Electronics - Bipolar Junction Transistors (BJT) - Discussion
Discussion Forum : Bipolar Junction Transistors (BJT) - General Questions (Q.No. 10)
10.
If VCC = +18 V, voltage-divider resistor R1 is 4.7 k
, and R2 is 1500, what is the base bias voltage?
, and R2 is 1500, what is the base bias voltage?Discussion:
19 comments Page 1 of 2.
Mazen Mohammed said:
3 years ago
Vbb = Vcc * (R2/R1+R2) = 4.35.
(2)
Shubham k said:
6 years ago
Vcc *R2\R1+R2.
=4.35.
=4.35.
(1)
Aneesh said:
6 years ago
Vb = Vcc x (R2/R2+R1).
Vb = 18*(1.5/1.5+4.7),
Vb = 4.37V.
Vb = 18*(1.5/1.5+4.7),
Vb = 4.37V.
(3)
Vaibhav said:
6 years ago
Vb = Vcc x (R2/R2+R1).
= 18 *(1.5e3/1.5e3+4.7e3),
= 4.35.
= 18 *(1.5e3/1.5e3+4.7e3),
= 4.35.
(2)
Parul said:
8 years ago
Then what is across r1 resistor?
(1)
Mallika said:
8 years ago
Base voltage is across r2 resistror, So we apply vB = Vcc(r2)/(r1+r2);.
(1)
Sravani said:
8 years ago
Apply the voltage divider rule.
(1)
Gyandeep said:
9 years ago
We will consider R2 because input is applied to base-emitter terminal. So we calculate voltage across R2. Now apply KVL.
Vcc*R2/R2+R1.
Vcc*R2/R2+R1.
(1)
RONI said:
10 years ago
Why not Vcc = (R1/R1+R2)?
Dydie said:
1 decade ago
VBB = VCC(R2/R1+R2).
(1)
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