1
\$\begingroup\$

I'm thinking of making my own electric bicycle, and I need help determining what the amp hour rating of my battery pack would be. I'm going for 36v and 8 or 9 amp hours. So, with 10 batteries in series, I can get a single, low capacity pack with 36v. If I were to make three of these, and put those in parallel, I'd have three times the capacity. But, with each battery providing 2500mah, that's 75 amp hours, and I know that's not correct. What am I forgetting? image according to my knowledge and to the picture, series doubles the voltage and the capacity stays the same

\$\endgroup\$
  • 1
    \$\begingroup\$ Two obligatory safety warnings: 1) You will need to ensure voltage balance between series connected batteries somehow. 2) Cell-integrated protection circuits cannot protect a battery pack due to too high reverse voltage, so you need a whole-pack protection circuit. \$\endgroup\$ – jpa Nov 25 '15 at 5:43
3
\$\begingroup\$

If you series connect the batteries the amp-hour rating stays the same. The voltage has increased though, so the power available has increased. That's why using watt-hours is more useful than amp-hours when comparing battery packs of different voltages.

So ...

  • If you're using 10 x 3.6 V batteries in a string you have a 36 V, 2.5 Ah battery.
  • That's a 36 V x 2.5 Ah = 90 Wh (watt-hour) battery.
  • Put three of those packs in parallel and you get a 36 V, 7.5 Ah battery.
  • That's a 36 V x 7.5 A = 270 Wh battery.

Now here's where watt-hours are so much more useful. Let's say you're going to need 135 W average power for your motor. Your battery duration will be 270 Wh / 135 W = 2 hours duration.

My electric bike uses about 150 W in normal cycling mode at almost 30 kph. I reckon my legs are putting in another 150 W.

Little technical point: In the SI system of units it's capital letters for abbreviated units named after someone but lower-case when spelled out: V for volt, A for amp, W for watt, etc. Yes, there will be exceptions ...

\$\endgroup\$
2
\$\begingroup\$

Showing my work:

2.5 amp hours (one cell) at 36 volts (10 cells in series) is 90 watt-hours.

Times 3 of these strings in parallel is still 36 volts (you knew that) and 7.5 amp hours or 270 watt hours.

So I think you forgot a decimal point. :-)


Side note about connecting batteries in parallel: If you can avoid it, don't! Use higher capacity batteries instead. The reason appears when one cell gets weak or fails short-circuit while the others are still good:

If you have one series string and that happens, the pack voltage drops and you have to replace that one cell. Inconvenient, but still the best of all scenarios.

If you have multiple strings in parallel, current can flow backwards through the string that contains the bad cell, potentially damaging the entire string.

If you tie across all the strings at each level, making one series string where each "cell" is actually a parallel combination of cells, then you're pretty much guaranteed to lose an entire "rung" at once because the bad cell will drag the rest down with it.

And there's the issue of how much current actually flows in each scenario. For example, when you first connect the batteries in parallel, whether single cells or strings of cells, they will be charged to slightly different voltages (nothing is ever equal, including end-of-charge from the same charger at different times), which will be imposed across the wires that connect the strings. How much current is that?

\$\endgroup\$
  • 1
    \$\begingroup\$ I think the 75 Ah came from multiplying the cell amp-hour rating by the number of cells. You correctly calculate the total as 7.5 Ah. \$\endgroup\$ – Transistor Nov 24 '15 at 23:31
  • \$\begingroup\$ However, if you isolate each string with a diode, you can avoid the problems you discuss. \$\endgroup\$ – WhatRoughBeast Nov 24 '15 at 23:33
  • \$\begingroup\$ electricbike.com/wp-content/uploads/2015/03/BatteryDIY11.png Why are the batteries welded like they are? \$\endgroup\$ – Hellreaver Nov 24 '15 at 23:33
  • 1
    \$\begingroup\$ @Hellreaver: You'd lose a lot of capacity by partial charging, and you wouldn't solve anything. The problem is the charger's cutoff voltage, which is different per charger, per temperature, per noise, per everything. Nothing is ever equal, regardless of whether you turn it down or use the full scale. \$\endgroup\$ – AaronD Nov 25 '15 at 0:29
  • 1
    \$\begingroup\$ @Hellreaver It would take 60V at 4.5Ah to get the same Wh as you had originally, which is over the 50V that is typically considered "safe" for muggles. (highly arbitrary, but you have to draw the line somewhere) Or if you're okay with 48V, you would have 216Wh in a single string instead of 270 across three strings. \$\endgroup\$ – AaronD Nov 25 '15 at 15:54
1
\$\begingroup\$

The same current goes through batteries in series, therefore the capacity in Ah is not the sum of the individual batteries, it's the same. The overall energy in Wh is increased because the voltage of the battery pack increases. If you want more capacity in Ah, you need to put several 36V stacks in parallel; 3 of them in parallel will give you 7.5Ah.

That's purely to answer your question, though you should watch out for cell failure modes and charge balancing.

\$\endgroup\$
  • \$\begingroup\$ The linked picture is correct... In what way do you think it is conflicting with what I wrote? \$\endgroup\$ – Mister Mystère Nov 25 '15 at 0:16
  • \$\begingroup\$ I was trying to make the distinction that I would be using many cells in series to increase the voltage, and three of those packs to increase the amp hour rating. \$\endgroup\$ – Hellreaver Nov 25 '15 at 0:18
  • \$\begingroup\$ Yes, but in that case you will only multiply the amp hour rating by 3, not 30 as your "75" figure suggests. \$\endgroup\$ – Mister Mystère Nov 25 '15 at 0:35

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.