I have a 12V 18Ah battery I wish to run through a boost converter at 40v and 2.5A to power a 100w led.

Could someone figure out how long my battery would power my led and show the calculatins/formulas.


  • 2
    \$\begingroup\$ Besides that its impossible to do accurately without knowing the efficiency of the boost converter, why didn't the gazillion questions that handle the same topic here answer the question already? \$\endgroup\$ – PlasmaHH Feb 10 '16 at 10:00
  • \$\begingroup\$ @PlasmaHH - that is essentially the same question and set of good answers. This question is a duplicate, and so should be closed. \$\endgroup\$ – gbulmer Feb 10 '16 at 10:44

Two methods:

Amp-hour calculation

Use formula

$$P = V \cdot I$$

For your load the power consumed is \$P = 40 \cdot 2.5 = 100~W\$ as you correctly identified.

If your boost converter is 100% efficient it will consume 100 W. Therefore \$I = \frac {P}{V} = \frac {100}{12} = 8.3~A\$.

Run time is then \$ \frac {18~Ah}{8.3~A} = 2.16~h\$.

Watt-hour calculation

This is simpler. Your battery can supply \$12~V \cdot 18~Ah = 216~Wh\$. You have a 100 W load so you could run it for 2.16 h (if you had a 100% efficient booster).

Correct these calculations with the efficiency of the boost converter and add in some safety factor for deterioration of battery over time.

See @SteveG's comment below regarding de-rating for fast discharge.

  • \$\begingroup\$ Presuming that this is a sealed lead acid battery, the 18Ah capacity is quoted for a 10h discharge. If you discharge it in 2h then the capacity is around 12.5Ah. The run-time becomes 12.5/8.3 = 1.5h. See the Discharge Characteristics in yuasabatteries.com/pdfs/NP_18_12_DataSheet.pdf \$\endgroup\$ – Steve G Feb 10 '16 at 10:28
  • \$\begingroup\$ I forgot to include that. I've added a pointer to your comment. Thanks. \$\endgroup\$ – Transistor Feb 10 '16 at 10:35

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