Answer to Question #136537 in Electric Circuits for Ankita

Question

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

Answer 1

Initialy we have base bias circuit with given quantities as

$R_B=15k\Omega,R_c=100\Omega,V_{cc}=9V$ ,$\beta$ at 25$\degree C$ = 110

Since the given circuit is a base bias circuit then

$V_{cc}=I_B\times R_B$

$9=I_B\times 15$

$I_B=0.6mA$

Case 1: At 0$\degree C$

$\beta$ reduce by 50%

so It will became $\frac{\beta}{2}$

Using formula of current gain we have,

$\frac{\beta}{2}=\frac{I_c}{I_B}$

$I_C=I_B\times \frac{110}{2}$

=$0.6\times 55$

=33mA=0.033A

Now using the equation

$V_{CE}=V_{CC}-I_CR_C$

=$9-0.033\times 100$

=9-3.3

=5.7V

Case 2: At $25\degree C$

$\beta=\frac{I_c}{I_B}$

$I_c=110\times0.6$

=66mA=0.066A

Again using the equation,

$V_{CE}=V_{CC}-I_CR_C$

= $9-.066\times100$

=9-6.6

=2.7V

Case 3: At $70\degree C$

$\beta$ increase by 70%

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

i.e.$\frac{7}{4}\beta$

using

$\frac{7}{4}\beta=\frac{I_c}{I_B}$

$I_c=I_B\times \frac{7}{4}\times 110$

$=0.6\times 7\times27.5$

=115.5ma=0.1155A

Again using,

$V_{CE}=V_{CC}-I_CR_C$

=$9-0.1155\times100$

=9-11.55

$=-2.55V$