Two-Port Parameters for
Single-Stage Amplifiers
Amplifier Type
Controlled Source
Input Resistance
Rin
Output Resistance
Rout
Common
Emitter
Gm = gm
rπ
ro || roc
Common
Source
Gm = gm
infinity
ro || roc
Common
Base
Ai = -1
1 / gm
roc || [(1 + gm(rπ||RS)) ro],
for gmro >> 1
Ai = -1
1 / gm, (vsb = 0)
-otherwise1 / (gm + gmb)
roc ||[(1 + gm RS)ro], (vsb=0)
-otherwiseroc || [(1+ (gm + gmb)RS) ro]
both for ro >> RS
Common
Collector
Av = 1
rπ + βο(ro || roc|| RL)
(1 / gm ) + RS / βο
Common
Drain
Av = 1 if vsb = 0,
-otherwisegm / (gm + gmb)
infinity
1 / gm if vsb = 0,
-otherwise1 / (gm + gmb)
Common
Gate
Note: appropriate two-port model is used, depending on controlled source
EE 105 Spring 2000
Page 1
Week 12, Lecture 27
Sinusoidal Function Review
Sinusoidal functions are important is analog signal processing
v ( t ) = v cos ( ωt + φ )
amplitude
(half of
peak-to-peak)
phase (degrees
or radians)
frequency
(radian) ... ω = 2π f = 2π (1/T)
v 1 ( t ) = v cos ( ωt )
v 2 ( t ) = v cos ( ωt – 45 )
2π
ω = -----T
T
1. EECS 20/120: periodic functions can be represented as sums of sinusoids
functions at different frequencies.
2. The response of a circuit to a sinusoidal input signal, as a function of the
frequency, leads to insights into the behavior of the circuit.
EE 105 Spring 2000
Page 2
Week 12, Lecture 27
Frequency Response
Key concept: small-signal models for amplifiers are linear and therefore,
cosines and sines are solutions of the linear differential equations which arise
from R, C, and controlled source (e.g., Gm) networks.
*
The problem: finding the solutions to the differential equations is TEDIOUS
and provides little insight into the behavior of the circuit!
vout(t)
vs(t)
EE 105 Spring 2000
R
C
Page 3
Week 12, Lecture 27
Phasors
It is much more efficient to work with imaginary exponentials as “representing” the
sinusoidal voltages and currents ... since these functions are solutions of linear
differential equations and
jωt
d jωt
----- ( e ) = jω ( e )
dt
How to connect the exponential to the measured function v(t)? Conventionally, v(t)
is the real part of the of the imaginary exponential
v(t) = v cos ( ωt + φ ) → Re ( ve
( jωt + φ )
jφ jωt
) = Re ( ve e
)
where v is the amplitude and φ is the phase of the sinusoidal signal v(t).
The phasor V is defined as the complex number
V = ve
jφ
Therefore, the measured function is related to the phasor by
v(t) = Re ( Ve
EE 105 Spring 2000
Page 4
jωt
)
Week 12, Lecture 27
Circuit Analysis with Phasors
*
The current through a capacitor is proportional to the derivative of the voltage:
i(t) = C d v(t)
dt
We assume that all signals in the circuit are represented by sinusoids.
Substitution of the phasor expression for voltage leads to:
v(t) → Ve
jωt
…
Ie
jωt
jωt
jωt
d
= C ( Ve ) = jωCVe
dt
which implies that the ratio of the phasor voltage to the phasor current through a
capacitor (the impedance) is
V
1
Z(jω) = --- = ---------I
jωC
*
Implication: the phasor current is linearly proportional to the phasor voltage,
making it possible to solve circuits involving capacitors and inductors as rapidly
as resistive networks ... as long as all signals are sinusoidal.
EE 105 Spring 2000
Page 5
Week 12, Lecture 27
Phasor Analysis of the Low-Pass Filter
*
Voltage divider with impedances --
Replacing the capacitor by its impedance, 1 / (jωC), we can solve for the
ratio of the phasors Vout / Vin
V out
1/jωC
----------- = -----------------------V in
R + 1/jωC
multiplying by jωC/jωC leads to
V out
1
---------- = ----------------------V in
1 + jωRC
EE 105 Spring 2000
Page 6
Week 12, Lecture 27
Frequency Response of LPF Circuits
The phasor ratio Vout / Vin is called the transfer function for the circuit
How to describe Vout / Vin?
complex number ... could plot Re(Vout / Vin) and Im(Vout / Vin) versus frequency
polar form translates better into what we measure on the oscilloscope ... the
magnitude (determines the amplitude) and the phase
*
“Bode plots”:
magnitude and phase of the phasor ratio: Vout / Vin
range of frequencies is very wide (DC to 1010 Hz, for some amplifiers)
therefore, plot frequency axis on log scale
range of magnitudes is also very wide:
therefore, plot magnitude on log scale
Convention: express the magnitude in decibels “dB” by
V out
V out
= 20 log -------------------V in dB
V in
phase is usually expressed in degrees (rather than radians):
Im ( V out ⁄ V in )
V out
---------= atan ----------------------------------∠
Re ( V out ⁄ V in )
V in
EE 105 Spring 2000
Page 7
Week 12, Lecture 27
Complex Algrebra Review
* Magnitudes:
2
2
X1 + Y1
Z1
Z1
------ = --------- = ----------------------- , where
Z2
2
2
Z2
X2 + Y2
Z 1 = X 1 + jY 1
Z2 = X2 + Y2
* Phases:
Z1
Y
Y
∠------ = ∠Z 1 – ∠Z 2 = atan -----1- – atan -----2Z2
X1
X2
* Examples:
EE 105 Spring 2000
Page 8
Week 12, Lecture 27