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I just want to clear my understanding about a basic source follower circuit. Following is the circuit Simple source follower

Here is my question:

Considering Vi > Vgs(th); Vo = Vi - Vgs. Also Vo = Id*Rs as shown in the picture. Now Id is defined by Vgs*gm. That much current MUST flow. If Rs is big then Vo will be high enough which will reduce Vgs to be less than Vgs(th) and drive the NMOS in cut off. Is my analysis correct?

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  • \$\begingroup\$ Think along the lines of Id=gm * (Vgs-Vgs(th)) and you're nearer the mark. \$\endgroup\$ – Brian Drummond Aug 15 '17 at 21:46
  • \$\begingroup\$ Yes, you are right. \$\endgroup\$ – RAN Aug 21 '17 at 11:12
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Your model (\$I_d = V_{gs}g_m\$) is just for small-signal currents. For large signal currents you should consider something like

$$I_d = K \left(V_{gs}-V_{th}\right)^2$$

If Rs is big then Vo will be high enough which will reduce Vgs to be less than Vgs(th) and drive the NMOS in cut off. Is my analysis correct?

It won't drive the NMOS totally into cut-off. It'll just drive it right to the edge (where \$I_d\$ is very small), which means \$V_{gs} \approx V_{th}\$.

Whether Rs is large or small (but not too small), the circuit will find a balance where the \$I_d\$ predicted by the MOSFET characteristic is equal to the \$I_d\$ required by the resistor characteristic.

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  • \$\begingroup\$ Hello, sorry for the delay in repsonse, i was travelling a bit, yes you are right, i did not consider the large signal model. I think i understand the working now \$\endgroup\$ – RAN Aug 21 '17 at 11:12

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