Solve MOSFET Confusion: Establish 0.8mA & 1.5V with Parameters VT, μpCox, L, λ

[vermin]

Homework Statement

The diagram:
http://images.fr1ckfr4ck.fastmail.fm/probC.jpg
This MOSFET has the following parameters:

• |V_T| = 0.6V
• μ_pC_ox = 100 (μa/V²)
• L = 0.25μm
• λ = 0

Establish a drain current of 0.8 mA and a voltage V_D of 1.5V.

Determine the transistor width W, and the resistance of resistor R, that meet the above drain current and voltage.

Homework Equations

see below

The Attempt at a Solution

First I dove into my bag of equations for the one involving MOSFETS which had both μ_pC_ox and W/L in it.

The best looking one I found was;
I_D=(1/2)μ_nC_ox(W/L)V²_OV

and when I fill in the appropriate values, I get a width that just doesn't make sense. I don't think I'm using the right equation but I don't know where to look.

 

[vermin]

Update, I think I figured out that I can use V_GS-V_T=V_OV

because V_GS should = V_DD = 2.5V

so V_OV=2.5-0.6 = 1.9V

Plugging that into the above equation for I_D gets me W=1.108

Also for the resistor I used
R_D=(V_DD-V_D)/I_D
= (2.5V-1.5V)/.08mA = 1250Ω

 

[rude man]

Update, I think I figured out that I can use V_GS-V_T=V_OV

because V_GS should = V_DD = 2.5V

Whence this? If the source voltage is +2.5V then Vgs can't be 2.5V.

 

[CWatters]

Establish a drain current of 0.8 mA and a voltage VD of 1.5V.

Also for the resistor I used
RD=(VDD-VD)/ID
= (2.5V-1.5V)/.08mA = 1250Ω

Perhaps I've missed something but if Vd = 1.5 the voltage drop across the resistor is 1.5V. The drain current flowing through R is 0.8mA so R is easy to calculate from Ohms law. It's not 1250Ω.

 

[CWatters]

Is the source voltage (the power supply voltage) specified in the problem statement?

 

[rude man]

Is the source voltage (the power supply voltage) specified in the problem statement?

R is obvious, as you say

In the OP's second post he mentions Vdd = 2.5V so I assume that's the source voltage.

 

[vermin]

Right sorry Vdd was given as 2.5V.

So. yeah. voltage across the resistor must be 1.5V, I_D is .8mA so R_D=V_D/I_D=1.5/.0008=1875ohms
That does seem obvious.

But now how do I go about solving for the Width? All I have are these equations I don't understand fully, and I know that I need V_OV to use the one I mentioned above. To get V_OV I need to know V_GS and I don't see how to find that...

 

[rude man]

Right sorry Vdd was given as 2.5V.

But now how do I go about solving for the Width? All I have are these equations I don't understand fully, and I know that I need V_OV to use the one I mentioned above. To get V_OV I need to know V_GS and I don't see how to find that...

You do know Vgs.

Vg = Vd (look at the scematic) and Vs = Vdd = 2.5V. And by now you know Vd.

 

[vermin]

so Vgs must = Vs-Vd = 2.5V-1.5V = 1V. Ok I will work with that and see where it gets me.

 

[rude man]

so Vgs must = Vs-Vd = 2.5V-1.5V = 1V. Ok I will work with that and see where it gets me.

Good move.