Question on Series Wired Cree R2's

[wtn]

I have been playing around with the LED series/parallel array wizard calculator and have a question. If my source voltage is 9.6V and diode vf is 3.7 and I want to drive each R2 at 1000ma, the wizard says I need a 2.2ohm resistor (this will be direct drive BTW). It goes on to state that the resistor will dissipate 2200mw, the diodes 7400mw, total power 9600mw, and that the array will draw 1000ma from the power source. Will each R2 be driven at 1000ma or 500ma??? I assume the answer is 1000ma, since the source voltage is greater than 2X3.7v, but I want to double check. Thanks.

 

[Mr Happy]

In a series circuit the current is everywhere the same (in simple terms you can say that all the current that goes in one end has to come out the other end). So the diodes will be passing 1000 mA, the resistor 1000 mA and the power supply 1000 mA.

 

[Illum]

2.2 ohm? so your planning to drive two R2s...
3.7x2 = 7.4V consumed, both getting 1000ma [series add voltage/parallel add current]

9.6V (Vin) - 7.4V (VLED) = 2.2V (Vdrop over resistor)
Resistance = Voltage/Amperage
R = (Vin - VLED)/I
R = (9.6-7.4)/1
R = (2.2)/1
R = 2.2 ohms

Power dissipated = Voltage drop x Current applied
Pd_resistor = 2.2 x 1 or 2.2 watts
Pd_LED = 7.4 (3.7x2) x 1 = 7.4 watts
Collectively, Pd_Resistor + Pd_LED = 9.6 watts

yeah...your fine on the math

I'd recommend Mouser Part MOSX3CT631R2R2J or Jameco part 660201

the LEDs will only receive 500ma if your driving 4 CREEs in 2s2p [or two parallel strings of two in series] if your total current consumption is limited to 1000ma

 

[wtn]

2.2 ohm? so your planning to drive two R2s...
3.7x2 = 7.4V consumed, both getting 1000ma [series add voltage/parallel add current]

9.6V (Vin) - 7.4V (VLED) = 2.2V (Vdrop over resistor)
Resistance = Voltage/Amperage
R = (Vin - VLED)/I
R = (9.6-7.4)/1
R = (2.2)/1
R = 2.2 ohms

Power dissipated = Voltage drop x Current applied
Pd_resistor = 2.2 x 1 or 2.2 watts
Pd_LED = 7.4 (3.7x2) x 1 = 7.4 watts
Collectively, Pd_Resistor + Pd_LED = 9.6 watts

yeah...your fine on the math

I'd recommend Mouser Part MOSX3CT631R2R2J or Jameco part 660201

each LED will be getting 1000ma, it will only receive 500ma if your driving 4 CREEs in 2s2p [or two parallel strings of two in series]

Well, this is exactly what I will be driving - 4 Cree's in 2S2P Now I am confused. You say in this configuration that each led will be driven at 500ma? In my mind I see one series string with 2 leds that attach to a power source (with resistor) and each gets 1000ma. Then I see another identical string that attaches to the same power source and each led gets 1000ma.
But you say each of the 4 will only get 500ma?

If this is true, then I need to run all 4 in series with a driver. I have a Fatman and George says it will drive 4 crees (Q5 or R2) at 1000ma each as long as I heatsink the driver.

 

[wtn]

Well, this is exactly what I will be driving - 4 Cree's in 2S2P Now I am confused. You say in this configuration that each led will be driven at 500ma? In my mind I see one series string with 2 leds that attach to a power source (with resistor) and each gets 1000ma. Then I see another identical string that attaches to the same power source and each led gets 1000ma.
But you say each of the 4 will only get 500ma?

If this is true, then I need to run all 4 in series with a driver. I have a Fatman and George says it will drive 4 crees (Q5 or R2) at 1000ma each as long as I heatsink the driver.

The reason I am confused is that I have this setup all wired up - 4 Cree's in 2S2P. While I was wiring up the resistors a connection broke to one of the series string, so 2 leds went out. So I figure out what broke and I am holding the broken wire in mid air. I power up the light - 2 of the 4 crees are lit - should be 1000ma to each. They are bright. I am quite familiar with what 900 to 1000ma looks like. I take the broken wire and touch it to where it broke off on the other string and now all 4 crees are burning. And they are all just as bright as the other 2 were. Now you say they are each getting 500ma. So they should have dimmed enough for me to notice, but they did not dim at all.

