In this heavily upvoted answer the answerer states that it is okay to supply a component with more current than what it's rated for. The analogy is that (paraphrasing here) "If Johnny wants to eat two apples, he'll only eat two regardless of whether you give him three or five, etc."

However, one of the most basic circuits you can possibly make is to power an LED from some power supply. Since most power supplies provide a current that is higher than what most LEDs can handle, you must put a resistor in front of the LED in order to not burn it out.

So which is it?!? Can someone explain to me when/where/how it is/isn't okay to provide higher (and lower, for that matter) current than what a component is rated for?

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    \$\begingroup\$ That question is talking about electronic devices/circuits. It is not talking about supplying singular components with unlimited current. A LED does not know how much current it wants by itself and will just keep pulling current until it blows. A LED with driving resistor is a circuit that knows how much current it wants and will only pull that much from the supply. \$\endgroup\$
    – I. Wolfe
    Commented Jun 12, 2015 at 18:26
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    \$\begingroup\$ You supply voltage (the apples), not current. What the component consumes (current) depends on the circuit (Johnny's appetite). Only then can you determine if the component can handle it or not. All of voltage/current/power need to be considered. \$\endgroup\$
    – user3169
    Commented Jun 12, 2015 at 22:46
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    \$\begingroup\$ Note that talking about components "knowing" the amount of current they need is a very simplified (and somewhat incorrect) view. The real answer is that a constant-voltage power supply supplies an amount of current that depends on the load resistance. \$\endgroup\$ Commented Jun 13, 2015 at 6:28
  • \$\begingroup\$ In addition to the analogy, while Johnny's appetite (LED by itself in a circuit) will let him eat a whole ton of apples, that doesn't mean he should - otherwise he'll end up with a stomach ache (blown LED). Putting in a resistor will limit the number of apples Johnny can eat to a safe amount. \$\endgroup\$
    – tangrs
    Commented Jun 13, 2015 at 11:18
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    \$\begingroup\$ An LED does not have a rated voltage; it has a rated current. So, you should not drive it using a constant voltage supply. (Yes, it has a forward voltage specification, but it is not stable or predictable enough to use it reliably with a constant voltage source.) The problem, I think, is confusion due to a senseless analogy. In reality, different loads have different I-V characteristics, which necessitates the use of constant-voltage, constant-current, or sometimes even constant-power sources, as appropriate. Constant voltage supplies are most common, but not the only type. \$\endgroup\$ Commented Jun 14, 2015 at 9:44

12 Answers 12


To answer the title of your question, the answer is no. It is not ok to supply more current to a component than its rated value.

However, it is ok to have a voltage power supply rated for more current than the components rated value because the component will draw as much as it needs. If you are pushing more current into (forcefully) the component, then the component will exceed its rated value, heat up and be destroyed. Such as if you use a constant current source or you use a large voltage (which will cause more current to flow). But if you use the rated voltage, then the load will only take what is required, regardless of how much current is available to be drawn from the source.

The difference is in how you word your question.

  • 4
    \$\begingroup\$ It is usually okay to have a supply which can output more current than devices expect, but some kinds of devices are only suitable for devices which have current limits. If a typical 0.25A fuse is fed by a supply that will current limit at 10A, for example, and its output is shorted, the fuse will interrupt the current. If the supply is capable of delivering 5,000A, however, and the shorted output can pass that much, the fuse might not be able to interrupt it. \$\endgroup\$
    – supercat
    Commented Jan 3, 2018 at 16:26
  • \$\begingroup\$ @supercat Why would the fuse be unable to interrupt it? Even if the shorted output (+ the rest of the circuit) can take as high as the delivered 5000A, the circuit is still only as strong as its weakest link i.e. the fuse, which should burn up quickly with all that dissipated heat. I suppose arcing could re-establish the circuit if the p.d. were high enough, but that would be afterwards. \$\endgroup\$ Commented Jul 7, 2021 at 1:54
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    \$\begingroup\$ @SoreDakeNoKoto: If a fuse vaporizes rather than melting, the metal vapor can draw an arc. If that arc is able to continue vaporizing more metal, it may take quite awhile before a fuse opens up. Arcs that are powered by DC can sometimes be very hard to extinguish, and the greater the current the lower the effective resistance. \$\endgroup\$
    – supercat
    Commented Jul 7, 2021 at 6:55

Your misconception stems from one incorrect statement: Since most power supplies provide a current that is higher than what most LEDs can handle, you must put a resistor in front of the LED in order to not burn it out.

The reason for the resistor in series with your LED is that if your power supply supplies a higher voltage than the LED requires, and your power supply is capable of supplying more current than the LED can handle, then you must limit the current your circuit draws from the power supply by using a suitable series resistor.

