# Can LDO Voltage Regulators be driven at max current output?

1. Jul 30, 2014

### aiq25

Hi. I was wondering if there is anything wrong with running LDO Voltage Regulators at max current output for extended periods of time (>12 hrs). The reason I ask is because I have a application where I have to drive a load that is less than 5Ω with a constant current of 500mA at 5V. I could use a resistor to limit the current but I was thinking if I could use a LDO Voltage Regulator and not use a current limiting resistor. With a 5Ω load at 5V I get a current of 1A but I need 500mA, so couldn't I use a LDO Volt Reg with a internal current limit of 500mA? I don't know much about voltage regulators, that's why I'm asking. The input voltage to my system is 6V, so I don't think the power dissipated in the voltage regulator would be that big of an issue. I'm willing to use a heatsink for the voltage regulator for better heat dissipation.

I could use a current source but I'm curious if this method will work or is there a problem with it.

2. Jul 30, 2014

### Staff: Mentor

Per Ohm's Law, you can set 2 out of the 3: V=IR. You cannot put 5V across a load of "less than 5 Ohms" and get a current of 500mA.

"So which is it you want, young feller?" (Quiz Question -- what movie is that line from?)

3. Jul 31, 2014

### Baluncore

500 mA through 5 ohm will drop only 2.5V across the load.

There may be no need for a regulator if your 6V is sufficiently stable.
You could drop the 3.5V with a (3.5V/0.5A) = 7 ohm series resistor. A difficult value to find.
It will need to be rated at 0.5A*3.5V = 1.75W.
You might just use 7 series 1 ohm ¼ watt resistors. Each will drop half a volt and dissipate 0.25W.

LDO linear regulators are only as efficient as a series resistor.
You could use a +5V LDO with a 5 ohm series resistor to your 5 ohm load.
From 6V only one volt will drop across the LDOR so it will dissipate 0.5A*1V=1W.

You could use a variable regulator such as a LM317. Two small resistors will set the Vout to 2.5V.
An LM317 will drop (6V – 2.5V) = 3.5V, it is linear so it will dissipate 0.5A*3.5V = 1.75W

Alternatively, use an adjustable switching regulator. It will be more efficient.
It will only draw about 250mA from the 6V rail to give 500mA out at 2.5V.
http://www.dx.com/ have a product SKU 280240 which will do the job efficiently without a heatsink.
http://www.dx.com/p/hzdz-adjustable-step-down-buck-module-blue-3a-280240# US$2.52 postage is free. Alternatively; SKU 255394 for US$1.99

4. Aug 20, 2014

### analogdesign

To directly answer you question running an LDO at maximum rated current indefinitely should be possible (unless it is explicitly forbidden in the datasheet)