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.
Dhayamurthy said:
1 decade ago
Ans
Vcc * R2 / R1 + R2
18 * (1500 / 6200) = 4.35V
Vcc * R2 / R1 + R2
18 * (1500 / 6200) = 4.35V
Atish said:
1 decade ago
Vbb={R2/R1+R2}* Vcc
=(1500/6200)* 18
= 4.35 V
=(1500/6200)* 18
= 4.35 V
Sivababu said:
1 decade ago
Vcc*(R2/R2+R1)
Pratul gupta said:
1 decade ago
Vb=Vcc*(r2/(r2+r1))
ANNA said:
1 decade ago
Since the voltage divides there we apply voltage divider rule to obtain the Vbb.
Niaz khan said:
1 decade ago
Apply V.D formula
vbb=vcc*(opposite resistance/sumof two resistance)
vbb=vcc*(opposite resistance/sumof two resistance)
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.
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?
Dydie said:
1 decade ago
VBB = VCC(R2/R1+R2).
(1)
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