I am very new to electrical engineering so I have very little experience. I am currently working on a project that has me using a 24 VDC 15 amp power supply. I was given a switch that is rated for 24 VDC 10 amps, but the switch has to power a circuit that requires 24 VDC 13 amps.

  1. Is the switch I was given able to handle the current?

  2. Do I have to find a way to reduce the current, so the circuit doesn't fry? If so how would I go about this?

  • \$\begingroup\$ Have a look at electronics.stackexchange.com/questions/624438/… because it answers question 1 (disclaimer: I wrote one of the answers) \$\endgroup\$ Jun 22, 2022 at 19:06
  • \$\begingroup\$ Exactly what is the circuit being powered, and which switch do you have? \$\endgroup\$ Jun 22, 2022 at 23:39
  • 7
    \$\begingroup\$ If the circuit you are powering is 13A then you are not running 15A through the switch. Instead you are using a power supply capable of supplying up to 15A and running 13A through the switch to power a 13A circuit \$\endgroup\$
    – slebetman
    Jun 23, 2022 at 1:52
  • \$\begingroup\$ Note that when you turn a switch off that's carrying current, you get some arcing between the switch contacts. The arcing creates plasma and damages (and cleans) the switch contacts. Putting 13A through a 10A switch, you will get more arcing and more plasma than the switch is designed for. This means more damage to the switch and a shorter lifespan. \$\endgroup\$ Jun 24, 2022 at 2:33

6 Answers 6


You need a better switch to handle the 13A.

That doesn't mean you can't also use the switch you are given. If for example it is the only one that fits on the front panel of your device (and fits stylistically) then absent any safety concerns1, you could let the user activate the 10A-limited switch, while the circuit actually passes through it only enough current to control a relay that switches the main power. Of course the relay should be rated above 13A (preferably including some safety margin, on both the maximum voltage and maximum current).

1 For example, latching emergency-off buttons often are part of a risk mitigation strategy that requires physically disconnecting power and not merely sending a "soft-off" logic signal. In such a case you cannot use a relay to abstract the user-activated button away from the switching action.


1: No. The switch says 10 amps; 10 amps is less than 13 amps.

2: It's hard to "reduce the current" going through the switch. It's not so much putting a flow restrictor washer in your showerhead to save water as much as it is putting a clamp on the fuel hose in your car and hoping you get better gas mileage; there's really no way to do so without lowering the voltage to the load which will cause it to operate incorrectly or not at all or doing some other kind of step-up step-down trickery that will certainly be more complex and costly than buying a switch rated for 15A (and can't be done with a switch rated for 24V anyways).

  • \$\begingroup\$ It's hard to reduce the current going to the load. It's easy to reduce the current going through the switch. \$\endgroup\$
    – Ben Voigt
    Jun 22, 2022 at 17:40
  • 1
    \$\begingroup\$ Yes, you could use a relay. \$\endgroup\$
    – vir
    Jun 22, 2022 at 17:41
  • \$\begingroup\$ In a constant-voltage supply system like 99% of electrical things, the load decides how much current to draw, and "getting in its way" is not the normal way to do things. There is no easy or elegant way to "hack" a load to draw less, except re-designing the load. \$\endgroup\$ Jun 23, 2022 at 4:17

Is the switch I was given able to handle the current?

It will work well for a while, then the contacts will die or weld themselves shorted.

Do I have to find a way to reduce the current, so the circuit doesn't fry? If so how would I go about this

Just put a 10A 250VAC switch on the mains side of your power supply.

If your DC load has a substantial amount of capacitance on the 12V rail, then this would be the best solution, as it will allow the power supply to soft-start into the capacitors. Otherwise the switch will sit between the power supply's output caps and the board's input caps, and it will take a huge inrush current at turn-on.

You don't mention the type of load in your question, but when picking a switch you should also consider the inrush current at turn-on, not just the average current while it is running.

  • 2
    \$\begingroup\$ Some switches and relays have separate specifications for the maximum current that can be actively switched, versus the amount of current that may be fed through them while they are closed (and which must fall below the first limit before the switch is open). In many cases, the latter limit may be much higher. If one has two devices which each draw 50 amps, but only one would be used at a time, one could use one SPST switch which is rated to reliably interrupt 50 amps, and one SPDT switch which is rated to pass more than 50 amps but might not be able to interrupt that much current. \$\endgroup\$
    – supercat
    Jun 23, 2022 at 20:42
  1. No. Switch is rated for 10A, not 13A.

