Electronics - Bipolar Junction Transistors (BJT) - Discussion

14. 

If an input signal ranges from 20–40 mu.gifA (microamps), with an output signal ranging from .5–1.5 mA (milliamps), what is the ac beta?

[A]. 0.05
[B]. 20
[C]. 50
[D]. 500

Answer: Option C

Explanation:

No answer description available for this question.

Rupali said: (Feb 26, 2011)  
Beta = (1.5-.5)mA/(40-20)micro Amp = 50

Fredy And Vimal said: (Jun 15, 2011)  
beta=d(Ic)/d(Ib)
typical value 50 - 250

Kosgei said: (Jul 27, 2011)  
Ensure you change the micro ampheres to milliamphere.

Vinod Kumar said: (Aug 30, 2011)  
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

Nidhi Patel said: (Jan 6, 2014)  
Beta = (1.5-0.5)*10^-3/(40-20)*10^-6.

you can get answer by solving above expression.

Jasmine George said: (May 28, 2014)  
Beta = current gain = output current/input current.

= ((1.5*10^-3)-(.5*10^-3)/(40*10^-6)-(20*10^-6)).

= 50.

Dishant Garg said: (Aug 11, 2014)  
Beta = (collector current gain)/(base current gain).
= (1*0.001)/20*0.000001.
= 50.

Jitesh said: (Mar 3, 2019)  
Beta is output /input.

i.e (1.5-0.5)/(40-20).

Tkay said: (Aug 12, 2019)  
1*10^-3/20*10^-6 = 50.

From equation B=collector current gain /base current gain.

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