IEEE Solid-States Circuits Magazine - Summer 2021 - 17

SHOP TALK: WHAT YOU DIDN'T LEARN IN SCHOOL
Chris Mangelsdorf
Stupid FET Tricks:
The Zero-Gain Amplifier
T
This all started with an interview
question: What is the gain of the circuit
in Figure 1? Of course, the problem
came with some " fine print. "
By gain, the interviewer meant
small-signal gain. It was implied that
Vin
had enough dc bias to move M1
into the active region. The key was
that the drain resistor R was equal
to 1/gm1
conductance of M1
, the inverse of the trans.
How would you
respond?
If you find yourself writing down
transcendental equations, you're
overthinking this. This is a bar napkin
" parlor trick, " not an exercise in
theoretical physics.
If you missed that, you failed the
interview. To work with industrial
circuit types, you need to be able
to shoot from the hip. Know the
answer yet?
The correct response is " zero. " Not
0 dB, just zero. None of the signal
makes it to the output. To see this,
break the circuit into two paths, as
shown in Figure 2. The top path, from
the input through the resistor to the
output has a small-signal gain of one.
The bottom path, through M1
, is just
-gm1 × R, or -1. The superposition of
the paths with gains of +1 and -1 is
zero. They cancel each other out.
Meet the zero-gain amplifier (ZGA).
OK, that was amusing, but that's
the end of it, right? Nobody needs
a signal-blocking amplifier. " That's
about as useful as a write-only memDigital
Object Identifier 10.1109/MSSC.2021.3089182
Date of current version: 25 August 2021
vin
R = 1/gm1
+
-
vin
vin
vout
M1
Bottom Path:
Gain = -1
+
-
FIGURE 1: What is the small-signal gain
of this circuit? M1
is biased in the active
region, and the conductance of the resistor
is equal to the transconductance of M1
.
FIGURE 2: Analyzing the gain as the superposition
of the signals from two paths.
IEEE SOLID-STATE CIRCUITS MAGAZINE
SUMMER 2021
17
vin
R = 1/gm1
vout
M1
ory. I can get the same performance
with a pair of wire cutters! " you protest.
Not so-and I can prove it!
It's Been There All Along
You may swear that you've never seen
anything so stupid in your entire
life, but I'll bet you have-and you've
probably even used it! Think of a
two-stage CMOS amplifier with Miller
compensation. How do you avoid the
right-half-plane zero caused by feedforward
through the compensation
capacitor? (No, the other method.)
That's right, you put a resistor in
series with the capacitor. What's the
value of that resistor? Hint: it's about
equal to 1/gm
of the gain transistor,
depending on where you want the
zero to land. Its purpose? To block
forward transmission at high frequency.
Zero gain! You owe me a beer.
But there is more. A lot more.
It turns out that the large-signal
behavior of the ZGA is what originally
put it on the map. To appreciate
this, we need to go back to 1966:
the Summer of Love was just about
ready to convene, silicon wafers were
1 inch in diameter, and transistors
were bigger than bond pads. (This
was even before my time!)
With such little silicon real estate
to go around and such visible-fromspace
lithography, designers were
careful to minimize the number of
active devices. The concept of throwing
great numbers of transistors at
a problem was still quite alien. Bias
circuits were particularly vexing
because you needed something to
light up your circuit but you didn't
want it to be the same size as the
amplifier you were trying to light.
Specifically, designers wanted
tiny currents of just a few microamperes
to bias input stages. (Bipolar
transistors draw a parasitic base
current, the equivalent of the gate
current in MOS, that is proportional
to their operating current,
and this would result in current
flowing in the input terminals of an
amp. Therefore, a tiny bias current
was a good thing.) You might try a
resistor across the supply, but you
Top Path:
Gain = +1
IEEE Solid-States Circuits Magazine - Summer 2021

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