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**[SOLVED] Simple resistor problem**

Edit: Oh bullocks! Just noticed the thread title is: "simple resistor problem" when it's obviously supposed to be

**transistor**. Can't seem to be able to edit the topic, though.

## Homework Statement

The problem statement is as follows:

Find the value of the resistance R and the channel width W1 and W2 of the following transistors.

http://img86.imageshack.us/img86/6320/transistornw3.jpg

This is what we know: The current Id over the resistor is 120 μA, and the voltage in a is 3.5v, and in b it's 1.5v. The rest of the information is given on the picture.

## Homework Equations

Relevant equations for this has to be the equations for current through a resistor in the triode-region and the saturation region, namely;

Id = K'n(W/L)((Vgs-vt)vds - 0.5vds^2))

and

Id = 0.5K'n(W/L)(Vgs-vt)^2

## The Attempt at a Solution

Finding the resistance for R was rather easy. Considering there's 3.5 volt in point

**a**, there has to have been a voltage drop of 1.5 volt over the resistance.

1.5volt/(120*10^-6) = 12500 ohms.

The real problem I'm having here is realizing how the voltage splits between Vgs and Vds when they're both from the same voltage supply, if you get my drift. What I can say however is this;

Vgs has to be greater than Vt considering a current is supposed to flow through the resistor, so we can't be in the cut off area.

Also, if transistor b only requires 1.5 volt to sustain the current ID, I guess we can conclude that the width of transistor b is larger than that of transistor a. After this however, I'm stumped.

I can't really use any of the formulas without knowing Vgs or Vds. I can say that Vgs + vds = 2V for the first transistor, but this leaves me with three unknowns (W, vgs and vds) and only two equations, and regardless if this I don't even know what region they're operating in.

If anyone could give me a little pointer to help me understand how voltage is divided in a transistor where both vgs and vds draws voltage from the same supply, it would be

**greatly**appreciated.

Thanks!

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