  2. The load takes 13A at 24V. It might not work properly if you limit current to 10A.

Depending on what the load is, it may not be possible to reduce current. If the load has a switch mode power supply, it takes the power it needs. Reducing voltage will make it take more current. And voltage can't be increased to reduce current, because switch is used at rated 24V already.

Also make sure you know the switch specs properly. It might be rated for 10A and 24V separately, but it might not be rated for 10A and 24V simultaneously.


A DC power supply often see a capacitive 0 charge on switching. Thus the ESR of the load cap and supply can be much less than <<0.1 Ohm and the switch must handle this contact turn on current even if it is only for microseconds and this momentarily arcs possibly with contact bounce and erodes the silver alloy contacts each time.

So for DC switching contact ratings are usually reduced for this reason. The better solution is to use a 25A switch and preferably a FET or a soft start inrush current limiter , ICL which is a special NTC thermistor with a 15A switch or without using a 25A relay which is less desirable but also reliable when the 1 mohm initial contact resistance is less than the net ESR of the capacitors.

Thus if you want a short lifespan, use a 10A relay otherwise, don't.

Power high side or low side FET power switches would be most reliable rated for 25A or more and cost less than a 25A relay. It won't need a PCB if you use short twisted magnet wires to Gate and Source with correct polarity as required , Pch for high side and Nch for gnd switched side.

If you do not have a C load and are just driving 12V LEDs then that is considered a resistive load and a 15A relay will do or a 25A FET without a heatsink on a through hole part.

All switches have resistance and the heat dissipation must be low for Pd= Isquared*R. Generally, you choose a switch for <<1% pref. 0.1% load loss so 15Ax12V= 180W x0.1% = 0.18W

  • that choice results in switch R= P/I^2 = 0.18W/225= 0.8 mohm . Relay contacts may start this low but erode, oxidize and rise sharply with aging then get hotter and burn out. FETs switches do not.
  • \$\begingroup\$ Note that OP can probably buy a "solid-state relay module" including small PCB already assembled, with the same pinout as a solenoid relay for probably not much more cost than sourcing the FETs separately. \$\endgroup\$
    – Ben Voigt
    Jun 22, 2022 at 19:20
  • \$\begingroup\$ Maybe, but important to be aware ti.com/lit/an/slva716a/… \$\endgroup\$ Jun 23, 2022 at 3:23

I have used a power transistor in the past to act as a switch or an scr rated for the current your circuit will draw. As a precaution, use a heat sink with either should you choose this method. With that, you can use 2 inexpensive momentary contact switches if they fit your panel properly. These switches come in different sizes. One switch will turn on the circuit, the other will turn it off.If you are not familiar with these components, many sites will show how to connect these devices properly. Just google or type in info you require and mate the result of the search with your circuit configuration. The idea for the SCR was obtained from the website Circuits Today. I will add an excerpt from that site:

Current ratings of an SCR. The current carrying capability of an SCR is solely determined by the junction temperature. Except in case of surge currents, in no other case the junction temperature is permitted to exceed the permissible value. Some of the current ratings used in industry to specify the device are given below.

(i) Forward Current Rating. The maximum value of anode current, that an SCR can handle safely (without any damage), is called the forward current rating. The usual current rating of SCRs is from about 30 A to 100 A. In case the current exceeds the forward current rating, the SCR may get damaged due to intensive heating at the junctions.

SSRs are I believe more expensive then scrs and the response time longer. It is possible that the op may be better off with an SSR since the circuit design was not shown. The better unit will be determined by what has been designed.I will edit again if it is necessary. For now, I will go with this.

  • \$\begingroup\$ This is a DC circuit, why do you recommend a SCR? \$\endgroup\$
    – Jens
    Jul 8, 2022 at 13:09
  • \$\begingroup\$ It is mentioned because it can take the place of a switch the op doesn't have and they can use a low rated mechanical switch to power the gate and latch the circuit on. Several comments have been made about arching. This will not happen with an SCR. \$\endgroup\$
    – j dorfman
    Jul 8, 2022 at 15:45
  • \$\begingroup\$ This may be a misunderstanding. For me an SCR is a component with TRIACs inside, only useful in AC circuits and with 1.6-2V saturation voltage, which would produce 20W heat here. An SSR typically contains MOSFETs with much lower saturation voltage and useful in DC circuits. \$\endgroup\$
    – Jens
    Jul 9, 2022 at 16:33

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