Page 295 - Mechatronics with Experiments

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JWST499-c05
JWST499-Cetinkunt
ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS 281
The input circuit of the transistor has the relationship (assuming the transistor is operating
in the linear active region), and V = 0.0,
in
1
i = ⋅ i C (5.178)
B

V R B
i = (5.179)
B
R B
V CC − V BE
= (5.180)
R
B
V CC
≈ (5.181)
R
B
Let use denote V = V to indicate the bias voltage, since V = 0. From the above
out 0 in
relations, we can determine the bias resistor needed to provide the desired bias voltage for
the given supply, load, and transistor,
V = V CC − R ⋅ i C (5.182)
L
0
= V CC − R ⋅ ⋅ i B (5.183)
L
V CC
= V − R ⋅ ⋅ (5.184)
CC L
R
B
( )
⋅ R L
V = V 1 − (5.185)
0 CC
R
B
The bias resistor is then
V CC
R = ⋅ ⋅ R (5.186)
B L
V − V
CC 0
It should be noted again that this is only one of many possible and simple ways of introducing
bias voltage to a transistor amplifier. In this design, the bias voltage is a linear function of
the amplifier current gain, , which is known to vary significantly due to manufacturing
variations and temperature. In a given initial design, if it turns out that the bias voltage (V )
0
is larger than anticipated as a result of variations in , then the designer may need increase
or reduce R .
B
Example Consider the bias voltage circuit for a transistor as shown in Figure 5.22. Let
the parameters of this circuit be given as follows,
V = 12 V, = 100, R = 10 KΩ (5.187)
CC L
Determine the value of the bias resistor, R , so that the bias voltage is V = 6V.
B 0
From the above equation, the bias resistor is
V CC
R = ⋅ ⋅ R L (5.188)
B
V CC − V 0
= 100 ⋅ 2 ⋅ 10 KΩ (5.189)
= 2MΩ (5.190)

After this initial design, if the circuit indicates that the V is different than the desired value
0
(due to the variation in the transistor gain ), then the value of the resistor R should be
B
modified according to the above relation until the desired bias voltage is obtained.