Good day friends,
Todays world of electronics finds me with a 48v, 20Ah LiFEPO4 battery pack for an electric bicycle which is a 16 cell pack which can charge to 3.65v per cell
I am trying to find a way to safely discharge that battery pack from full charge of 58.4v down to it's internally controlled BMS cut off voltage of 40v at 20Amps while using a PEMZ-15 or other battery / current monitoring device to measure the amount of energy withdrawn from that battery pack.
If I had a 48v inverter I would simply use that and an adequate load, but I do not have one.
I'm not sure if resistors, a shunt or something else would be best for this? Maybe 10 x 100watt light bulbs or some other load measured to draw 1168watts?
I don't want to lose that battery energy to heat without first measuring it during this discharge test. so I assume the resistors would be placed after the measuring device? But then, I'm not sure the measuring device can handle that much power either.
Since the pack starts out at it's full voltage, I will use that for the initial calculation
Watts = Amps x Voltage
Watts = 20A x 58.4v
Watts = 1,168
So I need to dissipate 1,168 watts (initially, since the voltage will drop constantly during the discharge)
I'm just not sure how to do it safely.
Thanks
Todays world of electronics finds me with a 48v, 20Ah LiFEPO4 battery pack for an electric bicycle which is a 16 cell pack which can charge to 3.65v per cell
I am trying to find a way to safely discharge that battery pack from full charge of 58.4v down to it's internally controlled BMS cut off voltage of 40v at 20Amps while using a PEMZ-15 or other battery / current monitoring device to measure the amount of energy withdrawn from that battery pack.
If I had a 48v inverter I would simply use that and an adequate load, but I do not have one.
I'm not sure if resistors, a shunt or something else would be best for this? Maybe 10 x 100watt light bulbs or some other load measured to draw 1168watts?
I don't want to lose that battery energy to heat without first measuring it during this discharge test. so I assume the resistors would be placed after the measuring device? But then, I'm not sure the measuring device can handle that much power either.
Since the pack starts out at it's full voltage, I will use that for the initial calculation
Watts = Amps x Voltage
Watts = 20A x 58.4v
Watts = 1,168
So I need to dissipate 1,168 watts (initially, since the voltage will drop constantly during the discharge)
I'm just not sure how to do it safely.
Thanks