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Circuit Diagram

This circuit looks relatively simple to build. I'm not going to bother with the transformer since I have a dc adapter. I was wondering how would I adapt this circuit to use a lm317 instead of a 7805. The maximum output is 1.5A. I presume I would need beefier transistors. However am unsure as the formulas at the bottom I never seen in my life. Could you point me in the right direction to understanding these formulas please? It's a 3.3v / 5v PSU selectable via SPDT switch.

Image from: http://apowersupply.com/short-circuit-protection-in-dc-low-voltage-systems-380.html

The reason I'm using this circuit is because it has an short circuit led. It's useful to have.

Am actually following spark fun schematic for a 5v breadboard supply. However I want a short circuit led as well as a power led. enter image description here

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  • \$\begingroup\$ Are you replacing the whole of the circuit above the emitter of T1 with your existing power supply? It sounds like your PSU performs the function of that part of the circuit admirably. Is it just short-circuit protection you want to add to it? \$\endgroup\$
    – Majenko
    Commented Jul 10, 2011 at 10:21
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    \$\begingroup\$ I deleted my incomplete answer. I think the PMOSFET is a good solution, but I've never used PMOSFETs before, and especially at these low voltages I have to study it more. \$\endgroup\$
    – stevenvh
    Commented Jul 11, 2011 at 5:06
  • \$\begingroup\$ Steven Could I not use an N channel mosfet? I thought they're the same only they use different material \$\endgroup\$
    – Ageis
    Commented Jul 11, 2011 at 13:36
  • \$\begingroup\$ The short circuit LED will only work if O/P2 is shorted. For normal usage you'd connect your load to O/P1 and i don't see how a short circuit there can trigger the LED to light up. \$\endgroup\$
    – Zy Gan
    Commented Sep 26, 2018 at 12:40

2 Answers 2

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If you want a LED lighted when there's a short circuit you can place the LED at the 7805's input, and control it with the 7805's output voltage. Use a PNP transistor to switch the LED, and drive it from the 5V output. Since the input voltage is a few volts higher than that, you'll need a voltage divider to the base to prevent the LED from lighting at 5V out. Placing the LED between emitter and input voltage also helps, though the divider is still required if the input voltage is several volts higher than 5V.
If the output is shorted it will go to 0V, and the PNP transistor will be switched on.

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dropping 5V to 4.5V or more at rate current seems like a big compromise for having short circuit sense LED. A Better way might be to choose an overcurrent protection resistor PTC driving the 7805 input selected just above needed current and then put LED across that device with series current limit R for LED. So the output is still regulated, LED indicator and OVC protection is added.

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    \$\begingroup\$ Wouldn't that oscillate? Short circuit causes PTC to open -> short circuit is gone -> PTC closes again -> short circuit, etc. \$\endgroup\$
    – Federico Russo
    Commented Apr 11, 2012 at 10:48

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