Battery Run Time - doing the math?

[deejayspinz]

Hey All,

Noob here.. I am in the midst of building a bike light for my night time MTB rides and my electronics education is foggy. I want to determine the run time that I will get.

Running a Romisen RC-U4 Rated at 3 - 8V. I have a 7.2 V 3300 mAh Li-ion battery pack. Not sure, but my assumption is that the light is pulling 3.7V. What I am unclear about is the math to determine how long it will run? I think somewhere in the formula I need to know what current the LED is drawing? Right?

BTW - Happy Independence Day from Canada eh!

 

[Mr Happy]

Let's see. In order of decreasing accuracy:

1. Do a test run with a clock and time it.

2. Measure the current and voltage at the battery to determine power consumption, then divide into battery capacity.

3. Estimate power consumption of light, then divide into battery capacity.

To calculate with methods 2 and 3, you need to estimate the power consumption of the light and the power capacity of the battery pack.

The power consumption of the light is voltage x current. So if the light draws 0.5 A at a voltage of 8.0 V the power consumption is 4 W.

The power capacity for the battery is average voltage times charge capacity. If you assume an average voltage of 7.2 V and a charge capacity of 3.3 Ah, the power capacity is 7.2 x 3.3 = 24 Wh.

Given these numbers, the estimated run time would be 24 Wh / 4 W = 6 h.

 

[TigerhawkT3]

Devices don't "pull" voltage, they pull current at a voltage. I'd recommend reading the WM.

Why does he need to calculate wattage? Just divide the battery's rated capacity in mAh by the current in mA pulled in that setup.

 

[Mr Happy]

Why does he need to calculate wattage? Just divide the battery's rated capacity in mAh by the current in mA pulled in that setup.

Because the driver is likely to pull more mA from the battery as the pack voltage declines during the run in order to maintain regulated current at the LED. Using mA and mAh will work to a first approximation, but using mW and mWh will be a bit more accurate if the driver has good regulation.

 

[TigerhawkT3]

You're quite right, that would be more accurate. However, I prefer dividing mAh by mA and then taking off around 20% for a decent (if rough) estimate, mostly because it's much easier.

Maybe measuring the current at the pack's midpoint voltage (say, 7.6V or 7.7V for a 7.2V Li-Ion pack) would be effective.

 

[AlexLED]

Devices don't "pull" voltage, they pull current at a voltage. I'd recommend reading the WM.

Why does he need to calculate wattage? Just divide the battery's rated capacity in mAh by the current in mA pulled in that setup.

Because, the voltage of the LED can differ from the voltage of the battery:

A 3,7 V and 1 A LED (3,7 W) would pull about 3 A from a 1,2 V NiMH (=3,6 W) and about 0,5 A from a 7,4 V (=3,7 W) Li-Ion 2S confiuration !
(Neglecting the loss of the converter.)

 

[deejayspinz]

Thanks for all the tips folks.

 

[TigerhawkT3]

Because, the voltage of the LED can differ from the voltage of the battery:

A 3,7 V and 1 A LED (3,7 W) would pull about 3 A from a 1,2 V NiMH (=3,6 W) and about 0,5 A from a 7,4 V (=3,7 W) Li-Ion 2S confiuration !
(Neglecting the loss of the converter.)

I know that. That's why we're talking about measuring current from the battery, not current through the emitter.