Using rechargeables in LED lights
- Thread starter jbrett14
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Charles Bradshaw
Flashlight Enthusiast
There are some you can use Lithium AAs in and some you can use Lithium AAAs in (due out holiday season 2004). Likewise with rechargables: some work, some don't, and some will overheat the LED(s). I am talking about the lights. It all depends oon the light in question.
gwbaltzell
Enlightened
[ QUOTE ]
Charles Bradshaw said:
... It all depends ...
[/ QUOTE ]
Amen!
Charles Bradshaw said:
... It all depends ...
[/ QUOTE ]
Amen!
The_LED_Museum
*Retired*
If the lights are resistored, the LEDs might run on the dim side.
If the lights are regulated, the LEDs should run correctly.
If the lights are direct-drive, you might let all the magic smoke out of them.
So I guess it just depends. /ubbthreads/images/graemlins/crazy.gif
If the lights are regulated, the LEDs should run correctly.
If the lights are direct-drive, you might let all the magic smoke out of them.
So I guess it just depends. /ubbthreads/images/graemlins/crazy.gif
BentHeadTX
Flashlight Enthusiast
What do adult diapers have to do with this? /ubbthreads/images/graemlins/jpshakehead.gif
gwbaltzell
Enlightened
[ QUOTE ]
The_LED_Museum said:
...
If the lights are regulated, the LEDs should run correctly.
...
[/ QUOTE ]
I, likely, would have said the same thing until I really thought about the consequenes. It was brought home to me when I was trying to find some answers for this thread about the Pelican Sabrelight. If you look in the reviews this light is very well regulated, but it uses a simple circuit rather than one of the many special ICs designed for this purpose. One of the facts of life is rechargeables can deliver very high current levels even when their voltage is extremly low. Alkalines cannot do this. With any regulated design we are trying to feed the LED with a constant amount of power. Power, in watts, is voltage times current, in amps. This particular flashlight uses a regular 1.2 W Luxeon and 3 "C" size cells. Lets call the efficientcy of the regulator 50%. So we would be trying to draw 2.4 W from the battery. OK, fully charged rechargeables (3 x 1.35 V) 4.05 V and we will be drawing (2.4 W / 4.05 v) 0.593 A or 593 mA. No problem! But a real flashlight has real resistance. In battery contacts, wiring, the switch, the traces on the circuit board. Also, since this is regulated, we don't know that our battery is running low. Let's call the resistance 1/2 an ohm (actual would just change the point were the problem occurs, acutal point in time wouldn't be much different). Another permutation of Ohm's law tells us that power is resistance times the current squared. So at full charge we are lossing 0.176 W in wiring. No problem. When rechargeables reach a point their voltage starts dropping fast, but remember they can still deliver high current. Lets pick a point in time were we have a voltage of 0.2 V per cell giving a current draw of (2.4 W / 0.6 V) 4 A! Wiring loss, which becomes heat, is (0.5 ohm * 4^2) 8 W. The LED is still fine, but likely the rest of the flashlight isn't! A requlator must be designed to shutdown when the current drawn reaches a certain point to allow it to be safely used with rechargeables. Another reason for moonlight mode!
George
The_LED_Museum said:
...
If the lights are regulated, the LEDs should run correctly.
...
[/ QUOTE ]
I, likely, would have said the same thing until I really thought about the consequenes. It was brought home to me when I was trying to find some answers for this thread about the Pelican Sabrelight. If you look in the reviews this light is very well regulated, but it uses a simple circuit rather than one of the many special ICs designed for this purpose. One of the facts of life is rechargeables can deliver very high current levels even when their voltage is extremly low. Alkalines cannot do this. With any regulated design we are trying to feed the LED with a constant amount of power. Power, in watts, is voltage times current, in amps. This particular flashlight uses a regular 1.2 W Luxeon and 3 "C" size cells. Lets call the efficientcy of the regulator 50%. So we would be trying to draw 2.4 W from the battery. OK, fully charged rechargeables (3 x 1.35 V) 4.05 V and we will be drawing (2.4 W / 4.05 v) 0.593 A or 593 mA. No problem! But a real flashlight has real resistance. In battery contacts, wiring, the switch, the traces on the circuit board. Also, since this is regulated, we don't know that our battery is running low. Let's call the resistance 1/2 an ohm (actual would just change the point were the problem occurs, acutal point in time wouldn't be much different). Another permutation of Ohm's law tells us that power is resistance times the current squared. So at full charge we are lossing 0.176 W in wiring. No problem. When rechargeables reach a point their voltage starts dropping fast, but remember they can still deliver high current. Lets pick a point in time were we have a voltage of 0.2 V per cell giving a current draw of (2.4 W / 0.6 V) 4 A! Wiring loss, which becomes heat, is (0.5 ohm * 4^2) 8 W. The LED is still fine, but likely the rest of the flashlight isn't! A requlator must be designed to shutdown when the current drawn reaches a certain point to allow it to be safely used with rechargeables. Another reason for moonlight mode!
