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In my school we used lab bench power supplies all the time as it is an electrical engineering school. Can someone point me in the direction of an implementation of an overcurrent protection found in such supplies?

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There are many hobbyist quality schematics out there, so it's harder than one might think to find the schematic for a commercial product. Below is taken from this website. It's a lower end commercial product from an Asian supplier. The overcurrent limit is via the differential comparator IC2D. Very crude current limiting.

enter image description here

The way it works is pretty straightforward- R3 carries about 20% of the current flowing to the output (R1~R5 are 0.5 ohm resistors), so the voltage across R9+M1+VR1 is proportional to the output current, and that voltage is divided down by R10/R11+VR2 for one side and R24/R25 for the other side. The frequency compensation pin of the LM723 is used for shutdown via D12.

Offhand, I'm not quite sure what the R23/D9/R23 network achieves. The R22/R23 divider yields nominally a bit under 0.6V unloaded so it would only have an effect if the output voltage was very close to 0V. It might be to prevent a problem with the case of an externally applied voltage to the output that causes it to go negative, which can happen (see D8).

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There are several ways this could be done and to list them all would be a book, not a quick answer.

But two of the more common are:

  1. Directly measure output current, or primary side current and either shut-down or go into burst mode (shutting-down and restarting), also called 'hiccup' mode. This is crude but does protect the power supply.

  2. Directly measure output current and add this as a second input to the feedback circuit implementing a true current limit.

I wont give circuits as there are lots of ways it can be done and you should be able to find several examples on the internet.

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The old, just barbarian solution that still may be applicable is some cases is to use incandescent light bulb rated for the voltage of your power source and operating current close to your protection current. It must be wired sequentially to your power consumer.

If the current is well below the protection level, the thread of the bulb is cold, and it has relatively limited resistance. If you short circuit your consumer, the bulb will light up warning about the event. You will not get higher current than the bulb is rated for, however. The resistance of the lighting bulb is much higher and it will restrict current more than when it was cold.

This works not so bad if the consumer normally needs much less current than the light current of the bulb, but the power source is powerful enough to support the bulb as well.

schematic

simulate this circuit – Schematic created using CircuitLab

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