# External battery for laptop with 18650

I would like to try to make an external battery for my laptop. I opted for 18650 batteries. (but feel free to tell me if there are better batteries for that).

I read this answer to a question here: How to build an external lithium battery pack for a laptop with 19V DC input?

Now, I have a question: Can I take eight 18650 batteries with cut-off at 2.5V?

Minimum: 8 x 2.5 = 20V => good to use only a buck converter! But if regulation fails, my laptop dies.

I'm surprised cause Russel said:

### 6 or more cells with a linear regulator:

• Not preferred but may be bearable.
• 75% to 98% efficient across falling battery range.
• Regulator failure MAY destroy laptop.

But I though it was the least efficient of the regulators!

So, which is better? Buck or boost regulator? The output current will be 2 to 3A.

Can I add a Schottky diode to every battery to decrease current drawn from each cell? Does anyone have a reference for a 2-3A diode with the smaller forward (?) voltage?

• "(but feel free to tell me if there are better batteries for that)" Considering that the old laptop batteries I chopped up gave me lots of those, I think you're fine. – Ignacio Vazquez-Abrams Aug 9 '14 at 23:53
• Ok, thank you ! I'll choose between 18650 or li po raw... – Alexis_FR_JP Aug 10 '14 at 5:08

2.5V is not a realistic cutoff voltage. A good 3.7V li-ion cell will stay above 3.2V until ~90% discharged, so 6 cells in series should be enough. At full charge a 6 cell battery puts out 25.2V, but under load it will rapidly drop below 24V. Therefore the practical range of voltages available is ~19-24V.

A linear regulator has to dissipate up to 15W with a fresh battery, so it will need a good heat sink. The power loss at half charge and average current may only be about 5W ((3.5V*6-19V)*2.5A) which corresponds to 90% efficiency. To get the full use of a 6 cell battery pack the regulator needs to be an LDO (Low Drop Out) type to minimize voltage drop towards the end of the discharge.

Putting diodes in series with the cells will reduce the pack voltage (not current) but wastes power just like a linear regulator. Better to just use a high power regulator and let it do all the work.

A step-down switching regulator should be more efficient when the battery is above half charge, and could allow use of a higher cell count to improve run-time (at higher voltage the switching regulator will draw less current from the battery). Since suitable Buck modules are readily available, this is probably your best solution.

• Example for the 2.5v cut-off : [industrial.panasonic.com/www-data/pdf2/ACA4000/… Yes but my experience shows that more the diode allows current, more the waste voltage is high. But ok, no diode. Ok, I think it will be more efficient with a buck/boost converter. (if I find the good one for my project). I've edited my post. – Alexis_FR_JP Aug 10 '14 at 4:54
• If each cell in that battery unit is 9000mAh then the total for two cells in series is 9000mAh @ 7.4V, not 18000mAh. When voltage is boosted to 19V the regulator must draw proportionally more current from the battery to get the same power, so this '18000mAh' battery is only equivalent to a 6000mAh 6 cell battery. To produce 19V at 3.5A it has to draw over 10A from its battery. A good boost regulator should be able to handle that without a heat sink, but larger inductors and thicker wires are required to handle the higher current. – Bruce Abbott Aug 10 '14 at 10:02