The proper way to figure a resistor for an LED string is to take the starting voltage - your example - 36 volts. Next you figure the total voltage drop across the LED's. Your case, 31 volts. Subtract the Vf from the supply voltage. 36 - 31 leaves you 5 volts to drop across a resistor. Since voltage is equal to resistance times current, and resistance is equal to voltage divided by the current (not a complete explanation of ohms law) then 5 volts divided by 600 mA (0.6 amps) means you'd need a resistor value of 5 ÷ 0.6 = 8.3 ohms. The calculated wattage of the required resistor would be the voltage times the current. This case, 5 x 0.6 = 3. You'd need at least a 3 watt resistor. Best practice is to always over rate your resistor. Most will say two times, I commonly say 1 and 1/2 times. By my practice, you'd need a 4.5 watt resistor. They don't make them. So a 5 watt resistor would likely do just fine. For each string.

Ok, thank you, so that would be with no buck controller in place.

 

Yes, without a buck.

That would be my approach. But that's just how I would do it. Personally. The final choice is yours.

 

If you want a stable current in each string, use 4 constant current sources.

Bob

We have come full circle again.. ok would cheap current regulator as suggested in post 49 be able to limit the current of each string better?

 

We have come full circle again.. ok would cheap current regulator as suggested in post 49 be able to limit the current of each string better?

Yes, it regulates the current perfectly in each string.

 

Tony is certainly correct! The resistor approach will be more reliable because of only having one extra part, which you are in control of. And the resistor for each string does not need to be nearly that precise. To determine the resistor value, first calculate the nominal forward voltage of the string, and decide the current value that you want. For best LED life this would be between half and 3/4 of the specified maximum current. Next, with the known supply voltage, determine the difference in voltage between the LED string and the power supply voltages. This is the needed voltage drop across the resistor. It should be at least a bit greater than twice the change in the LED string's change in voltage drop as they get warm during operation. So if the change were 0.1 volt for each of ten LEDs, the total change would be 1 volt, and so you would select the voltage drop across the resistor to be at least two volts. But this is a minimum value, not an exact requirement. using ohms law, E/I=R, and so 2 volts/0.6A = 3.33 ohms. But that is a minimum value, and not standard or easy to get. The next standard value of resistor would be 4.7 ohms, which is quite common. This value will work well, although you may need to slightly adjust the supply voltage to get the current you want, or the brightness that you want.
I hope that this explanation makes sense.

 

Tony is certainly correct! The resistor approach will be more reliable because of only having one extra part, which you are in control of. And the resistor for each string does not need to be nearly that precise. To determine the resistor value, first calculate the nominal forward voltage of the string, and decide the current value that you want. For best LED life this would be between half and 3/4 of the specified maximum current. Next, with the known supply voltage, determine the difference in voltage between the LED string and the power supply voltages. This is the needed voltage drop across the resistor. It should be at least a bit greater than twice the change in the LED string's change in voltage drop as they get warm during operation. So if the change were 0.1 volt for each of ten LEDs, the total change would be 1 volt, and so you would select the voltage drop across the resistor to be at least two volts. But this is a minimum value, not an exact requirement. using ohms law, E/I=R, and so 2 volts/0.6A = 3.33 ohms. But that is a minimum value, and not standard or easy to get. The next standard value of resistor would be 4.7 ohms, which is quite common. This value will work well, although you may need to slightly adjust the supply voltage to get the current you want, or the brightness that you want.
I hope that this explanation makes sense.

Thanks it Does, My concern with using straight resistors is the heat generated and that they will go poof lol, also the effeciency of running the leds 12 hours a day minimum. I've actually just ordered more buck converters that i will collect shortly. Pricey but atleast it gives me bit more peace of mind and control over the leds

 

If you are plugging the circuit in then you can just add resistors in each branch for the array. Jsut as you show. You can sill use your constant current device for brightness. I would pick a resistor

Thanks it Does, My concern with using straight resistors is the heat generated and that they will go poof lol, also the effeciency of running the leds 12 hours a day minimum. I've actually just ordered more buck converters that i will collect shortly. Pricey but atleast it gives me bit more peace of mind and control over the leds

Resistors won't go "poof" as long as they are properly sized.

 

Hello,

If the resistor has enough power there should be no fear that the resistor will go poof.
When higher power resistors are needed, I often take the aluminium housed resistors.
Here is an example of those resistors
https://www.rapidonline.com/arcol-hs10-22r-j-10w-aluminium-clad-resistor-62-8078
You can mount them on the heatsink to keep them cool.

Bertus

 

Hello,

If the resistor has enough power there should be no fear that the resistor will go poof.
When higher power resistors are needed, I often take the aluminium housed resistors.
Here is an example of those resistors
https://www.rapidonline.com/arcol-hs10-22r-j-10w-aluminium-clad-resistor-62-8078
You can mount them on the heatsink to keep them cool.

Bertus

Power in watts equals voltage times voltage divided by resistance. For a resistor that means voltage measured across the resistor. AND power also equals current times current times resistance. BUt note that if the current is in milliamps then the power is in milliwatts. so if the current is 600 milliamps and the resistance is 10 ohms, then the power is (600mA) x (600mA) x 10ohms, = 360000x10 =3600,000 MICROwatts, =3.6 watts.
So a 5 watt resistor would be totally adequate.

 

My concern with using straight resistors is the heat generated and that they will go poof

Resistors don't go poof. They overheat and burn out. The reason why they burn out is because they are not properly rated for the wattage you're going to pull through them.

