A common-emitter amplifier has R_C = 2 kΩ and an emitter resistor R_E = 500 Ω with a bypass capacitor. When the bypass capacitor is removed, the voltage gain magnitude drops dramatically. What is the primary reason?
AThe transistor's transconductance g_m decreases when the bypass capacitor is removed
BR_E appears in series with r_e in the gain denominator, reducing gain to R_C/(r_e + R_E)
CThe coupling capacitors can no longer pass the AC signal effectively
DThe DC operating point shifts when the bypass capacitor is removed
With the bypass capacitor, R_E is shorted at signal frequencies and gain is −g_m·R_C = −R_C/r_e. Without it, R_E is in series with r_e, giving gain −R_C/(r_e + R_E) — much smaller when R_E >> r_e. The Q-point, g_m, and coupling capacitors are unaffected by the bypass capacitor.
Question 2 Multiple Choice
In a common-emitter amplifier, the input voltage rises (becomes more positive). What happens to the collector voltage, and why?
AIt rises, because the transistor acts as a voltage follower
BIt falls, because increased base-emitter voltage increases collector current, increasing the voltage drop across R_C
CIt remains constant, because the voltage divider bias stabilizes it
DIt rises, because the emitter bypass capacitor inverts the signal back
A rising input increases v_be, which increases the controlled current g_m·v_be flowing through R_C. Greater current means greater voltage drop across R_C, so the collector voltage (V_CC − I_C·R_C) falls. This 180-degree phase inversion is fundamental to the CE configuration — it is why the gain expression A_v = −g_m·R_C carries a negative sign.
Question 3 True / False
The emitter bypass capacitor in a common-emitter amplifier is optional and primarily affects the DC bias point.
TTrue
FFalse
Answer: False
The bypass capacitor has no effect on the DC bias — it is an open circuit at DC. Its critical role is at signal frequencies, where it shorts R_E, preventing the emitter resistance from appearing in the gain expression. Without it, gain drops from −g_m·R_C to −R_C/(r_e + R_E). Far from optional, it is responsible for most of the amplifier's voltage gain.
Question 4 True / False
Removing the emitter bypass capacitor from a common-emitter amplifier reduces voltage gain but improves linearity, input impedance, and bandwidth.
TTrue
FFalse
Answer: True
When R_E is un-bypassed, it introduces series-series feedback. This reduces gain but stabilizes the operating point against signal swings (improving linearity), increases the input impedance seen by the source, and extends bandwidth. This is the classic gain-stability tradeoff: the same resistor that fights AC gain is responsible for DC stability and improved small-signal linearity.
Question 5 Short Answer
Why does the common-emitter amplifier produce a 180-degree phase inversion between input and output? Explain using the transistor's action and the role of R_C.
Think about your answer, then reveal below.
Model answer: When the input rises, it increases the base-emitter voltage, which increases the collector current. Greater collector current causes a larger voltage drop across R_C, which lowers the collector voltage. The output therefore moves opposite to the input — 180-degree inversion.
The key is V_out = V_CC − I_C·R_C. Since I_C increases when v_in increases, V_out decreases. The small-signal model captures this as v_out = −g_m·v_be·R_C, where the negative sign encodes the inversion caused by the resistive load at the collector.