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What happens if my circuit draws more current than my source is able to supply? For example, if I have a source that can supply 1 V and 1 A and I attach it to a resistor that is 0.5 ohms, the circuit will try to draw 2 A but my source is only rated to supply 1 A. What happens here?

Would the source supply the maximum rated current of 1 A and only be able to output 0.5V ? Or would the source begin to overheat?

How would this be different from connecting the 1 V 1 A source to a short circuit? In this case, I am trying to draw near infinite current but I can only supply 2 A.

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  • \$\begingroup\$ The power supply current compliance rating is just that, a rating. It's not a precision statement about physical law. It's just that it will perform properly up to that load current. After that? Who knows? But a simple power supply, those without specialty concepts like current-foldback or a settable current-limit mode, will probably have their output voltage go lower until the load stops drawing more and the supply can manage what's being taken. \$\endgroup\$
    – jonk
    Sep 17, 2021 at 3:18
  • \$\begingroup\$ If the load draws more current than the supply can manage, the voltage will dip or sag or collapse. The exact behavior beyond that basic statement depends on too many things. But for sure, if the load draws more than the supply can put out, the voltage will go down. That much is safe to say. \$\endgroup\$
    – mkeith
    Sep 17, 2021 at 3:42
  • \$\begingroup\$ Varies. Shuts down, foldback, hick-up or constant current limit. Do you have a datasheet for it or have you asked the manufacturer? \$\endgroup\$
    – winny
    Sep 17, 2021 at 8:51

2 Answers 2

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Depends on the power supply.

If it's a simple thing, like a transformer + rectifier + capacitor, the power supply will go "Okey dokey!" And give you the current you are requesting. Then, you are in a "race condition" as to whether the power supply will destroy itself, wiring will set the building on fire (e.g. an AC mains source), or an overcurrent protective device (fuse, breaker) will interrupt.

Modern switching supplies will detect the overload and "crowbar" - interrupt power, take some time to reset, power up again, crowbar, rinse wash repeat.

A few power supplies have high internal impedance - think, batteries. So as you increase current draw, their internal impedance will make voltage sag. for instance, this is how cheap solar path lights work with NiCd batteries - if you try to upgrade them to "better" NiMH batteries, the NiMH have too low internal impedance, and the light uses it too quickly.

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Assuming no protections, one of two things will happen:

  1. if your output resistance/impedance is the bottleneck, then as you increase output current, your output voltage will drop because more and more voltage is lost across the resistance which is inherently in series with the output (via components). At some point, the output voltage has dropped so low that it is no longer within specification or tolerance and is therefore no longer considered to be properly producing the rated output voltage. This is similar to you trying to run while carrying something that is too heavy; Beyond a certain weight, you run so slowly that you can on longer be considered to be running.

  2. If heat is the bottleneck then your power supply will overheat and melt/and or explode. This is like you actually having the strength to run at speed, but your body structurally cannot handle it so you break bones or tear muscles.

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  • \$\begingroup\$ Useful analogies! \$\endgroup\$
    – Ahmed Tawfik
    Sep 17, 2021 at 15:59

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