Page 294 - Mechatronics with Experiments

JWST499-Cetinkunt
JWST499-c05
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Notice that as long as the Zener diode is conducting, the voltage across the base and
ground will be V = V , and the maximum base current is
B
Z
i = (V CC1 − V )∕R 1 (5.170)
B
Z
The voltage between the emitter and ground is
V = V − V BE = V − 0.7 (5.171)
E
Z
Z
Hence, current across the resistor R is
E
i = V ∕R = (V − 0.7)∕R E (5.172)
E
E
Z
E
The collector current which also is the load current is

i = i − i = i E (5.173)
E
C
B
+ 1
Notice that i is very small compared to i .The i is constant as long as the V > V and
B E E CC1 Z
the R is constant. Therefore, the i will remain a constant current despite the variations in
E C
the load resistance R .
L
Bias Voltage So far we have considered transistor amplifiers that amplify current or
voltage and that when the input signal is zero, the output signal is zero as well, that is
V out = K ⋅ V in (5.174)
v
Many applications require offset voltage in the input and output voltage relationship as
follows,

V out = V + K ⋅ V in (5.175)
v
0
where V is the bias voltage. There are many different versions of circuit designs to introduce
0
bias voltage to the amplifier, one of which is shown in Figure 5.22. When the input voltage
(V ) is zero, there is still a base current due to the power supply voltage (V CC ) and the
in
resistor R . Given the supply voltage, load resistance, amplifier current gain, and the desired
B
bias voltage: V , R , , V , determine the bias resistance (R ) needed.
CC L 0 B
The output circuit relation of the transistor is
V = R ⋅ i + V = R ⋅ i + V (5.176)
CC L C CE L C out
V CC − V out
i = (5.177)
C
R L
V
CC
R R
B L
v out = V o
i
B
i c
v in

FIGURE 5.22: Bias voltage circuit for an
amplifier.