0
\$\begingroup\$

I have a project that requires an odd voltage of 15VDC. The power pack that I plan to build will be two 12VDC 7AH SLA batteries with a DC-DC converter.

So I know that my starting conditions will be:

  • Input volts: 24V
  • Input capacity: 7Ah
  • DC-DC converter (LM2596): 24VDC to 15VDC
  • Load: intermittent (6-12 times/day) 15VDC, ~1 amp, ~30 second duration

How does the DC-DC converter factor into the load calculations to figure out estimated battery life?

How do I figure in losses from the converter?

Supplemental information:

I know that running the same system at a reduced 12VDC results in about a 48 hour runtime on a 12V SLA 1.3Ah battery.

\$\endgroup\$
  • \$\begingroup\$ Consider watt-hours, rather than amp-hours, and think about the efficiency of your buck converter. \$\endgroup\$ – Hearth Feb 8 at 14:18
1
\$\begingroup\$

There is an efficiency curve in the datasheet that will give you a general idea of what to expect: enter image description here

At 24V in and 15V out with a 3A load, it looks like the DC to DC will be about 92% efficient. You can directly multiply the available Wh rating of the battery to 0.92 to get the number of watt hours you will deliver to the load.

If you are using a cheap Chinese board, the parts might be more or less ideal than the sample circuit in the datasheet, so you will need to measure the input and output current to get the exact efficiency for your system.

Note that the LM2596 is a non-synchronous buck converter, you can probably get another 5% or so efficiency by switching to a synchronous converter.

Also remember that if you are 92% efficient, you are converting 8% of your input energy into heat at the converter (24V * 3A = 72W, 72W * 0.08 = 5.8W of heat).

Edit: BTW, since your device is operating at a low duty cycle, you need to consider the Quiescent (idle) or disabled power consumption. Like most DC/DC converters, the LM2596 has an on/off (or enable) input, shutting down the converter using this input will use significantly less idle power than simply disconnecting the load from the output. This diagram (from the datasheet) shows the difference between the two options:

enter image description here

If you are operating at less than 100% duty cycle, then you are consuming the idle/quiescent power when off and consuming load + switch dissipation power when on.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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