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I have some trouble understanding why M1 is in saturation/active mode.

Circuit

According to Wikipedia a MOSFET is in saturation mode if \$V_{GS} > V_{th}\$ and \$V_{DS} \ge (V_{GS} – V_{th})\$.

However as drain and gate are tied together \$\implies V_{DG} = 0 \implies V_{DS} = V_{GS}\$. Therefore \$V_{DS} \ge (V_{GS} – V_{th})\$ can't be true (\$V_{th} > 0\$)? What am I missing?

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  • \$\begingroup\$ I don't think wikipedia explicitly says that Vgs>Vth is necessary for saturation. It isn't. Saturation and linear regions also exist in subthreshold. \$\endgroup\$ – HKOB Mar 6 '15 at 23:02
  • \$\begingroup\$ "Therefore VDS≥(VGS–Vth) can't be true". Let X=Vds=Vgs and Vth is a positive number. Then X-Vth is less than X. Check your logic. \$\endgroup\$ – Austin Mar 16 '15 at 8:49
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If \$V_{GS}=V_{DS}\$, and \$V_T>0\$, you can change the saturation requirements of \$V_{DS}\ge V_{GS}-V_T\$ to \$V_{DS}\ge V_{DS}-V_T\$. Subtracting \$V_{DS}\$ from both sides gives you \$0\ge-V_T\$, which can also be written as \$V_T\ge0\$. This is why this configuration is always in saturation as long as you meet the other saturation criteria of \$V_{GS}>V_T\$.

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  • \$\begingroup\$ Thank you, in retrospect, I don't even know why I asked. Just some simple math... \$\endgroup\$ – Bob Oct 10 '14 at 17:46
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I think was is happening is that you are looking at \$V_{DS} \ge (V_{GS} – V_{th})\$ wrong. That equation is saying that the \$V_{DS}\$ has to be greater than the overdrive voltage. Putting some numbers in there will be helpful. Say that \$V_{th} = 0.5\$ and \$ V_{GS} = 0.7\$ Volts.

So we get.

\$ 0.7 \ge (0.7 -0.5) \$

i.e.

\$ 0.7 \ge 0.2\$

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