IEEE Solid-States Circuits Magazine - Spring 2021 - 15

This result is indeed similar to
ix
Req
v1 +- v2
vx
gm, ro
(a)
ix
v1
gm, ro
(b)
FIGURE 6: (a) Measuring the output resistance
between the two nodes of a current
mirror using a test current source. (b) Splitting
the test current source into two singleended
current sources.
vv ()11 22
xl vv vrl r
=+ -+
.
=+
ir rxo o12
Step 4: Find Req
^h
as /:
vi
Rr .rooeq 12
=+
xx
+
vx
-
ix
v2
gm, ro
gm, ro
the result we found in relation to the
circuit of Figure 1.
As our final example, we determine
the output resistance of a current mirror
as shown in Figure 6(a). At a first
glance, the output resistance may
appear to be the sum of ///,
1 gr
mo
which is the resistance looking into
the left node, and
r ,o which is the re -
sistance looking into the right node.
However, using the step-by-step procedure
we followed for the differential
pair, one can verify that
vr i2xo x
=
Rr .2 o
eq =
This result, being exact (that is, no
approximation), may seem surprising.
(assuming
/gr1
% ). Howo
ever,
the resistance looking into
v2
is 2r .o How could this be? The
answer lies in the act of mirroring,
which effectively doubles the short
circuit current [1] of node 2: one
explicit ix
mirrored from the left side. When
added and multiplied by ,ro
produce a voltage that is twice as
References
[1] A. Sheikholeslami, " Looking into a node
[Circuit Intuit ions], " IEEE Solid State
Circuits Mag., vol. 6, no. 2, pp. 8-10,
Spring 2014. doi: 10.1109/MSSC.2014
.2315062.
on the right side and one
they
[2] A. Sheikholeslami, " Source degeneration
[Circuit Intuitions], " IEEE Solid State
Circuits Mag., vol. 6, no. 3, pp. 8-10,
Summer 2014. doi: 10.1109/MSSC.2014
.2329233.
Indeed, the resistance looking
into v1 alone can be approximated
with /g1 m
large or, equivalently, a resistor that
is r2 o
(given that we see only one ix
leaving node 2). This example also
illustrates how a differential current
going into the current mirror does
not produce a differential (complementary)
voltage at its two nodes.
The reader can verify that the amplitude
of v2
is approximately gr2 mo
times that of v .1
In summary, to determine the
resistance between two nodes in an
LTI circuit, we apply a set of differential
currents (ix
and ix
^hv ,x
vi
- ) to the
two nodes, measure the resulting
voltage difference between the two
nodes
xx
sure the current ix
and find /. Alternatively,
we can apply a voltage source
vx
between the two nodes, meathat
flows from
one node to the other through the
voltage source, and find /.
vi
xx
THE ANALOG MIND (continued from p. 11)
the transmission line, thereby drawing
a frequency-independent input
current,
in=+ It follows
VVV IR RST
that /( /)() ,1=+ -
is given by (19).
out2
IV RR ).
out2 in
in /( ST
in
where VI /out2 in
The input interface therefore does
not exhibit
the effect depicted in
Figure 9(c).
Figures 11(b) and 11(c) present
this design's received eye diagram
and ;;S ,11
indicating satisfactory performance.
The latter remains below
- 10dB up to 28 GHz.
References
[1] D. Feucht, Handbook of Analog Circuit Design.
New York: Academic, 1990.
[2] S. Galal and B. Razavi, " 10-Gb/s limiting
amplifier and laser/modulator driver in
0.18um CMOS technology, " IEEE J. SolidState
Circuits, vol. 38, pp. 2138-2146,
Dec. 2003. doi: 10.1109/JSSC.2003.
818567.
[3] J. Paramesh and D. J. Allstot, " Analysis
of the bridged T-coil circuit using the
extra-element theorem, " IEEE Trans. Circuits
Syst. II, vol. 53, pp. 1408-1412, Dec. 2006.
doi: 10.1109/TCSII.2006.885971.
[4] M. Kossel et al., " A T-coil enhanced 8.5-Gb/s
high-swing SST transmitter in 65-nm bulk
CMOS with ¡-16 dB return loss over 10-GHz
bandwidth, " IEEE J. Solid-State Circuits,
vol. 43, pp. 2905-2920, Dec. 2008. doi:
10.1109/JSSC.2008.2006230.
[5] B. Razavi, " The bridged T-coil, " IEEE SolidState
Circuits Mag., vol. 7, pp. 10-13, Fall
2015. doi: 10.1109/MSSC.2014.2369332.
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IEEE Solid-States Circuits Magazine - Spring 2021

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