Semiconductor Devices (8hrs) · Basics of Electrical & Electronics Engineering

BJT Applications: CE configuration as amplifier

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BJT Applications: CE configuration as amplifier

1. What CE Configuration Is

CE (Common-Emitter) = emitter common to input and output: input at base-emitter, output at collector-emitter (emitter grounded for AC via capacitor). It is the most used BJT amplifier configuration because it gives high voltage, current, and power gain, and phase inversion (180°180°).

Alternatives: CB (common-base, voltage gain but α≈1α≈1, low input ZZ, high output) and CC (emitter follower, ββ current gain, voltage gain ≈1, buffer) — but CE is standard.

2. Circuit (bias + AC)

Vcc (12V) → Rc (collector resistor, e.g., 2kΩ) → Collector → Emitter → Re (emitter resistor with bypass Ce) → Ground
Base → voltage divider R1,R2 (e.g., 40k/10k) setting Vb≈2V ⇒ Ve=1.3V ⇒ Ie≈1.3mA, Q-point → coupling capacitors Cin, Cout block DC, pass AC.

Bias (Q-point): sets ICI_C and VCEV_{CE} in active region (e.g., VCE≈6VV_{CE}≈6V) so signal swing not clipped.

3. Working as Amplifier

  • Small AC input vbev_{be} (e.g., 10mV) modulates IBI_B:
ib=vbe/rπi_b = v_{be}/r_π

where

rπ=VT/IE≈25mV/IEr_π = V_T/I_E≈25mV/I_E

(e.g., 1mA ⇒ 25Ω? Actually rπ=β/gmr_π=β/gm, gm=IC/VTgm=I_C/V_T).

  • Collector current
ic=βib=gmvbei_c = β i_b = g_m v_{be}

with transconductance

gm=IC/VT≈40⋅IC(mA)g_m = I_C/V_T≈40·I_C(mA)

mS (e.g., 1mA ⇒40mS).

  • Voltage gain:
Av=−gm⋅(RC∣∣RL)=−RC/reA_v = -g_m·(R_C||R_L) = -R_C/r_e

where

re=VT/IE≈25Ωat1mAr_e = V_T/I_E≈25Ω at 1mA

, negative sign = 180° phase inversion (when base goes up, collector goes down due to VRcV_{Rc} drop).

  • For RC=2kΩR_C=2kΩ, IE=1mAI_E=1mA ⇒ Av≈−80A_v≈ -80 (40mS·2k).

4. Characteristics

Feature CE Amplifier
Voltage gain AvA_v High (-50 to -500)
Current gain High β≈100β≈100
Power gain High
Input ZZ Moderate ≈R1∣∣R2∣∣rπ≈R1||R2||r_π (~kΩ)
Output ZZ Moderate-high ≈RC≈R_C
Phase 180° inverted (input up ⇒ output down)

5. Worked Design

Need Av=-100 with Ic=2mA: gm=80mS ⇒ Rc=Av/gm=1.25kΩ ⇒ choose 1.2kΩ.
Input rπ=β/gm≈100/80mS=1.25kΩ, with bias divider 10k ⇒ Zin≈1.1kΩ.

6. Exam Points

  • Draw CE amplifier with voltage divider bias, RCR_C, Cin/outC_{in/out}, RER_E — 7 marks, very frequent (almost certain to ask CE circuit).
  • Explain IC=βIBI_C=β I_B, VCEV_{CE} bias, and Av=−gmRCA_v=-g_m R_C with 180° — 7-10 marks, very frequent (the most common amplifier question).
  • Mention high voltage/current/power gain — 3-5 marks, frequent.
  • Distinguish CE vs CC vs CB (gain, ZZ) — 5 marks, frequent.
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