volts..." Vgg=5 volts. Wouldn't Vgs=Vgg-V2?
Molecular
Molecular is offline
#3
Nov19-07, 07:41 AM
P: 29
Aha, I see now, of course. Thanks!
I managed to solve it now by using four different equations, was quite a hustle but I've never figured it out if not for this!

Thanks again!


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