George
Rothrandir
Flashaholic
[ QUOTE ]
BentHeadTX said:
What do adult diapers have to do with this? /ubbthreads/images/graemlins/jpshakehead.gif
[/ QUOTE ]
once you kill a led by using lithiums, you'll know... /ubbthreads/images/graemlins/icon15.gif
BentHeadTX said:
What do adult diapers have to do with this? /ubbthreads/images/graemlins/jpshakehead.gif
[/ QUOTE ]
once you kill a led by using lithiums, you'll know... /ubbthreads/images/graemlins/icon15.gif
gwbaltzell
Enlightened
[ QUOTE ]
Rothrandir said:
[ QUOTE ]
BentHeadTX said:
What do adult diapers have to do with this? /ubbthreads/images/graemlins/jpshakehead.gif
[/ QUOTE ]
once you kill a led by using lithiums, you'll know... /ubbthreads/images/graemlins/icon15.gif
[/ QUOTE ]
ROFLOL /ubbthreads/images/graemlins/grinser2.gif /ubbthreads/images/graemlins/bowdown.gif
Rothrandir said:
[ QUOTE ]
BentHeadTX said:
What do adult diapers have to do with this? /ubbthreads/images/graemlins/jpshakehead.gif
[/ QUOTE ]
once you kill a led by using lithiums, you'll know... /ubbthreads/images/graemlins/icon15.gif
[/ QUOTE ]
ROFLOL /ubbthreads/images/graemlins/grinser2.gif /ubbthreads/images/graemlins/bowdown.gif
[ QUOTE ]
gwbaltzell said:
Lets pick a point in time were we have a voltage of 0.2 V per cell giving a current draw of (2.4 W / 0.6 V) 4 A! Wiring loss, which becomes heat, is (0.5 ohm * 4^2) 8 W.
[/ QUOTE ]
I know 'numbers don't lie', but let's keep it between the ditches. You're only going to get a maximum current of .6/.5 or 1.2 Amps through that .5 ohms (using *all* your .6 Volts, nothing for the LED). Under a watt maximum available, total.
And as suggested, real world converters are probably shut down long before they look like dead shorts.
Doug Owen
gwbaltzell said:
Lets pick a point in time were we have a voltage of 0.2 V per cell giving a current draw of (2.4 W / 0.6 V) 4 A! Wiring loss, which becomes heat, is (0.5 ohm * 4^2) 8 W.
[/ QUOTE ]
I know 'numbers don't lie', but let's keep it between the ditches. You're only going to get a maximum current of .6/.5 or 1.2 Amps through that .5 ohms (using *all* your .6 Volts, nothing for the LED). Under a watt maximum available, total.
And as suggested, real world converters are probably shut down long before they look like dead shorts.
Doug Owen
gwbaltzell
Enlightened
[ QUOTE ]
Doug Owen said:...
I know 'numbers don't lie', but let's keep it between the ditches. ...
Doug Owen
[/ QUOTE ]
OK, keeping it "between the ditches", the CMOS chip is rated for 2 V (there is nothing else I see in the circuit to shut it down). If you'll agree to the 50% figure we would be drawing 1.2 A. Likely the transistor would have to be at a almost full on time. The mostly likely candidate for this "N15D" has a limit of 1 A. Even if the oscillation stops both the CMOS and the output transistor will be left at some state were they could draw excesive current and the LED will be dark. One would almost hope the wiring resistance was high. There were warnings about not using rechargeable with the Sabrelight long before I came along.
