What Do I Do If My Power Supply’s Voltage Drops Under Load?( 'Takes about 9 minutes to read. )

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(This article has been edited and re-published, having been first published on my old website.)

You may be asking: “Why is this geeky article on a musician/entertainer’s website?”

Well musicians & entertainers use computers, and computers belonging to musicians & entertainers get as many faults as any others. – ‘No more need be said.


I assume that either you’ve checked your power supply’s output in one of or both of these two ways. – Those being: –


  • a) While the PSU’s outputs are unconnected, and also while the PSU is running under an artificial load.


OR you preferred to leave the PSU fitted and tested it: –


  • b) While the PSU is connected to the computer’s components in a normal way.


The question “What do I do if my power supply voltage drops under load?” is a bit of a multi-pronged question. I’ll try in this article to cover a number of those points in the hope that you can come to a conclusion based upon the information and guidance herein.

I’ll start by saying that, in the case of any supply of power, the voltage will always drop to some extent under sufficient load. To clarify that point somewhat, I’ll draw your attention back to Ohm’s Law and the relationship between wattage (Power), current (Amperage), voltage, and resistance (The load):

If the wattage or power (P) = the current (I) squared multiplied by the load resistance (R) in a DC circuit; then it follows that R divided by I squared = P.

If we give I the fixed value of 10 amps in this example, then we can see that as R decreases, so does P. : Translated into English, this shows that as the load increases and therefore the parallel load resistance decreases, so the amount of unused power available from the supply also decreases. Why does this occur? Because as the load increases so the resistance decreases, therefore more of the available electrical power in watts is used by the load and transformed into other kinds of energy.


Ohm's Law


We know from Ohm’s Law that V = IR (Voltage = current multiplied by resistance.). Therefore, if we use the fixed figure of 10 amps as the value for I again, we note that as the value of R (The resistance) increases then so does the value of V (The voltage.). Inversely as the (load) resistance drops as extra load is added to a DC circuit, so the voltage drops proportionally.

For those of you who don’t understand the connection between load and resistance in a DC circuit; let me put it to you this way, in terms of extremes:

Connect wires to the terminals of a DC power source and just leave it at that… That is theoretically zero load because there is infinite resistance between both poles. (In fact it’s not infinite resistance; it’s probably somewhere in the region of half a billion ohms or more; but that’s as good as… ) As a result there is no load and no electrical power is used.
Now let’s go to the other extreme; (*Don’t actually do this.*) if you touch the wires together – meaning there is ZERO resistance between the poles of the DC power source – then all of the available power flows in the circuit at once – and as a result the circuit (consisting of the two wires only ) can get very hot and catch fire, usually something goes bang if there is a decent amount of electrical current available, the power-source gets overloaded and drains very quickly if it’s a battery, or malfunctions and/or explodes if it’s another type of power source.(Unless it has some kind of semiconductor-based short-circuit protection on its output.)

All electrical circuits, no matter what type of electrical circuit it is, have a resistance to the flow of electricity, which is measured in OHMS and denoted by the Greek letter Omega. The lower the resistance of the circuit in ohms the greater the load, the more electrical current flows – therefore the greater the potential voltage drop. – Does that make more sense to you now?

The higher the amount of current (I) available in the circuit; the less the decrease in R (Increasing the load.) affects the decrease of V. (Drop in voltage.)

We can prove that by increasing the value of I to 20 amps: Do the math again and you’ll see that this is true.


Conclusion: The greater the electrical load on a given supply; the greater the voltage drop on said supply.


This is always true to a given extent; no matter how much current is available.

Bearing this in mind, then; how much voltage drop is too much?

In a perfect world; any voltage drop is too much. – But as we’ve seen above; there can never be zero voltage drop unless there is zero load. As we know; if there is zero load then there is zero computer, so the point is moot. The objective of the exercise then is to minimise the voltage drop to the greatest extent possible.

In the examples above we saw that the more current we had at our disposal, the smaller the voltage drop under any given load: In Joule’s Law; I x V = P (Current multiplied by voltage = power (wattage).). We know that the voltage outputs of a computer power supply are fixed at 12v, 5v, and 3.3v. therefore the available current (Amperage) is dependent upon the available power (Wattage): The more wattage the supply can output the more current is available; therefore the less the voltage drop will be under any given load.


Conclusion: To minimise any voltage drop under any given load conditions; make more wattage (power) accessible.


In other words; if the voltage on your supply rails drops substantially under the load that your computer’s components put on those supply rails, then your power supply isn’t outputting the required wattage.

In the face of that; you have two choices. Either: –


  • 1) Keep your existing power supply unit and decrease the load on it. (- By not running so many components: In other words removing something from the circuit, such as a hard-drive, graphics card, whatever.)


  • 2) Fit a new power supply capable of delivering the required wattage.


At this point you may say:

“I’ve checked everything and worked out that the load on whatever rails is X amps; that’s Y watts at Z volts. My power supply’s output rating on whatever rails is greater than that: Therefore something’s wrong.”

Correct. Something is indeed wrong. There is an inconsistency somewhere. The problem now is to discover exactly what that inconsistency is.


It could be three main things: –


  • 1) The equations are wrong.


  • 2) Your calculations have arrived at incorrect figures.


  • 3) The PSU ratings are wrong.


Firstly there is no chance that the equations are wrong. The equations used form the basis for modern electronics theory and have done so for more than 100 years. They are an integral part of basic electronic calculus and have been proven to be correct time and time again.

Secondly, check that your calculations are correct and that you have indeed arrived at the correct figures.

If there is no fallacy in your calculations then there is now only one possibility left – That being that the PSU ratings are wrong. How can this be?

Some manufacturers; especially the manufacturers of cheapo power supplies, overrate their product. This isn’t actually lies though. It’s an overstatement:


Computer Power Supply UnitA PSU rated at 500 watts is indeed capable of supplying 500 watts; but not necessarily 500 watts of continuous power. In tests done by a leading UK computer magazine in the last 2 years, a number of 500 watt PSUs failed when fully loaded. One, the second cheapest under test, even detonated! There is an old saying that you get what you pay for. Sure; a $15 PSU built in China and rated at 500 watts will supply a computer that needs 300 watts to run efficiently without (m)any problems. It’s when you add a powerful graphics card that drains an extra 100 watts, plus a RAID array = another 30 watts, extra optical disk =  +10 watts, and upgrade to a more powerful processor = +35 watts that things start going askew: BSOD s occur, the graphics card doesn’t perform properly, the power supply starts making strange noises…

– That’s only 475 watts total, and the PSU is rated at 500 watts, so theoretically it should work; but it doesn’t. – Because the PSU can provide 500 watts peak, – but not 500 watts continuous. – In fact probably only 400 watts or less continuous. At peaks where the power requirement is at its greatest there may be a voltage drop of over a volt causing memory failures and processor outages.

Maybe it’s not that serious in your case? Maybe you’re on the edge of the precipice; but are still noticing a voltage drop on measurement under load, even though the PSU is still holding up, the CPU’s not having outages, and the RAM is just managing on the reduced voltage?

That’s a good thing inasmuch as your computer is still working. – But it’s so close to the wire that it’ll only take a feather to tilt the balance. – So don’t expect your computer to carry on without problems much longer in those circumstances. All you need is for the PSU to get too hot, the CPU to use a few watts more, the graphics card to draw that extra piece of polygonal shading, and…

Get a better quality and / or higher-rated power supply unit: The sooner the better. Once it’s fitted, measure the voltage drop: There will still be a voltage-drop; but it’ll be smaller and your computer will be happier as a result.

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