Common-Emitter Bias Shift and Asymmetric Clipping
A student designs a single-stage common-emitter amplifier with VCC = 12 V, R1 = 47 kΩ from VCC to base, R2 = 10 kΩ from base to ground, RC = 3.3 kΩ, RE = 1 kΩ, and a silicon NPN transistor with hFE ≈ 200. The design target is VE ≈ 1.4 V, VC ≈ 7.4 V and VCE ≈ 6.0 V. After building it, the student measures VB = 1.50 V, VE = 0.80 V and VC = 9.40 V. The transistor is not hot. With a small sine input the amplifier works, but when the input is increased the output clips on its positive peaks much earlier than expected. The student has already checked VCC (12.0 V) and the transistor pinout. The resistors are carbon film 5%. The student suspects the transistor has low hFE and wants to replace it.
1. Calculate the intended bias values and compare them with the measurements. Does a low hFE explain the measured VB, VE and VC?
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Using the unloaded divider and neglecting base current for a first estimate:
VB ≈ 12 × 10 / (47 + 10) = 2.105 V. VE ≈ VB − 0.7 = 1.405 V. IE ≈ VE / RE = 1.405 / 1 kΩ = 1.405 mA. VC ≈ 12 − IC × RC ≈ 12 − 1.405 × 3.3 = 7.36 V. VCE ≈ 7.36 − 1.405 = 5.96 V.
So the design values are close to the stated targets. The measured VB of 1.50 V is far below 2.105 V, while VE and VC have moved in a consistent direction: lower emitter current and a higher collector voltage. A low hFE cannot explain this. The divider current is about 12 / 57 kΩ = 210 µA, much larger than the base current of roughly 7 µA, so the base voltage is set mainly by R1 and R2, not by hFE. Replacing the transistor would not move VB from 2.1 V to 1.5 V.
2. Use the measured VB to infer the actual value of R2, assuming R1 is 47 kΩ and VBE ≈ 0.7 V. Check consistency with the measured VE and VC.
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The divider relation can be rearranged:
VB = VCC × R2 / (R1 + R2) R2 = R1 × VB / (VCC − VB) R2 = 47 kΩ × 1.50 / (12.0 − 1.50) R2 = 47 kΩ × 1.50 / 10.50 = 6.71 kΩ.
The nearest standard 5% value is 6.8 kΩ, not 10 kΩ. This suggests R2 was fitted with the wrong resistor, possibly through a colour-code misread.
Check with R2 = 6.8 kΩ: VB = 12 × 6.8 / (47 + 6.8) = 1.52 V. VE ≈ 1.52 − 0.7 = 0.82 V. IE ≈ 0.82 / 1 kΩ = 0.82 mA. VC ≈ 12 − 0.82 × 3.3 = 9.29 V.
These are close to the measured 0.80 V and 9.40 V within resistor tolerance and VBE spread. The measurements are consistent with a 6.8 kΩ R2.
3. Explain why the amplifier clips first on positive output peaks and propose the minimum correct fix. Include the approximate unclipped positive swing before and after the fix.
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The quiescent collector voltage is too high. Before the fix, VC ≈ 9.4 V, so the output can rise only about VCC − VC = 12 − 9.4 = 2.6 V before the transistor cuts off. Downward, it can swing until saturation, roughly VC − (VE + 0.2 V) = 9.4 − 1.0 = 8.4 V. Because a common-emitter stage inverts, the output positive peak corresponds to the transistor turning off, so the positive peak clips first at only about 2.6 V. The negative peak has much more room.
After replacing R2 with 10 kΩ, the bias returns to approximately VC = 7.4 V. The positive swing to cutoff becomes about 12 − 7.4 = 4.6 V, and the negative swing to saturation is about 7.4 − 1.6 = 5.8 V. The clipping is much more symmetric and the maximum useful output is larger. The minimum correct fix is to restore R2 to 10 kΩ, not to change the transistor. The input drive should also be kept below the level that demands more than this swing.