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So I bought battery brick from Amazon.ca it was a 20,000 mAh, it's a 5v battery pack. I also just bought a Milwaukee 9 Ah 18v battery for my tools and beside it I saw an adapter that let me click onto the top of the battery the ability to use it to charge my phone. I could not tell what one was better when it comes to capacity. Because one is an 18V battery and the other is 5V. How can I calculate what is a better battery to use to charge my battery if my phone requires 5V 3A to charge. If I only had room for 1 battery on a trip what one should I bring?

Update: My phone battery is 4000 mAh so lets assume that every time i charge it I want to fill it up.. I really don't care what I am charging though but what battery listed above would last longer... therefore charge my phone more if you assume 5V 3A when charging it!

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    \$\begingroup\$ Total energy stored = Ah rating times voltage. \$\endgroup\$
    – Eugene Sh.
    Jun 4, 2018 at 18:01
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    \$\begingroup\$ How long is the trip? How often will you recharge your phone and for how long? Why can't you use a cigar lighter in your vehicle? Which battery is lightest? Do you need to take tools with you? Are you allowed to take the adapter with you? Have you done any research on what ampere hour capacity means? \$\endgroup\$
    – Andy aka
    Jun 4, 2018 at 18:07
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    \$\begingroup\$ Note that neither pack is actually a battery with the listed specs; the former is a Li-ion or Lipo battery with a boost circuit, and the other is a battery with 18V nominal voltage but whose voltage will actually vary with time. \$\endgroup\$ Jun 4, 2018 at 18:16
  • \$\begingroup\$ This is just a hypothetical question not real life, I have a cig lighter charger I was just curious how you would compare the two devices to each other, different battery types, different mAh different voltage rated for.. If a 5V 3A load was placed on both till they were dead which one would last longer.. and how would you calculate that? \$\endgroup\$
    – Steve Main
    Jun 4, 2018 at 18:30
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    \$\begingroup\$ It doesn't make sense to specify one battery in mAh, and the other in Ah, especially when the first is a higher value than the second. We do engineering here. Learn to use grown-up units if you want to be taken seriously. \$\endgroup\$ Jun 4, 2018 at 20:08

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If a 5V 3A load was placed on both till they were dead which one would last longer.. and how would you calculate that?

  • Load power consumption:
    5 V * 3 A = 5*3 V*A = 15 W.

  • Amazon power bank energy:
    While 20 Ah (20,000 mAh = 20 Ah) at 5 V would store 5*20 V*Ah = 100 Wh of energy, I really doubt that the 20 Ah battery within really has a nominal voltage of 5 V. Most likely 5 V is the (boosted) output voltage of the thing, and the actual battery has a nominal voltage around 3.7 V. A 20 Ah 3.7 V battery stores 3.7*20 V*Ah = 74 Wh of energy. In Joules, that is 74 Wh * (3600 s)/(1 h) = 74*3600 Wh*s/h = 266400 Ws = 266400 J = 266.4 kJ.

  • Milwaukee tool battery energy:
    9 Ah at 18 V stores 18*9 V*Ah = 162 Wh of energy. Again, in Joules, 162 Wh * (3600 s)/(1 h) = 162*3600 Wh*s/h = 583200 J = 583.2 kJ.

Assuming a 100 % efficient DC/DC conversion to 5 V:

  • The power bank will supply the load for 266.4 kJ / 15 W = 266.4 kWs / 15 W = 17.46 kWs/W = 17.46 ks = 17460 s (slightly under 5 hours).
  • The tool battery will supply the load for 583.2 kJ / 15 W = 38.33 kWs/W = 38330 s (almost 11 hours).

For a more realistic figure, multiply those estimated durations by 0.85 to account for inefficiencies in the power conversion.

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