Blue_Shift
Newly Enlightened
Does anyone know the Voltage Drop for a 1W luxeon star?
luxeon star 1W question
- Thread starter Blue_Shift
- Start date
Blue_Shift
Newly Enlightened
Ok, so what I really need to know is how much resistance I need. I am running the white 1W LS with two Alk. AA's.
Blue_Shift
Newly Enlightened
thank you! /ubbthreads/images/graemlins/grin.gif
INRETECH
Flashlight Enthusiast
It depends on the Bin-Code
Do not try to design your circuit thinking of the LED as a "voltage device" and trying to calculate the res using Ohms law, the voltage drop changes as the current increases
Think of it as a "current device", you can use Ohms law to "get close" to the resistor value, but not exact
Properly heat-sink the LED, and then try different resistors until you get the CURRENT to the proper amount (350ma)
Do not try to design your circuit thinking of the LED as a "voltage device" and trying to calculate the res using Ohms law, the voltage drop changes as the current increases
Think of it as a "current device", you can use Ohms law to "get close" to the resistor value, but not exact
Properly heat-sink the LED, and then try different resistors until you get the CURRENT to the proper amount (350ma)
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PhotonBoy
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So, 5 ohms for 2 AA alkaline cells.
If you switch to NiMH cells, does the LED get brighter (lower voltage, 1.2 instead of 1.5, but also lower internal resistance), or dimmer due to lower voltage alone?
What's the optimal resistance for NiMH?
I prefer to use NiMH since you don't have to limp along on reduced output, trying to milk your $$$ out of the dying alkalines, you can simply top 'em up and away you go at full brightness again.
If you switch to NiMH cells, does the LED get brighter (lower voltage, 1.2 instead of 1.5, but also lower internal resistance), or dimmer due to lower voltage alone?
What's the optimal resistance for NiMH?
I prefer to use NiMH since you don't have to limp along on reduced output, trying to milk your $$$ out of the dying alkalines, you can simply top 'em up and away you go at full brightness again.
How electrons work...
[ QUOTE ]
Blue_Shift said:
Does anyone know the Voltage Drop for a 1W luxeon star?
[/ QUOTE ]
Depending on how tightly you want to define it, sure. It's 0 volts in the "first order approximation", 3 in the second, 3 plus one Ohm in the third. These are *approximations* used for circuit design, your individual device will differ a bit. How much such difference mean determines which of the three approximations an engineer will use in a 'proper design' (what ever that is....).
Although the actual "forward voltage drop" varies from device to device (and seems, at least to me, to be slightly higher with more efficient LEDs??? That is high output bin number run with high Vf bin numbers), as does the effective internal resistance that makes up that one Ohm, the third order approximation seems the correct call. This means:
" the Voltage Drop for a 1W luxeon star" is three Volts plus one 'Volt per Amp'.
That is, if we're running 350 mA (.35 Amps) we have 3.0 + .35 or 3.35 Volts. For 100 aM it's 3.0 + .1 for 3.1 Volts. However, as other posters have correctly pointed out, this is not really a good way to go about this.
Let's go through it step by step 'the other way around', shall we?
Let's take the four cell case, at 6 Volts, for simple "resistored drive" (IMO highly preferable over 'direct drive', and voltage driven circuits). Six minus three is three 'Volts across the (total) resistor'. We want (let's say) full blast, .35 Amps, so we have three Volts divided by 3.5 Amps. This is by Ohm's Law the *total* resistance we need, 8.6 Ohms of which one is internal (already there), We need 7.6 (more or less) additional to control current at 350 mA *while the battery is really 6.0 Volts*.
Changing the above system to NiMH (or simply running alkaline cells to *half used*) lowers the starting voltage to 4 X 1.2 Volts, 4.8. Less three is 'only' 1.8 Volts now. Divided by our 8.6 Ohms gives us a more modest 209 mA, nearly half the light output. A resistor adjustment is called for. 1.8 / .35 gives us 5.1 Ohms for this lower voltage. The danger is, of course, absent mindedly putting fresh alkalines (1.57 Volts) or 'hot off the charger' (often as high) cells in for over 640 mA (and no doubt a bright, but possibly becoming permanently dark, Luxeon).
