A technician replaces a BJT in a fixed-base biased amplifier (single resistor from VCC to base) with a transistor of the same type but double the beta. What happens to the collector current?
ANothing changes — the base resistor sets the collector current independent of beta
BThe collector current approximately doubles, potentially pushing the transistor into saturation
CThe base current doubles while collector current remains constant
DThe voltage divider compensates to maintain the original Q-point
In fixed-base bias, IB ≈ (VCC − VBE) / RB, set by the resistor. Then IC = β × IB. Because β appears as a direct multiplier, doubling β doubles IC. There is no feedback mechanism to compensate — the base voltage and base current are fixed by the resistor, and the collector current scales linearly with whatever β the transistor happens to have. This is why fixed-base bias is unsuitable for production circuits where transistors from the same batch can have 2:1 or greater beta spread. Option D incorrectly describes voltage-divider bias, not fixed-base bias.
Question 2 Multiple Choice
In a voltage-divider biased amplifier with emitter resistor RE, the junction temperature rises, causing IC to increase. Trace through the negative feedback mechanism that limits this increase.
ARising IC → rising VE (= IC × RE) → falling VBE (= VB − VE, with VB held fixed by stiff divider) → IC decreases back toward original value
BRising IC → rising VC → reduced RB equivalent → feedback reduces base current
CRising IC → rising temperature → thermal runaway unless heat sinking is adequate
DRising IC → rising VCC drop → Thevenin equivalent adjusts base voltage downward
This is the core stabilization mechanism: RE acts as a negative feedback element in the DC bias path. VB is held approximately constant by the stiff voltage divider (if the stiff divider condition is met). As IC rises, VE = IC × RE rises with it, reducing VBE = VB − VE. A lower VBE means the transistor is driven less hard, so IC decreases back toward its original value. The feedback gain is negative, so it is self-correcting. Without RE (or if RE is bypassed at DC), this stabilizing path is absent and thermal runaway becomes possible.
Question 3 True / False
Voltage-divider bias eliminates the need for the stiff divider condition because the emitter resistor RE alone provides most of the stabilization needed, regardless of how large the base current is relative to the divider current.
TTrue
FFalse
Answer: False
The emitter resistor provides negative feedback only if VB stays approximately constant when IC changes. VB stays constant only if the divider current is much larger than the base current (the stiff divider condition, typically ID ≥ 10 × IB). If the divider current is comparable to the base current, changes in IC cause changes in IB which change the current through R2, which changes VB — the base voltage is no longer fixed, and the feedback mechanism is undermined. The stiff divider condition and the emitter resistor work together; neither alone is sufficient.
Question 4 True / False
The emitter bypass capacitor CE is placed in parallel with RE so that RE provides DC bias stabilization while CE short-circuits RE at AC signal frequencies, restoring full voltage gain for the amplified signal.
TTrue
FFalse
Answer: True
This is the elegant design separation: at DC (and very low frequencies), CE is essentially open-circuit (its impedance 1/(2πfC) is large), so the full RE is present in the DC bias loop providing Q-point stability. At AC signal frequencies, CE is essentially short-circuit, bypassing RE entirely. Without RE in the AC signal path, the full transconductance drives the output and voltage gain is maximized. This allows the engineer to design the bias network for stability and the signal path for gain independently, without compromise.
Question 5 Short Answer
Explain why fixed-base bias is unsuitable for production BJT amplifiers, and describe how voltage-divider bias with an emitter resistor solves this problem at the circuit level.
Think about your answer, then reveal below.
Model answer: Fixed-base bias sets IC = β × IB where IB is determined by a single resistor. Since β varies 2:1 or more across transistors of the same type and also drifts with temperature, IC shifts proportionally — the Q-point is unreliable. Voltage-divider bias solves this in two ways. First, the voltage divider (R1, R2) with a stiff divider current sets VB approximately independent of the transistor's β, so the base voltage is a stable reference regardless of which transistor is installed. Second, the emitter resistor RE creates a negative feedback loop: any increase in IC raises VE = IC × RE, which reduces VBE = VB − VE, which in turn reduces IC. This self-correcting loop stabilizes the Q-point against beta variation and temperature drift. The result is a circuit whose operating point is determined primarily by resistor values and supply voltage — components that are stable and well-controlled — rather than by the transistor's unpredictable β.
The key insight is substituting a reliable feedback mechanism for direct dependence on an unreliable parameter. The same principle applies broadly in circuit design: feedback is more robust than open-loop precision.