Drawing enough current?

[Strydur]

Maybe we are talking about different versions of the MR, all I know is the 3w work just fine and look great. And if the mr is only supplying 500mA then evidently that is all that is needed to drive them to full brightness since they run just as bright as driving them off of 4aa at 1400mA. If you want to take pictures of your multimeter hooked up and testing it while running off the mr and again while running off of 4aa then I would love to see the results.

Tim
The Custom Saber Shop

[Strydur]

Ok I got my hands on a fluke meter and went a testing... Depending on how you tap into the mr power/sound module is what decides how much current you get.
[img]../scripts/mrmod.jpg[/img]
Ok in this picture. If you hook you negative to pin1 and your positive to vdd you will only get about 500mA which is the way I have done it so far and seems to be plenty of current for the star III. From what I have read up to about half the rated current can be a visible difference but after half it is hardly noticeable. Ok, but if you want to get the full 1400mA then hook your positive to vdd and you negative to all the pins 1-6 and you will get pretty much exactly 1400mA. If you need 1000mA then use a 2.2ohm resistor in the circuit.

If any of you can confirm that this is the same for your modules please do so.

So I guess we where all right, sorta.. The mr board using the wiring I showed only gave you 500mA or less but if you do it this new way then you can get the 1400mA no problem.

Dont forget while testing you must take the LED out of the circuit and put your meter in place of it in order to properly calculate the mA going to the LED. Otherwise you are just measuring how much current is left over after the LED takes what it needs.

This is just the info I have learned and I of course could be totally wrong. So if you have other info or think I am figuring something wrong then please let me know so I can block you from posting..er I mean look into it.


Tim
The Custom Saber Shop

[Bobby Zero]

Sweet! Thanks for the info. I'll have to try that out.

[MythranDir]

Thanks for the info!! I will definitely give this a try.

MythranDir

[Strydur]

While this shows we have 1400mA of current it doesnt take into account that whole forward voltage calculation the led needs. I still dont fully understand how this all works. Someone get a rep from luxeon to explain it all before I rip my hair out[!]I need to look more into how that calculation works. If someone can shed more light on how this all works please do so.

Tim
The Custom Saber Shop

[MythranDir]

Tim, for a given desired current, the required value of resistance is obtained by dividing the voltage difference between the battery and diode by the desired current. For example, for a battery voltage of 4.5 V, and a diode forward-voltage of 3.9 volts, and a desired current of 1 Amp, the required resistance is (4.5 - 3.9)/ 1 = 0.6 Ohms. Some of the on-line resistance calculators round this up to a 'standard' value of resistance.

I tried hooking up the LED to the MR electronics as you suggested (hooking the negatives to pins 1-6) and was able to get more current, although for the blue LED I can still only get a maximum of ~ 500 mA. For the cyan and green I can easily get over 700 mA. However, I am not happy with the quality of my connections, and once I have solved this issue I will repeat these measurements. Thanks again for the info on the MR electronics.

MythranDir

[Strydur]

Do you measure the amperage while the led is connected?

Tim
The Custom Saber Shop

[MythranDir]

Tim,

You can measure the current when the LED is in the circuit. The multimeter is put into the circuit in series with the LED. In this way, you will be able to tell what the current actually is with the LED in place. When I do this with the blue LED using 3 AA's and no resistor, the best I can do is ~ 450 mA, using lithium ion batteries.

BTW, there is a huge difference in current depending on the type of battery. When I switch from the Energizer lithium ion batteries to Radio Shack alkaline batteries the best I can do is 350 mA.

I have also confirmed that when I use all the ground wires from the MR board I can get 450 mAmps through the blue LED, again without using a resistor. With red, green, or cyan LED's I can get upwards of 800 mA from the MR board. I think the issue with the blue LED is that it probably has a higher forward voltage than the other colors. This makes sense because blue photons are more energetic than the other colors.

Of course, 4 AA's can provide all the current you want for any color.

Thanks again for the info on the MR board. Connecting all the grounds has definitely increased the current.


MythranDir

[Strydur]

The thing I dont get is...If you measure the current using 4aa's and the resistor that the calculator says to use for 6v and a blue LED then you get the following measurements which dont make much sense.

[img]../scripts/ledtesting1.jpg[/img]
no resistor..5770mA
[img]../scripts/ledtesting2.jpg[/img]
with resistor..1663mA
[img]../scripts/ledtesting3.jpg[/img]
with blue led and resistor..425mA

Which leads me to believe you cant accurately measure these fricking things!!!!


Tim
The Custom Saber Shop

[MythranDir]

I think your measurements are ok. If your question is why does the current go down to 450 mA when the LED is in the circuit, the reason is there is a ~ 3.9 V drop across the LED, which means the voltage drop across the resistor is only about 6 - 3.9 = 2.1 V. The current through the circuit is 2.1 V divided by the value of the resistor. Without the LED in the circuit, the entire battery voltage, 6 V, is dropped across the resistor, which means the current in the circuit will be 6 V divided by the value of the resistor. So, the current with the LED in the circuit will be at least 3 times smaller than when there is no LED in the circuit.

One weird thing I noticed is that when I first turn on the power (using the MR electronics) the current starts off pretty high and then drops and after a few minutes stabilizes at 350 mA. I think one problem is the AA batteries are not really intended to operate at these high current draws. I am thinking that as the battery discharges the battery resistance or voltage changes, thus changing the current delivered to the diode. I think that is why I observe higher and more stable currents when I use lithium ion batteries.

MythranDir