Razavi stated that

the maximum output swing at X or Y is equal to \$V_{DD}−(V_{GS}−V_{TH})\$

as shown in the following picture, the second picture is the Fig.4.3

Razavi Chapter 4
Figure 9.4 I don't understand why the equation for maximum output swing is written as..
$$ V_{DD}-(V_{GS}-V_{TH}) \tag{1} $$ ?
For a common source amplifier with a resistor load at drain, the output voltage should be $$V_{out}=V_{DD}-I_{DS}R_{D} \tag{2}$$ Due to equation 2, maximizing the output voltage swing entails maximizing the Drain-source current. I understand that the current maximizes when the drain-source voltage equals the overdrive voltage. If the above reasoning is correct, shouldn't the maximum output voltage be represent as
$$V_{out}=V_{DD}-\frac{1}{2}K_n{V_{ov}}^2R_{D} \tag{3}$$ instead of equation 1?

Reference: Razavi, Design of analog CMOS integrated circuits


2 Answers 2


Maybe you skipped the importance detail of chaptor 3, single-stage amplifier. let me recap for you.

First, look at this figure. you see the single stage amplifier only behave close to linear if \$V_{out}\$ is between \$V_{dd}\$ and \$V_{gs}-V_{th}\$.

Note:The term \$V_{gs}-V_{th}\$ came form saturation region.

That why the maximum voltage range (with linear behavior) become.

$$ V_{DD}-(V_{GS}-V_{TH}) \tag{1} $$

  • \$\begingroup\$ I just figured it out , thanks \$\endgroup\$
    – Tong Su
    Feb 21 at 17:40

I have just figured it out.
First, my previous understanding is somehow swapped; I thought that Vdd is the minimum and \$V_{GS}-V_{TH}\$ is the maximum, which should be exactly the opposite.
Secondly, for a CS stage like figure 4.3, the output voltage is exactly the Drain-Source voltage. The minimum requirement for the transistor to operate in the saturation region is when the Drain-Source voltage is equal to the overdrive voltage. That is to say: $$V_{outmin}=V_{DS}=V_{OV}=V_{GS}-V_{TH}=V_{DD}-\frac{1}{2}K{V_{OV}}^2R_{d}$$ And since the maximum voltage is just the supply voltage \$V_{dd}\$, we indeed have the maximum swing $$V_{Omax}-V_{Omin}=V_{DD}-V_{OV}$$


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