The AC current gain in a common base configuration is_________
The AC current gain is denoted by α_{ac}. The ratio of change in collector current to the change in emitter current at constant collector base voltage is defined as current amplification factor.
The value of αac for all practical purposes, for commercial transistors range from_________
For all practical purposes, αac=αdc=α and practical values in commercial transistors range from 0.90.99. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
A transistor has an IC of 100mA and IB of 0.5mA. What is the value of αdc?
Emitter current I_{E}=I_{C}+I_{B} =100+0.5=100.5mA
α_{dc}=I_{C}/I_{E}=100/100.5=0.995.
In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.
Here, I_{C}=4.9/5K=0.98mA
α = I_{C}/I_{E} .So,
I_{E}=I_{C}/α=0.98/0.98=1mA.
I_{B}=I_{E}I_{C}=10.98=0.02Ma.
The emitter current I_{E} in a transistor is 3mA. If the leakage current I_{C}BO is 5µA and α=0.98, calculate the collector and base current.
I_{C}=αI_{E} + I_{C}BO
=0.98*3+0.005=2.945mA.
I_{E}=I_{C}+I_{B} . So, I_{B}=32.495=0.055mA=55µA.
Determine the value of emitter current and collector current of a transistor having α=0.98 and collector to base leakage current I_{C}BO=4µA. The base current is 50µA.
Given, I_{B}=50µA=0.05mA
I_{CBO}=4µA=0.004Ma
I_{C}=α/(1 α)I_{B}+1/(1 α)I_{CBO}=2.45+0.2=2.65Ma
I_{E}=I_{C}+I_{B}=2.65+0.05=2.7mA.
The negative sign in the formula of amplification factor indicates_________
When no signal is applied, the ratio of collector current to emitter current is called dc alpha, α_{dc} of a transistor. α_{dc}=I_{C}/I_{E}. It is the measure of the quality of a transistor. Higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
The relation between α and β is _________
The terms Alpha and Beta refer to BJTs, Bipolar Junction Transistors. In these devices current leaves the emitter and crosses to the base and most is diverted towards the collector. The remainder leaves the base terminal.
Using a convenient notation, Ie = Ic + Ib
So Ib = Ie  Ic
Alpha is the proportion of Ie that flows to the collector,
ɑ = Ic/Ie ————— Definition
So Ic = ɑ Ie
Also Ib = Ie  ɑ Ie = Ie(1  ɑ)
Beta is the current gain, the ratio of Ic to Ib.
β = Ic/Ib ————— Definition
Substituting in this definition for Ib and Ic
β = (ɑ Ie)/ Ie(1  ɑ)
Cancelling Ie leaves:
β = ɑ/(1  ɑ)
A transistor has an I_{E} of 0.9mA and amplification factor of 0.98. What will be the I_{C}?
Given, I_{E} = 0.9mA, α=0.98
We know, α= I_{C}/I_{E}
So, I_{C}=0.98*0.9=0.882mA.
The collector current is 2.945A and α=0.98. The leakage current is 2µA. What is the emitter current and base current?
(I_{C} – I_{CBO})/α=I_{E}
= (2.9450.002)/0.98=3mA.
I_{E}=I_{C}+I_{B} . So, I_{B}=32.495=0.055mA=55µA.
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