DIY Led help

[weasel440]

No, current doesn't double from the driver - it is cut in half. So if the driver gives 700mA, each LED string, if voltage matches exactly, will receive 350mA.

My point is make sure the driver has enough amperage to drive the total mA of the strings paralled to driver.
3 strings @ 700mA is 2100mA total or 3 strings @ 350mA is 1050mA total.

Taken from this site... http://www.astrodynetdi.com/power_supplies/guides/led-drivers/faq/

Series

Perhaps the simplest configuration is to connect all of the LEDs in series, the anode of the second LED connected to the cathode of the first. A single, constant current source can then illuminate the entire string. This works very well with a limited number of LEDs in the string. However, as the string voltage is proportional to the number of LEDs in the string, long strings can generate rather high voltages. Assuming a forward voltage of 3.5V, a string of 24 LEDs would generate a voltage of about 84V. If any given LED fails short, there is limited impact on the operation of the lamp. However, if any LED fails open, the entire lamp will fail. Despite the high voltage, this is perhaps the most energy efficient way to power a lamp.

Parallel Strings

In order to minimize the operating voltage, multiple strings can be connected in parallel. Using the same 24 LEDs, one could form four series strings of six LEDs each and then connect these strings in parallel. The total voltage would now be only about 21V but it would require four times as much current to power the lamp. If any given LED failed open, one of the strings would fail but the other strings would remain lit. If any LED failed short, that string would carry much more current than the remaining strings. These situations will likely result in lower reliability as the remaining LEDs in the string with the failed LED are subject to significantly increased stress.

 

[jedimasterben]

Yes it does double. 2 strings of 700mA (or whatever the mA for his specific leds are) parallel to driver would be 1400mA and so on.

No, it doesn't. On a constant current driver, the current is precisely that - constant. Take a 48v, 700mA LED driver. It is capable of supplying about 34 watts of power (Volts times Amps). If you have one string of 12x LEDs at 3.0v each drop at 700mA, then each LED gets 3.0v and 700mA, totals to about 25 watts.

What you're saying is that if I added another string of LEDs, 12x with 3.0v drop at 700mA, that the current to each string would be the same as what the driver provides. Going to 24x LEDs with 3.0v drop and 700mA each is just over 50 watts. Please explain where the driver gets this extra power from.

 

[weasel440]

I don't mean the driver doubles. Driver is what it is. You purchase accordingly.
I'll use mine as example. I have 9 strings of leds parralel to my power supply / driver. My total max draw is 6300mA (6.3 amps) so i bought a 8 amp power supply. If I hook up my amp gauge to main plug to leds i get a very close reading to 6 amps.

Each string of 6 leds on my set up is running @ 19.9 volts @ 668mA

 

[jedimasterben]

I don't mean the driver doubles. Driver is what it is. You purchase accordingly.
I'll use mine as example. I have 9 strings of leds parralel to my power supply / driver. My total max draw is 6300mA (6.3 amps) so i bought a 8 watt power supply. If I hook up my amp gauge to main plug to leds i get a very close reading to 6 amps.

Each string of 6 leds on my set up is running @ 19.9 volts @ 668mA

An 8 watt power supply is giving out about 120 watts of power the LEDs?

19.9v x .668 gives over 13 watts. To get 6.3A from an 8 watt power supply the voltage output would need to be 1.26v, not enough to get even a single one of your LEDs to light up, much less strings of six.


But again, what you're saying does not apply to a constant current LED driver. You are not using LED drivers.

 

[weasel440]

Sorry I meant 8 amp - not 8 watt.

No i'm not using store bought drivers. All 100% diy for efficiency. Arduino controlled mosfets and resistors. And it is set up / wired for constant current.
And yes each string of leds, if tested with volt meter is running at my set voltage @ set % by arduino.

 

[oreo57]

Series

Perhaps the simplest configuration is to connect all of the LEDs in series, the anode of the second LED connected to the cathode of the first. A single, constant current source can then illuminate the entire string. This works very well with a limited number of LEDs in the string. However, as the string voltage is proportional to the number of LEDs in the string, long strings can generate rather high voltages. Assuming a forward voltage of 3.5V, a string of 24 LEDs would generate a voltage of about 84V. If any given LED fails short, there is limited impact on the operation of the lamp. However, if any LED fails open, the entire lamp will fail. Despite the high voltage, this is perhaps the most energy efficient way to power a lamp.

