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Design CE amplifier (BC547B, no bypass)

Short answer

a) Yes, Q-point at VCE=4.5V, IC=1.5mA is mid-load-line for maximum symmetrical swing. b) Zin≈22.3kΩ, Av≈-3.9. c) vce/vi = -4.9, so max input peak ≈ 0.88V (limited by saturation).

Design voltage divider CE amplifier (without emitter by pass capacitor). Given: Transistor BC 547B having β = 295, Ic = 1.5 mA and Vcc = +9V. a) Is this the best Q point? Why? b) Calculate its input impedance and voltage gain. c) What is the maximum peak voltage of the signal that can be applied to the input of this amplifier to ensure the transistor is always in active region?

Design (VCC = 9 V, IC = 1.5 mA, β = 295, RE not bypassed)

Guidelines: VE ≈ VCC/10, VCE ≈ VCC/2, divider current I2 = 10 IB, VB = VE + 0.7 V.

IB = IC/β = 1.5 mA/295 = 5.08 µA          IE = 1.505 mA
VE = 0.9 V      RE = 0.9/1.505 mA = 598 Ω  (use 600 Ω)
VCE = 4.5 V     RC = (9 - 0.9 - 4.5)/1.5 mA = 2.4 kΩ
VB = 1.6 V      I2 = 10 IB = 50.8 µA
R2 = 1.6/50.8 µA = 31.5 kΩ                 R1 = (9 - 1.6)/55.9 µA = 132 kΩ

Choose R1 = 130 kΩ, R2 = 31.5 kΩ (30 kΩ), RE = 600 Ω, RC = 2.4 kΩ.

a) Is this the best Q-point?

Yes, for the unloaded amplifier. The dc load line equation is VCC = IC(RC + RE) + VCE. The Q-point is at VCE = 4.5 V and IC(RC + RE) = 1.5 mA × 3 kΩ = 4.5 V, so it lies at the middle of the load line (VCEQ = VCC/2 and ICQ = half of the saturation current of 3 mA). Since RE is not bypassed and there is no load, the ac load line is the same as the dc load line, so the Q-point is also at the centre of the ac load line. The output can swing about equally towards cut-off and towards saturation, giving the maximum symmetrical undistorted swing. (The actual swing is slightly less than 4.5 V because VCE(sat) ≈ 0.2 V.)

b) Input impedance and voltage gain

re = VT/IE = 25 mV/1.505 mA = 16.6 Ω
Rib = (β+1)(re + RE) = 296 × (16.6 + 598) = 181.9 kΩ
Zin = R1 || R2 || Rib = 132 k || 31.5 k || 181.9 k = 22.3 kΩ

Av = -RC/(re + RE) = -2400/(16.6 + 598) = -3.9

Zin ≈ 22.3 kΩ, Av ≈ -3.9.

c) Maximum input peak for active operation

Because RE is unbypassed, the signal current ic flows through both RC and RE, so the collector-emitter voltage changes by ic(RC + RE), not just ic RC:

vce/vi = -(RC + RE)/(re + RE) = -3000/614.6 ≈ -4.9
  • Saturation side: VCE can fall from 4.5 V to VCE(sat) ≈ 0.2 V, a swing of 4.3 V, so vi(max) = 4.3/4.9 ≈ 0.88 V.
  • Cut-off side: ic can fall by at most ICQ = 1.5 mA, so vi(max) = 1.5 mA × (re + RE) ≈ 1.5 mA × 614.6 Ω ≈ 0.92 V.

The smaller value (saturation side) sets the limit:

vi(max) ≈ 4.3/4.9 ≈ 0.88 V (peak)

The input peak must be below about 0.88 V (assuming a signal source with negligible resistance). The large unbypassed RE keeps the stage linear, which is why such a large input is possible.

✓ Answer checked by Claude Opus 5.5 (high effort)

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