If a laptop forced to drag out more current than the power supply can provide at the given voltage (i.e. 20 volts), the voltage does decrease, usually approximately anti-proportionally if I am correct.

For a 100W (20V 5A) power supply, it is U[V]=(20×5)/I[A] starting at 5 ampères:

  • 5A 20V
  • 6.25A 16V (voltage dropped to 16V)
  • 8A 12V
  • 10A 10V (might be unrealistic if the conductors of the cable are not strong enough. Maybe it's just 10V 8A or even 10V 5A if 5A is the supplier's ampèrage limit despite voltage drop).
  • 20A 5V (again hypothetical, just for the anti-proportional concept.)
  • 40A 0V (by that time, the short circuit protection should kick in, thus 0V and testing if the short circuit has already vanished by sending short power impulses every few seconds.)

If the limit of 5A at 20V gets exceeded, the voltage will drop for one of these reasons:

  • Voltage drops naturally and the laptop power supply just keeps on supplying 100W.
  • Laptop power supply reduces voltage deliberately instead of naturally, to avoid exceeding 100W of output power.

Which one is the correct reason?

Which one is the reason for the drop of the voltage?

  • 1
    \$\begingroup\$ Under what specific theory do you get these thoughts? The current limit is a design compliance figure. It's not as though the power supply measures the power delivered to the load and uses that measurement to adjust its output voltage. So what's your mechanism underlying your hypothesis? \$\endgroup\$
    – jonk
    Sep 16, 2018 at 21:21
  • 1
    \$\begingroup\$ The power supply has a limit of 5A...it will not provide more current at a lower voltage. If you try to draw more current the voltage may drop or the power supply may simply shut down. The supply is designed to provide a voltage close to 20V and a current up to 5A but it is not designed to provide 100W regardless of voltage or current. \$\endgroup\$ Sep 16, 2018 at 23:28
  • \$\begingroup\$ @ElliotAlderson I have a mobile phone charger rated for 5V 2A, which provides actually 5V 2.6A but also provides 4V 3.1A on too much power draw. Beyond that, not tested yet. \$\endgroup\$
    – neverMind9
    Sep 18, 2018 at 18:16
  • 1
    \$\begingroup\$ The mobile charger you describe behaves as I suggested. You have been exceeding the design limits of the charger. You may be reducing the usable life of the device. If you can draw 10A at an output voltage of 1V for 24 hours, get back to us. \$\endgroup\$ Sep 18, 2018 at 18:31
  • \$\begingroup\$ @ElliotAlderson I actually have a working 5V 1A charger which is nearly 10 years old, and it was often at full load and still works. Right now, I do not have the testing equipment for 10A 1V, but if I do, I might test it on a charger which is exclusively for testing in a secured environment. \$\endgroup\$
    – neverMind9
    Sep 18, 2018 at 19:48

2 Answers 2


Which one is the reason for the drop of the voltage?

This question is a conceptual mix-up. Nothing happens "naturally" without getting out of specifications. Power supplies are not "constant power supplies", they are either "constant voltage", or "constant current". It is difficult to design "constant power" supply because output rectifying elements are limited in current-carrying capability, so supplying high current at low voltages (to maintain constant power) is problematic. And, as explained by user287001, switching-mode PSU don't convert energy 100%, they have timing and other component limits, and "natural" conservation of energy doesn't fully apply.

Practically there are at least four categories of power supplies and "charger adapters":

  1. Constant-voltage power supply with overcurrent protection. This PS keeps its output voltage constant (example: 20V @ 5A ) until the load draw exceeds 5 A. To get 5 A the load must be 4 Ohms. If the load gets smaller, the PS will simply shut down, and a power OFF - ON cycle will be needed to reset the output. There are several lines of protection against burning into smokes, like thermal shutdown etc.

