what size resister

[joohn]

I am putting together a torch, I have fitted two sst90 in series and the batteries are two 5000mAh 7.4V 30C Lipo's in parallel, the build is going ahead well. What I am stumped on is the size of resister needed? Do you have any idea? Would it be a 4.7K or 5K. thanks.

 

[AnAppleSnail]

I think you have some confusion going on. A 4.7 kilo-ohm resistor would make those SST-90s about as bright as a mag lite solitaire.

1. I don't suggest using a resistor for high-power LEDs. If you run a 1-ohm resistor at 1 amp, you'll be wasting a watt of power.
2. You might try just wiring it up as-is, and hope that the wire resistance is enough. But if the LEDs go angry blue, pull the plug.
3. I suggest using an LED driver for high-power LEDs.

 

[joohn]

Thanks for the reply I thought a resister would work, where do I find a driver? is a driver better than a resister? don't want to wire it up bare need to know it will last, any help would be good?

 

[joohn]

Can anyone help me with the calculations for a resister? The build will be finished by the weekend apart from the resister. I have used 1.2mm wire. I want to see it shine!
Thanks Joohn

 

[USSR]

The resistor value is found by subtracting the LED forward voltage drop from the supply voltage, and then dividing by the desired LED operating current . if its led ... forgive me if my answer is irrelevant..

 

[iron potato]

As AnAppleSnail suggested, google for a dedicated driver for your project, its safer & easier.

Or u may read this for a Current Regulator > http://www.reuk.co.uk/Using-The-LM317T-With-LED-Lighting.htm

Also, get a decent heatsink for the led as well (salvage Pentium 4 heatsink or equivalent), it will get pretty hot :devil:

Hope u enjoy your project :thumbsup:

 

[joohn]

The resistor value is found by subtracting the LED forward voltage drop from the supply voltage, and then dividing by the desired LED operating current. If its led ... forgive me if my answer is irrelevant..

Thanks for your reply, from what I read above my Led forwards max voltage is 3.87 each led = 2 x 3.87 that’s 7.74V for the Led's taken from my supply of 7.2 volts = -0.54V. Is a resister necessary? the Led's are two sst-90 drawing 9amps and the supply is from a 7.2V Lipo.

 

[USSR]

from what i have red and found use of resister is necessary only when your supply voltage is greater then the forward voltage ,, to drop the rest of voltages.. in case you are lucky and have exactly same source voltage as the forward voltage then you do not need a current limiting resistor.. still some people say its necessary.. i have tried with and without resistor..and found no difference..



now in you case u do not need a resistor but keep in mind using one is good .. in that case assume only .1 volts accross your resister .. that will not drop too much volts to affect your leds performance.. from what u specified your voltage is already low but by very small quanty.. though your led may glow but u shold try at correct voltages to verify if the amount of light is not affected by small voltages..

i ,m not sure if your lipo can sustain such a huge currents..

u may stack two or three lipos to reduce he current drawn from each .. at low discharge currents batteries give good capacity and runtime as well as life...

u may calculate the Resistance ( assuming ur case where the although supply voltage is low compared to needed volage but still for safty adding a resistor is mneeded) as .. assume only .1 v drop accross Resistor

so v=i * r

r = v / i

r= .1/ 9 [ if both leds are in series .]

i m not sure if such a small resistane is available ..

and power ratting of resistor should be more then p = i * v accross the resistor
i.e

=.1 * 9

or more then at least 1 watt ,, u can use 2 watt rating resistor..

but 1 watt is huge powe loss ..

u may search for some driver ckt

 

[alpg88]

just to be safe, use 0.1ohm resistor, i'd use 5w resistor so it doesn't get too hot.

 

[alpg88]

but 1 watt is huge powe loss ..
u may search for some driver ckt

1w is not huge loss, i've seen drivers that loose as much on diode alone. when you loose 1w in a set up that has 1w led, than yes it is huge loss, when you loose 1w for 2 30W leds, than it is not, you'll pbly waste as much with driver.

 

[joohn]

I am using two 7.4V 30C 5000mAh Li-poly batteries parallel; the heat sink will be ok as the torch will be in water. Can I go without a resister? What is the danger without one; I want this torch to last a while.

 

[joohn]

It’s built and under charge, I can observe the voltage with the Lipo charger attached while I turn it on, big day for it tomorrow.
Thanks for your help!
Joohn

 

[USSR]

that led has typical forward cutrrent of 3 amps and absolute max of 9 amp..
so u may fry it at its absolute ratings,,,
try first 3 amps by using resistor...
then try incremrenting it to see if with increae in amperage produces any increase in brightness..

u may buy few suitable resistors for different currents then use one that suits most...

for such high power led i would recomend using resistor..

 

[joohn]

just finished the charge of the Liop 7.2V but it's not fully charged it's 8.4V so a set back, still on the hunt for the best resister or ? to controll the Led's, what do you think?

 

[jpou]

You are walking on a thin ice with power supply voltage being lower than forward voltage drop for LEDs. Once your supply voltage drops below turn on threshold of your LEDs, you torch will cease to shine.