Imagine you have a 1 watt resistor and are pulling 1 watt through it. It's GOING to run hot. And eventually go "Poof" as you say. A 2 watt resistor will handle the load much better. Here's where I might start an argument among other members, but drawing 1 watt though ANY resistor is going to produce 1 watt of heat. A 2 watt resistor can handle it better. A 5 watt resistor will do better still. Yet, the amount of heat produced is the same. Given a large enough resistor (say a 100 watt resistor) will feel cool to the touch because of such an extensive surface area for the heat to dissipate. Still, 1 watt of heat energy is 1 watt of heat energy.

So to avoid resistors that go poof in the night it's common among my contemporaries to double the size of the resistor (wattage rating wise), and I can find no problems with that. Me - personally - I'd go for anywhere from 1 1/2 times to 2 times. If I'm pulling 750 milliwatts through a 1 watt resistor - it can handle it, but I'd opt for using a resistor ((in theory)) of 1 and 1/2 times that size. So I'd need a 1 and 1/8 watt resistor. Since nothing like that exists, I'd opt for the next size - 2 watts. But if I were pulling 650 mW ((again, in theory)) I'd need a 975 mW resistor. Since nothing like that exists, a 1 watt resistor would suit me just fine. 650 mW times 1.5 = 975 mW, and I'm happy. Others will still go for the 2 watt resistor. AND THERE'S NOTHING WRONG WITH THAT.

Standard engineering practices among commercial products is to build to an over-rating of 133% normal expected operation. Aerospace, automotive and medical, it's common for 150% over-rating that of normal expected operation. Higher degree of reliability. 200% is, in my opinion, overkill. It's like building a pedestrian bridge that can support a herd of elephants where none can fit. SURE it can handle the load, but is it really needed to be that strong?

My opinion.

 

Tony, it would be difficult to put it any better than you have done in this explanation. Thanks for that! Considering a resistor's rating is just as important as calculating the value, or choosing the value if that is the process. For finding the value to use as the current limiting resistor for an LED string, the first step is to find the nominal (typical) forward voltage drop of the string, which is simply the sum of the published voltage drops of all the LEDs in that string. This voltage is then subtracted from the planned supply voltage, with the difference being the voltage to be dropped across the resistor at the planned LED current. Use ohms law to solve for the resistance, R= V/I. If the current is in milliamps then the result is in Kilohms. Pick the closest value of resistance from the 10% values table, that will be the resistor to use. Then calculate the power in the resistor, P=IxV, and if the current is in milliamps then the resulting power is in milliwatts.So the resistor wattage must be greater than that value, which the math is easy for this part. The one exception is if the resistor is to also serve as a fuse, going "poof" in the event of excess current. Seldom do they fail that way, unless something goes terribly wrong.

 

I'll add something. The color of the LED changes the Vd range.

For uniformity purposes, it's better to SELECT the LEDS in a particular string by Vd. This will minimize intensity differences.

Flame proof resistors will disintegrate and act like a fuse. Metal film resistors puddle and carbon composition change value (usually get larger). There are Flanged resistors too when you have a substrate to sink too.

You might in the long run, use a large resistor and a few smaller ones.

let's say the variation is 1.8 to 2.2 for Vd. Your big resistor pretends that all LEDs drop 2.2 V.
Then you size an array of smaller resistors that you clip/open to adjust the current slightly for the string.

So, you populate everything, then measure and clip the appropriate resistors.

e.g. suppose you get 9-12 mA using all Vmin and all Vmax. So maybe create a 9, 10, , 11, and 12 mA operational points. Use a resistor that would drop an appropriate voltages to get 9-12 mA.

You might have to say the string consists of 9 LEDs and 1 fictitious LED, and a current limiting resistor.
You ficticious LED is basically a few resistors.

it would take some experimentation to get it to work because of the eye's sensitivity to different colors.

It is typical to do that sort of thing. You do a trim in steps with fixed resistors in series and parallel. You cut the ones you don't need.

 

@KeepItSimpleStupid Not sure if I follow you on your proposed cutting of resistors. What I'm imagining (probably wrongfully so) is you have a few resistors in parallel acting as a single resistor of smaller value (as parallel resistors do). Proposed is to cut one at a time to tune in the brightness desired. Assume 5 parallel resistors, each 20 ohms: That's 4 ohms. Cut one resistor out of circuit and the resistance goes up to 5 ohms. Higher resistance will reduce the current, and thus the brightness. If I'm wrong in my assumptions - excuse me. Is this what you're suggesting? Start with the highest current and dim the LED's to desired brightness?

Stepping out for lunch. Be back in a bit.

 

If you design for a bright "typical" forward voltage then some strings will look fine, some strings will be dim and some strings will burn out.
If you do not want dim and burn out then feed the LED string with a constant current source and use enough supply voltage for worst case.

 

few resistors in parallel acting as a single resistor of smaller value (as parallel resistors do). Proposed is to cut one at a time to tune in the brightness desired. Assume 5 parallel resistors, each 20 ohms: That's 4 ohms. Cut one resistor out of circuit and the resistance goes up to 5 ohms. Higher resistance will reduce the current, and thus the brightness. If I'm wrong in my assumptions - excuse me. Is this what you're suggesting?

I never said exactly how to do it. May 2-3 sets of parallel resistors with one being a short. I have no idea.
So with 10 || 10 || 0 ohms + 10 || 10 || 0 ohms + 10 || 10 || 0 + (1 fixed value)

You could make 0, 5, 10 with each set. So 0-30 ohms in steps of 5 by cutting added to the fixed value.

There are 0 ohm resistors or jumpers in a resistor base. "+" is add and || is "in parallel with". Be careful calculating resistor wattage.

I can think better now the rain passed.