Voltage drop on AA alkaline
- Thread starter nsmith
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SilverFox
Flashaholic
Hello Nsmith,
On alkaline's the voltage is constantly dropping. I think for figuring you can use 1.2 volts per cell. That would give you 4.8 volts for 4 cells.
Tom
On alkaline's the voltage is constantly dropping. I think for figuring you can use 1.2 volts per cell. That would give you 4.8 volts for 4 cells.
Tom
jtr1962
Flashaholic
Short answer:
After about the first few milliseconds you should get no voltage drop and no current drain. A fresh AA alkaline battery has a resistance of about .15 ohms. Four in series have a total resistance of about 0.6 ohms. An LIII has a typical forward voltage drop of 3.90V at 1A. The four alkalines will be able to put out 1A at maybe 5.4V, 2A at 4.8V, 3A at 4.2V, and 4A at 3.6V. This means that the LIII will probably initially draw pretty close to 3A immediately before it burns out.
Long answer:
Don't direct drive, especially with 4 cells. If you won't use regulation and must direct drive then use 3 cells. This should keep you almost within the LIII's specs of 1A or less. By my calculations you should draw roughly 1.1 or 1.2A on fresh cells but you can still burn out the LIII even with 3 cells if you get one with a low Vf. Direct drive is never a good idea even in the best of circumstances. You get varying output as the batteries die, and a chance of easily blowing the LED with fresh batteries. Also, should you ever use rechargeables you'll risk blowing the LED even if the light works fine with AA alkaline. And using disposable batteries in a high-current drain device like a flashlight really isn't a good idea anyway. Using maybe 4 NiMH with a step-down current regulator makes the most sense for this situation. You'll get two or three times the run time of alkaline, won't risk buring out the LED, and won't be adding to the landfill by constantly throwing batteries away.
After about the first few milliseconds you should get no voltage drop and no current drain. A fresh AA alkaline battery has a resistance of about .15 ohms. Four in series have a total resistance of about 0.6 ohms. An LIII has a typical forward voltage drop of 3.90V at 1A. The four alkalines will be able to put out 1A at maybe 5.4V, 2A at 4.8V, 3A at 4.2V, and 4A at 3.6V. This means that the LIII will probably initially draw pretty close to 3A immediately before it burns out.
Long answer:
Don't direct drive, especially with 4 cells. If you won't use regulation and must direct drive then use 3 cells. This should keep you almost within the LIII's specs of 1A or less. By my calculations you should draw roughly 1.1 or 1.2A on fresh cells but you can still burn out the LIII even with 3 cells if you get one with a low Vf. Direct drive is never a good idea even in the best of circumstances. You get varying output as the batteries die, and a chance of easily blowing the LED with fresh batteries. Also, should you ever use rechargeables you'll risk blowing the LED even if the light works fine with AA alkaline. And using disposable batteries in a high-current drain device like a flashlight really isn't a good idea anyway. Using maybe 4 NiMH with a step-down current regulator makes the most sense for this situation. You'll get two or three times the run time of alkaline, won't risk buring out the LED, and won't be adding to the landfill by constantly throwing batteries away.
I will be using a 1 ohm series resistor for current limiting. So with the .6 ohm internal resistance of the batteries and 5.5V, I will have about 1A through the LED for first few minutes or so. As the batteries discharge to 1.1V per cell, I will have about 600ma through the LED. I wanted to use NIMH batteries and a current regulator, but the person I am making this for, does not want to mess with chargers and stuff. Also the current regulator ($30-40 buck puck, etc)nearly doubles the cost of the device.
gadget_lover
Flashaholic
I'm not sure why you are starting with 5.5 volts, since the internal resistance is applied to the entire cell voltage. According to the data sheet for the eveready e91 alkaline battery, you'll have more than 1.4 volts for about 5 minutes at a 750ma draw. The initial current through a K binned lux III will be around 1.3 amps with the 1 ohm resistor. I'm guessing that's a K bin if the 1 amp Vf is 3.9 volts.
Of Course, I could be all wet /ubbthreads/images/graemlins/smile.gif
Daniel
Of Course, I could be all wet /ubbthreads/images/graemlins/smile.gif
Daniel
