Question

find vc (assume a stiff bias circuit).
answer:

Explanation:

Step1: Calculate the base voltage \( V_B \)

Using the voltage - divider formula \( V_B=\frac{R_2}{R_1 + R_2}\times V_{CC} \), where \( R_1 = 10\space k\Omega=10\times10^{3}\space\Omega \), \( R_2 = 1.8\space k\Omega = 1.8\times 10^{3}\space\Omega \), and \( V_{CC}=40\space V \)
\( V_B=\frac{1.8\times 10^{3}}{10\times 10^{3}+1.8\times 10^{3}}\times40=\frac{1.8}{11.8}\times40\approx6.1\space V \)

Step2: Calculate the emitter voltage \( V_E \)

Assume \( V_{BE}=0.7\space V \) (for a silicon transistor). Then \( V_E=V_B - V_{BE} \)
\( V_E=6.1 - 0.7=5.4\space V \)

Step3: Calculate the emitter current \( I_E \)

Using \( I_E=\frac{V_E}{R_E} \), where \( R_E = 150\space\Omega \)
\( I_E=\frac{5.4}{150}=0.036\space A = 36\space mA \)

Step4: Assume \( I_C\approx I_E \) (since \( I_B\) is very small in a stiff - bias circuit)

\( I_C\approx36\space mA \)

Step5: Calculate the voltage drop across \( R_C \)

Using \( V_{R_C}=I_C\times R_C \), where \( R_C = 820\space\Omega \)
\( V_{R_C}=36\times10^{- 3}\times820 = 29.52\space V \)

Step6: Calculate \( V_C \)

Using \( V_C=V_{CC}-V_{R_C} \)
\( V_C=40 - 29.52 = 10.48\space V \)

Answer:

\( 10.48\space V \)