CE amplifier saturates on one board: transistor batch or base divider?
Students build a single-stage common-emitter amplifier on breadboard: Vcc = 12 V, R1 = 47 kΩ from Vcc to base, R2 = 10 kΩ from base to ground, Rc = 2.2 kΩ, Re = 470 Ω bypassed by 100 µF, input 10 mV peak-to-peak at 1 kHz through a 10 µF capacitor, transistor BC547 with β ≈ 150. The lab sheet expects Vb ≈ 2.1 V, Ve ≈ 1.4 V, Vc ≈ 5.4 V and a clean inverted output of about 1 V peak-to-peak. One board measures Vb = 3.2 V, Ve = 2.5 V, Vc = 0.2 V. The output is a badly clipped waveform of only 150 mV peak-to-peak. The student has already replaced the BC547 twice with the same result and now blames a bad batch of transistors. A second identically built board works exactly as the lab sheet predicts. The supply is 12.0 V and the 470 Ω emitter resistor reads 468 Ω in circuit.
1. Calculate the expected DC values from the resistor network and compare them with the measured set. What operating condition do the measurements show?
Read the worked answer
Ignoring base current: Vb = 12 × 10/(47 + 10) = 2.11 V. Then Ve = 2.11 − 0.7 = 1.41 V, Ie = 1.41/470 ≈ 3.0 mA, and Vc = 12 − (3.0 mA × 2.2 kΩ) = 12 − 6.6 = 5.4 V. This matches the lab sheet.
Measured: Vb = 3.2 V (above the 2.11 V the divider alone can produce), Ve = 2.5 V, so Vbe ≈ 0.7 V and the emitter current is 2.5/468 ≈ 5.3 mA. But Vc = 0.2 V instead of 5.4 V, i.e. the collector is only 0.2 V above ground while 5.3 mA flows - the transistor is hard in saturation. A saturated stage cannot swing symmetrically, which is exactly why the output is clipped and small.
2. Identify the most probable faulty component and rule out the alternatives the student is considering.
Read the worked answer
R2 is open (or has risen to a very high value). With R2 open, the only path into the base node is R1: base current = (12 − 3.2)/47 kΩ ≈ 187 µA. That is far more than the 5.3 mA/150 ≈ 35 µA the stage needs, so the transistor is driven into saturation, Vc falls to 0.2 V, and the base node floats up to roughly Ve + 0.7 ≈ 3.2 V - precisely the measurements.
Ruling out alternatives: a faulty transistor would not give three identical results with fresh devices and a normal 0.7 V Vbe; a collector-emitter short would make Vc ≈ Ve = 2.5 V, not 0.2 V; an open Rc would remove collector current and pull Ve to 0 V; a shorted Re would also give Ve ≈ 0 V; a leaking coupling capacitor would shift bias but not produce a clean, repeatable saturation with a raised Vb.
3. Describe a confirmation test and state the effect of this fault on the AC performance.
Read the worked answer
Confirm by isolating R2: lift one leg or measure it out of circuit; a cracked carbon-film resistor often reads open only when voltage is applied, so also measure the voltage across R1 (should be ≈8.8 V for 187 µA) and across R2 (should be ≈3.2 V but with almost no current - an inconsistent V/I pair confirms an open path). Refit a 10 kΩ resistor and recheck Vb ≈ 2.1 V, Ve ≈ 1.4 V, Vc ≈ 5.4 V.
AC effect: the quiescent point sits at the saturation boundary, so on one half-cycle the collector cannot move below ≈0.2 V and the waveform is flattened; the stage also has almost no available swing, hence the 150 mV peak-to-peak instead of 1 V. After repair, gain ≈ Rc/re′ (re′ ≈ 8.7 Ω at 3 mA) is high, so a 10 mV input may still clip - verify with a smaller input before declaring the amplifier linear.