2026-08-22
You picked a nice LM317 or LT1085 for your bench supply, but the load turned out to be 5 A and your regulator maxes out at 1.5 A. Rather than switching topologies, you can bolt on an external pass transistor and let the regulator IC handle the loop while a beefy PNP or P-channel MOSFET carries most of the current.
The classic circuit: Insert a small "sense" resistor RSC between VIN and the regulator's input pin. Tie a PNP power transistor (say a TIP42 or MJ2955) with its emitter to VIN, collector to the regulator's output, and base to the junction of RSC and the regulator input. When regulator input current rises high enough to drop ~0.6 V across RSC, the PNP turns on and shunts additional current around the IC to the load.
How the loop stays honest: The regulator still senses its output pin voltage and adjusts its internal pass element. If the load pulls harder, the regulator tries to source more current, RSC drops more voltage, and the PNP delivers proportionally more. The regulator effectively becomes the error amplifier for the composite pass device.
Sizing RSC: Rule of thumb — pick RSC so the PNP starts conducting at roughly half the IC's rated current. For a 1.5 A LM317, target turn-on at 750 mA:
Concrete example: A 12 V → 5 V @ 5 A supply. Voltage across the PNP is 7 V, so at 4.25 A it dissipates ~30 W. That transistor needs a serious heatsink (θJA under ~3 °C/W with a 25 °C ambient to stay below 125 °C junction). The LM317 dissipates only about 5 W — a modest TO-220 heatsink handles it.
Gotchas:
