I have a 5V power supply with DC output rating: 5V @ 1500mA. It could power my High Torque servo very well.

I, however, have a fairly used 9V Alkaline battery. I shorted this battery and I measured the current to be ~3 Amps using a multimeter.

Later on, I connected this battery to a 5V regulator to power my High Torque servo, and it doesn't seem to be drawing as much power as my 5V DC power supply.

My question is, what does 5V @ 1500mA mean exactly? Is 1500mA the maximum current it draws or the minimum? If it's the maximum then my 9V battery seems to produce more current than the power supply, but that's not the case.

And why are power supplies not equally as powerful even though the voltage measure maybe similar.

My guess is that it all boils down to less internal resistance produces more current, and my 5V DC power supply has less internal resistance.

Is the internal resistance of my 5V DC power supply then: 5V / 1.5 A ?

Or maybe this doesn't even have anything to do with internal resistance.

  • \$\begingroup\$ Is 1500mA the maximum current it draws ... .... a power supply does not draw current, it supplies current ..... in your case, the power supply will supply up to 1500mA (1.5A) before it starts to overload and shuts down or the output voltage starts to drop \$\endgroup\$
    – jsotola
    Jul 12, 2018 at 23:33

2 Answers 2


Your battery when shorted produced zero volts and thus zero power, but internal losses were 27 Watts but suggests it has an ESR of Voc/Isc= 3 Ohms. So at 1.5A it will have an output of 4.5V

Meanwhile your DC supplies 5V @ 1.5A or 7.5W but you don't say what the voltage load regulation error is. Let's say it is 1% or 50mV . This would suggest the regulator output impedance is 50mV/1.5A= 33 milliohms .

Any questions.

So what did you learn?

ESR= ΔV/ΔI For both linear unregulated batteries and regulated sources in the linear range ( not short circuit protected range)
P max = I max @ Vout.
A 9V battery can generate 27 Watts of short circuit internal heat but zero output power. (warning)

But now you can estimate how much current any battery supply with say a 10% drop in voltage using the ESR of the battery.

What you didn't ask is that every battery has memory, some less than others. As if a bigger capacitor with a larger ESR restores the battery voltage after a temporary short circuit. (double layer electric effect) So in effect R1C1//R2C2 storage in the battery above the 0% SoC voltage. ( look up what you dont understand )

Also the max a battery "should" supply depends on ESR and more importantly the junction internal temperature rise. THis is often defined by the C rate for some temperature rise.

Any questions?


The 5V power supply is regulated, meaning that its internal circuits will hold the output voltage at about 5V for any output load current up to 1500mA. It's really not a matter of having less internal resistance, it has feedback circuits that maintain the desired output voltage.

The 9V battery is unregulated. Sure, you may have measured 3A when you shorted the battery with your meter, but the output voltage from the battery was very small. The internal resistance of the battery is a factor, but physical size is also important. A battery provides power from a chemical reaction, and the more chemicals you have available to react the more current you will get.

  • \$\begingroup\$ To put it more bluntly, with the battery shorted, its output voltage was zero. Since power is volts times amps, it was providing zero power to the load (the short). \$\endgroup\$ Jul 12, 2018 at 23:51
  • \$\begingroup\$ @WhatRoughBeast Agreed, but I didn't want to complicate matters by getting into a discussion of the burden voltage of the meter. I was afraid that if I said the voltage was zero someone would complain that it wasn't exactly zero. \$\endgroup\$ Jul 12, 2018 at 23:55

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