# LED determining Series / Parellel configurations for most efficent design

My circuit has an input range of 11.8 - 28 vdc. I have 2 types LEDs that I would like to place in this circuit.

The first has a VF Typ of 3.4v and an Idrawnom of .150 A The second has a VF Typ of 2.2v and an Idrawnom of .150 A

(Link to data sheet) http://www.optekinc.com/datasheets/OVS5MABCR4.pdf

I have 2 sections in my PCB.

In one section I would like to put 4 of each color. In the other section put 20 of each of these.

These LEDs could be considered to be on all of the time after power up.

I am trying to figure out what would be the most efficient use of power and determine the series parallel configuration. PWM is a possibility if needed.

The large swing in the input voltage is why I am asking this question. If it were only 12v or 24v then the answer would be straight forward. Also if I was not concerned about efficiency / power distribution it would easy as well.

Also I am trying to stay away from the use of LED drivers vs PWM because of the code rewrite involved.

• Are you regulating the input voltage down at all? It might be best to run the LEDs off of that. Aug 15 '13 at 20:27
• I was planning on using an LDO for the electronics only (ie straight bus for the LEDs). But I could move to a switcher and use that at say 14vdc and put 4 LEDs in series. That was deff one idea that I had. I dont know if that is the best solution though? Thanks for the suggestion. Aug 15 '13 at 20:29
• Yes sorry Ill fix that now. Aug 15 '13 at 20:41
• There are led drivers with a single input/enable pin, which you can effectively use pwm on. You don't have to have an addressable/spi/i2c led driver. Aug 15 '13 at 21:08
• I have found this one from TI ti.com/product/tlc5926 (allows up to .120 A per channel and has 16 channels) I would assume then that I would limit the voltage to ~11v (3x 3.4v which is the highest VF) and then set the current limiting resistor on the chip. to understand the function correctly it can draw up to .120 A for each channel so that could be up to 16 x .120 A or 1.92A? Aug 15 '13 at 21:23

LEDs are current driven devices, not voltage driven. Hence, as long as the drive current requirements are met, the voltage involved is merely a dependent variable, so to speak.

For both types of LEDs mentioned, a constant current buck LED driver would probably provide the lowest cost Bill of Materials for a very high efficiency design.

For instance, the Texas Instruments LM3407 constant current, floating buck converter will allow a supply of 4.5 to 30 Volts, and a LED string forward voltage of anywhere between 10% and 90% of the supply. Taking the voltage numbers provided in the question:

• Vin(min) = 11.8 Volts ==> maximum Vf of string = 0.9 x 11.8 = 10.62 V
• Vin(max) = 28 Volts ==> minimum Vf at full input voltage = 0.1 x 28 = 2.8 V

So, as long as each string is between 2.8 and 10.62 Volts Vf, this regulator can support the application. Typical implementation, taken from the datasheet:

By using switching current regulators, the issue of getting rid of surplus voltage as heat is pretty much eliminated. As can be seen from the example of the LM3407 above, switching regulators aren't too fussed even if 90% of the supply voltage needs to be dropped within the regulator: The device does cycle-by-cycle current regulation of the load (the LEDs), and simply switches off the rest of the time, hence not needing to dissipate the excess voltage as heat.

This simplifies the series / parallel LED decision significantly: Simply put as many LEDs in series as you like, up to the maximum determined by the supply voltage lower limit (10.62 Volts as calculated above).

Thus, 3 LEDs of 3.4 Vf, or 5 LEDs of 2.2 Vf can be supported per string, in this particular application.