It should be clear, that the modest internal resistance of typical cells is not a significant factor here (not in the face of several ohms), given reasonable margins. In 'direct drive' circuits, always a bit dicey, they are often the only other factor besides internal LED resistances holding current in check. In such systems (and other 'voltage regulator' circuits), another important factor is often in play. The 'nominal three Volt' value goes down with temperature, typically .002 Volts per degree C. Modest, but in already marginal designs (from a safety standpoint) this can be a very big deal. A 50 degree rise in such systems lowers Vf by .1 Volt. This causes a 100 mA *increase* in current, which further raises the temperature......
So, the short of it is, no. Five Ohms on a three Volt system won't realistically light the LED. Five Ohms is 'about right' for three cell systems (the minimum for white or blue LEDs), using fresh alkalines or four NiMH cells. Otherwise, the math is fairly easy. If it doesn't make sense, speak up and will 'run it a few more times'.
Otherwise, you can easily customize the circuit for your needs. It's actually easy to measure the true Vf of the diode junction itself. Measure the apparent Vf (including all drops from internal resistances) at two different currents (say 100 mA and 200 mA). Let's say 3.15 Volts and 3.25 Volts respectively. The additional .1 Volts (3.25 - 3.15) divided by the typical one Ohm is the result of the 100 mA increase. If we subtract that same (100 mA caused) change from he 100 mA number we get 3.05 as 'the real' value for *that* LED (what it's Vf would be at zero current). Predicting the exact total battery voltage is not so easy. Conservative design will combine with real shifts (discharging battery and contact resistance in the batter pack or other connections for instance) that lower current generally to keep us safe.
A good lesson here is that while DD systems are 'on the edge', even more stable 'resistored' ones have highly variable performance for small changes. Optimizing for this makes us vulnerable to overdriving in other reasonable conditions. This tends, at least to me, to favor conservative design.
The overall solution is, of course, a proper current regulated supply. A "smart" series resistor if you will. Solves a multitude of sins, as the saying goes.
Cheers.
Doug Owen
[ QUOTE ]
Blue_Shift said:
Does anyone know the Voltage Drop for a 1W luxeon star?
[/ QUOTE ]
Depending on how tightly you want to define it, sure. It's 0 volts in the "first order approximation", 3 in the second, 3 plus one Ohm in the third. These are *approximations* used for circuit design, your individual device will differ a bit. How much such difference mean determines which of the three approximations an engineer will use in a 'proper design' (what ever that is....).
Although the actual "forward voltage drop" varies from device to device (and seems, at least to me, to be slightly higher with more efficient LEDs??? That is high output bin number run with high Vf bin numbers), as does the effective internal resistance that makes up that one Ohm, the third order approximation seems the correct call. This means:
" the Voltage Drop for a 1W luxeon star" is three Volts plus one 'Volt per Amp'.
That is, if we're running 350 mA (.35 Amps) we have 3.0 + .35 or 3.35 Volts. For 100 aM it's 3.0 + .1 for 3.1 Volts. However, as other posters have correctly pointed out, this is not really a good way to go about this.
Let's go through it step by step 'the other way around', shall we?
Let's take the four cell case, at 6 Volts, for simple "resistored drive" (IMO highly preferable over 'direct drive', and voltage driven circuits). Six minus three is three 'Volts across the (total) resistor'. We want (let's say) full blast, .35 Amps, so we have three Volts divided by 3.5 Amps. This is by Ohm's Law the *total* resistance we need, 8.6 Ohms of which one is internal (already there), We need 7.6 (more or less) additional to control current at 350 mA *while the battery is really 6.0 Volts*.
Changing the above system to NiMH (or simply running alkaline cells to *half used*) lowers the starting voltage to 4 X 1.2 Volts, 4.8. Less three is 'only' 1.8 Volts now. Divided by our 8.6 Ohms gives us a more modest 209 mA, nearly half the light output. A resistor adjustment is called for. 1.8 / .35 gives us 5.1 Ohms for this lower voltage. The danger is, of course, absent mindedly putting fresh alkalines (1.57 Volts) or 'hot off the charger' (often as high) cells in for over 640 mA (and no doubt a bright, but possibly becoming permanently dark, Luxeon).
