# LED lighting/driver question

Discussion in 'General Electronics Chat' started by brad1138, Jun 20, 2017.

Jun 20, 2017
14
0
This is the driver that came with the LED light I bought on amazon. I am installing 5-10 of them, and would like to use 1 Driver/PS. It says "DC3-9V" I used an ohm meter and it puts out 9.4V (~16V when not under load). I can't find any "3-9V" drivers on line, or any at or above 2 amps. Can I just use a generic 12DC power supply? I read how "drivers" are supposed to back the voltage down when the LED heats up (or something like that), but after 1/2 hour of being on, it still reads 9V.

Thanks

Jun 20, 2017
14
0
Seems I can not edit it anymore. I mean, 1 PS/driver to run all of the Lights, probably 7.

3. ### EM Fields Active Member

Jun 8, 2016
583
154
What you have there is a constant-current power supply which is designed to connect directly to 100 to 220 volt RMS 50 to 60 Hz mains and pump 300 milliamperes into a load by automatically adjusting its output voltage over the range of 3 to 9 volts DC with loads than can vary from:

$R = \frac {E}{I} =\frac{3V}{300mA} = 10 \text{ ohms, to} R = \frac{9V}{300mA} = 30 \text{ ohms.}$

That is, if you have a 10 ohm load on the supply's output it'll adjust its output voltage until it gets to be 3 volts, if you have a 30 ohm load on the supply it'll adjust its output until it gets to be 9 volts, and if you have any load between 10 and 30 ohms on the supply, it'll adjust its output voltage until there's 300 mA through the load.

Last edited: Jul 15, 2017
#12 likes this.

Nov 30, 2010
18,076
9,686

Jun 20, 2017
14
0
6. ### #12 Expert

Nov 30, 2010
18,076
9,686
It will if the LEDs actually need 3.00 volts, but you never said what the required voltage is.
I can't guess any more accurately than that so I'll just give you the math:
(Vsupply-Vled)/0.3 amps = resistance in ohms
Power rating = at least 2 times 0.09 x R

Jun 20, 2017
14
0
Thanks, I don't know anything more than what is printed on the driver picture above.

Jun 20, 2017
14
0
I just want to double check, I read that as needing 7 resistors, 1 for each LED. Also, I may add an 8th, what value resistor would I need in that situation?

You said 5 V PS, I assume at least 2.1 amp (or 2.4 if I add an 8th), would any 5v 3amp PS work, like this one? https://www.amazon.com/dp/B006QYRVU6

I am feeling dumb, I took calculus in school, but the math here is giving me a headache. A lot to do with not having enough time to study it...

Thanks for the help.

9. ### #12 Expert

Nov 30, 2010
18,076
9,686
The same as all the rest of the resistors.
The math for this barely qualifies as Algebra, and I did the Algebra for you. Now you're working on fifth grade Arithmetic.
Yes, if the LEDs need 3 volts.
How about you connect the driver to an LED and measure the voltage?

Jun 20, 2017
14
0
I thought the number might vary depending on number of LEDs..

You mean like in the OP where I said "I used an ohm meter and it puts out 9.4V (~16V when not under load)"?

11. ### KeepItSimpleStupid AAC Fanatic!

Mar 4, 2014
3,448
632
I'll try. LEDs have a forward voltage that varies with the color of the LED and temperature. It also has some manufacturing variability. The manufacturer provides min,max and typical. Temperature is a small component.
A really good design makes sure the system operates over all of the tolerances, temperature etc.

To get uniformity, you have to BIN your bunches of LEDs by Vf. Your series resistor will be a little different for each BIN; R=(6-Vled)/300 mA.

The 6 is chosen because it's the center of the range. A 300 mA LED may exist, but you can sure have ten LED's in parallel operating at 30 mA. A "Parallel" LED in this case, is a resistor in series with an LED.
So, you can have 30 of these in parallel. Each with a slightly different resistor based on your BIN, so that the LEDS look like they have the same brightness.