A 5A power supply will not force 5A through anything you connect to it. It will only allow up to a maximum of 5A to flow and how much actually flows depends on the power supply voltage and the effective total resistance of the circuit connected to it.


A simple example: You can have a power supply rated for 5V at 1 billion Amps. Now say you attach a resistor to this supply, lets say 5 Ohms. How much current will it draw? (a) 1A, or (b) 1 billion A?

The answer is (a). Ohms law says that I = V/R. Therefore, if you have a 5V supply across a 5 Ohm resistor, you get 1A current flowing? But what happened to the other 999 million or so Amps? Well there wasn't enough voltage to drive that through the circuit. Now if you had a 5e-9 resistor then you would get your 1billion Amps flowing.

In an LED circuit, the diode is non-linear. This means that as voltage increases, current doesn't increase with Ohms law. In fact it is exponential - an LED could conduct 10mA at 2V, but be able to conduct 1A at 2.1V for example - not usually quite that extreme, but you can see that if we don't limit the current, the LED will undoubtedly blow up. How does the resistor help? Well you can consider the LED to be like an ideal voltage source (not quite true, but bear with me). This example LED is essentially dropping roughly the same voltage at 10mA as it is at 1A, so we say, well hey it always has the same voltage, so if we add a resistor, then the voltage over that will be the supply minus what the LED drops. We can then use ohms law to select a resistor which will drop that voltage at the required current level.

Now the point where current rating of a supply becomes important is this. Say you have a supply which is rated at 5V at 10mA. You connect a 5 Ohm resistor to it. What is the current? (a) 1A or (b) much less?

The answer would be (b). Why? Well the supply simply cannot drive that much current - it could be because of its internal resistance, it could be a current source type supply. Whatever. So what happens is either the voltage at the terminals of the supply decreases (because of say, more voltage being dropped across the internal resistance) or (and) it blows up, melts, burns out however you want to phrase it. The key thing here is if the supply survives and the voltage has dropped, then there is less voltage across the resistor which means there will be less current required to satisfy Ohms law - now this all happens in a very quick transient, so essentially all you see is a 5Ohm resistor with a very low voltage across it.

In terms of the direct answer to the question title, the answer is in most cases No. The rated current is what the manufacturer of the component says it will work correct at.

In many cases it could be a component like an LED or resistor (usually limited by power rating not current, but still...) which given a lack of current limiting or the right supply voltage could easily conduct much higher current than its rated for resulting in excessive heating and/or damage.

In other cases if you apply the correct supply voltage, the device will operate at its required current even if you have a supply capable of sourcing much more than that. This is because all devices are in the end just resistors be it fixed value ones or ones which change resistance with voltage (e.g. semiconductors, transistors, etc.). At the given supply voltage the arrangement of these resistances will operate at a current level they are designed for.


Going with the "eating apples" analogy, an LED will consume every apple you give it until it explodes. It is not capable of limiting itself. You need to provide the LED with a reasonable current (aka, quantity of apples). It's fine to have a million apples stashed away somewhere (the power supply), but you (the resistor) need to stand in the way of the LED to prevent the LED from self-destructing.

A power supply has a voltage and current rating (amongst other ratings). The power supply will normally supply the rated voltage up to the rated current. Just because a 12v power supply can supply 10 amps, doesn't mean that the power supply will force 10 amps through the circuit.

It is, generally, not okay to supply a component with more current than its rated for. LEDs (and all diodes) act like shorts once their forward voltage requirement is met. They will sink all the amps the power supply can give it, just for a very, very short time, and then they become open circuits.


The question you linked is about fixed voltage power supplies. These supplies are typically specified with a fixed output voltage and a maximum load current.

If you use a well chosen current limiting resistor, your power supply can be rated for any maximum current you like and it won't destroy the LED.

For example, you have a 5 V fixed-voltage supply, and an LED that drops 2.0 V.

If you use a 300 ohm current-limiting resistor, the LED-resistor circuit will draw (about) 10 mA, regardless of whether the supply is rated for 100 mA or 100 A.


First let's address supplying more current than a device is rated for. Suppose you have a very simple circuit, with one resistor (100ohm) and one voltage source (5V). Then, by

V=IR 5=I(100) I=0.05A

This circuit will always draw 0.05A of current. It does not matter if the power supply you are using to generate the supplied 5V is capable of supplying 0.05A or 5A of current; because the voltage and resistance are fixed, the current through the circuit will also be fixed. (Though note that it is better practice to have slightly larger power supply than the 0.05A required so you are not running your supply at 100% all the time)

Now specifically for the LED - this component is a diode, and will permit current to flow through it very easily in the forward direction. This means very little resistance to the current. Observe from V=IR that as resistance gets very small, I gets very large. This means that placing only an LED in series with your voltage source will result in a current that is large - probably likely much higher than the LED is rated to handle. This is the reason that a resistor is placed in the circuit - to reduce the amount of current drawn by adding some resistance to the circuit.