 

[Illum]

post reworded to remove the misunderstanding, my mistake...I thought your power supply was limited to 1000ma, which was a faulty assumption

9.6V input
3.7V per LED...
2S2P would make two strings of 7.4V corrected parallel to your input. If you are drawing a total of 1000ma then it is shared between the two strings, 1000/2=500ma under theory.

Due to LED variations [including those within the same batch/bin/company] the current sharing will not be 1:1 but instead be like 0.7/0.3, etc...that is, one string may be severely over-driven while the other under-driven, for this reason its strongly recommended that you place current limiting resistors in series to each string.

+----|>|----|>|---/\/\/----+  R = 2.2 ohms
+----|>|----|>|---/\/\/----+  R = 2.2 ohms
+                          +
+ 9.6V                     + Gnd

If your intending for your CREEs to draw 1000ma each you would be drawing as total of 2000ma from your source [before losses]

 

[mdocod]

visually you'd be surprised how hard it is to see the difference between 500mA and 1000mA on a modern cree. Remember it takes about a 30% difference in output to see at all, since LEDs increase in efficiency when under-driven, the output at 500mA would not be half of 100mA, but probably more like 65%...


Now... This all assumes a bench regulated power supply at 9.6V. If your power source is a 9.6V battery pack, the the actual operating voltage under a load can be more or less than 9.6V by enough to make all these numbers far more difficult to theorize on paper. What might actually be happening is, on 2 LEDs, you are getting 1200mA per LED, and when you hook up the other set you are getting 700mA per LED, this would make the difference in output even more difficult to see...

I like Illums suggestion to use a resistor on each string, but I think in order to make that actually work, you would be approaching from 2 angles:
1. Determine the Vf of each LED, put together 2 series pairs that add up to as close a total Vf as possible.
2. Start with the 2.2ohm per string, and then adjust with much lower value resistors (fractions of an ohm or less) in series on the string that's being driven hard until both strings test as having the same running current.

Just FYI,
a pair of 2.2ohm resistors in this configuration is very much like just using a 1.1ohm resistor with the LEDs still in the 2S2P configuration. I'm pretty sure it would work fine either way, however you would still need to use low value resistors and Vf pairing to get each string of LEDs operating close to the same current.

-Eric

 

[Mr Happy]

Just FYI,
a pair of 2.2ohm resistors in this configuration is very much like just using a 1.1ohm resistor with the LEDs still in the 2S2P configuration. I'm pretty sure it would work fine either way, however you would still need to use low value resistors and Vf pairing to get each string of LEDs operating close to the same current.

I think the pair of 2.2 ohm resistors would give much better balancing and would definitely be the better way to go.

 

[wtn]

post reworded to remove the misunderstanding, my mistake...I thought your power supply was limited to 1000ma, which was a faulty assumption

9.6V input
3.7V per LED...
2S2P would make two strings of 7.4V corrected parallel to your input. If you are drawing a total of 1000ma then it is shared between the two strings, 1000/2=500ma under theory.

Due to LED variations [including those within the same batch/bin/company] the current sharing will not be 1:1 but instead be like 0.7/0.3, etc...that is, one string may be severely over-driven while the other under-driven, for this reason its strongly recommended that you place current limiting resistors in series to each string.

+----|>|----|>|---/\/\/----+ R = 2.2 ohms
+----|>|----|>|---/\/\/----+ R = 2.2 ohms
+ +
+ 9.6V + Gnd

If your intending for your CREEs to draw 1000ma each you would be drawing as total of 2000ma from your source [before losses]

This is exactly how I have them hooked up - using two 2.2 ohm resistors. Thanks for the correction. I thought I was losing my mind. And yes, 2A will be drawn from the power source.
BTW - Mouser has the new Arcol 5W Silicone Coated Axial Resistors. These are 5W, but come in the same size as 3W resistors. They are $1 each, but if you need a 5W in a small footprint these are the answer. The part number is 284-ACS5SW-2.7

 

[Illum]

thats alotta heat though

if your planning to mount this on your bike mount the resistors on your handlebars, that'll keep the dew off of it or under your seat for "bun warming" capability

 

[mdocod]

A decent quality boost circuit with the string of LEDs wired series would be a ~85%+ efficient solution that would guarantee even current flow through all the LEDs.

A pair of 1A buckpucks or other decent buck regulators would operate right at ~95% or better efficiency in your application.

If you have the flexibility to incorporate true regulation of some sort, that's always the best solution.

 

[Illum]

unless the op favors the heat...as in cold locations heat is invaluable