IEEE Solid-States Circuits Magazine - Spring 2021 - 13

Req
12 12
i1
R1 R2
v1
+
-
R1 R2
Req
i2
+
-
R3
(a)
(b)
(c)
v v1
2
+
-
ic
Req
id
R1 22ic
R2
R3
ic
+
-
v2
+
-
ic
+
-
R1 R2
R3
vd
Req
-vd
(d)
FIGURE 1: (a) Finding the equivalent resistance between nodes 1 and 2. (b) An attempt to find the equivalent resistance using two singleended
voltage sources connected to nodes 1 and 2. (c) Splitting the currents into common-mode and differential. (d) Applying complementary
voltages to nodes 1 and 2 for the purpose of measuring the equivalent resistance.
In other words, we can fix only the
voltage difference between the
two nodes but not the individual
node voltages; otherwise, the re -
sulting currents flowing through
the two terminals will not be differential
(equal magnitudes and
opposite directions). Similarly, applying
differential currents to two nodes
gives rise to a particular set of v1
v .2 These are in contrast with our
initial attempts in which we applied
two independent v1
and v2
to nodes
1 and 2, violating the current constraint.
Even applying complementary
voltages ( /v 2d
and -v /2d
) to
the two nodes may violate the current
constraint. Let us now use
the proper methods of finding the
equivalent resistance.
Between the two methods illustrated
in Figure 2(a) and 2(b), we prefer
the latter as it explicitly shows
that the current entering a node
is equal to the current leaving the
other node. Using this circuit, it is
now easy to see that no current goes
through R3
, and hence v 03 = The
.
voltage difference between nodes
1 and 2 is then
vR iR i xx x12
xx 12
,
=+ or
Rv /. eq iR R== + For the numerical
example we provided earlier,
this will amount to an equivalent
resistance of
6k Ω, which is the correct
answer.
For further intuition into this
answer, we will use superposition.
The current leaving node 2 and entering
node 1 can be split into two current
sources as shown in Figure 2(c):
(a)
(b)
(c)
FIGURE 2: (a) Measuring the equivalent resistance seen between the two nodes of a circuit
using a test voltage source. (b) Using a test current source to find the equivalent resistance.
(c) Splitting the test current source into two single-ended current sources.
IEEE SOLID-STATE CIRCUITS MAGAZINE
SPRING 2021
13
and
one leaving node 2 and entering the
ground and one leaving the ground
and entering node 1. Since the current
that enters the ground is equal to
the current that leaves the ground,
there is no impact on ground (KCL
remains satisfied). Now, we can
use superposition: keep one current
source on at a time, find its contribution
to
v ,x and then add the contributions
from the two sources to find the
final value of
v .x
that vx, due to the left ix
due to the right ix
vR ,Ri
xx
result for R .eq
=+
12
, is Rix1
, is Ri .x2
It is easy to verify
and,
Hence
^h giving us the same
The reader should attempt to
solve the same problem by applying
a voltage source vx
between the two
nodes, as shown in Figure 2(a), and
confirm that the same answer can
be found.
vx
+-
v1
ix
R1
v3
R3
R2
v2
v2
ix
ix
v1 ++vx
v1
R1 R2
v3
R3
R1
v3
R3
ix
vx-- v2
ix
R2
To recap, the only way not to violate
the current constraint in calculating
the equivalent resistance is
by either applying a voltage source
as shown in Figure 2(a), without
specifying v1
and v ,2 or applying
a current source as shown in Figure
2(b), both ensuring equal but
opposite currents through node 1
and node 2.
We are now ready to apply this
method to small -signal ana lysis
of several examples of transistor
circuits. In Figure 3, we wish
to find the resistance looking into
the gate-drain pair of a commonsource
amplifier. This resistance,
denoted by
R ,gd
plays an important
role in determining the time constant
associated with the gate-drain
capacitance Cgd
and in estimating
the amplifier's bandwidth. In this

IEEE Solid-States Circuits Magazine - Spring 2021

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