Now if you decide to increase the size of your battery stack, you will have to calculate resistor value using this formula:
R = (Power supply voltage - LED voltage) / desired current

Power rating of your resistor will have to be AT LEAST:
P = I^2 R

Now let's say you have a battery with 12V output, your LED's will need 7.74V and you want to drive your LEDs at 3 AMPs:
R = (12-7.74) / 3 = 1.42 Ohm resistor

Power rating = 3^2 * 1.42 = 12.78W

Now for comparison, your LEDs power will be (U * I) = 7.74 * 3 = 23.22 W

Now it's up to you to decide if you want to waste 12.78W heating the water around your torch....

 

[USSR]

You are walking on a thin ice with power supply voltage being lower than forward voltage drop for LEDs. Once your supply voltage drops below turn on threshold of your LEDs, you torch will cease to shine.

Now if you decide to increase the size of your battery stack, you will have to calculate resistor value using this formula:
R = (Power supply voltage - LED voltage) / desired current

Power rating of your resistor will have to be AT LEAST:
P = I^2 R

Now let's say you have a battery with 12V output, your LED's will need 7.74V and you want to drive your LEDs at 3 AMPs:
R = (12-7.74) / 3 = 1.42 Ohm resistor

Power rating = 3^2 * 1.42 = 12.78W

Now for comparison, your LEDs power will be (U * I) = 7.74 * 3 = 23.22 W

Now it's up to you to decide if you want to waste 12.78W heating the water around your torch....

or to further reduce the wasted power u can add another led in series ..

Thanks for your reply, from what I read above my Led forwards max voltage is 3.87 each led = 2 x 3.87 that’s 7.74V for the Led's taken from my supply of 7.2 volts = -0.54V. Is a resister necessary? the Led's are two sst-90 drawing 9amps and the supply is from a 7.2V Lipo.

bro u have quoted the absolute max ratings.. avg/nominal/optimal working conditions are at some other points..
from the following source
http://www.luminus.com/products/SST-90.html

the specs are
Typical Voltage, Current
3.2V, 3.2A
now you have 2 leds
two lipo packs voltage is 7.4 and in parralleld to increase runtime

your options are
connect both the leds in series with 1 ohm resistor but that will waste 10 Watts of power.. [ and from where u got the specs that u earlier pointed out those were wrong.. if the specs of the above sit are considered correct ]

and combined power consumption of your led will be 20 watts..
total power consumption 30 watts..

 

[arek98]

connect both the leds in series with 1 ohm resistor but that will waste 10 Watts of power.. [ and from where u got the specs that u earlier pointed out those were wrong.. if the specs of the above sit are considered correct ]

and combined power consumption of your led will be 20 watts..
total power consumption 30 watts..

1 ohm is way too much; at 9A it would drop 9V which is more than input voltage.

0.1 ohm that alpg88 suggested would be about right. Power lost on it would be about 1W which is very little considering that LEDs itself are about 20W. Depending how much LiPos drop under load little bigger may be better, like 1.5 ohm.

Adding resistor gives you protection for LEDs. You wrote 7.4V LiPo but they are 4.2V when freshly charged. 30C are high current cells so they may drop very little under load. You can have more than 8V to LEDs. This most likely will destroy them.


I would use a constant current driver (9A capable) but if you want resistor, then 0.15 ohm, 2W resistor is where I would start. Check current thru LEDs to see how much room you have and change resistor to lower value if needed.

 

[CKOD]

1 ohm is way too much; at 9A it would drop 9V which is more than input voltage.

0.1 ohm that alpg88 suggested would be about right. Power lost on it would be about 1W which is very little considering that LEDs itself are about 20W. Depending how much LiPos drop under load little bigger may be better, like 1.5 ohm.

Adding resistor gives you protection for LEDs. You wrote 7.4V LiPo but they are 4.2V when freshly charged. 30C are high current cells so they may drop very little under load. You can have more than 8V to LEDs. This most likely will destroy them.


I would use a constant current driver (9A capable) but if you want resistor, then 0.15 ohm, 2W resistor is where I would start. Check current thru LEDs to see how much room you have and change resistor to lower value if needed.

Quick note here, with the high currents we are talking here, and slight voltage changes causing large changes in current, even 0.05ohms from a multimeter current shunt would really throw off the reading. Read the voltage across the resistor (V=I*R) to determine the current. I.E. 9A across the 0.15 ohm resistor would be 1.35V.

I would suggest that 9A is fine with an SST-90, and a 3A current drive on that would be like running an XM-L at 700mA because thats what all the ratings are at.

 

[USSR]

i think the best solution is to use a constant current driver ..
best place to start with is use LM338 with proper heatsink . and use as a constant current source..

 

[alpg88]

no it is not best place to start.
lm338 is good for 5A, voltage drop across it is around 1,2v.
i have tried lm's, they can be used as cc regulators on paper, but in reality you don't wanna use them as such.
the best place would be amc7135 chips, you can either solder one driver for each led, or use 1 driver for both leds, with leds before and after the driver.