Sure, it you wanted to illuminate 1 standard LED using this power supply where Vf = 2V @ 20 mA. First, we design for 6V at 20 mA and use R=(6-2)/20e-3; But we need 280 mA That we need to WASTE across a resistor.
So, we can use 6/280 mA; So, we need a 21.4 ohm resistor. We don;t not pick. We find the closest resistor we need. Maybe 5%, Maybe 1%; Let's say it's 20 Ohms.

Now we have to check the power. P=I*I*R or ((280 mA)^2 * 20) or 1.56 W. So, we need a resistor that's bigge rthan 1.56 W; We could use 3 or 5W. The larger wattage resistor might run cooler.

Reality says, we want to put that current into Light, not heat.

it's not algebra in the sense of exact answers, but it's to get things to fill a creiteria.

If i had one small LED, then that's not the power supply for me,

All sorts of series/parallel combinations of the LEDs can be made. Can you attain exactly 6V, NO, but that may be your starting point which says the LEDS can get colder or hotter and you've compensated for the varying Vf.

Doing lab with tolerences just makes the lab much more fun.

e.g. 300 ma =
Your 6V isn't 6V; It can vary.
Your LED operates at some current with some tolerance
The LED's Vf determine relative brightness. Binning makes more sense that using a different resistor for each LED.

Transistors can be classified by the gain range, because you can't make every transistor have a gain of 100.0000000000000

When they are all on the same substrate, they are much closer. Resistors can be laser trimmer in IC packages.

If = I = V/R; then make V=6V because it's in the center of it's operating range. 300 mA is what it will deliver.
So, if R = 6/0.3 or 200 ohms (let's not use a LED) that's what you will see across that resistor. If you make the resistor 270 or 330 ohms, you will have 300 mA through that resistor.

Jun 20, 2017
14
0
Thanks, I didn't think the LED type was relevant, but you were mentioning different colors. This is standard "white" spot light.

I don't know if that changes anything you said above.

I appreciate the help, I didn't expect this to be anywhere near this complicated. I bought LED 12V light strips (like pic below) years ago, hooked them to generic 12V Power supplies I picked up at Goodwill, or had laying around, and they have worked fine ever since(no resistors or anything). I thought this would be more or less that same.

I have 7 or 8 of these(above pic) in my home theater ceiling, and want to use a dimmer. They are a bit bright as is, especially after watching a movie or lights out listening to music. If I don't consolidate to 1 power supply, I would need 7 or 8 dimmers... #12's first post was useful, told me what kind of resistor to get, but he kept mentioning 3V, when my original post says twice, they are drawing ~9V, even after a few hours. So I don't know if his resistor value is correct or not. Then Keep it simple stupid said I should be using 1 resistor total, not 1 per LED.... So not sure what is right now...

Jun 20, 2017
14
0
OK, I tried the math here, and if you put 9V in for "Vsupply-Vled" the math works out to a ~5 V PS, but if you put in 3V, it works out to 1.8V. But you said I need 5V PS, "if the LEDs need 3 volts." Am I missing something, or did you make a mistake?

I am also assuming "Vsupply-Vled" does not mean "voltage of power supply" minus "voltage of LED".

I don't see any calculations in here for figuring the value of the resistor, is that just a standard value resistor used with LEDs?

14. ### #12 Expert

Nov 30, 2010
18,076
9,686
That's exactly what it means.
"Not under load" means with no LED connected.
If you want to know the voltage the LEDs require, you have to connect one and measure the voltage with a VOLT meter, not an OHM meter.
I said you could use a 5V power supply. You can use any voltage greater than what the LED needs when it's actually connected to the power supply, then you "waste" the excess voltage with a resistor. The maximum voltage of the power supply is not what the LED needs, but that's the voltage you will see on a meter if you don't connect the LED.

Jun 20, 2017
14
0
Well, that creates a problem, if the 2 are the same value you get 0/x...

OK, I guess my bad, I was using that as a generic term, it is a volt/ohm meter. I said it reads 9 volts, that IS with the LED connected, I threw in the "not under load" in case it was useful to anyone. Which also infers that the other (9V) reading would be under load.

