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.
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)
Sravya said:
1 decade ago
Vbb = Vcc*R2/R1+R2.
Use this formula and substitute the values. Finally you got answer.
Use this formula and substitute the values. Finally you got answer.
RAVAL ROSHNI said:
1 decade ago
BY Apply voltage divider rule.
Vcc*R2/R1+R2.
You can get true answer 4.35 volt.
Vcc*R2/R1+R2.
You can get true answer 4.35 volt.
ANNA said:
1 decade ago
Since the voltage divides there we apply voltage divider rule to obtain the Vbb.
Mallika said:
8 years ago
Base voltage is across r2 resistror, So we apply vB = Vcc(r2)/(r1+r2);.
(1)
Niaz khan said:
1 decade ago
Apply V.D formula
vbb=vcc*(opposite resistance/sumof two resistance)
vbb=vcc*(opposite resistance/sumof two resistance)
Suleman said:
1 decade ago
Why not use R1 Vcc = R1/R1+R2?
Why use the R2 Vcc = R2/R1+R2?
Why use the R2 Vcc = R2/R1+R2?
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)
Atish said:
1 decade ago
Vbb={R2/R1+R2}* Vcc
=(1500/6200)* 18
= 4.35 V
=(1500/6200)* 18
= 4.35 V
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)
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