Shunt resistors can compensate for a "fail open" condition and many LED's incorporate such.
Fail short will probably shut off a good driver or go into "limp mode".
At $4-$7 per LDD breaking large series strings into ind. series strings is economical. (Each string has it's own driver..string size of course being limited to voltage available)

Parallel Strings

In order to minimize the operating voltage, multiple strings can be connected in parallel. Using the same 24 LEDs, one could form four series strings of six LEDs each and then connect these strings in parallel. The total voltage would now be only about 21V but it would require four times as much current to power the lamp. If any given LED failed open, one of the strings would fail but the other strings would remain lit. If any LED failed short, that string would carry much more current than the remaining strings. These situations will likely result in lower reliability as the remaining LEDs in the string with the failed LED are subject to significantly increased stress.

fuses would help this situation and a current balancing circuit is usually recommended due to differences in the characteristics of the individual diodes.

 

[mcgyvr]

Sorry I meant 8 amp - not 8 watt.

No i'm not using store bought drivers. All 100% diy for efficiency. Arduino controlled mosfets and resistors. And it is set up / wired for constant current.
And yes each string of leds, if tested with volt meter is running at my set voltage @ set % by arduino.

Your point/warning is valid for some situations but the OP is using LED drivers which do not require a separate power supply. So it doesn't apply here.. But still good for others to know..

 

[weasel440]

Your point/warning is valid for some situations but the OP is using LED drivers which do not require a separate power supply. So it doesn't apply here.. But still good for others to know..

Agreed. But it's still the same math (voltage & current) that is needed to drive the leds. I just chose diy driver over store bought driver for efficiency.

 

[oreo57]

Agreed. But it's still the same math (voltage & current) that is needed to drive the leds. I just chose diy driver over store bought driver for efficiency.

Hmm. I curious.. Why do you think it is more efficient than say an LDD???
LDD's are listed at 95% efficient..

 

[weasel440]

Hmm. I curious.. Why do you think it is more efficient than say an LDD???
LDD's are listed at 95% efficient..

You have total control on how you power them. I use mine as example... I do all the math for a particular power supply I pick for the number of leds per string. Which I did 6, so I got a 20 volt power supply. Then you determine how many strings you are going to use, and add up the total mA per string. Which I have 9, so I made sure I found a 20v power supply that had 8 amps output.

Now I done mosfets and resistors to set current. I used some online led resistor calculators to determine which current (mA) at what voltage for 6 leds would give me the best current (mA) with the lowest "waste".
In my case it was 19.9 volts @ 668 (mA) current set with a 1/8th watt 1 ohm resistor. But you can see in photo all I had on hand was 1 watt resistors. Which makes no difference.
All resistors & leds are cool - luke warm. Virtually no heat. And they are about 2 inches off water sandwiched in between plexi cover and aluminum plate. I'm also running 3 different optics. 60,90,120

Just to explain whats in photo. The 1 (far right) buck converter is no longer used. It used to power 2 - 12 volt mini pumps. The other buck converter with gauge is used to drop voltage from 20 volts to 10 volts. It powers arduino, fans and cabinet lights. I use 1 power supply to run everything.

 

[oreo57]

You have total control on how you power them. I use mine as example... I do all the math for a particular power supply I pick for the number of leds per string. Which I did 6, so I got a 20 volt power supply. Then you determine how many strings you are going to use, and add up the total mA per string. Which I have 9, so I made sure I found a 20v power supply that had 8 amps output.

Now I done mosfets and resistors to set current. I used some online led resistor calculators to determine which current (mA) at what voltage for 6 leds would give me the best current (mA) with the lowest "waste".
In my case it was 19.9 volts @ 668 (mA) current set with a 1/8th watt 1 ohm resistor. But you can see in photo all I had on hand was 1 watt resistors. Which makes no difference.
All resistors & leds are cool - luke warm. Virtually no heat. And they are about 2 inches off water sandwiched in between plexi cover and aluminum plate. I'm also running 3 different optics. 60,90,120

Just to explain whats in photo. The 1 (far right) buck converter is no longer used. It used to power 2 - 12 volt mini pumps. The other buck converter with gauge is used to drop voltage from 20 volts to 10 volts. It powers arduino, fans and cabinet lights. I use 1 power supply to run everything.

Understood. A couple of gotchas though:
1) all strings need to be of equal length and electrical characteristics to match the single power supply.
2) By using a variable voltage power supply to match LDD's/LED parameters you achieve "close" to the same efficiency. If I know the cumulative V(f) at 1000mA and can adjust the power supply to match
3)In your case the MOSFETS still burn (if I did this right) 1/2W of power..Which is close to minimum achievable but looks to be better than an LDD
http://www.pcbheaven.com/userpages/LED_driving_and_controlling_methods/?topic=worklog&p=3
http://www.pcbheaven.com/userpages/LED_driving_and_controlling_methods/?topic=worklog&p=5

I'm not arguing your point, tailoring a circuit usually leads to more efficiency than "one size fits all".. just pointing out a few things I think are worth thinking about..

The 500mA vs say 655mA "efficiency" is generally not very large..

The LDD is pretty much based on the same circuitry as the MOSFET piecework.. I think.