That makes sense, but I don't know how to figure what size resistor to "waste" the appropriate amount of excess voltage.[/quote][/QUOTE]

16. ### #12 Expert

Nov 30, 2010
18,076
9,686
Now that you know the LED needs 9 volts, you can get a power supply of more than 9 volts and use the formula to find the resistance.

Jun 20, 2017
14
0
Thank you, and sorry for coming across short. I am getting to cantankerous....

I found this site http://led.linear1.org/led.wiz which has a calculator for just that. However, when I plug in the original numbers you were using, 5V PS, 3V FV(LED Voltage correct?), 300mA, it say 6.8 ohm resistor, not 0.68. I get 6.6667 when I do (5-3)/300mA.

I just want to get the right ones.

Thanks again

http://led.linear1.org/led.wiz

18. ### #12 Expert

Nov 30, 2010
18,076
9,686
Yep. I slipped a digit.
Of course, now that you have measured the LED running at 9 volts DC, a 5V power supply won't work. I'd go for 12V because they're easy enough to find.
Then 3 excess volts/0.3A =10 ohms
and power = 0.3 squared times R
P=0.9 watts, buy 2 watt resistors.

19. ### KeepItSimpleStupid AAC Fanatic!

Mar 4, 2014
3,448
632
Here's http://www.lumex.com/article/led-color-guide a table of the typical voltage drops for LEDS. Note, the white LEDS can have a 4 Volt drop. So, you can use two white LED's in series which will drop a max of 8 volts. This happens to be close to what your measuring.

To make it brighter we add multiple 2 LED + resistor combinations in parallel.

When we buy a replacement 12 V LED for our automotive lamp, that's what we're doing, using series/parallel combinations of multiple LED's and resistors to get the brightness we need.

These http://www.mouser.com/ProductDetail/Bivar/PM5-S17R28V are much wierder. They allow an AC and DC voltage to work, but fundamentally they are a mixture of LEDs, resistors and possibly diodes and maybe an IC chip.

Here's http://www.mouser.com/ProductDetail/Dialight/557-1605-203F a 24 V LED, but again, they are fundamentally made up of basic LEDs in a series or parallel combination.

==

Since everything isn;t created the same, I bought a non-dimmable light bar which works from about 8 to 32 Volts. The LED combinarions see about 6 Volts. there are 5 pairs of( 2 LEDS in series and one resistor). I can't dim this bar with input voltage. There is a DIM pin on the IC used, but I have been unable to investigate it fuller.

AC dimmable bulbs are really complicated, so I want to stay out of that for now.

LED's like to be current controlled. The easiest way to get current is add a resistor. the better way is to place a resistor inside a feedback loop and keep a certain voltage across that resistor.

LED's have a polarity. They don;t like more than about 5V reverse voltage. Some LAMPS, not LEDs may add a diode.

A linear power supply, a LED and a Resistor gets you dimmable LED. A linear power supply and a LED LAMP won't necessarily give you a dimmable LAMP. With my example of 8 to 32 V is full brightness,

Linear power supplies are not efficient, so we tend to pulse width modulate the LED's. This is a square wave with a variable duty cycle.

So, just as you have the power supply you have, there should be others that are dimmable. Dimmable, might mean a 0-10V control signal, a 0-100% PWM signal or a potentiometer.

Replacing LEDS for incandescent bulbs in cars are problematic because of polarity and lamp monitors. I replaced the interior #194 lamps with LEDs without issue, but the trunk lamp required a minimum load or a parallel resistor, otherwise the trunk lamp would be dim until a door was opened. There's a complex dimmable lighting controller in the car. Brightness ramps up, ramps down and will automatically turn off the lamps if they are on too long.

I need to find some example datasheets for the IC;s that may be used in dimmable controllers.

==

Your LAMP operates at 9V and probably 12 V just fine. Parallel lamps may also act just fine although there "COULD" be slight intensity variations between the lamps. A variable power supply should work too.

PWM is generally the right way to dim the lamps. So, a fixed 12 V supply that can handle the current and a PWM module 0-100% that also can vary the current.

Now, it you want your automation system to be able to dim, that's another story. Search for DALI as an example.