Question

find ie (assume a stiff bias circuit). express answer in ma
40v
10 kω
820 ω
c
c
1.8 kω
250 ω
c
choose...
ma
a
answer:
choose...

Explanation:

Step1: Calculate the voltage at the base (\(V_B\))

For a stiff - bias circuit (voltage - divider bias), the voltage at the base \(V_B\) is given by the voltage - divider formula \(V_B=\frac{R_2}{R_1 + R_2}\times V_{CC}\). Here, \(R_1 = 10\ \text{k}\Omega=10\times10^{3}\ \Omega\), \(R_2 = 1.8\ \text{k}\Omega = 1.8\times 10^{3}\ \Omega\), and \(V_{CC}=40\ \text{V}\).

$$V_B=\frac{1.8\times 10^{3}}{10\times 10^{3}+1.8\times 10^{3}}\times40=\frac{1.8}{11.8}\times40\approx6.1\ \text{V}$$

Step2: Calculate the voltage at the emitter (\(V_E\))

Assume \(V_{BE}=0.7\ \text{V}\) (for a silicon BJT). Using the relation \(V_E=V_B - V_{BE}\), we get \(V_E=6.1 - 0.7=5.4\ \text{V}\)

Step3: Calculate the emitter current (\(I_E\))

Using Ohm's law \(I_E=\frac{V_E}{R_E}\), where \(R_E = 550\ \Omega\) (assuming a typo in the problem statement, if \(R_E = 550\ \Omega\)).

$$I_E=\frac{5.4}{550}\approx 10\ \text{mA}$$

Answer:

\(10\ \text{mA}\)