IEEE Solid-States Circuits Magazine - Winter 2020 - 7

of this circuit is identical to that of the
differential one since both the input
and the output are divided by two in
the half circuit. We observe that the
half circuit is an amplifier with source
degeneration [3]. In general, the gain
of a common-source amplifier is
reduced when a resistor is added to
its source terminal. In this case, at low
frequencies, when C s is considered
open, a nonzero R s does reduce the
voltage gain of the circuit. At higher
frequencies, however, the capacitor
(C s) begins to short R s, reducing the
degeneration and increasing the gain.
So, in effect, this capacitor produces
a frequency-dependent gain where
the gain increases with frequency (as
the capacitor becomes a short circuit
with increasing frequency). This is the
exact effect we wished to produce.
Let us now find Vo in the half circuit using the method described in
[2]. We find the short circuit current
at the output node (I scd) and multiply it by the total impedance (Z eqd)
seen at the output.
Figure 4 presents the steps toward
finding I scd . In the first step [Figure  4(b)], we find Vs by multiplying
its short circuit current (I scs) and its
equivalent impedance Z eqs . We then
use Figure 4(c) to find I scd .
The short circuit current at the
source node can be written as
I scs = g m Vi,
where g m is the transistor's transconductance. Note that this is the short
circuit current at the source node
while the drain is also shorted to
ground. The equivalent impedance at
this node, while the drain is shorted,
can be written as
Z eqs =

1
,
g m + 2/R s + 2sC s

where we have ignored the body
effect and the channel-length modulation (that is, we have assumed
g me = g m and g m ro & 1). The voltage
at the source node, while the drain
is shorted, can be written as the
product of I scs and Z eqs:
Vs =

g m Vi
.
g m + 2/R s + 2sC s

RL

CL
Vo

Zeqd
Iscd

M1

Vi

RS
2

Vo
Vi

RS
2

2CS

(a)

Vi

M1

VS
Zeqs
2CS

(b)

Iscs

M1

Iscd

VS

(c)

FIGURE 4: (a) A small-signal half circuit for the differential equalizer in Figure 3. We use this
circuit to derive an expression for Vo . (b) A circuit model to derive an expression for Vs while
the output node is shorted to ground. (c) Calculating the short circuit current at the output
node given Vi and Vs.

We now use Figure 4(c) to find the
output short circuit current
I scd = g m (Vs - Vi).
If we further divide I scd by Vi, we
find the short circuit transadmittance
of this amplifier, which is
G (s ) =

- g m 1 + sC s R s
,
a 1 + sC s R s /a

where a = 1 + g m R s /2 is the degeneration factor.
We note that G (s ) does not include
the load. This is simply because
we have shorted the output node to
ground. This transfer function exhibits a gain at dc (-g m /a) and has one
zero at 1/2rC s R s and one pole at
a/2rC s R s . The high-pass characteristic of this equalizer is due to the fact
that the zero frequency is lower than
the pole frequency (since a 2 1).
We now multiply this transfer
function by the equivalent output
impedance (Z eqd) to find the overall
voltage gain of this amplifier:
Z eqd =

RL
.
1 + sC L R L

In writing this equation, we have
ignored the impedance looking down
into the drain of the transistor as it is
assumed to be much larger than the
load. Finally, we can write
Vo -g m R L 1 + sC s R s
1
=
.
Vi
a
1 + sC s R s /a 1 + sC L R L
This expression clearly identifies
one zero and two poles of the trans-

fer function. More explicitly, in connection to the parameters shown on
Figure 2, we have
fz = 1/2rC s R s,
fp1 = a/2rC s R s,
fp2 = 1/2rC L R L .
The reader can verify that, by
changing C s, we can move fz and fp1
without changing fp2 or the equalizer's dc gain. Indeed, by making C s
tunable, this circuit can equalize a
wider variety of channels, although an
equalizer with only one zero may not
be able to compensate for higher-order
channel attenuations. In these cases,
we may require two to three stages of
equalization or a more sophisticated
equalizer altogether.
In summary, an equalizer provides a
transfer function in the signal path that
is the inverse of the transfer function
introduced by the wire connecting the
transmitter and the receiver. In doing
so, the equalizer compensates for the
frequency-dependent loss caused by
the wire and restores the original spectrum of the transmit signal.

References

[1] A. Sheikholeslami, "Circuit intuitions: The
electrical length of a wire," IEEE Solid-State
Circuits Mag., vol. 11, no. 3, pp. 7-9, 2019.
doi: 10.1109/MSSC.2019.2922885.
[2] A. Sheikholeslami, "Circuit intuitions: Looking into a node," IEEE Solid-State Circuits Mag.,
vol. 6, no. 2, pp. 8-10, 2014. doi: 10.1109/
MSSC.2014.2315062.
[3] A. Sheikholeslami, "Circuit intuitions:
Source degeneration," IEEE Solid-State Circuits Mag., vol. 6, no. 3, pp. 5-6, 2014. doi:
10.1109/MSSC.2014.2329233.

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IEEE Solid-States Circuits Magazine - Winter 2020

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