I have a DC-DC buck step down converter module LM2596 voltage regulator set to supply 3.0 volts to an atomic clock using a 8 Ah 12 V battery. The clock normally uses 2 AA batteries.


The battery started out at 12.07 V. After 10 days, it's at 8.45 Volts.

It does seem like it should not have dropped that much.

It claims static power: 20 ma. Does that mean power used?

The converter has a digital voltage readout. Does it use a lot of current?


closed as off-topic by Marcus Müller, RoyC, Finbarr, Bimpelrekkie, Nick Alexeev Apr 29 at 21:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Marcus Müller, RoyC, Finbarr, Bimpelrekkie
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    \$\begingroup\$ It should be quite obvious that if a device has 3-digit digital readout with 7-segment LEDs, it should draw quite measurable current. Assume 1 mA per LED, and all 21 segments can easily eat 20 mA from input power source. \$\endgroup\$ – Ale..chenski Apr 27 at 22:33
  • \$\begingroup\$ You should post this question to the eBay seller. EE.SE isn't his volunteer application support. \$\endgroup\$ – Nick Alexeev Apr 29 at 21:57
  • \$\begingroup\$ Rather than guessing, measure the current. Also realize you've killed your battery, potentially permanently. \$\endgroup\$ – Chris Stratton Apr 30 at 15:39

It looks like you have bought a converter that uses at least 20mA, regardless of what load you put on it. Some part of that will be driving the voltage display.

Roughly speaking, that converter would drain an 8AH battery in 400 hours, or 16 days, even if nothing else was connected.

  • \$\begingroup\$ Thanks. I found a switch that turns the led display off. That should extend my runtime. :-) \$\endgroup\$ – fixit7 Apr 27 at 22:48
  • 2
    \$\begingroup\$ The display and A/D chip are likely fed by a linear regulator too. \$\endgroup\$ – Jack Creasey Apr 27 at 22:49
  • \$\begingroup\$ @Jack Creasey How would that apply to my situation? \$\endgroup\$ – fixit7 Apr 28 at 0:03
  • 2
    \$\begingroup\$ @fixit7 They draw constant power not related to the DC-DC convertor task. I imagine your display has at least 11 segments on for 3V. Even turning off the display may not remove as much of the current as you think. The LM2596 alone requires a quiescent current of 5mA+ in addition to requiring 3A pulse current, and then whatever chip and linear regulator is used for the display. \$\endgroup\$ – Jack Creasey Apr 28 at 2:21
  • \$\begingroup\$ Good info. It seems like converting 12 V to 3 V is not very efficient with what I have. I did notice one of the components got pretty warm, so a lot of loss is through heat. \$\endgroup\$ – fixit7 Apr 28 at 2:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.