# How can current-limiting supplies be possible?

Knowing Ohm's law, the current drawn is directly proportional to the resistance. In some power supplies, you can limit the current. Let's say the voltage is 12 volts, it supplies 1 amp, and the resistance is 8 ohms. The current going through should be 1.5 amps, but the supply can only deliver 1. How can this defy Ohm's law, and limit the current?

What you do with a current limited power supply is set to maximum properties. The power supply will regulate its output voltage in such a way that the lowest condition is met.

I drew a graph for how a current limited power supply will act with varying resistance connected. Consider a 12V power supply limited to 1A.

• The X-scale varies the load resistor from 1 to 30Ω;
• The left Y-axis and red line represents the ouput voltage of the power supply. Clearly when the resistor is too low, the power supply output voltage drops, to fulfill the 1A max output;
• The right Y-axis and green line represents the output current. Once the maximum output voltage is reached the output current will drop inversely proportional with the resistor value.
• You can clearly see voltage regulation taking over from current regulation at R=12Ω

The name current limited power supply is a little misleading, because such a supply is also voltage limited. It delivers as much power as possible, without exceeding either the voltage limit or current limit.

If it's connected to an open circuit, it will be limited by voltage, because it takes no current to develop a voltage over an open circuit.

If it's connected to a short, it will be limited by current, because it takes to voltage to develop current through a short.

If it's connected to something else (0 < $R_L < \infty$), then it could be limited by current or voltage, but it's much more common to design circuits to be powered by voltage sources than current sources.

• Nearly all practical power supplies are not only going to have some limit to the voltage they can output, but will also be constructed such that the voltage under zero-current conditions will be somewhat close to the voltage under expected-current conditions, and certainly not so great as to damage the supply. Thus, "voltage limited" is a bit redundant. By contrast, many power supply designs will output many times more current under short-circuit conditions than under recommended operating conditions--quite possibly enough current to overhead and/or damage the supply in short order. – supercat Jun 22 '13 at 18:45
• The fact that a supply claims to be current-limited doesn't just mean that there's some finite limit to the amount of current it will supply (just about any supply could claim that), but also that the current limit is predictable, and that the supply is designed so that it can be operated continuously under current-limiting conditions without overheating or damaging itself. – supercat Jun 22 '13 at 18:47
• @supercat sure, i'm just pointing out that a supply that has adjustable voltage and current knobs, like most lab supplies, aren't voltage supplies any more than they are current supplies. There is a symmetry to them that isn't always obvious, mostly because of the terminology we use. It's a current-limited voltage supply as much as it is a voltage-limited current supply, depending on how you turn the knobs, and what load you connect. – Phil Frost Jun 22 '13 at 19:33

You can't violate Ohm's law. Under that situation, the voltage at the output would drop to 8 volts. If you were to continue reducing the load impedance, at some point something will go pop, and it will fail. A specification like 12V output is only valid under certain operating conditions, like a load of less than 1 amp. Outside that range, the voltage will not (and cannot) stay constant.

Power supplies often include "foldback" current limiting, so that with a low-impedance load, the voltage will drop enough (in a controlled way) so that the output power is reduced enough so that nothing pops. When the load goes back within the specified range, the supply will operate normally again.

Measure the output voltage: you'll find the supply has reduced the output voltage such that the output current is at or below the target value.

So the output voltage should be:

$$1A \cdot 8 \Omega= 8V$$