--- title: "Series regulator current limiting; LM317 design 7-21 V" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:series-regulator-current-limit-lm317" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/series-regulator-current-limit-lm317 site: BCT Engineering --- # Series regulator current limiting; LM317 design 7-21 V Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2082 Baisakh, 2+4+1+3 marks - 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-current-limit-lm317 - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Draw series voltage regulator with current limiting circuit and explain how this protection circuit works. Design a voltage regulator to give output voltage from 7 V to 21 V using LM317. ## Short answer Current limiting uses a sense resistor and transistor to clamp output current to 0.6V/Rsc. For LM317 7–21V supply: R1=240Ω, R2 = 1.1kΩ fixed + 2.7kΩ pot, input ≈ 24V. ## Answer #### Series voltage regulator with current limiting A series regulator with current limiting protects the pass transistor and the load from damage in the event of an overload or short circuit. A sense resistor Rsc and a transistor Q2 limit the maximum output current. #### Circuit diagram (current limiting) ```text Vi o----+-----------------------+ | | C [R] | | B | +---------+-------------Q1 (series pass) | | | (error amplifier | | E / Zener drive | +-----+---[Rsc]---+-----+----o Vo to Q1 base) | | | | | C | | [RL] Q2 B---------------+ | | E | GND | | +-------------------------------+ ``` Connections: Q1 is the pass transistor (collector to Vi). Its emitter goes through the small sense resistor Rsc to the output Vo. The base of Q2 is connected to the Q1 emitter side of Rsc, and the emitter of Q2 to the output side, so VBE2 = IL Rsc. The collector of Q2 is connected to the base of Q1. Q1 is driven by the regulator (error amplifier or Zener) in the usual way. #### Working 1. **Normal load:** IL Rsc < VBE2(on) ≈ 0.6 V, so Q2 is OFF and has no effect. Q1 is controlled by the regulator and keeps Vo constant. 2. **Overload or short circuit:** IL rises. When IL Rsc reaches about 0.6 V, Q2 turns ON. 3. Q2 now takes (shunts) the base current away from Q1 (Q2's collector is connected to the base of Q1). Q1 conducts less, so IL cannot rise further. 4. Thus the output current is clamped to **IL(max) = VBE2 / Rsc ≈ 0.6 V / Rsc** 5. In a short circuit the output falls and the current stays at IL(max), so the power dissipation of Q1 is limited. This is called constant-current limiting. **Example:** For IL(max) = 0.5 A: Rsc = 0.6/0.5 = 1.2 Ω. A disadvantage is the drop IL·Rsc in the output path (about 0.6 V at full current), which reduces the efficiency. Foldback limiting reduces the current further as the output voltage falls, giving lower dissipation in a short circuit. #### Design of a 7 V to 21 V regulator using LM317 **LM317** is a three-terminal adjustable regulator. It keeps a constant reference voltage Vref = 1.25 V between its output and adjust terminals. The output voltage is Vout = 1.25 (1 + R2/R1) + Iadj · R2 with Iadj ≈ 50 µA (small, usually neglected here). #### Circuit ```text +--------------+ Vin o--+---| IN OUT |---+------+--o Vout | | LM317 | | | [C1] | ADJ | [R1] [C2] | +------+-------+ | | GND +-----------+ GND | [R2 = fixed + pot] | GND ``` R1 = 240 Ω (the usual 120 to 240 Ω range), so that 1.25 V / 240 Ω = 5.2 mA always flows, enough for the LM317's minimum load current. C1 = 0.1 µF (input bypass), C2 = 1 µF (output). #### Design for 7 V to 21 V - Take R1 = 240 Ω. - For Vout = 7 V (minimum), R2 = R1 (Vout/1.25 − 1) = 240 × (7/1.25 − 1) = **1104 Ω** - For Vout = 21 V (maximum), R2 = 240 × (21/1.25 − 1) = **3792 Ω** A potentiometer alone cannot give a minimum above 1.25 V, so a fixed resistor is connected in series with the pot: - Fixed resistor Rfixed = 1104 Ω → standard value **1.1 kΩ** - Potentiometer = 3792 − 1104 = 2688 Ω → **2.7 kΩ** if available, otherwise a 5 kΩ pot set so that the maximum is 21 V (a 2.5 kΩ pot is too small: it reaches only 20 V) Check (neglecting Iadj): Vout(min) = 1.25 (1 + 1100/240) = 6.98 V; Vout(max) = 1.25 (1 + 3800/240) = 21.04 V. Iadj = 50 µA adds Iadj·R2 ≈ 0.06 V at the low end and about 0.19 V at the high end. #### Input voltage The LM317 needs a dropout of about 3 V, so Vin ≥ Vout(max) + 3 = 24 V. Use an unregulated supply of about **26 V** (below the 40 V device limit). The pass dissipation is (Vin − Vout) × IL, so a heat sink is needed for large currents (up to 1.5 A). **Answer: R1 = 240 Ω, R2 = 1.1 kΩ fixed + 2.7 kΩ pot (or 5 kΩ set to 21 V) in series, Vin about 26 V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Q2 connected in the current-limiting circuit? A. Collector to the load, emitter to the supply B. Base-emitter across Rsc, collector to the base of Q1 (pass transistor) C. Base to ground, emitter to Vi D. In parallel with RL Answer: B. VBE2 = IL Rsc, and Q2 can steal base drive from Q1. (Key point: Current limiter circuit) 2. At what current does the limit act? A. When IL Rsc ≈ 0.6 V, i.e. IL(max) = 0.6/Rsc B. IL = Rsc x 0.6 C. IL = Vo/Rsc D. When Vo = 0.6 V Answer: A. Q2 conducts when its VBE reaches about 0.6 V. (Key point: Limit current) 3. For IL(max) = 0.5 A, what Rsc is needed? A. 0.3 Ω B. 12 Ω C. 3 Ω D. 1.2 Ω Answer: D. Rsc = 0.6 / 0.5 = 1.2 Ω. (Key point: Rsc example) 4. For 7 V to 21 V with R1 = 240 Ω, what R2 range? A. 720 Ω to 3600 Ω B. 1104 Ω to 2688 Ω C. 1104 Ω to 3792 Ω D. 240 Ω to 3792 Ω Answer: C. 240(7/1.25 - 1) = 1104 Ω and 240(21/1.25 - 1) = 3792 Ω. 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