IEEE Solid-State Circuits Magazine - Spring 2014 - 10

it turns out that it is easier to find v oc
first. This is because with the circuit
being open, the current that flows
through R S will be zero and this in
turn makes v s (and hence g m v s and
g m v bs) zero. As a result, the equivalent circuit will have an open-circuit
voltage that is -g m ro v in . Given this
and the value of R eq found in Element
5, we can find i sc as -g m ro v in /R eq .
Finding i sc proves to be easier when
we look into the source of a transistor.
This is illustrated in Element 8. Here,
by shorting the source to ground, we
effectively zero g m v s and g mb vbs .
The transistor current g m v in now goes
through a current divider consisting of
ro and R D . Once we find i sc, it is easy
to find v oc as i sc times R eq (as found
in Element 6).
Now, we will use these elements
to analyze two circuits shown in
Figures 3 and 4. In Figure 3 (cascode
configuration), we are interested in
finding v out as a function of v in . We do
this in two steps:
■■ We replace M 1 with its Thevenin
equivalent circuit (using Elements
2 and 7) and replace M 2 and R L

contributors

The problem
of -finding the
-small-signal voltage
of a node in a circuit
including transistors reduces to the
-problem of finding
the Thevenin and
Norton equivalent
circuits at that node.

with its equivalent resistance (using
- lement 6).
E
■■ Using these equivalent circuits,
we find the current going to
ground (which is equal to the current going through R L ) and hence
the voltage across R L .
Figure 4 shows a differential pair
with diode connected loads. Here, we
are interested in determining the voltage gain from the input to the common node ^v d5h and to the output
voltage ^v outh . The circuit consists of
five transistors, and as such it would
be time-consuming and cumbersome
to draw the small-signal models for all

Maurits Ortmanns
is a full professor at
the University of Ulm,
Germany.

Tianyi Liu is a research assistant/
Ph.D. candidate at the
Institute of Microelectronics, Faculty of Engineering and Computer
Science, University of Ulm, Germany.

Hanspeter Schmid
is the professor for
analog microelectronics at IME/FHNW and
is also a part-time
senior lecturer at ETH
Zürich.

s p r i n g 2 0 14	

References
For Thevenin and Norton equivalent circuits,
refer to R. E. Thomas, et al., The Analysis &
Design of Linear Circuits, 7th ed. New York:
Wiley, 2012, pp. 109-122.
For small-signal analysis of transistor
circuits, refer to A. S. Sedra and K. C. Smith,
Microelectronic Circuits, 6th ed. London, U.K.:
Oxford Univ. Press, 2010.
B. Razavi, Fundamentals of Microelectronics. New York: Wiley, 2008.

(Continued from p. 3)

Jonas Handwerker is working toward
his Ph.D. degree at the
Institute of Microelectronics at the University of Ulm, Germany.

10	

the transistors. To find v d5 , we use the
short-circuit current ^i scl h at this node
along with the three resistances that
are connected to this node in parallel
^R eq1, R eq5, and R eq2h . v d5 can then be
found as a simple product of i sc and
R eq1 < R eq5 < R eq2 . Once we know v d5 ,
we find the current that goes through
R eq2 (and hence through R eq4 ). v out is
then equal to v d5 /R eq2 times R eq4 .
In summary, using elements introduced in this article, the problem of
finding the small-signal voltage of a
node in a circuit including transistors
reduces to the problem of finding the
Thevenin and Norton equivalent circuits at that node.

IEEE SOLID-STATE CIRCUITS MAGAZINE	

Alex Huber is with
the Institute of Microelectronics of the
University of Applied
Sciences Northwestern
Switzerland, Windisch.





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