I used an LM317 in my circuit to get a regulated 1.5 volts output from a 3.7 volt battery using two resistors (220 and 60 ohm.)

It worked fine, but the circuit dissipated too much power even when not in use. After a day, the battery was empty (with no load.)

What is wrong with my circuit or the LM317 IC? If this is natural, what alternatives can I use to get a regulated 1.5 volts output from a 3.7 volt battery with no or very less power dissipation?

The project is simple, an input of around 4 volts to the LM317 which uses R1 and R2 (220 and 60 ohms respectively) to step down the voltage to 1.5 volts. On the other end, there is a small load that uses about 100 mA. The battery was empty the next day while the load was not yet connected.

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    \$\begingroup\$ Usually you'd use a switching regulator with a battery since that lets you step down voltage without having to waste it as heat. For a small load a linear regulator might be ok. How big is your load? How big is your battery? \$\endgroup\$ Commented Aug 22, 2021 at 15:44
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    \$\begingroup\$ 1.5V/60Ohm= 22.5 mA. Just divider consumption. Not good choice for battery operated device. \$\endgroup\$
    – user263983
    Commented Aug 22, 2021 at 15:56
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    \$\begingroup\$ Minimal load required 3.5 mA. You may increase voltage divider resistances. \$\endgroup\$
    – user263983
    Commented Aug 22, 2021 at 16:04
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    \$\begingroup\$ If you wire it up like that, R1+R2 are part of the load, so use large values. With 220 and 60 ohms, you're using ~130 mah per day just heating the voltage divider. Make them at least 10 times bigger. \$\endgroup\$ Commented Aug 22, 2021 at 17:07
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    \$\begingroup\$ Please don’t use LM317 for battery applications in 2021. There are LDOs with four orders of magnitude lower self-consumption these days for pennies. \$\endgroup\$
    – winny
    Commented Aug 22, 2021 at 18:48

2 Answers 2


With a 220/60 ohm voltage set network, you are pulling about 5.3 mA with no load. Add in another 0.1 mA for the current from the ADJ terminal, and you are talking 5.4 mA. Over 24 hours, this will draw a total of about 130 mA-hr from the battery. If your battery was not fully charged, or possesses a very low effective capacity (this happens as batteries approach the end of their lifetime), this would explain things.


LM317 are linear voltage regulators.

This type of regulators are good at:

  • Stepping down voltages
  • Being simple to use
  • Having a low voltage ripple output

The problem with linear regulators is that they have a terrible efficiency. This is because the current input = current output.

If you want to use a regulator with greater efficiency, then you need to use a switching regulator (probably a buck converter in your case.)

They can:

  • Step down and step up the output voltage depending on what kind of regulator you are using
  • Achieve a better efficiency (somewhere to around 70% to 80% on average)

These are better suited for projects where battery life matters, but these converters are noisy. So you have to be careful on which projects you use these on. If you're doing something like powering a microcontroller, then you'll be fine.

I would also like to know more about your project, and if there is a way to reduce the current consumption of your circuit.

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    \$\begingroup\$ Linear voltage regulators don't have a terrible efficiency when running light compared to a switching converter. In fact, when running into a light load they may sometimes have a better efficiency. \$\endgroup\$
    – Andy aka
    Commented Aug 22, 2021 at 16:08
  • \$\begingroup\$ @Andyaka It is depending of application. Minimum load is required. \$\endgroup\$
    – user263983
    Commented Aug 22, 2021 at 16:22
  • \$\begingroup\$ See update to the question. [Edited by a moderator.] \$\endgroup\$
    – Asmat Ali
    Commented Aug 22, 2021 at 16:42

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