Current limited power supply and Ohm's law

[uby]

Suppose I have some resistor with resistance R. Using a typical power supply, application of a voltage difference V across the resistor leads to a current draw of I = V/R according to Ohm's law (so long as the only circuit element is a resistor without any complex impedances).

Now, suppose I have a current limiting power supply that can generate the same voltage difference V, but whose maximum output current is set to some fraction of I (let's call it f*I, where 0<f<1). Does this continue to obey Ohm's law? Can the full voltage potential V be placed across the resistor using such a power supply, or will the maximum voltage that can be applied be limited by the same ratio f*V in order to satisfy Ohm's law (f*I = f*V/R)?

Thanks!

 

[Drakkith]

I'm not sure, but I don't think it works quite that way. The circuit involving a current limited power supply and a resistor needs to take into account the power supply as well. Also, the only reason you should reach the limit to on the current in your circuit is if a short develops and bypasses the resistor, or if the resistor was simply to low in resistance compared to what the circuit was designed for.

But I really don't know, that's just from a few minutes looking it up here: http://en.wikipedia.org/wiki/Current_limiting

 

[berkeman]

Suppose I have some resistor with resistance R. Using a typical power supply, application of a voltage difference V across the resistor leads to a current draw of I = V/R according to Ohm's law (so long as the only circuit element is a resistor without any complex impedances).

Now, suppose I have a current limiting power supply that can generate the same voltage difference V, but whose maximum output current is set to some fraction of I (let's call it f*I, where 0<f<1). Does this continue to obey Ohm's law? Can the full voltage potential V be placed across the resistor using such a power supply, or will the maximum voltage that can be applied be limited by the same ratio f*V in order to satisfy Ohm's law (f*I = f*V/R)?

Thanks!

Drakkith's wikipedia link is good. There are several types of current limiting strategies that are used in different power supplies. The simplest is that the voltage only rises to where the maximum output current is flowing, and does not rise any higher.

But that strategy can lead to some over-power problems for a linear regulator based power supply, so linears often use "fold back" current limiting, where when the limit output current is reached, the current limit is decreased to a small value, meaning that the voltage drops even more. The overcurrent condition has to be released before the linear will raise its output voltage again.

And with DC-DC switching power supplies, you can have time-dependent current limits, where the supply tries to "burp" the output voltage until the overcurrent condition is released.

Probably more info than you wanted, though. The basic answer is that Vout = I(limit) * R for a simple linear regulator that is in current limit.

 

[Antiphon]

I do this every day. Ohms law will be obeyed.

The current limiting supply will reduce it's normal output voltage for the limited current.

A resistor has no choice; ohms law is always obeyed without exception.

 

[berkeman]

I do this every day. Ohms law will be obeyed.

The current limiting supply will reduce it's normal output voltage for the limited current.

A resistor has no choice; ohms law is always obeyed without exception.

<hijack>

Never anthropomorphise resistors. They hate it when you do that.

</hijack>

 

[Drakkith]

<hijack>

Never anthropomorphise resistors. They hate it when you do that.

</hijack>

Yeah they really get heated when you do that!

 

[berkeman]

Yeah they really get heated when you do that!