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Hi, this is from Razavi's Analog CMOS IC Design book (2nd edition) - Pg 49. He says that " We observe that the circuit provides a gain of about 3 in the input range of 0.4V to 0.6V"

But if you look at the graph, at the input of 0.5V, the output voltage is approx 0.55V. 0.55/0.5 = 1.1. Where is he getting the gain of 3?


2 Answers 2


The author is talking about small-signal gain, not large-signal. A small change in the input of 0.01V will change the output by about 0.03V. Although it's actually a negative gain here, as an increase in the input means a decrease in the output.

This is an estimate, though; just visually looking at the graph I would estimate a gain more around -2.5 or perhaps a bit more, probably lower magnitude than the 3 stated in the text.

  • \$\begingroup\$ Sorry, would you mind just explaining a bit more on differential gain. I usually thought that was in reference to differential amplifiers? How exactly is this differential gain? \$\endgroup\$ Commented Aug 18, 2019 at 15:31
  • \$\begingroup\$ Sorry, I used the wrong term there. Early-morning sloppiness. Fixed it. \$\endgroup\$
    – Hearth
    Commented Aug 18, 2019 at 15:32
  • \$\begingroup\$ That makes more sense now. Small-Signal gain is the derivative of large signal gain (slope) in the linear region of an amp. \$\endgroup\$ Commented Aug 18, 2019 at 17:03
  • \$\begingroup\$ @Hearth -- in some universe it's called differential gain. I knew what you meant, but from context, not from the word you used. \$\endgroup\$
    – TimWescott
    Commented Aug 18, 2019 at 17:11
  • \$\begingroup\$ @TimWescott Mornings tend to make me even worse with words than I normally am! \$\endgroup\$
    – Hearth
    Commented Aug 18, 2019 at 17:39

The gain comes from the small-signal gain, first you have to bias (put a dc voltage into) \$V_{in}\$ so that the \$|\frac{d}{dV_{in}}V_{out}|\$ is big, in a simpler way, use a \$V_{in}\$ that is located at a very steep position of the graph. In the example the author mentions

input in the range of 0.4V and 0.6V.

Lets use \$V_{in}=0.5V\$, at the very center of the range he suggests (that leads to \$V_{out} \approx 0.55)\$.

Now, if you have a signal added to that constant part, say, \$V_{in}=0.5+0.1\$, the output is \$V_{out} \approx 0.3 = 0.55 - 0.25\$.

From this, it can be determined that \$\frac{d}{dV_{in}}V_{out} \approx -0.25/0.1 = -2.5\$ if you bias \$V_{in}\$ with 0.5V. This gives you the gain for small signals applied to device while biased.

Remember that, by the first terms of the Taylor series $$f(x_0+\delta x) \approx f(x_0) + \frac{d}{dx}f(x_0)\delta x$$


$$V_{out}(0.5+\delta x) \approx V_{out}(0.5) -2.5 \cdot \delta x = 0.55 -2.5 \delta x $$

Where is he getting the gain of 3?

The author approximated 2.5 to 3 and ignored the fact that this device is an inverting amplifier (has a negative gain of 2.5, which is almost 3).


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