Example-1
Find the transistor current in the circuit shown in fig. 5, if ICO= 20nA, β =100.
Solution:
For the base circuit, 5 = 200 x IB + 0.7
Therefore,
Since ICO << IB, therefore, IC = β IB = 2.15 mA
From the collector circuit, VCE = 10 - 3 x 2.15 = 3.55 V
Since, VCE = VCB + VBE
Thus, VCB = 3.55 - 0.7 = 2.55 V
Therefore, collector junction is reverse biased and transistor is operating in its active region.
Fig. 5
Example - 2
If a resistor of 2K is connected in series with emitter in the circuit as shown in fig. 6, find the currents. Given ICO= 20 nA, β =100.
Solution:
IE = IB + IC = IB + 100 IB = 101 IB
For the base circuit, 5 = 200 x IB + 0.7 + 2k x 101 IB
Therefore,
Since ICO << IB, therefore, IC = βIB = 1.07 mA
From the collector circuit, VCB = 10 - 3 x 1.07 - 0.7 - 2 x 101 x 0.0107 = 3.93 V
Therefore collector junction is reverse biased and transistor is operating in its active region.
Fig. 6
Example - 3
Repeat the example-1 if RB is replaced by 50k.
Solution:
The circuit is shown in fig. 7.
Since the base resistance is reduced, the base current must have increased and there is a possibility that the transistor has entered into saturation region.
Assuming transistor is operating in its saturation region,
VBE (sat)= 0.8 V and VCE (sat) = 0.2V
Therefore,
and
The minimum base current required for operating the transistor in saturation region is
Since IB > IB(min), therefore, transistor is operating in its saturation region.
Fig. 7
Example - 4
Repeat the example-2 if RB is replaced by 50k.
Solution:
The circuit is shown in fig. 8.
Since the base resistance is reduced, the base current must have increased and there is a possibility that the transistor has entered into saturation region.
Assuming transistor is operating in its saturation region,
Solving these equations, we get,
IC = 1.96mA and IB = 0.0035mA
The minimum base current required for operating the transistor in saturation region is
Since I B < I B(min) , therefore, transistor is operating in its active region and not in saturation. The base and the collector currents can be recalculated assuming the transistor to be in active region.
For the base circuit, 5 = 50 x IB + 0.7 + 2k x 101 IB
Therefore,
From the collector circuit, VCB = 10 - 3 x 1.71 - 0.7 - 2 x 101 x 0.0171 = 0.716 VIC = 1.71mA
Fig. 8
