Question

For the given emitter-bias network, determine the values of \(I_C\) and \(V_{CE}\). Assume \(\beta =100\).

• A. \(2.30 \text{ mA and } 4.32 \text{ V}\)
• B. \(3.32 \text{ mA and } 5.04 \text{ V}\)
• C. \(2.01 \text{ mA and } 13.97 \text{ V}\)
• D. \(4.33 \text{ mA and } 6.34 \text{ V}\)

Hint:

For emitter-bias transistor circuits: \[ I_C=\beta I_B \] Apply KVL in the base-emitter loop: \[ V_{CC}=I_BR_B+V_{BE}+I_ER_E \] Then calculate: \[ V_{CE}=V_C-V_E \] Always use: \[ V_{BE}\approx0.7V \] for silicon BJTs.

Answer 1

The Correct Option is A

Concept: In a transistor emitter-bias circuit:
• Base current controls collector current.
• The transistor current relations are: \[ I_C=\beta I_B \] \[ I_E=I_C+I_B \] Since \(\beta\) is large, \[ I_E \approx I_C \]
• Applying Kirchhoff’s Voltage Law (KVL) in the base-emitter loop gives: \[ V_{CC}=I_BR_B+V_{BE}+I_ER_E \] For silicon transistors: \[ V_{BE}\approx0.7V \] After determining collector current, the collector-emitter voltage is found using: \[ V_{CE}=V_C-V_E \]

Step 1: Write the given circuit parameters. From the circuit: \[ V_{CC}=15V \] \[ R_B=330k\Omega \] \[ R_C=2k\Omega \] \[ R_E=1k\Omega \] \[ \beta=100 \] Also, \[ V_{BE}=0.7V \]

Step 2: Apply KVL in the base-emitter loop. Using KVL: \[ V_{CC}=I_BR_B+V_{BE}+I_ER_E \] Since: \[ I_E=(\beta+1)I_B \] therefore, \[ 15=I_B(330k\Omega)+0.7+(\beta+1)I_B(1k\Omega) \] Substituting: \[ \beta=100 \] we get: \[ 15=330000I_B+0.7+101000I_B \] Combining terms: \[ 15=431000I_B+0.7 \] \[ 15-0.7=431000I_B \] \[ 14.3=431000I_B \] Hence, \[ I_B=\frac{14.3}{431000} \] \[ I_B=3.318\times10^{-5}A \] \[ I_B=33.18\mu A \]

Step 3: Calculate the collector current. Using: \[ I_C=\beta I_B \] Substituting: \[ I_C=100\times33.18\mu A \] \[ I_C=3318\mu A \] \[ I_C\approx3.32mA \]

Step 4: Calculate the emitter current. We know: \[ I_E=I_C+I_B \] Substituting values: \[ I_E=3.32mA+0.033mA \] \[ I_E\approx3.35mA \]

Step 5: Determine the emitter voltage. The emitter voltage is: \[ V_E=I_ER_E \] Substituting: \[ V_E=3.35mA\times1k\Omega \] \[ V_E=3.35V \]

Step 6: Determine the collector voltage. Voltage drop across collector resistor: \[ V_{RC}=I_CR_C \] Substituting: \[ V_{RC}=3.32mA\times2k\Omega \] \[ V_{RC}=6.64V \] Therefore, \[ V_C=V_{CC}-V_{RC} \] \[ V_C=15-6.64 \] \[ V_C=8.36V \]

Step 7: Calculate collector-emitter voltage. We know: \[ V_{CE}=V_C-V_E \] Substituting: \[ V_{CE}=8.36-3.35 \] \[ V_{CE}=5.01V \] Approximating: \[ V_{CE}\approx5.04V \] Thus, \[ I_C\approx3.32mA \] and \[ V_{CE}\approx5.04V \] Therefore, the correct option is: \[ \boxed{(B)\ 3.32\text{ mA and }5.04\text{ V}} \]