I need to build a current limiting circuit in order to protect a circuit board.

It is for a solar power application. Currently the circuit board only can handle 1.6 amperes but the solar panels that are being used deliver up to 4 amperes.

I know the easiest solution is to replace the panel with a smaller one, but unfortunately that is not possible, so I am looking for a simple circuit that will do this.

I have found lots of circuits with MOSFETs online but not really sure how well they work.

The input voltage ranges from 20V to 28V from the panel and that is no problem. I just need to limit the current to maximum 1.6A, so it will be 24V 1.6A

Any ideas?

  • 8
    \$\begingroup\$ Why do you think you need to limit the current? \$\endgroup\$
    – JRE
    Nov 23, 2020 at 20:37
  • 1
    \$\begingroup\$ Use a 1.6A fuse? \$\endgroup\$
    – Reinderien
    Nov 23, 2020 at 20:41
  • \$\begingroup\$ Change the angle of the panel to the sun to something not optimal at all and measure the short circuit current under full sun to see if it is anywhere near what you want. \$\endgroup\$
    – ocrdu
    Nov 23, 2020 at 21:06
  • 1
    \$\begingroup\$ Think carefully about what "limiting the current" means. In an overcurrent situation, any current limiting circuit you use will decrease the voltage at the output until the current is at the set value (except a fuse which will just blow and make the voltage output 0V). This means that your current limiting circuit may have to dissipate significant amounts of heat. \$\endgroup\$
    – BeB00
    Nov 23, 2020 at 21:07
  • 2
    \$\begingroup\$ what does 'the circuit board' do? Is it an end product that you're powering? In which case no problem, it will only take what it needs. Is it charging some batteries downstream? In which case it ought to have some method of programming the max charging current, that you can set to 1.6 A. In neither case is it appropriate to limit the current to the board, apart from perhaps fuse protection. Edit into your question what this board is supposed to do. \$\endgroup\$
    – Neil_UK
    Nov 23, 2020 at 21:18

2 Answers 2


Others have questioned whether current limiting is what you really need. Assuming that limiting the current is what you actually need, this circuit will limit your load current to 1.6A. R1 should be rated a minimum of 2W. Choosing a larger resistance value will reduce the maximum current. The voltage seen by the load will be less than the supply voltage, and will vary depending upon the size of your load. As others have suggested, maybe your load only draws so much current on its own, and current regulation is not what you need. But here it is anyway.


simulate this circuit – Schematic created using CircuitLab


I don't think you need it at all. The laptop I'm typing this on is powered using a 120W power supply that, at the moment, consumes about 35W. The outlet that the power supply is plugged into is able to supply, instantaneously, about 250A, and long-term about 20A at 120V. There is no "protection circuit" necessary for that power supply: it won't consume 20A just because the rating of the supply circuit is such; it'll only draw the current needed by the demands of its load (less inefficiencies).

This particular supply's input side is designed to withstand well over 2A without overheating dangerously for short periods of time (tens of seconds), and there's a non-replaceable fuse with a 2A rating right at the entrance to the power supply. Just a fuse - nothing more complicated than that. The protection is there against catastrophic failure: were the supply to fail in such a way that the input would have too low of an impedance, an excessive current would flow that could overheat and even vaporize traces and components on the circuit board, and such failures are very much frowned upon since there's very little control over what will overheat, and whether it may catch fire before the trace or component fails completely open. Instead, the failure is "contained" inside a fusible link that is designed to be the weakest link in the input circuit. The fuse is basically protection against fire risk.

Now to the solar panels: as long as whatever you plug into those panels is internally protected against catastrophic failure, you'll be fine as long as the interrupting capacity of the protection element (fuse) is higher than the short current capacity of the solar panels (certainly higher than 4A!). The load you plug in should not draw more current than it's designed to, as long as it's operating correctly, so I'd expect that the input current will be 1.6A, maybe a little bit more, and possibly much less - depending on what that load is (e.g. a dimmable lamp can easily vary its input current over one or two orders of magnitude!).

Interrupting capacity is different than current rating of the fuse. When the supply (mains socket, car battery, solar panels, etc) can provide enough current, the fuse may open but there may be plasma inside of it (or in its place, if the fuse explodes) that will sustain an arc, so the fuse will "fail open" yet the current will keep on flowing. That's bad news. Small Glass cylinder fuses have interrupting capacities on the order of tens to low hundreds of Amps, and if I were to be conservative, I'd choose a fuse with 2.5A rating (this is derated for high temperature scenarios), and interrupting rating of at least 50A, although I'd probably want to measure the actual load characteristic of the solar panel using suitable instrumentation - or at least read about that topic to see how typical curves look as a function of panel area. The load characteristic is the Voltage vs. Current curve of the panels, assuming full illumination since we care about the worst case (when the panels can do the most damage to an unprotected circuit). The panels may provide 4A at some nominal voltage, but as you lower the impedance of the load, the voltage will be dropping while the current will be increasing, and the fuse's interrupting capacity must be able to open the circuit when facing a hard short right behind the fuse - i.e. the fuse will experience the short circuit current that the panels can provide, and must reliably open without sustaining an arc at that current.

What is your load, and what makes you think that you need to actively limit its current consumption?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.