--- title: "Current limiting circuit in dc regulator" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:current-limiting-circuit-draw" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/current-limiting-circuit-draw site: BCT Engineering --- # Current limiting circuit in dc regulator Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2070 Chaitra, 2 marks - Video: [Voltage regulator short circuit protection (7 - Regulators)](https://www.youtube.com/watch?v=03KeyVS0LAo) — Aaron Danner - 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/current-limiting-circuit-draw - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Draw the circuit of current limiting circuit in dc voltage regulator. ## Short answer A current limiting circuit uses a series-pass transistor Q1 with a current-sense resistor Rsc in its emitter path. A limiting transistor Q2 has its base-emitter junction across Rsc and collector tied to Q1’s base. When load current causes Rsc voltage drop to reach ~0.6 V, Q2 turns on, stealing Q1’s base drive and limiting output current to IL(max) ≈ 0.6 V / Rsc. ## Answer ```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 | | +-------------------------------+ ``` - Q1 is the series pass transistor of the regulator, and Rsc is the small current-sense resistor in series with the load. - Q2 is the limiting transistor. Its base-emitter junction is connected across Rsc (base on the Q1 side, emitter on the output side), and its collector is connected to the base of Q1. - Normal operation: IL Rsc < 0.6 V, so Q2 is OFF. - Overload: when IL Rsc ≈ 0.6 V, Q2 turns ON and takes away the base drive of Q1. The output current is limited to **IL(max) ≈ 0.6 V / Rsc** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the current limiting circuit, how is the limiting transistor Q2 connected? A. Base-emitter across the sense resistor Rsc, collector to the base of Q1 B. Base-emitter across the load RL, collector to the output C. Collector across Rsc, emitter to the base of Q1 D. Base-emitter across the Zener diode, collector to ground Answer: A. Q2's base-emitter junction senses the voltage across Rsc, and its collector is tied to Q1's base so it can divert Q1's base drive. (Key point: Connection of limiting transistor Q2) 2. During normal load current (below the limit), what is the state of Q2? A. ON, sharing load current with Q1 B. OFF, because IL x Rsc is less than about 0.6 V C. In breakdown, clamping the output D. ON, supplying the base current of Q1 Answer: B. The drop across Rsc is below 0.6 V, so Q2's base-emitter junction is not forward biased and Q2 stays OFF. (Key point: Normal operation) 3. What happens when the load current rises until IL x Rsc is about 0.6 V? A. Q1 is driven harder to supply more current B. Q2 turns OFF and Q1 saturates C. Q2 turns ON and takes away the base drive of Q1, limiting the output current D. The Zener diode shorts the output to ground Answer: C. At about 0.6 V across Rsc, Q2 conducts and diverts base current away from Q1, which stops the load current from increasing. (Key point: Overload action) 4. If Rsc = 0.5 ohm, what is the approximate limiting current IL(max)? A. 0.3 A B. 0.6 A C. 3 A D. 1.2 A Answer: D. IL(max) = 0.6 V / Rsc = 0.6 / 0.5 = 1.2 A. 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