# How to find W/L of this CMOS??

Discussion in 'Homework Help' started by jean28, Jan 24, 2013.

1. ### jean28 Thread Starter Member

Sep 5, 2012
76
0
1. The problem statement, all variables and given/known data
The exercise already tells me that the value of W/L is 2. However, I can't seem to understand where that result came from. I'd like to know how I can get to that conclusion.

Here is the exercise with the graph of the inverter:
Inverter image:
http://i1226.photobucket.com/albums/...ps9e7f8724.png

Exercise, part 1:
http://i1226.photobucket.com/albums/...ps136b69f7.png

Exercise, part 2:
http://i1226.photobucket.com/albums/...ps32b9a8ee.png

2. Relevant equations
Here is a picture of relevant formulas of MOSFETS:
http://i1226.photobucket.com/albums/...pse3d8245e.png

There is also the formula in the hint that is given in the exercise:
http://i1226.photobucket.com/albums/...ps32b9a8ee.png

3. The attempt at a solution

So, using the formula given in the hint, I substitute the values (I already know the 48 KΩ part):

48k = 1/((125μ)(W/L)(2.5 - 0.5))

Solving for W/L, I get:

12 = 1 / (W/L)

W/L = 1 / 12 = 0.08333

However, the answer says that W/L is supposed to be 2. What am I doing wrong here???

Thanks a lot.

P.S.

It might be worth noting that, since this is an exercise working with CMOS circuits, then the relevant state that it should be analyzed in is in Triode and Cutoff modes.

2. ### toffee_pie Active Member

Oct 31, 2009
162
7
them photos dont work

3. ### Audioguru New Member

Dec 20, 2007
9,411
896
Post your pictures HERE attached to your reply, not over at PhotoBucket where we cannot open them.

4. ### jean28 Thread Starter Member

Sep 5, 2012
76
0
Here are the pictures.

First one is the circuit, second and third are the exercise with the answer, and the fourth is a page with possibly useful formulas.

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Oct 31, 2009
162
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