Doug Owen said:...
I know 'numbers don't lie', but let's keep it between the ditches. ...
Doug Owen
[/ QUOTE ]
OK, keeping it "between the ditches", the CMOS chip is rated for 2 V (there is nothing else I see in the circuit to shut it down). If you'll agree to the 50% figure we would be drawing 1.2 A. Likely the transistor would have to be at a almost full on time. The mostly likely candidate for this "N15D" has a limit of 1 A. Even if the oscillation stops both the CMOS and the output transistor will be left at some state were they could draw excesive current and the LED will be dark. One would almost hope the wiring resistance was high. There were warnings about not using rechargeable with the Sabrelight long before I came along.
[ QUOTE ]
jbrett14 said:
A more specific question: Can we use rechargeables in the PT Surge without harm? Anyone
[/ QUOTE ]
Yes. At least I have for the year or so I've owned mine. Not that I have all that many hours on it, but I've run several cycles on it. Not quite as bright as with *fresh* alkalines, but almost.
Doug Owen
jbrett14 said:
A more specific question: Can we use rechargeables in the PT Surge without harm? Anyone
[/ QUOTE ]
Yes. At least I have for the year or so I've owned mine. Not that I have all that many hours on it, but I've run several cycles on it. Not quite as bright as with *fresh* alkalines, but almost.
Doug Owen
[ QUOTE ]
gwbaltzell said:
[ QUOTE ]
Doug Owen said:...
I know 'numbers don't lie', but let's keep it between the ditches. ...
Doug Owen
[/ QUOTE ]
OK, keeping it "between the ditches", the CMOS chip is rated for 2 V (there is nothing else I see in the circuit to shut it down). If you'll agree to the 50% figure we would be drawing 1.2 A. Likely the transistor would have to be at a almost full on time. The mostly likely candidate for this "N15D" has a limit of 1 A. Even if the oscillation stops both the CMOS and the output transistor will be left at some state were they could draw excesive current and the LED will be dark. One would almost hope the wiring resistance was high. There were warnings about not using rechargeable with the Sabrelight long before I came along.
[/ QUOTE ]
I think I failed to make my point. With .5 ohms contact and other resistances in circuit at one Amp .5 of the total .6 Volts will be dropped elsewhere, the converter will 'see' .1 Volts (and is unlikley to notice it). A *dead short* will draw only 1.2 Amps. It's impossible to draw any more under any conditions with any sort of converter.
Doug Owen
gwbaltzell said:
[ QUOTE ]
Doug Owen said:...
I know 'numbers don't lie', but let's keep it between the ditches. ...
Doug Owen
[/ QUOTE ]
OK, keeping it "between the ditches", the CMOS chip is rated for 2 V (there is nothing else I see in the circuit to shut it down). If you'll agree to the 50% figure we would be drawing 1.2 A. Likely the transistor would have to be at a almost full on time. The mostly likely candidate for this "N15D" has a limit of 1 A. Even if the oscillation stops both the CMOS and the output transistor will be left at some state were they could draw excesive current and the LED will be dark. One would almost hope the wiring resistance was high. There were warnings about not using rechargeable with the Sabrelight long before I came along.
[/ QUOTE ]
I think I failed to make my point. With .5 ohms contact and other resistances in circuit at one Amp .5 of the total .6 Volts will be dropped elsewhere, the converter will 'see' .1 Volts (and is unlikley to notice it). A *dead short* will draw only 1.2 Amps. It's impossible to draw any more under any conditions with any sort of converter.
Doug Owen
Bob Snow
Newly Enlightened
I have been using rechargeables in a Surge for a long time with no problems. I also use three lithium AA batteries and a dummy cell in a Streamlight 4AA 7LED light. I use litiums in the Streamlight because it becomes dim fairly quickly with alkalines due to the fact that it is unregulated. With 3 AA lithiums, it stays bright much longer because the lithiums continue to put out high current and maintain voltage until they are nearly used up.
Bob: How long have you been using the 3 Lithiums in your SL? Was this your own idea? Is the light output significantly less? What is your dummy cell made of? Perhaps this would solve one of my issues for what lights to keep in gloveboxes. I already own 3 of the SL's but don't really want to rely on alkalines in glovebox.
