I'm trying to determine the max amp output of a battery pack that has 28 cells in it. To be more specific towards what I'm doing: I'm trying to size up a diode for the charge port; the battery charges at 29.4V max. On the website where I bought the e-bike it says it's a 466Wh battery. So, doing conventional math 466 / 29.4 = 15.85A. But, theoretically if a cell says it's rated at 7.56Wh, and the voltage drops from 4.2v to 3.7v, the battery would have to increase amps to maintain 7.56Wh. In my case my battery may be putting out 15.85A at 29.4V, but when the battery is low at 25.9V does it put out 17.99A? One website recommended using a diode that's twice the rating as the output to reduce the need for a heat sink. In this case I would be looking at a diode rated at 58.8v and 35.98A.

Update 10:44AM: I need a diode between the battery harness and battery charger; the charger gives me an error that says it's receiving current from the battery, so it's unwilling to charge the battery pack. It makes sense as it would be better to not have current flowing from the charge port.


closed as off-topic by Voltage Spike, Daniel Grillo, Sparky256, Dave Tweed Jul 23 '16 at 16:04

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  • \$\begingroup\$ Why would you need a diode? That is only going to increase your losses, you didn't say where you planned on putting the diode in the circuit or drawing a circuit diagram. The major problem with this question is its about an ebike and product questions are discouraged. Please read the guidelines about posting questions electronics.stackexchange.com/help/how-to-answer electronics.stackexchange.com/help/on-topic \$\endgroup\$ – Voltage Spike Jul 20 '16 at 15:24
  • 1
    \$\begingroup\$ "charger gives me an error that says it's receiving current from the battery" - odd, some chargers won't charge unless they can detect current, or at least a voltage, from the battery. Are you sure this is normal operation? \$\endgroup\$ – pjc50 Jul 20 '16 at 15:50
  • \$\begingroup\$ The current through the diode is only the charging current. What's the max. A the charger will provide? \$\endgroup\$ – JimmyB Jul 20 '16 at 17:29
  • \$\begingroup\$ Are you using the correct charger? If it outputs a lower voltage than the battery there's current flowing into the charger but a diode won't help with charging. \$\endgroup\$ – JimmyB Jul 20 '16 at 17:32
  • \$\begingroup\$ "One website recommended using a diode" - which website? \$\endgroup\$ – Bruce Abbott Jul 20 '16 at 22:30

Dividing energy [Watt-hours] by voltage [V] does not give amps [I], you have to divide power by voltage to get amps. The power output is V*I with voltage determined by the battery and current output determined by the load. The current capacity of the battery pack is not specified by the energy capacity but you can calculate it if you have the C rating.

A 100Whr battery can deliver 100 watts for 1 hour or 10 watts for 10 hours, it depends on the load. The voltage will decrease as the battery discharges and for a fixed resistive load the current will also decrease according to Ohm's law.

As far as needing a diode, it sounds like your charger is not correctly configured for the battery pack. The only way for the battery to source current into the charger is if the charger voltage is lower than the battery voltage. No charging will ever happen if that is true.


The maximum current that a battery is designed to safely discharge is known as the C rating. I believe usually a laptop or cell phone battery is considered 1C where rc batteries (more ideal for ebikes) are usually 10C+ with 40 and 50C not being uncommon. Now in your example you divided 466Wh/29.4V = 15.85 Ah not A so since the capacity with respect to the C rating is based on Ah the maximum current you should be able to pull from (or push into) the battery safely is a multiple of that. The current that actually gets pulled is up to the way you try to discharge the battery so with great power comes great responsibility and that's up to you.


What you're getting confused here is that WattHours is not Volts x Amps.
WattHours are synonomous with Joules, with the exception of the scale factor. what's you're looking at is the wattage rating, or the C rating.

Watts are simple enoug to calculate, it's simply the Veff X A. Voltage, being measured in Joules/culomb, and amperage, being measured in culombs/second, multiply out to give the units Joules/second, or Watts.

C rating
the C rating of a battery is usually only used for LI-Po batteries, because of their composition. The basic unit of a 'C' is Joules/(Culomb hours).This seems simmilar to voltage, with the exception of hours being on the bottom. Following good ol' PEMDAS, we find that we multiply this by (Culomb hours/second), which is also known as AmpHours. For example, if a battery is rated for 20C, aand can hold 5Ah, then the maximum discharge rate is 100A.

To answer the first question, No, the amperage does not increas as the voltage drops, it decreases, as shown here:

enter image description here


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