Image Processing Reference
In-Depth Information
I ds
I ds
I ds
V dd
Drive
transistor
Drive
transistor
Drive
transistor
V in
Load
transistor
Drive
transistor
Constant
current source
Resistor
C in
V out
V out
V ds
V out
V dd
V ds
V out
V dd
V ds
C out
Load
device
Constant
current source
Resistor
Load transistor
(a)
(b)
(c)
(d)
FSA: A buffer circuit for impedance conversion
Voltage gain: G V < 1
Charge amount gain: G Q C out / C in
FIGURE 2.24
Source follower amplifier: (a) general configuration; (b) constant current source; (c) resistor; (d) load transistor.
transistor and load device is shown in Figure 2.24. As the currents through the drive
transistor and load device are the same, the intersection point of the two voltage-current
curves is the operating point. If the signal voltage of the input V in in SFA changes, the oper-
ating point moves. The solid curved lines of the voltage-current characteristic of the drive
transistor in Figure 2.24b-d are the characteristics before the input voltage of V in changes,
that is, a no-signal situation. The broken lines show those after the change, that is, the
signal charge quantity information is added to input V in . Therefore, the voltage difference
between two intersection points on the voltage-current characteristic curve indicates the
output voltage amplitude of SFAs. Regarding the output change in contrast with the input
voltage change, the constant current source is the highest of the three. Although it has a
good characteristic, the area of circuitry tends to be large, and therefore the load transistor
is used in most cases because of characteristic—circuit size trade-offs.
An ideal characteristic is that the output voltage V out is equal to the input voltage V in ,
however, output voltage is obtained by subtracting the threshold voltage V th of the drive
transistor from the input voltage and multiplying by SFA voltage gain G V (<1, around
0.7 - 0.9). This means the actual output voltage is lower than the input one, as expressed by
the following equation:
(
)
VVVG V
=−
(2.5)
out
in†
th
Despite the name “amplifier,” the voltage gain is less than unity, as mentioned above.
Their respective load capacitors, C in and C out , are at the input part and the output part of
the SFA. Under a very simplified condition, G V = 1, V th = 0, and V in = V out . Then the amount
of charge accumulated in C out is C out / C in times the charge amount in C in because of the
same voltages. Although the voltage gain of SFA is less than unity, the amount of charge is
greatly multiplied and this is important to drive the later circuit.
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