# Homework Help: Find K and VTR for MOSFET

1. Feb 21, 2013

### hogrampage

1. The problem statement, all variables and given/known data
A MOSFET has the set of v-I characteristics shown in Fig. P5.5.

What are the values of K and VTR?

2. Relevant equations
iD=K(VGS-VTR)2

3. The attempt at a solution
I have looked in the book and all of the example problems give VTR, but it doesn't show how to actually get the K and VTR values based on the graph. I really have no idea what to do.

The answer in the book says K=0.25 mA/V2 and VTR=2V. Is VTR 2V because VGS increases by 2V?

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2. Feb 21, 2013

### rude man

First job for you is to figure out where on the Vds axis (the x axis) you're operating at.
Hint: notice that Vds does not appear in your equation. What does that suggest?

3. Feb 21, 2013

### hogrampage

The transition from triode to constant current uses the equation VDS=VGS-VTR. So, substituting into the equation I gave in my first post:

iD=KVDS2

I also noticed that I can find VTR using the point where the graph goes horizontal. For VGS=4V, the point would be at 2V. Then 4-2=2V is what I want (verified this by looking at an example in the book where I was given VTR). This seems to work fine, but not sure if it is the "correct" way.

4. Feb 21, 2013

### rude man

OK, there are 2 ways to go:
1. Use VT = Vgs - Vds. So look where the curves bend flat. As you point out, using the Vgs = 4V curve it looks like VT ~ 2V.

But if you used the Vgs = 8V curve you'd get more like VT = 3V (the curve bends at about Vds = 5V so 8 - 5 = 3). The Vgs = 6V and 11V curves would also compute to about VT = 3.

A better way to go is to use your original equation of Ids = k(Vgs - VT)^2. Use it twice, on two different curves, to solve simultaneously for k and VT.

But as I said, first you have to decide where on the Vds axis you're sitting. That equation of yours holds only if Vds > (Vgs - VT).

5. Feb 21, 2013

### hogrampage

Do I just pick a region (i.e. constant current, saturation, triode) and use the corresponding equation?

6. Feb 22, 2013

### rude man

Yes, but remember what I said about the region where that equation is valid.

There are separate equations fdepending on the region of the i-V characteristics you're looking at.

For example, in the triode region your equation would not be appropriate.

Last edited: Feb 22, 2013
7. Feb 22, 2013

### hogrampage

Ah, gotcha. I figured it out after using two different curves, as you said.

Thank you for the help.