Question
Question: For a common-emitter transistor, input current is \[5\,\mu A\]. Circuit is operated at load resistan...
For a common-emitter transistor, input current is 5μA. Circuit is operated at load resistance of 10kΩ. The voltage across collector emitter will be
A. 5 V
B. 10 V
C. 12.5 V
D. 7.5 V
Solution
Using the relation between amplification factor, collector current and base current calculate the collector current. Use the formula for voltage across the collector and emitter to determine the collector emitter current.
Complete step by step answer:
The current amplification factor is given as,
β=IBIC
Here, IC is the collector current and IB is the base current.
The voltage across collector emitter is given as,
VCE=RLIC
Here, RL is the load resistor across the output of the collector emitter.
We know that the current amplification factor is defined as the ratio of collector current to the base current.
β=IBIC
Here, IC is the collector current and IB is the base current.
We know that in common emitter configuration, the input current is given to the base of the transistor. Therefore, the base current is IB=5μA.
From the above equation, we can calculate the collector current as follows,
IC=βIB
We substitute 100 for β and 5μA for IB in the above equation.
IC=(100)(5×10−6A)
⇒IC=5×10−4A
⇒IC=0.5mA
We have the voltage across collector emitter is given as,
VCE=RLIC
Here, RL is the load resistor across the output of the collector emitter.
We substitute 10kΩ for RL and 0.5mA for IC in the above equation.
VCE=(10×103Ω)(0.5×10−3A)
⇒VCE=5V
So, the correct answer is “Option A”.
Note:
The amplification factor or current gain β represents the increase in collector current in terms of base current. While solving the question relating the output of transistors, students should know the relation between collector current, base current and emitter current, IE=IC+IB. The voltage across the collector-emitter is the change in the output voltage between collector and emitter.