PMOS Saturation Condition

Question

we know for nmos works in active region, we must have Vgs-Vth>0 and Vds>Vgs-Vth.

For PMOS can I write like this | Vgs|-|Vth|>0 , |Vds|>|Vgs|-|Vth|.

Please correct me if I am wrong

Answer 1

The terms \$V_{\text{GS}} \$ and \$V_{\text{DS}} \$ are polarity sensitive, so you cannot just take the absolute values. The requirements for a PMOS-transistor to be in saturation mode are $$V_{\text{GS}} \leq V_{\text{t}} \: \: \text{and} \: \:V_{\text{DS}} \leq V_{\text{GS}}-V_{\text{t}} \tag1$$ where \$V_{\text{t}} \$ is the threshold voltage for the transistor (which typically is \$-1 \; \text{V} \$ for a PMOS-transistor).

---

OP's Conjecture

The inequality \$V_\text{GS} - V_\text{t} \leq 0 \$ is equivalent to \$|V_\text{GS}| - |V_\text{t}| \geq 0 \$.

Counterexample

We set \$V_\text{t} = - 1 \; \text{V}\$ and \$V_\text{GS} = 1 \; \text{V} \$ and check if the inequalites are either both violated or both satisfied. The first inequality yields

$$\begin{align*} V_\text{GS} - V_\text{t} &\leq 0 \; \text{V} \tag2 \\\\ 1 \; \text{V} + 1 \; \text{V} &\leq 0 \; \text{V} \tag3 \\\\ 2 \; \text{V} &\leq 0 \; \text{V} \; \; \; \color{red}{\text{violated}} \tag4 \end{align*}$$

The second inequality yields

$$\begin{align*} |V_\text{GS}| - |V_\text{t}| &\geq 0 \; \text{V} \tag5 \\\\ |1 \; \text{V}| - |-1 \; \text{V}| &\geq 0 \; \text{V} \tag6 \\\\ 1 \; \text{V} - 1 \; \text{V} &\geq 0 \; \text{V} \tag7 \\\\ 0 \; \text{V} &\geq 0 \; \text{V} \; \; \; \color{green}{\text{satisfied}} \tag8 \end{align*}$$

The two inequalities are not equivalent.

---

Special case \$V_\text{GS} \leq 0 \$ and \$V_\text{t} \leq 0 \$

In the special case (usually the case in textbook problems for PMOS) where \$V_\text{GS} \leq 0 \$ and \$V_\text{t} \leq 0 \$ the two inequalities \$V_\text{GS} - V_\text{t} \leq 0 \$ and \$|V_\text{GS}| - |V_\text{t}| \geq 0 \$ are equivalent.

Proof

If \$V_\text{GS} \leq 0 \$ then \$|V_\text{GS}| = -V_\text{GS}\$, and if \$|V_\text{t}| \leq 0 \$ then \$|V_\text{t}| = -V_\text{t} \$.

$$\begin{align*} |V_\text{GS}| - |V_\text{t}| &\geq 0 \tag{10} \\\\ -V_\text{GS} + V_\text{t} &\geq 0 \tag{11} \\\\ V_\text{GS} - V_\text{t} &\leq 0 \tag{12} \end{align*} $$

which shows that in this special case the two inequalities are equivalent. Additionally, if \$V_\text{DS} \leq 0\$ then \$V_{\text{DS}} \leq V_{\text{GS}}-V_{\text{t}} \$ is equivalent to \$|V_{\text{GS}}|-|V_{\text{t}}| \leq |V_\text{DS}| \$.