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?