IEEE Solid-State Circuits Magazine - Summer 2015 - 10

(A > 1)

ν

AV1

V1
r
R1
R2

and the fact that R 1 + R 2 = R, we will
arrive at the same equations as in
(1) and (2).
It would be interesting to use this
graphical approach to gain intuition
in the case when A is positive and
greater than one. In this case, both
ends of the resistive string have voltages of the same sign, and hence

R
1

R1 + R2 = R
R1 < 0, R2 > 0

2

In either case,
the sum of the
two resistances in
series is equal to the
original resistance.

Figure 3: When A > 1, R can be split into
one negative (R 1) and one positive resistance (R 2) such that R = R 1 + R 2 .

¥
(3"1)*$
450$,

splits the total resistance R into
R 1 and R 2, which are the string
resistances from Nodes 1 and 2 to
the 0 V location, respectively. This
is shown pictorially in Figure 2(c),
where we have assumed V1 to be
a positive voltage and V2 = AV1 to
be a negative voltage. The voltage
along the resistive string decreases
linearly from V1 (at Node 1) to AV1
(at Node 2). The intersection of this
line with the resistance axis breaks
the resistor into two pieces, R 1 and
R 2 . One can easily verify through
the two similar triangles that the
ratio of R 2 to R 1 is | A | . Given this

10

there exists no intermediate location
with 0 V. This is shown in Figure 3
for the case where both nodes have
positive voltages. If we connect V1
and V2 via a line, we will find that
the location with 0 V lies outside the
[0 R] region at a negative resistance
(R 1) with respect to Node 1 and at a
positive resistance (R 2) with respect
to Node 2. This makes intuitive sense
because if we apply a positive voltage
to Node 1, given V2 > V1, there will
be a current moving toward Node 1
(not leaving Node 1 as we expect with
a positive resistance). For this reason,

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IEEE SOLID-STATE CIRCUITS MAGAZINE

Node 1 experiences a negative resistance (R 1) . From the perspective of
Node 2, the current always leaves the
node (when V2 is positive) indicating
a positive resistance. The resistance,
however, between this node and
ground (R 2) is now larger than R as
indicated in Figure 3. Similar to the
previous case, given the two similar
triangles in this figure, we can verify
that | R 2 /R 1 | = A and R 1 + R 2 = R to
arrive at the same equations as (1) and
(2). In other words, we can still split
R into R 1 and R 2 but with R 1 being
negative for the common node of the
two resistors to be at 0 V.
To summarize, in applying Miller's theorem, we are essentially
splitting the resistor between two
nodes as two resistors in series
such that their common node will
have 0 V! If the two node voltages
have opposite signs, we will end up
with two positive resistors. If they
have the same sign, we will end up
with one negative and one positive
resistor. In either case, the sum of
the two resistances in series is equal
to the original resistance.

References

[1] A. S. Sedra and K. C. Smith, Microelectronic Circuits, 7th ed. London, U.K.: Oxford
Univ. Press, 2014.
[2] B. Razavi, Fundamentals of Microelectronics. Hoboken, NJ: Wiley, 2008.

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Table of Contents for the Digital Edition of IEEE Solid-State Circuits Magazine - Summer 2015

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