Question

find vrc (assume a stiff bias circuit).
40 v
10 kω
820 ω
c_c
c_a
q1
1.8 kω
150 ω
c_e
answer:

Explanation:

Step1: Calculate the base voltage \(V_B\)

For a stiff - bias circuit (voltage - divider bias), the base voltage \(V_B\) is given by the voltage - divider formula. The two resistors in the voltage divider are \(R_1 = 10\space k\Omega\) and \(R_2=1.8\space k\Omega\), and the supply voltage \(V_{CC}=40\space V\).

$$V_B=\frac{R_2}{R_1 + R_2}\times V_{CC}$$

$$V_B=\frac{1.8\times10^{3}}{(10 + 1.8)\times10^{3}}\times40\space V$$

$$V_B=\frac{1.8}{11.8}\times40\space V\approx6.10\space V$$

Step2: Calculate the emitter voltage \(V_E\)

Assume \(V_{BE}=0.7\space V\) (for a silicon BJT). Using the relationship \(V_B=V_E + V_{BE}\), we can find \(V_E\)

$$V_E=V_B - V_{BE}$$

$$V_E=6.10\space V- 0.7\space V = 5.4\space V$$

Step3: Calculate the emitter current \(I_E\)

The emitter current \(I_E\) is given by \(I_E=\frac{V_E}{R_E}\), where \(R_E = 150\space\Omega\)

$$I_E=\frac{5.4\space V}{150\space\Omega}=0.036\space A = 36\space mA$$

Step4: Assume \(I_C\approx I_E\) (since \(\beta\gg1\) for a BJT in active region)

The collector - emitter current relationship for a BJT in active region is \(I_C=\alpha I_E\), and when \(\beta\gg1\), \(\alpha\approx1\), so \(I_C\approx I_E = 36\space mA\)

Step5: Calculate the voltage across the collector resistor \(V_{RC}\)

The voltage across the collector resistor \(R_C=820\space\Omega\) is given by \(V_{RC}=I_C\times R_C\)

$$V_{RC}=36\times10^{- 3}\space A\times820\space\Omega$$

$$V_{RC}=29.52\space V$$

Answer:

\(29.52\space V\)