Question #136537
The base bias circuit is subjected to a temperature variation from 0°C to 70°C. The ß decreases by 50% at 0°C and increases by 75% at 70°C from it's normal value of 110 at 25°C. What are the changes in IC and VCE over the temperature range of 0°C to 70°C Given: RB = 15kΩ, Rc = 100Ω, Vcc = 9V
1
Expert's answer
2020-10-05T15:17:55-0400

Initialy we have base bias circuit with given quantities as


RB=15kΩ,Rc=100Ω,Vcc=9VR_B=15k\Omega,R_c=100\Omega,V_{cc}=9V ,β\beta at 25°C\degree C = 110


Since the given circuit is a base bias circuit then

Vcc=IB×RBV_{cc}=I_B\times R_B


9=IB×159=I_B\times 15

IB=0.6mAI_B=0.6mA


Case 1: At 0°C\degree C

β\beta reduce by 50%

so It will became β2\frac{\beta}{2}


Using formula of current gain we have,

β2=IcIB\frac{\beta}{2}=\frac{I_c}{I_B}


IC=IB×1102I_C=I_B\times \frac{110}{2}

=0.6×550.6\times 55

=33mA=0.033A

Now using the equation

VCE=VCCICRCV_{CE}=V_{CC}-I_CR_C

=90.033×1009-0.033\times 100

=9-3.3

=5.7V

Case 2: At 25°C25\degree C

β=IcIB\beta=\frac{I_c}{I_B}


Ic=110×0.6I_c=110\times0.6

=66mA=0.066A


Again using the equation,

VCE=VCCICRCV_{CE}=V_{CC}-I_CR_C

= 9.066×1009-.066\times100

=9-6.6

=2.7V

Case 3: At 70°C70\degree C


β\beta increase by 70%

i.e. β\beta will became β+75100β\beta+\frac{75}{100}\beta

i.e.74β\frac{7}{4}\beta

using

74β=IcIB\frac{7}{4}\beta=\frac{I_c}{I_B}


Ic=IB×74×110I_c=I_B\times \frac{7}{4}\times 110

=0.6×7×27.5=0.6\times 7\times27.5


=115.5ma=0.1155A

Again using,

VCE=VCCICRCV_{CE}=V_{CC}-I_CR_C

=90.1155×1009-0.1155\times100

=9-11.55

=2.55V=-2.55V


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