It should be clear, that the modest internal resistance of typical cells is not a significant factor here (not in the face of several ohms), given reasonable margins. In 'direct drive' circuits, always a bit dicey, they are often the only other factor besides internal LED resistances holding current in check. In such systems (and other 'voltage regulator' circuits), another important factor is often in play. The 'nominal three Volt' value goes down with temperature, typically .002 Volts per degree C. Modest, but in already marginal designs (from a safety standpoint) this can be a very big deal. A 50 degree rise in such systems lowers Vf by .1 Volt. This causes a 100 mA *increase* in current, which further raises the temperature......
So, the short of it is, no. Five Ohms on a three Volt system won't realistically light the LED. Five Ohms is 'about right' for three cell systems (the minimum for white or blue LEDs), using fresh alkalines or four NiMH cells. Otherwise, the math is fairly easy. If it doesn't make sense, speak up and will 'run it a few more times'.
Otherwise, you can easily customize the circuit for your needs. It's actually easy to measure the true Vf of the diode junction itself. Measure the apparent Vf (including all drops from internal resistances) at two different currents (say 100 mA and 200 mA). Let's say 3.15 Volts and 3.25 Volts respectively. The additional .1 Volts (3.25 - 3.15) divided by the typical one Ohm is the result of the 100 mA increase. If we subtract that same (100 mA caused) change from he 100 mA number we get 3.05 as 'the real' value for *that* LED (what it's Vf would be at zero current). Predicting the exact total battery voltage is not so easy. Conservative design will combine with real shifts (discharging battery and contact resistance in the batter pack or other connections for instance) that lower current generally to keep us safe.
A good lesson here is that while DD systems are 'on the edge', even more stable 'resistored' ones have highly variable performance for small changes. Optimizing for this makes us vulnerable to overdriving in other reasonable conditions. This tends, at least to me, to favor conservative design.
The overall solution is, of course, a proper current regulated supply. A "smart" series resistor if you will. Solves a multitude of sins, as the saying goes.
Cheers.
Doug Owen
INRETECH
Flashlight Enthusiast
Re: How electrons work...
Try experimenting with some values, but remember - a diode has a non-linear E/I waveform and will change its operating voltage as the current changes - measure the current with a meter, and adjust the resistor
"a current meter is your friend"
Try experimenting with some values, but remember - a diode has a non-linear E/I waveform and will change its operating voltage as the current changes - measure the current with a meter, and adjust the resistor
"a current meter is your friend"
eluminator
Flashlight Enthusiast
Re: How electrons work...
Ohm's law is your friend also. If you have a known resistor in the circuit, just measure the voltage across it and divide by the resistance. If that doesn't give you the current, then the resistor isn't really a resistor.
Current meters work the same way. They put a small resistance in series with the circuit. They use a smaller resistance at higher current ranges, so you should use the highest range that will give you the required precision. If you use too low a current range, the resistance the meter puts in the circuit can reduce the current significantly and give incorrect results.
It can be misleading to say an alkaline cell produces 1.5 volts. Because of the relatively high internal resistance (compared with NiMH) the voltage drops noticeable as the current draw increases. In my experience, alkalines do quite well at 350 ma though. It's only when the current draw exceeds 1 amp, that the voltage drops below the voltage from an NiMH cell. To make a long story short, alkalines would give a brighter light at 350 ma than NiMH.
Ohm's law is your friend also. If you have a known resistor in the circuit, just measure the voltage across it and divide by the resistance. If that doesn't give you the current, then the resistor isn't really a resistor.
Current meters work the same way. They put a small resistance in series with the circuit. They use a smaller resistance at higher current ranges, so you should use the highest range that will give you the required precision. If you use too low a current range, the resistance the meter puts in the circuit can reduce the current significantly and give incorrect results.
It can be misleading to say an alkaline cell produces 1.5 volts. Because of the relatively high internal resistance (compared with NiMH) the voltage drops noticeable as the current draw increases. In my experience, alkalines do quite well at 350 ma though. It's only when the current draw exceeds 1 amp, that the voltage drops below the voltage from an NiMH cell. To make a long story short, alkalines would give a brighter light at 350 ma than NiMH.
