# Why is small-signal output resistance not dependant on device width

For a MOSFET in deep triode region, we can approximate it as a resistor with the following:

$$\frac{1}{{\mu}_n C_{ox}(W/L)(V_{GS} - V_{Th})}$$

However, in small-signal for a MOSFET in saturation, I know that $$\R_{out} = \frac{1}{\lambda \cdot I_d}\$$, where lambda is channel-length modulation factor and that is proportional to inverse of length. So increasing length increases $$\R_{out}\$$. But why isn't it dependent on width? Surely, if I make the transistor very wide, resistance should drop.

• Take a look at your equation...
– Chu
Jun 14 '19 at 14:48
• @Chu I don't understand your comment. Which equation are you talking about? If you are talking about the first equation, this is not the point of the question. Jun 14 '19 at 14:58
• If the equation is Rout = ....., and if W = width, and L= length, then Rout is proportional to L an inversely proportional to W. (BTW it's 'dependent' not 'dependant')
– Chu
Jun 14 '19 at 14:58
• @ElliotAlderson, OK, I can't see an entire equation on my laptop, perhaps there's a problem.
– Chu
Jun 14 '19 at 15:03
• @Chu There are two equations provided, one as a graphic and the other in the text of the question. The equation in question does not have W or L as factors. I will grant you that the first equation is botched, as the RHS is in the graphic but the LHS is in the text. Jun 14 '19 at 15:19

The small signal output resistance depends on W because $$\I_D\$$ depends on it.

After all, you have:

$$R_o=\dfrac{1}{{\lambda}I_D}$$

With $$\I_D\$$ being the current in the saturation region for the MOSFET. In saturation, $$I_D=\dfrac{1}{2}\text{K}\frac{W}{L}(V_{GS}-V_T)^2$$

So yeah, if you increase W, the $$\R_o\$$ does decrease because $$\I_D\$$ is in its denominator.

$$R_o=\dfrac{1}{\dfrac{1}{2}{\lambda}\text{K}\frac{W}{L}(V_{GS}-V_T)^2}$$

• Ah Yes. Dependance on W/L in the Id equation. Forgot about that. Thanks Jun 15 '19 at 17:31