TV0J and 3 alkaline cells

I'd expect about 1.5 amps DD with a 3.4v Vf on alkaline batteries. A 1 ohm, 1 watt resistor would get you down to about 1 amp. Better would be a Shottky diode, which would add about a .3v drop, effectively giving you a 'k' Vf with very little loss. A 2 watt Shottky is smaller than a 1/2 watt resistor. DD on alkalines wouldn't be too bad, tho, since they would drop to about 1 amp after maybe 1/2 hour of use. HTH, -RussH
 
Cool, just the info I needed. I am going to use the 2 watt Shottky.
Thanks.
Yaesumofo

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RussH said:
I'd expect about 1.5 amps DD with a 3.4v Vf on alkaline batteries. A 1 ohm, 1 watt resistor would get you down to about 1 amp. Better would be a Shottky diode, which would add about a .3v drop, effectively giving you a 'k' Vf with very little loss. A 2 watt Shottky is smaller than a 1/2 watt resistor. DD on alkalines wouldn't be too bad, tho, since they would drop to about 1 amp after maybe 1/2 hour of use. HTH, -RussH

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[ QUOTE ]
RussH said:
I'd expect about 1.5 amps DD with a 3.4v Vf on alkaline batteries. A 1 ohm, 1 watt resistor would get you down to about 1 amp. Better would be a Shottky diode, which would add about a .3v drop, effectively giving you a 'k' Vf with very little loss. A 2 watt Shottky is smaller than a 1/2 watt resistor. DD on alkalines wouldn't be too bad, tho, since they would drop to about 1 amp after maybe 1/2 hour of use. HTH, -RussH

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Actually, for a Luxeon and 3 alkaline cells, resistive current limiting gives a flatter discharge curve than the use of a diode. A diode gives an almost constant drop across it while the drop across a resistor decreases as the current decreases.
 
I have an SX1J I'm direct driving, Its definately over an amp when I first turn it on. I love it! I don't need no Resistors! I keep forgetting to measure it on fresh cells but it definately gets a bit warm. I think its probably over 1.2 amps..

Im running on C Cells which have lower internal resistance than D's I believe.
 
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