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Series regulator with current limiting

Short answer

A sense resistor Rsc in series with the load develops IL·Rsc. When it reaches ~0.6 V, Q2 turns on and shunts base current from the pass transistor, limiting output current to IL(max) ≈ 0.6 V / Rsc.

Draw series voltage regulator with current limiting circuit and explain how this protection circuit works.

A series regulator with current limiting protects the pass transistor and the load from damage during an overload or short circuit. A sense resistor Rsc and a transistor Q2 limit the maximum output current.

Circuit diagram

 Vi o----+--------------------+
         |                    | C
        [R]          B1 +-----|  Q1 (pass)
         |              |     | E
         +--------------+     +------+---[Rsc]---+------+----o Vo
         |              |            |           |      |
         |              |            |           |     [RL]
         |              +---+ C      |           |      |
         |                  |--B-----+           |     GND
         |                  | E   Q2             |
         |                  +--------------------+
  [Zener / error amp]

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 (node B1). 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 diverts (shunts) base current away from Q1, because 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

  1. In a short circuit the output voltage falls while 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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