However, note also that with an LED and resistor in series with a voltage source, the current through that specific circuit will always be a constant value, regardless of how much additional current the power source is capable of providing.


yes and no.

Some components like LED's allow for pulsed current higher than a continuous rating. An LED rated for 50ma continuous may allow 100ma at a 50% duty cycle at 1khz but not at 0.1 hz .( ie 10secs on 10 secs off). or not 1A at %5 duty cycle.

All cases have the same total average power consumption. Some will work , some will not.


Let's say you have an incandescent lamp rated for 100 watts when it's connected to 120 volt mains.

Since power (P) equals voltage (E) times current, (I) we can write:

$$ P = E \times I, $$

and to find out how much current the lamp is drawing from the mains we can rearrange the formula like this:

$$ I = \frac PE. $$

Then, if we plug in what we know, we can solve for the current through the lamp like this:

$$ I =\frac {100W}{120V} \approx0.83 \ \text {amperes} $$

and we can calculate the lamp's resistance like this:

$$ R = \frac EI =\frac {120V}{0.83A} \approx145 \ \text {ohms} $$

So now we know pretty much everything about the lamp that we need to for this exercise.

\$ \ \$

Next, let's say you've got one of those 120 volt 1000 watt gasoline powered generators.

Since \$\ I = \frac{1000W}{120V}\ \$ it'll be rated to put out a maximum of 8.3 amperes, and if you plug the 100 watt lamp into it the lamp will only take 0.83 amperes because the lamp's 145 ohm resistance will limit the current through it [the lamp] to 0.83 amperes with 120 volts across it.

So now, if you've got a 15 ampere circuit breaker on one of the 120 volt circuits in your house, that circuit can supply 1800 watts without the breaker opening up, so your hair dryer, which dissipates 1200 watts, will draw 10 amperes from the line because it has a resistance of 12 ohms.

Plug your lamp in, though, and because the lamp's resistance is 145 ohms it'll only draw 0.83 amperes from the line even though the line is rated to supply 15 amperes if needed.

Bottom line then, is that even though a source may provide the capability of supplying a great deal of current at a particular voltage, a load will only take what its [the load's] resistance will allow.

  • \$\begingroup\$ Not all loads are resistors or light bulbs. For example, I had an aircraft as a load. It was full of regulated switching power supplies that delivered the same power to internal loads even when there were wide variations in supplied voltage. At low supply voltage the current would INCREASE. \$\endgroup\$ Commented Dec 2, 2021 at 19:55
  • \$\begingroup\$ And another issue is the nature of the power supply: An ordinary power supply can be modeled as a constant voltage source with a maximum current capability, or a constant current source with a maximum voltage capability, or something between those extremes. A smart regulating power supply can NOT be so modeled. Exercise due diligence! \$\endgroup\$ Commented Dec 2, 2021 at 20:01

There are 2 types of power supply. The first, and by far the most common is-

Constant Voltage Source. This will provide, for example, 5 volts. If you short this out with an insulated wire, you [usually] get a spark, followed by a Rather Hot Wire. Oodles of current will flow, and you may even smell burning insulation. Putting a resistor in series will limit the current, just as with your LED example.

Constant Current Source. This much rarer beastie will provide, for example, 1 amp. If you hook this up to a 1000 ohm resistor, you would [in theory] have 1000 volts across the resistor. IN THEORY, having nothing attached [open circuit] would require the power supply to produce enough volts to cause an arc, just to make that 1 amp flow.

Back in the real world, most supplies emulate a constant voltage source. If you try to draw too much current, the output voltage tends to drop down.

In conclusion: Yes, it is OK to have a supply capable of supplying more current than you need. If you are building something yourself, however, make sure it isn't a dead short before you connect it - or you will smell that burning insulation.

  • 1
    \$\begingroup\$ Best answer so far, but you need to extend this idea to loads. Not all loads follow ohm's law. For example, an NE-2 neon bulb in series with a resistor, an electronic system with a switching regulated power supply, or a saturable reactor. Back in the 19th century (I was not alive then) FUZES were invented for situations where the available current was too great. \$\endgroup\$ Commented Dec 2, 2021 at 20:12
  • \$\begingroup\$ richard1941 you can indeed have fancy loads that do odd things. Negative resistance, ionising gas in a flourescent tube, etc. Heck, even lithium batteries get hot and grumpy if you throw too much at them. For those starting out however, worrying about a 2A power supply for a simple blinking LED drawing 20mA, some basic reassurance seems appropriate. \$\endgroup\$ Commented Dec 4, 2021 at 10:36

If it was up to me, I'd supply ALL my circuits with a billion amps, providing I fuse them appropriately. It's up to (my design of) the circuit how much it will actually require. And if it requires too much (think short circuit, bad design, etc.) then that's where the fuse comes in. So, the answer is YES, it's OK to supply more current than what the component is rated for.