  2. A power supply with constant-voltage and constant-current mode. Say, the same 20V@5A. Again, if the load is light, under 5 A, the PS output is 20V (plus-minus some variation called "voltage regulation"). If the load gets under 4 Ohms the PS enters CC (constant current mode), so the voltage drops but load current stays constant at 5 A. So if the load is as low as 1 Ohm, the output voltage will be just 5V, no more, and the power will be just 25W. Even if you short the load completely, the PS will still drive 5 A at zero voltage on its output terminals. This is how most laboratory benchtop power supplies are designed. Maintaining full-range CV-CC regulation is expensive, so the benchtop PSU are pricey.

  3. A power supply that has some range of output voltages when it turns into CC mode. This is due to design simplifications and by deliberate choice. Most frequent use of this kind of PSU is called "LED Driver". The example PSU will deliver 20V until the load is below 5A, and enters CC mode after that, just as #2. However, these kind of PS have a limited range of output voltages (typically 50% of nominal) where it can maintain the steady current. After that it either shuts down as per #1, or behaves erratically-intermittently. In applications, for most efficient operation the set of LEDs must be rated at 5A total (say, 5 1-A LEDs in parallel), with sum of forward voltages BELOW 20 V, but above 10 V (50% CC range assumed).

  4. A power supply that is designed for supplying/charging mobile devices as laptops and smartphones. They are more like #3, and start dropping their voltage when slightly overloaded. This is a design feature mandated by battery charging specifications. Many smart devices are designed to sense this voltage drop and reduce the consumption, so the PS keeps outputting reasonable voltage. Smart devices can reduce the consumed current down to the point when their battery no longer charges (they can control the charge rate in wide range), and can reduce it further by managing the processor power at the expense of device performance, or balance the two processes (charging battery while having some reduced functionality).

So, to answer your question as you framed it, yes, laptop power supply reduces voltage deliberately, likely with some help from "natural causes".

  • \$\begingroup\$ You explained what I knew much better. Thank you. I marked this answer as accepted. I am very grateful for this answer. \$\endgroup\$
    – neverMind9
    Nov 7, 2018 at 14:56
  • \$\begingroup\$ Dell laptop chargers are type 1 and Lenovo laptop chargers are type 2. \$\endgroup\$
    – neverMind9
    Nov 7, 2018 at 14:59
  • \$\begingroup\$ To me, type 3 and 4 just seem like variations of type 2. \$\endgroup\$
    – neverMind9
    Nov 7, 2018 at 15:01

Laptop chargers seem to use the flyback principle. That is: The voltage in the output capacitor is observed. Every time it seems to be too low (there's a drop treshold well below 1V, say 100mV) a new current pulse is injected to the output capacitor to rise the voltage back to the acceptable level.

Injecting a pulse is a two phase operation:

  • a switch lets an inductor to accumulate current from the rectified mains voltage
  • the switch is turned off, the inductor current continues where ever it founds a way (it will make the way by rising the voltage until insulations break, if the route isn't prepared).

In power supplies the prepared route is to the output capacitor which gets all energy which was accumulated to the inductor during the on-period of the switch.

This system has natural protection against overloading. Too high output current simply causes the output voltage to drop. Your guess "It pushes out max. 100W and drops the voltage if the current is too high" is not at all bad. The switch system cannot inject the pulses too frequently, accumulating the energy to the inductor takes time.

Flyback principle has a drawback. When the output current is low, the needed injections can be so sparse that the pulse frequency is in audio range. That can be heard because the inductors vibrate slightly. One solution is to observe the output current and make the injected pulses smaller to get a reason to output them more often, at an unhearable frequency. Direct current measurement is not a must. Smart controller can decide that now the injections are so sparse that they can be made smaller.

There's also another reason to observe the output current. A short circuit causes all injected energy to be dissipated inside the charger which will cause overheating. Again a direct current measuring is not a must. Indirect reasoning can reveal a short circuit. If repeating injections will not rise the output voltage high enough, the controller can decide "there's a short circuit" and halts the operation.


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