Question

Consider the voltage divider circuit shown in the figure below.


The value of cut in voltage for the amplifier circuit $V_{BE\left(on\right)}=0.7V$, and the value of $\beta =140$, then the value of power dissipated by the transistor is equal to

Answer 1

$V_{Th}=\frac{R_2\times V_{CC}}{R_1+R_2}=\frac{3.9}{39+3.9}\times 22=2V$

$R_{Th}=R_1||R_2=3.55k\Omega$

$\therefore I_B=\frac{V_{Th}-0.7}{R_{Th}+(\beta +1)R_E}=\frac{2-0.7}{3.55+141\times 1.5}$
$=\frac{1.3}{3.55+211.5}=6.05\mu A$
$I_C=\beta I_B=140\times 6.05\times {10}^{-6}=0.847mA$ and $I_E=141\times 6.05\times {10}^{-6}=0.853mA$
$\therefore V_{CE}=22-10\times 0.847-1.5\times 0.853$
$V_{CE}=12.25V$
$\therefore P_{dis}=V_{CE}{I}_C=10.376mW$