BJT biasing and amplifiers (CE amplifier analysis)
Analog electronics · Physics
DC circuit
Interactive 3DElectrons drift around the loop — raise the voltage and watch the current obey Ohm's law.
Controls
Drag the sliders or type a value — the simulation updates live.
Study notes
**Worked Example: Common‑Emitter Amplifier** Given: - Vcc = 12 V - R1 = 100 kΩ, R2 = 10 kΩ (bias network) - RE = 1 kΩ, RC = 2 kΩ - Transistor β = 100 - Assume VBE ≈ 0.7 V **Step 1 – Find Base Voltage (Vb)** Vb = Vcc × (R2 / (R1 + R2)) Vb = 12 × (10k / (100k + 10k)) = 12 × (10/110) ≈ 1.09 V **Step 2 – Find Emitter Voltage (Ve)** Ve = Vb – VBE = 1.09 – 0.7 ≈ 0.39 V **Step 3 – Find Emitter Current (Ie)** Ie ≈ Ve / RE = 0.39 / 1k ≈ 0.39 mA **Step 4 – Find Collector Current (Ic)** Ic ≈ Ie (since base current is small) ≈ 0.39 mA **Step 5 – Find Collector Voltage (Vc)** Vc = Vcc – (Ic × RC) = 12 – (0.39 mA × 2 kΩ) ≈ 12 – 0.78 ≈ 11.22 V **Step 6 – AC Voltage Gain (Av)** First find transconductance gm = Ic / Vt, where Vt ≈ 25 mV. gm = 0.39 mA / 25 mV ≈ 0.0156 S Av ≈ -RC / (RE + 1/gm) = -2k / (1k + 1/0.0156) ≈ -2k / (1k + 64) ≈ -2k / 1.064k ≈ -1.88 So the amplifier inverts the signal and amplifies it by about 1.9 times.