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# How to calculate the power needed for 750 LEDs?

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#### Maverickmax

Hi

I am looking for someone's advice in regard to power approximately 750 LEDs. I am thinking of using 24V powering 10 arrays of 15 LEDs with resistor in series on each board. Is it feasible?

I look forward to hear from you asap

Maybe I should power 750 LEDs with circuit via AC voltage

MM

#### gszczesz

##### Full Member level 2
Re: Powering LEDs

Firstly, you mention 750 LED's, but 10 Arrays of 15 is only 150 LEDs. So the question is, is it 150 LED's or 750 LED's?

If it's 150 LED's, then the next question is how much current?

If you are using 20-30mA of current, then parts like the Allegro A8500/A8504 may be used. They provide excellent current matching, accuracy and include a Boost converter so that your part only uses as much power as it needs. For 150 LEDs, you would only use 2 IC's.

When putting more then 3 or 4 LED's in series, you have to be very concious of mis-match and voltage variation. Take for example a typical Nichia white LED. with 12 LED's in series, the mis-match between the strings can be as much as 3.5V. If you are going to use a resistor to set your current, then the current variation between each string will be tremendous unless you use 10V or so for the resistor. This will eat up a lot of power. This is why for large strings of LED's it is highly advisable to use a driver IC that will guarantee good current matching and minimize power loss.

The other thing to consider is the Forward Voltage variation over temperature, or batches of LED's. This also can be very large, and using a driver IC that only uses as much power as is required is very advantagous.

For a typical Boost converter, Power In = Power Out * 1.1, assuming 90% efficiency. So for 150 LEDs at 3.6V forward voltage and 20mA of current, the power out is 10.8 Watts. Given a 5V supply, that equates to 2.16A of current. For a 12V supply, this equates to 0.9A of current, and for 24V this equates to 0.45A of current.

For the A8500, it needs 5V supply for the IC, but can take almost any voltage up to 40V on the inductor side. So you could always drop a simple 5V regulator from your 24V supply to generate the 5V IC voltage. I would recommend using 12V.

If what you meant was 750 LEDs, then you would need to use 8 of the A8500. The more the LED's, the more important it is to use a power converter and active current sources so that wasted power is at a minimum.

Greg
Greg

#### Maverickmax

Re: Powering LEDs

gszczesz said:
Firstly, you mention 750 LED's, but 10 Arrays of 15 is only 150 LEDs. So the question is, is it 150 LED's or 750 LED's?

If it's 150 LED's, then the next question is how much current?

If you are using 20-30mA of current, then parts like the Allegro A8500/A8504 may be used. They provide excellent current matching, accuracy and include a Boost converter so that your part only uses as much power as it needs. For 150 LEDs, you would only use 2 IC's.

When putting more then 3 or 4 LED's in series, you have to be very concious of mis-match and voltage variation. Take for example a typical Nichia white LED. with 12 LED's in series, the mis-match between the strings can be as much as 3.5V. If you are going to use a resistor to set your current, then the current variation between each string will be tremendous unless you use 10V or so for the resistor. This will eat up a lot of power. This is why for large strings of LED's it is highly advisable to use a driver IC that will guarantee good current matching and minimize power loss.

The other thing to consider is the Forward Voltage variation over temperature, or batches of LED's. This also can be very large, and using a driver IC that only uses as much power as is required is very advantagous.

For a typical Boost converter, Power In = Power Out * 1.1, assuming 90% efficiency. So for 150 LEDs at 3.6V forward voltage and 20mA of current, the power out is 10.8 Watts. Given a 5V supply, that equates to 2.16A of current. For a 12V supply, this equates to 0.9A of current, and for 24V this equates to 0.45A of current.

For the A8500, it needs 5V supply for the IC, but can take almost any voltage up to 40V on the inductor side. So you could always drop a simple 5V regulator from your 24V supply to generate the 5V IC voltage. I would recommend using 12V.

If what you meant was 750 LEDs, then you would need to use 8 of the A8500. The more the LED's, the more important it is to use a power converter and active current sources so that wasted power is at a minimum.

Greg
Greg

Yes, I want to power 5 x150 LEDs (750 LEDs) together. The A8500 sounds interesting device. I assume it is PDIP package? Is there any more similar driver that allows me to drive the 750 LEDs?

MM

#### gszczesz

##### Full Member level 2
Re: Powering LEDs

I believe that one is QFN. If you can't use QFN, then are alternative parts both from Allegro and other companies (such as Linear, Maxim, etc...).

If I go to www.allegromicro.com, I notice that the A8501 is out as a TSSOP, but that parts has only 4 strings and is intended for higher currents (100mA) and is automotive grade... I also don't see it available anywhere, so I imagine it just came out.... Alternativelly, you can try linear's portfolio:

www.linear.com

Their parts are o.k., but none of them drive a whole lot of LEDs (25 at most, unless you use the 40 one but that one is pretty complicated with a lot of external components.

At Maxim's site (www.maxim-ic.com), you can find the MAX17061 which is a like the A8500 but with serial-bus interface. That part accepts a wider input voltage range so you don't need the regulator like the A8500, it has slightly lower efficiency though and you need a micro-processor to program it. Also, it looks new so I doubt it is easily available.

Greg

Points: 2