--- title: "Series regulator efficiency; bandgap-reference regulator" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:series-regulator-low-efficiency-bandgap" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/series-regulator-low-efficiency-bandgap site: BCT Engineering --- # Series regulator efficiency; bandgap-reference regulator Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2078 Bhadra, 2+5 marks; 2072 Kartik, 2+4 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/series-regulator-low-efficiency-bandgap - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Why transistor series regulator has lower efficiency? Explain the operation of voltage regulator using band gap voltage reference. ## Short answer Series regulator efficiency is low because pass transistor drops (Vin-Vout), wasting power as heat; efficiency ≈ Vout/Vin. A bandgap-based regulator compares a resistor divider feedback to a stable 1.25V bandgap reference; the error amplifier adjusts the pass element to keep output constant against line and load variations. ## Answer #### Why a transistor series regulator has low efficiency In a series regulator, the pass transistor is in series with the load and carries the whole load current IL. It acts like a variable resistor that drops the extra input voltage: - Power delivered to load: Po = Vo IL - Power drawn from the source: Pi = Vi (IL + Iq), where Iq is the small bias/control current - Power wasted in the pass transistor: PD = (Vi − Vo) IL η = Po/Pi ≈ Vo/Vi So if Vi = 12 V and Vo = 5 V, the efficiency is at most 5/12 = 42 %; the rest, 7 V × IL, is lost as heat in the transistor. The efficiency is poor when Vi is much larger than Vo, or when Vo is adjusted to a low value. The wasted heat also needs a heat sink. The transistor also works as a linear (class A like) element all the time, so the power loss continues at every load. (Switching regulators avoid this by using the transistor as a switch.) #### Voltage regulator using a bandgap reference ```text Vin o----+--------[ Q1 pass ]--------+----o Vo | ^ | | | [R1] | +----+----+ | | | Error |<------+--- feedback +--------| amp | | | +----+----+ [R2] | ^ | | +------+------+ GND +------| Bandgap Vref| | ~1.25 V | +-------------+ ``` 1. **Bandgap reference** supplies a stable Vref ≈ 1.25 V, independent of temperature and Vin. It is formed by adding VBE (negative temperature coefficient, about −2 mV/°C) to a PTAT voltage K·VT (about +0.086 mV/°C × K). With K ≈ 23, the drifts cancel. 2. **Feedback divider** R1, R2 takes a sample of the output: Vf = Vo R2/(R1 + R2). 3. **Error amplifier** compares Vf with Vref and drives the base of the pass transistor Q1. 4. If Vo rises (higher Vin or lighter load), Vf rises above Vref, the error amplifier reduces Q1 drive, and Vo falls back. If Vo falls, the opposite happens. 5. In steady state Vf = Vref, so **Vo = Vref (1 + R1/R2)** For example, Vref = 1.25 V, R1 = 3 R2 gives Vo = 5 V. The output is stable against temperature, line and load changes, as the reference does not depend on them. The LM317 and most IC regulators work on this principle. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does a transistor series regulator have low efficiency? A. The pass transistor is in parallel with the load B. The pass transistor drops (Vi - Vo) while carrying the full load current, wasting power as heat C. The reference draws most of the current D. The transistor is saturated and has no drop Answer: B. Wasted power is (Vi - Vo) IL, so efficiency is about Vo/Vi. (Key point: Low efficiency) 2. For Vi = 12 V and Vo = 5 V, the maximum efficiency is about? A. 42 % B. 58 % C. 95 % D. 240 % Answer: A. η ≈ Vo/Vi = 5/12 ≈ 42 %. (Key point: Efficiency value) 3. How does the regulator correct a rise in Vo? A. The error amplifier increases the pass transistor drive B. The bandgap voltage rises to match Vo C. R1 is switched out of the divider D. The sample Vf rises above Vref, the error amplifier reduces drive to the pass transistor, and Vo falls back Answer: D. Negative feedback compares Vf with Vref and reduces conduction when Vo is too high. (Key point: Feedback operation) 4. In steady state, what is Vo? A. Vo = Vref R2/(R1 + R2) B. Vo = Vref - Vf C. Vo = Vref (1 + R1/R2) D. Vo = Vref (R1/R2) Answer: C. Vf = Vo R2/(R1+R2) = Vref, giving Vo = Vref(1 + R1/R2). 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