Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 1 (Q.No. 20)
20.
The transistor of following figure in Si diode with a base current of 40 μA and ICBO = 0, if VBB = 6V, RE = 2 kΩ and β = 90, IBQ = 20 μA then RB =
Answer: Option
Explanation:
.
Discussion:
13 comments Page 2 of 2.
Arnab Bhattacharya said:
10 years ago
Ic = βIB + (β+1) ICBO => Ic = βIb [Since, ICBO = 0; given].
SO, Ic = 90*20 uA = 1.8 mA.
Voltage drop across RE = IE*RE = (IC+IB)*RE = IC*RE (approx) = 1.8*2 = 3.6 V.
Now, VBB = RB*IB + VBE + voltage across RE.
6 = RB*20 uA + 0.7 + 3.6 = RB*20 uA + 4.3.
RB*20 uA = 1.7 => RB = 1.7/20 uA = 85 KOhm.
So, my result is none of the above options. Please mention if I am missing something. Thanks in advance.
SO, Ic = 90*20 uA = 1.8 mA.
Voltage drop across RE = IE*RE = (IC+IB)*RE = IC*RE (approx) = 1.8*2 = 3.6 V.
Now, VBB = RB*IB + VBE + voltage across RE.
6 = RB*20 uA + 0.7 + 3.6 = RB*20 uA + 4.3.
RB*20 uA = 1.7 => RB = 1.7/20 uA = 85 KOhm.
So, my result is none of the above options. Please mention if I am missing something. Thanks in advance.
Balu said:
1 decade ago
Why we not consider loop for i/p side?
Atul said:
1 decade ago
What about Re?
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