Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 10)
10.
A certain transistor has IC = 15 mA and IB = 167 
A; 
DC is:
A; DC is:Discussion:
6 comments Page 1 of 1.
Abhishek mishra said:
10 years ago
Beta dc is DC current amplification factor for common emitter mode beta dc = ic/ib.
= 15*10^-3/167*10^-6 = 89.820 approximately 90.
= 15*10^-3/167*10^-6 = 89.820 approximately 90.
Abhishek Mishra said:
10 years ago
Beta dc is DC current amplification factor for common emitter mode.
Beta dc = Ic/Ib.
= 15000/167 = 89.82.
Beta dc = Ic/Ib.
= 15000/167 = 89.82.
Mohammed Arif said:
1 decade ago
We know that.
Ic = Beta*Ib.
So we can write,
Beta = Ic/Ib.
Now put the value of Ic &IB in this formula,
b = 15000/167 = 89.820 or 90(approx equal).
Where the value of collector current is mA so we changed mA in uA.
Ic = Beta*Ib.
So we can write,
Beta = Ic/Ib.
Now put the value of Ic &IB in this formula,
b = 15000/167 = 89.820 or 90(approx equal).
Where the value of collector current is mA so we changed mA in uA.
Sri said:
1 decade ago
Beta dc is current amplification factor ic/ib = 15*10^-3/167*10^-6 = 89.820
i.e 90.
i.e 90.
Amala said:
1 decade ago
Pleas give the correct explanation.
Komal said:
1 decade ago
Since Beta = ic/ib
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