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I have a battery setup of 3s4p (12V) with the batter of 20A high drain, it has bms with it. It works perfectly if i put it on my inverter (2500w, 12v) and power laptops or monitor screen. But when i use it for computer or vacuum, the inverter doesnt seem to handle it. Am i missing something? could it be my battery cant handle the surge power?

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closed as off-topic by Chris Stratton, RoyC, Dave Tweed Jan 16 '18 at 5:53

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." – Chris Stratton, RoyC, Dave Tweed
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    \$\begingroup\$ This battery pack is far too small to allow your inverter to run anywhere near full power. Work the math: 3s is anywhere from 10.8 Vdc to about 12 Vdc. Assume the worst: call it 10.8 Vdc. Multiply that times the maximum current: 10.8 * 20 or about 210 Watts. Then multiply that by the efficiency of the inverter - I'm guessing about 75% to 80%. That gives you a maximum of about 160 Watts. Not enough to run a vacuum cleaner or even a modern desktop computer. \$\endgroup\$ – Dwayne Reid Jan 11 '18 at 3:45
  • \$\begingroup\$ Why did you think this would work to begin with? You are asking for an enormous amount of power (up to 2500w) and clearly didn't do even the most basic bit of math behind it. \$\endgroup\$ – Bryan Boettcher Jan 11 '18 at 17:51
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A vacuum could be drawing 15A or more at 120v (ac) from your inverter, that's 15A*120V = 1800W. The inverter must draw 1800W or more from your 12v battery to supply that. To pull 1800W from your 12V battery, it will have to supply 1800W/12V = 150A! This is probably well outside of it's capabilities.

So how many 18650s would you need to power a vacuum cleaner? Lets start with the figure of 1800W. These inverters appear to have an efficiency of around 85%. So we're actually going to need 1800W/0.85 = 2117W of input power. The 18650's also will not actually supply 12v, under load it's going to be more like 11v. That won't cause any issues, but the inverter will have to draw more current to compensate. This puts us at 2117W/11V = 192A. 18650's vary in max current rating, but lets say 7A continuous. I've seen ratings between 0.6A and 10A so definitely check that for your specific batteries. 192A/7A = 28 cells, aka 3s28p. If you add a 10% margin onto that you're at around 30 cells.

This 30 cell figure is based on the max current that each cell is rated to produce. Now what happens if you try to draw more than that? If you only use 14 cells you will be drawing around twice the rated current. This can causes a couple of things to happen.

  1. If your battery has some kind of protection circuit built in it may trip, causing everything to shutdown.

  2. The batteries may max out, causing the voltage to drop and the inverter to shutdown or hiccup.

  3. The batteries may overheat or otherwise be damaged by the high current.

Note: we've only determined how many cells are required to run a vacuum at all. How long they will last is a whole other matter.

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  • \$\begingroup\$ Does that mean if i have 3s4p it wont even start it because there's not enough current? I thought it would start it just wont last long. \$\endgroup\$ – monmonja Jan 11 '18 at 7:06
  • \$\begingroup\$ Added some more detail at the end of my answer. We've only figured out how many batteries are needed to supply the instantaneous current. We haven't even gotten to how long they will last once they're doing that. \$\endgroup\$ – Drew Jan 11 '18 at 17:20

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