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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.

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  • \$\begingroup\$ Take a look at your equation... \$\endgroup\$ – Chu Jun 14 at 14:48
  • \$\begingroup\$ @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. \$\endgroup\$ – Elliot Alderson Jun 14 at 14:58
  • \$\begingroup\$ 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') \$\endgroup\$ – Chu Jun 14 at 14:58
  • \$\begingroup\$ @ElliotAlderson, OK, I can't see an entire equation on my laptop, perhaps there's a problem. \$\endgroup\$ – Chu Jun 14 at 15:03
  • \$\begingroup\$ @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. \$\endgroup\$ – Elliot Alderson Jun 14 at 15:19
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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.

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

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  • \$\begingroup\$ Ah Yes. Dependance on W/L in the Id equation. Forgot about that. Thanks \$\endgroup\$ – AlfroJang80 Jun 15 at 17:31

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