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Emitter feedback bias, IC and VCE

Short answer

Use base loop KVL: VCC − IB RB − VBE − (β+1)IB RE = 0, VBE=0.7 V. IB = (12−0.7)/(430k+51×1k) ≈ 23.5 μA. Then IC = βIB = 1.175 mA, VCE = VCC − IC RC − IE RE = 12 − 2.349 − 1.198 = 8.45 V. So IC ≈ 1.17 mA, VCE ≈ 8.45 V.

Draw emitter feedback bias circuit of BJT by labeling all the circuit components. Find IC and VCE in the circuit if VCC = +12 V, RB = 430 kΩ, RC = 2 kΩ, RE = 1 kΩ and β = 50.

Emitter feedback (emitter-stabilised) bias circuit

            +VCC = 12 V
             |          |
            [RB]       [RC] 2 kΩ
           430 kΩ       |
             |          +----o Vo (collector)
             |        |/ C
             +--------|   NPN (β = 50)
                  B   |\ E
                        v
                        |
                       [RE] 1 kΩ
                        |
                       GND

This is a fixed-bias circuit with an emitter resistor RE added. RE provides negative feedback, which makes IC more stable against changes in β and temperature.

Calculation (assume VBE = 0.7 V for silicon)

Base loop (KVL): VCC − IB RB − VBE − IE RE = 0, with IE = (β + 1) IB:

IB = (VCC − VBE)/(RB + (β + 1) RE) = (12 − 0.7)/(430 k + 51 × 1 k) = 11.3/481 k = 23.49 µA

Collector current: IC = β IB = 50 × 23.49 µA = 1.175 mA

Emitter current: IE = (β + 1) IB = 51 × 23.49 µA = 1.198 mA

Collector-emitter voltage (collector loop): VCE = VCC − IC RC − IE RE = 12 − (1.175 mA × 2 k) − (1.198 mA × 1 k) = 12 − 2.349 − 1.198

VCE = 8.45 V

Since VCE is well above about 0.2 V, the transistor is in the active region.

Answer: IC ≈ 1.17 mA, VCE ≈ 8.45 V

✓ Answer checked by Claude Opus 5.5 (high effort)

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