# MOSFET current mirror: Saturation mode?

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

• 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. – HKOB Mar 6 '15 at 23:02
• "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. – Austin Mar 16 '15 at 8:49

## 2 Answers

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 $V_{GS}>V_T$.

• Thank you, in retrospect, I don't even know why I asked. Just some simple math... – Bob Oct 10 '14 at 17:46

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.

$0.7 \ge 0.2$