  • \$\begingroup\$ You will need some very expensive HRC fuses to guarantee they will interrupt the current if you have a billion amps at your disposal! Normal fuses will just arc over with that level of current and the power won't be interrupted. \$\endgroup\$
    – Malvineous
    Commented Dec 12, 2018 at 11:44
  • \$\begingroup\$ OP stated component not circuit. \$\endgroup\$
    – Geordie
    Commented Jan 15, 2019 at 0:50
  • \$\begingroup\$ @Malvineous Respectfully disagree. It could be a billion amps at 1 mV. Of course that requires a mighty fat wire.... \$\endgroup\$ Commented Dec 2, 2021 at 20:14
  • \$\begingroup\$ @richard1941 Fair point, it may not arc over at that low voltage, but I suppose another issue is that even at 1 mV the fuse could easily explode sending shrapnel everywhere if it blows at well above its rated maximum rupture current, causing injury to anyone nearby. HRC fuses are designed to contain the explosion but finding one that will safely blow at a billion amps without outright exploding will be quite a challenge. \$\endgroup\$
    – Malvineous
    Commented Dec 3, 2021 at 9:15
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    \$\begingroup\$ @Malvineous Ah, thanks for pointing that out. Looks like I need to learn more about HRC fuses. \$\endgroup\$
    – Noji
    Commented May 16, 2022 at 11:54

As an electronics enthusiast for over 60 years (started at 5!), and one who recently did a STEM teaching stint, I think the water analogy nearly perfectly answers this.

Voltage is like water pressure - a Water-Pik has higher "voltage" (pressure) than a huge shallow kiddie pool.

Current is like water flow rate - The kiddie pool has way more water flow per second.

Voltage pressure is applied. Current flow results. The word "supply" is vague; both the water-pik and the kiddie pool "supply" both; Let's take the common used meanings: Supplying a specific voltage means immediately applying it, and supplying a specific current means having the ability to provide that high a current flow rate.

Now to the question under discussion. "Is it really ok to supply more current than what the component is rated for.

Translating, "Is it really ok to have the ability to provide more current capability than what the component is rated for?"

That answer is yes. The extra capability is not used immediately. You can connect a 12v flashlight bulb to a car battery just fine. It may even stay on for a year but it will not burn out.

Final answer: Yes!

  • \$\begingroup\$ Except even if you string a few LEDs together in series to get them up to 12 V, they won't work as well on the car battery as the light bulb does... So it's not always safe to provide more current than the component is rated for! \$\endgroup\$
    – Malvineous
    Commented Dec 12, 2018 at 11:49
  • \$\begingroup\$ Best not to advertise that you were a STEM teacher, serving the needs of the corporate HR people for an expanded army of industrial reserve. \$\endgroup\$ Commented Dec 2, 2021 at 20:05

A more complete analogy --

Jonny likes Apples. Volts = the size of the Apples Amps = the speed at which Jonny eats apples Resistence = The body's natural limitations on how quickly Jonny can eat the apples.

Jonny loves to eat honey crisp large apples Jonny the LED has no Resistence to apples Jonny eats the apples too quickly because he has no resistence and he dies.

Jack loves to eat smaller red delicious apples Jack the body builder has an iron core that forces him to eat his apples slowly No matter how many apples you put on Jacks plate, he will only eat one per minute Jack lives a long life because his resistance protects him.

Jonny should have been given/taught some resistance and his mother Mrs Fuse should have been in the loop to keep jonny safe.

The end.

P.S. LED resistence gets lower as the LED heats up leading to exponential energy consumption until the LED burns up. Even if your LED voltage matches your supply voltage you still need an ohm or two to protect "Jonny"

  • \$\begingroup\$ Awful analogy. But, can you at least spell "resistor" and "resistance" correctly? \$\endgroup\$ Commented Feb 4, 2022 at 15:18
  • \$\begingroup\$ Thank you. It was intended to be awful. Now, feel free to correct if fundimentally wrong. The Boring translation -- Voltage comes from the electrical source. The consumer must be able to handle the voltage. Amperage is not "pushed from the source but is limited by the source. The consumer may consume at any amperage up to the limit of the breaker, fuse, or natural limit of the source. If a consumer like an LED array tends to act like a short because it has no natural resistance, resistance must be added. Don't bother make cracks about spelling I've better things to do with my time. \$\endgroup\$ Commented Apr 27, 2022 at 19:11

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