Digital Signal Processing Reference
In-Depth Information
n
V
dd
I
bias
I
bias
I
bias
i
in
i
out
1
2
3
g
m
r
DS
C
GS
G
m, virtual
Figure A.1.
A basic multistage transconductance amplifier. The gray-shaded area
represents a virtual current-gain amplifier which includes the parasitic
output resistance of the previous stage and the gate capacitance of the
current stage. For reasons of simplicity, the dc-operating point of the
transistors is ignored.
amplifier. Suppose that the transistors are biased in their saturation region, and
the output impedance of the ideal current source is sufficiently high so that it
be neglected.
For low frequencies, the small-signal output current
i
out
of each stage is
absorbed by it's own output resistance
r
ds
. At the same time, the voltage drop
over this resistor is also the small-signal input voltage of the following stage.
The current gain of each virtual intermediate stage (highlighted by the shaded
area in Figure A.1) is thus given by
g
m
r
ds
. For increasing frequencies, an in-
creasing portion of the input current is dumped in the parasitic gate-source
capacitor
C
gs
, until the cut-off frequency
f
T
has been reached. Apart from
consuming power, the intermediate stage does not provide any additional ben-
efit beyond this frequency point.
Back to the original question posed in the beginning of this section: should
the transconductance
g
m
be increased by reducing the overdrive voltage or by
increasing the bias current? As an experiment consider the case where
V
gs
−
V
T
is reduced. If the bias current
I
bias
of the transistors remains constant, if follows
from Equation (A.3) that
g
m
is inverse proportional to the overdrive voltage:
1
V
gs
−
g
m
∝
V
T
,
for fixed
I
bias
(A.8)
thus:
g
m
if
V
gs
−
V
T
On the other hand, it follows from Equation (A.2) that the width of the transis-
tor needs to be increased in order to maintain a fixed bias current. As a result,
the gate-source capacitance
C
gs
of the transistor is increased by the same ratio:
