Question

1) Find the Q-point value
2) Find the minimum power rating of transistor. Given : ß = 50, Vcc = -12V, Rc = 1.8kΩ, R1= 33kΩ, R2 = 5.6kΩ, RE = 560Ω.

Answer 1

i)To find Q-value we need to calculate,

$V_{BB}=V_{cc}\frac{R_2}{R_1+R_2}$

=-12$\times\frac{5.6}{5.6+33}$

=-1.74V

Resistance

$R_B=R_1||R_2$

=$\frac{R_1R_2}{R_1+R_2}$

=$\frac{5.6\times 33}{5.6+33}$

=4.78K$\Omega$

$I_{E}=\frac{V_{BB}-V_{CC}}{\frac{R_B}{\beta+1}+R_E}$

$=\frac{-1.74-(-12)}{\frac{4.78}{51}+0.56}$

=$\frac{10.26}{0.653}$

=15.71mA

$V_{CE}$ =$V_{CC}-I_E(R_C+R_E)$

=$-12-15.71(1.8+0.56)$

=-12-37.0756

=-49.07V

Hence Q point is (-49.07V,15.71mA)

ii) Power of a transistor is given by

P$=V_{CE}I_{CE}+V_{BE}I_{B}$

=V$_{CE}I_{CE}$

=$-49.07\times 15.71$

=-0.77 Watt