yaesumofo
Flashlight Enthusiast
Hey guys I want to direct drive a TV0J emitter in a 3"D" body. I can't use NiMh cells for this project. Do I need to put a resistor in line? What value. Any other ideas?
Yaesumofo
Yaesumofo
TV0J and 3 alkaline cells
- Thread starter yaesumofo
- Start date
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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
yaesumofo
Flashlight Enthusiast
Cool, just the info I needed. I am going to use the 2 watt Shottky.
Thanks.
Yaesumofo
[ 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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Thanks.
Yaesumofo
[ 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
[/ QUOTE ]
Doug S
Flashlight Enthusiast
[ 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
[/ QUOTE ]
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.
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
[/ QUOTE ]
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.
idleprocess
Flashaholic
Hmm... wonder where those emitters are coming from...
Justintoxicated
Flashlight Enthusiast
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.
Im running on C Cells which have lower internal resistance than D's I believe.
yaesumofo
Flashlight Enthusiast
I am going to build up a light and try it both ways. I have been colecting emitters from all over the place. I am gearing up to build a few lights for some buddies.
Yaesumofo
Yaesumofo
