Sunday 14 April 2013
TRANSISTOR PROBLEM 5
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a)if V1 = 1V find V0
b)if ICBO = 10nA at 250C.find maximum temperature
at which the transistor is in cut-off.
SOLUTION
a)Given V1 = 1V
Solving above equation we get VTH = -0.695V
From the circuit VTH = VBE
As VBE ≤ 0.6V
the transistor is in cutoff.
So,there is no IC ie.,IC = 0
Therefore V0 = 12V
b)
IB = I2 – I1
As the transistor is in cut-off IB is
approximately equal to -ICBO
-ICBO = I2 – I1
ICBO = I1 – I2
I1 = 0.6 – (-12 ) / 100K = 0.126mA
I2 = 1 – 0.6 / 15k = 0.4 /15k =0.026mA
ICBO = 0.126mA – 0.026mA
=0.1mA
ICBO (T) = ICBO (250C ) * 2 (
T – 25 )/10
0.1mA = 10nA * 2 ( T – 25 )/10
Applying logarithm
( T – 25 ) /10 = 13.28
T –25 =132.8
T = 157.80C
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