# Does the LM317 voltage regulator have a minimum current output of 1.5 A?

The LM317 voltage regulator is supposed to have a minimum current output of 1.5 A. What does that mean? I am only familiar with the idea of a maximum current output. With this minimum current, will it be safe to use this for replacing AA batteries in a wall clock?

• Where did you find this information about a minimum current? What is very probably meant is that it can handle at least a max. current of 1.5 A. Commented Jan 18, 2023 at 10:15
• You'll find the quiescent current of the 317 and its divider resistors is far greater than that of the wall clock, so it will eat batteries. Only use it with a power supply, or rechargables. Commented Jan 18, 2023 at 10:31
• @cjs that's possible. But I find if I assume stuff, I make an ASS out of U and ME. Commented Jan 19, 2023 at 15:22

The LM317 datasheet doesn't say that the minimum output current is 1.5A, though I can see how you'd read it that way.

Those words are trying to express what the specifications table says:

What that means is that the maximum current you can usually get from the device will be 2.2 A. Due to variations in the parts, you may get up to more or less from any particular device, but you will always be able to get 1.5 A from it.

It does not mean that you would have to at all times draw at least 1.5A from the LM317.

• Your assessment is correct. The recommended operating conditions also hint in this direction: i.sstatic.net/3N2P1.png Commented Jan 18, 2023 at 10:24
• Due to variations in the parts, the average is around 2.2A. You may get even more. But it will be at least 1.5A if the part isn't faulty. Commented Jan 18, 2023 at 13:37
• the "Description" part on the upper right says it quite clearly even though the "Features" list is briefer: it's "capable of supplying more than 1.5 A..." (too bad the red line in the image here partly strikes over that part) Commented Jan 19, 2023 at 6:27
• As one can see from that same chart, the actual minimum current (to maintain regulation) is typically only 3.5 mA. Commented Jan 19, 2023 at 18:32

The minimum current for the LM317 to stay in regulation is about 10mA under all conditions. If you roll the dice, as many do, half that (~5mA) is possible, and it's reasonably safe at relatively low input voltages since typical is 3.5mA. That can be achieved by using relatively low-value divider resistors (the current drawn from the regulator 'OUT' pin with no load is 1.25V/240Ω = 5.2mA):

Thus you can draw from 0mA to 1.5A or so (briefly in the case of high current) from the Vout pin. The schematic below shows resistor selection suitable for a single AA cell output voltage. If you have more than one in series adjust R2 according to the datasheet to get more output voltage (though the LM317 has a pretty high dropout voltage - 3V is recommended though it will work with about 1.5V under typical conditions) so you may have to increase the input voltage.

The 'minimum' 1.5A on the datasheet tells you that's the minimum current the LM317 is guaranteed be able to supply under the stated conditions. It's a capability, not a requirement.

simulate this circuit – Schematic created using CircuitLab

Of course you should measure the output voltage before connecting it to something you care about, and if something in your circuit fails you may end up destroying the clock electronics. That's not possible with AA batteries unless you find a way to put more in series than the designer intended.

You can set the output voltage by resistors. It's not a constant current source, it's a constant voltage source. 1.5 A is max. Maybe you have an old version of the datasheet.

• The datasheet shows how to configure the LM317 as a current source. Commented Jan 19, 2023 at 7:27