I need to protect my device from the flowing through currents above let's say 30 Amps.

So I need a black box which will generate a digital signal if the current will be greater than specified (e.g. +/- 30 Amps). Rest things (circuit breaking) will be done elsewhere.

First I think about hall-effect sensor with comparator. Except that this is a branchy scheme with a separate isolated power supply this approach is slow (typically about 5 us).

Isn't there any off the shelf solutions for protection?


I need to protect my own device primarily (load protection is also welcome). Mainly from a short circuit. My estimation is that I need to break the circuit within about 1,5 us: if I wrong with this estimation please let me know your consideration.


The load most likely will be LED drivers. So it will not likely to be inductive. Probably capacitive in some degree. Nominal voltage is 230 VAC, nominal current estimated to 20 Amps.


If asking about how I'm going to break the circuit that fast - the answer is that I have MOSFETs in series so I can switch them all off pretty fast:

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  • \$\begingroup\$ You're missing some important information: What is the normal operation voltage/current? What type of load are you driving? Highly inductive, capacitive, resistive, etc.? \$\endgroup\$ – helloworld922 Dec 14 '15 at 17:01
  • \$\begingroup\$ @helloworld922, please see my update ABOUT THE LOAD in the original question for details. \$\endgroup\$ – Roman Matveev Dec 14 '15 at 17:38
  • \$\begingroup\$ How are you going to break a 30A AC load in 1.5 us? Nothing mechanical will do that, I don't think. Maybe crowbar it and let something else open the circuit? I know this is not what you are asking about, but I cannot restrain my curiosity on this point. \$\endgroup\$ – mkeith Dec 14 '15 at 17:49
  • \$\begingroup\$ @mkeith I have MOSFETs in the schematic which can do this pretty fast. The part of the schematic is here: goo.gl/yT4YZb \$\endgroup\$ – Roman Matveev Dec 14 '15 at 18:56

5us is slow? All others solutions I think they will be ultra slow, then. You can use a reed switch instead of hall sensor, but you'll get an aproximative threshold with long delay (compared to 5us).
Anyway, what is the purpose of this ovecurrent detector, I mean what is the voltage source, because if you are trying to detect SC over mains power supply the current will be far more than 30A, and what load you are trying to protect (resistive, inductive)?

  • \$\begingroup\$ Thank your for your answer! Please see SOME MORE DETAILS in the original question for answering your questions and for clarifications. \$\endgroup\$ – Roman Matveev Dec 14 '15 at 10:00
  • \$\begingroup\$ One more comment: resistor shunt with a comparator can be pretty fast I hope. \$\endgroup\$ – Roman Matveev Dec 14 '15 at 10:52
  • \$\begingroup\$ Yes, but 1st: you can't disconnect the load in few us. 2nd: if your comparator circuit will have no low pass filter it will trip at every noise spike. 3rd: In a SC event the current will rise at much higher level - domestic is rated at 25kA. The only thing that can limit the slope is a choke in a series. Usualy the SCR and triacs have only fuses to prevent fire, other protection is more expensive that SCR itself (well you didn't mention what value part you need to protect). \$\endgroup\$ – Marko Buršič Dec 14 '15 at 18:03
  • \$\begingroup\$ Thank you! This thing appears to be much more tricky than I think it in the beginning. Regarding the circuit breaking - see my EDIT 2 \$\endgroup\$ – Roman Matveev Dec 14 '15 at 19:00
  • \$\begingroup\$ @Roman Go easy with all capitals, please. \$\endgroup\$ – Nick Alexeev Dec 14 '15 at 19:08

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