LED-WATER-HEATSINK

[LedElk]

I just checked and I-max is indeed at 1A.

No need to risk more problems, right? I'll probably re-route one of the LEDs to a third 35w driver. That makes room for 30w extra worth of LEDs

[Weezard]

I just checked and I-max is indeed at 1A.

No need to risk more problems, right? I'll probably re-route one of the LEDs to a third 35w driver. That makes room for 30w extra worth of LEDs

Quite right!
There's some interpretation needed with the spec. sheets.
If you read them carefully, you will find some interesting gotchas.

Fer instance, it may not be possible on some emitters to even sneak up on I-max.
The nominal junction drop is is just that, nominal.
There's plenty of wiggle room around that target voltage and the actual voltage drop will vary with the emitters, and over time.
I measured all my emitter drops and designed for the weakest link.

Current regulator regulate current by adjusting the voltage applied to your load.

Here's the fun part:
I might not be possible to drive some emitters to I-max, because doing so would exceed V-max.
My 15 Watt leds have an I-max of 1.5A.
That's 10 Volts to keep the math happy.

One of my emitters was hitting 10V. across the junction at 1.4A.
So, I locked 'em all down at ~1.2 Amps.
That seems to be the "sweet spot" for photon efficacy.

My 150 Watt array became a 120 Watt array.
Driving it "balls to the wall" would take another 25% in power.
But would deliver less than 10% more photons.
And it would drastically reduce the useful life of the assembly.

Why am I bothering you with this?

"That makes room for 30w extra worth of LEDs"

No headroom!
Max Headroom says, "ya gotta have it."
So, maybe another 25W. yah?

Meanwhile, your plants are looking good.

Aloha,
Weeze

[LedElk]

For sure, Max Headroom would be frowning hard if he came here I need to listen more to Max, cause he's right.

But, seriously. you got dimmable drivers or what? I'm counting on my newbie-drivers to feed the LEDs what they need, maybe sometime in the future I'll be adjusting drivers but that's a few steps ahead of me just yet

One question though, what happens if the LEDs aren't getting the V they need? I don't really have a grasp of how this works, you know, but would a lack of V restrict the "uptake" of A's and vice versa?
If you meant that setting your driver to 1.2A would give the LEDs <10V (V-max)?

And I'm a little curious on life-expectancy of these LEDs. I hope they last a while, but I guess it depends on giving them the right conditions.

The plants are loving the lights, definitely. As I've said I don't have much experience with growing, but I'm sure a green plant growing fast is a happy plant

And by all means bother me with technical posts, I'm trying to understand them and learning is the by-product

[Weezard]

Be happy to help.

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current,voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.
[align=center]I = V / R
[/align]

Where:
[align=right]I = [/align]	Electrical Current (Amperes)
[align=right]V = [/align]	Voltage (Voltage)
[align=right]R = [/align]	Resistance (Ohms)

[align=left]Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

Using this simple formula I trick a dirt cheap 3 terminal voltage regulator into regulating the current for my emitters.
I tie a 1 ohm 10W. resistor to the regulator's output pin and draw current through it to the leds.
Then I tie the sense lead of the regulator to the other end of the resistor.
That sense lead, often labeled adj ,is used to set the output voltage and is ordinarily tied to a potentiometer to vary the out put voltage.
Because of it's internal construction it needs 1.4V. on the sense lead to start regulating.
So, it will try to change it's output voltage to keep the math happy.
As we draw more current the voltage across that resistor will try to climb.
The chip will throttle it's output to keep it from doing so.

So, my "driver" is a 40 cent regulator and a 10 cent resistor that regulates current flow of up to 1.5A.
If you need to control higher current you could splurge and spend 2 bucks on a Mosfet and 2 more resistors to hang off the end.

Attachment 291333Attachment 291334

For power, it's:
Watts equals voltage times current. W = VA in a purely resistive circuit
So if we drop 2 volts across the regulator and draw 1.2 amps through it, it will have to dissipate 2.4W. (Won't even need a heat sink)
And the resistor will be a little over 1 ohm at the same current. so ~ 1.3 Watts.
Why do I use a 10W.?
Headroom!
Resistance increases with temperature and we want a cool and stable supply.
I strap the regulator to a heatsink for the same reason.
The cooler your electronics run, the longer they will live.

That's is why we overspec drivers too. If you want to drive 40 Watts worth of emitters, it's best to use a 50-60 Watt driver.
Same reason you should never try to draw more than 80 amps from a 100 amp service panel.

It's foolhardy to load any electrical circuit to 100% of it's capacity.
A very tiny voltage spike can cost you the whole game.
If you google ohm's law you will get a much clearer explanation.
I'm just a talented amateur at electronics, and not very good at teaching.


Aloha,
Weeze
[/align]

[Shovelhandle]

Good job, Weezard! I am an old electrician/technician.

As you explained, If your fuse or circuit breaker is rated to 'blow' or 'trip' at a given value you wouldn't want to be too close to that value when sizing your circuit. Otherwise the circuit breaker may open inconveniently. 80% is the maximum value you want to put on any fuse/breaker.

[Shovelhandle]

retired local 300 Vermont ibew, Weezard