- 2070 Ashad · 2+4+2 marks
Series regulator with error amplifier
Short answer
Circuit: series pass Q1, error amp Q2, Zener reference, divider R1,R2. Output Vo = (Vz+VBE2)(1+R1/R2). Stability factor Sv = (1+R1/R2)·rz/(R+rz). Improve by using op-amp error amp, Darlington pass, constant-current Zener bias, low rz Zener, temperature compensation, and output capacitor.
Draw the circuit diagram of variable series voltage regulator with transistor as error amplifier. And find its voltage stability factor and explain how change in output voltage can be improved.
Circuit diagram: variable series regulator with a transistor error amplifier
Vi o----+------------------------+
| | C
[R] |
| B |
+-----------+------------Q1 (series pass)
| |
| C | E
Q2 B---+ +------+-------+-----o Vo
E | | |
| | [R1] [RL]
(Dz) | | |
| +-----------+ GND
GND |
[R2] (variable, pot)
|
GND
Connections: Q1 is the series pass transistor (collector to Vi, emitter to Vo). Q2 is the error amplifier: its collector is connected to the base of Q1 (and to Vi through R), its emitter is held at the Zener voltage Vz, and its base is connected to the divider R1-R2 across the output. Changing the variable resistor R2 changes the fraction of Vo that is sampled, and so the output voltage.
Operation
VB2 = Vz + VBE2 = Vo R2/(R1 + R2) (R2 = the lower part of the divider)
Vo = (Vz + VBE2)(R1 + R2)/R2 = (Vz + VBE2)(1 + R1/R2)
Varying R2 changes the ratio R1/R2, so Vo can be set to any value above the minimum (Vz + VBE2).
If Vo rises: VB2 rises, Q2 conducts more, its collector voltage falls, the base of Q1 falls, Q1 conducts less and Vo falls back to its set value. The opposite happens when Vo falls.
Voltage stability factor
Sv = ΔVo/ΔVi at constant load. When Vi rises by ΔVi, the base of Q1 stays at about Vo + VBE1, so the current in R rises by about ΔVi/R. This extra current flows through Q2 into the Zener and changes the Zener voltage by
ΔVz ≈ ΔVi · rz/(R + rz) (≈ rz ΔVi/R, since rz ≪ R)
and Vo = (Vz + VBE2)(1 + R1/R2), so with VBE2 constant,
Sv = ΔVo/ΔVi = (1 + R1/R2) · rz/(R + rz)
Example: rz = 10 Ω, R = 1 kΩ, R1 = R2: Sv = 2 × 10/1010 = 0.0198 (about 2 %, that is 19.8 mV per volt).
Improving the change in output voltage
- Use a higher-gain error amplifier (a differential pair or an op-amp in place of the single transistor), which reduces the variation of Vo through the large loop gain.
- Use a Darlington pair as the pass element, so that the base current drawn from the error amplifier is smaller (less load regulation error).
- Make the reference more constant: feed the Zener from a constant-current source or from the regulated output (bootstrap) instead of a resistor, and choose a Zener with low rz (about 6 V, near zero temperature coefficient). Raising R or lowering rz reduces Sv.
- Use a temperature-compensated or bandgap reference.
- Add an output capacitor and current limiting for stability and protection.
✓ Answer checked by Claude Opus 5.5 (high effort)
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