Electronics - Bipolar Junction Transistors (BJT) - Discussion
Discussion Forum : Bipolar Junction Transistors (BJT) - General Questions (Q.No. 14)
14.
If an input signal ranges from 20–40 
A (microamps), with an output signal ranging from .5–1.5 mA (milliamps), what is the ac beta?
A (microamps), with an output signal ranging from .5–1.5 mA (milliamps), what is the ac beta?Discussion:
9 comments Page 1 of 1.
Rupali said:
1 decade ago
Beta = (1.5-.5)mA/(40-20)micro Amp = 50
Fredy and vimal said:
1 decade ago
beta=d(Ic)/d(Ib)
typical value 50 - 250
typical value 50 - 250
Kosgei said:
1 decade ago
Ensure you change the micro ampheres to milliamphere.
Vinod kumar said:
1 decade ago
Beta=D(output)/D(input)
=> Beta= (1.5-0.5)*10^-3/(40-20)*10^-6
=> Beta= 1*10^3/20
=> Beta= 1000/20
=> Beta= 50
=> Beta= (1.5-0.5)*10^-3/(40-20)*10^-6
=> Beta= 1*10^3/20
=> Beta= 1000/20
=> Beta= 50
(2)
Nidhi patel said:
1 decade ago
Beta = (1.5-0.5)*10^-3/(40-20)*10^-6.
you can get answer by solving above expression.
you can get answer by solving above expression.
(2)
Jasmine George said:
1 decade ago
Beta = current gain = output current/input current.
= ((1.5*10^-3)-(.5*10^-3)/(40*10^-6)-(20*10^-6)).
= 50.
= ((1.5*10^-3)-(.5*10^-3)/(40*10^-6)-(20*10^-6)).
= 50.
(1)
Dishant garg said:
1 decade ago
Beta = (collector current gain)/(base current gain).
= (1*0.001)/20*0.000001.
= 50.
= (1*0.001)/20*0.000001.
= 50.
(2)
Jitesh said:
6 years ago
Beta is output /input.
i.e (1.5-0.5)/(40-20).
i.e (1.5-0.5)/(40-20).
TKay said:
6 years ago
1*10^-3/20*10^-6 = 50.
From equation B=collector current gain /base current gain.
From equation B=collector current gain /base current gain.
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