how do you keep I at V

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Volter
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how do you keep I at V

Post by Volter »

I got a wallwart at 6.3Vdc/1000mA but I need to power a couple of LEDs at 5Vdc/700mA. Would you guys have any suggestion what do to get it work without frying the LEDs? :???:
Thank you much.
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dyarker
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Re: how do you keep I at V

Post by dyarker »

Got a part number for the LEDs ?
Dale Y
Volter
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Re: how do you keep I at V

Post by Volter »

Dale, I got that working. I measured the Vout and found it was more than it said and so I hooked up 3LEDs and it just works correctly!? It has the right V and I. What I would be interested to know if those wallwarts are self-regulated depending on the V and/or I? :idea:
I did not get any notification and so I did not know when you replied.
Thank you,
George
dyarker
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Re: how do you keep I at V

Post by dyarker »

What I would be interested to know if those wallwarts are self-regulated depending on the V and/or I?
Not usually.

Not enough info to say anything else. Not even wild-a guesses.

Kind of LEDs? Hooked serial or parallel? Any other parts?
Dale Y
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haklesup
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Re: how do you keep I at V

Post by haklesup »

A wall wart may be regulated or not. If it is regulated, it would only regulate the voltage. The load decides for itself (by way of ohms law) how much current it needs and the rest is overhead the supply could deliver but does not need to. For the most part, one only needs to match the voltage and make sure the current rating of the supply is higher than the load (device) you are powering. (an exception is LED current sources which work backward to that explanation)

Using a high current supply won't necessarily damage the load if the V is correct but if the load fails for another reason like a short, then the higher current from the oversized supply may cause more secondary damage than one that is better matched (good reason to use a properly sized fuse). Furthermore, an overpowered supply costs more than you need to spend most times.

Wall warts and low cost switchers are not current limited exactly at the rated value. In most cases they will supply at least that much current then voltage will drop off if you try to load it more but that's just running out of power, not necessarily a true current limit circuit at work.

Older ones based on transformers and linear supply architecture have more leeway above the ratings but the newer switcher based wall warts are generally better regulated and cut off closer to their rated power ratings. There are many designs and performance will vary.

a typical LED will work over a range of V and I operating points just like any other diode. running at a slightly high voltage will usually result in higher brightness but sometimes reduced lifetime (often not an issue in a project but a concern for production). So running a LED at 6V instead of 5V probably only resulted in a slightly higher current and brightness, putting 2 or 3 in series will reduce the V drop across each one thereby bringing it closer to your ideal operating point.
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Lenp
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Re: how do you keep I at V

Post by Lenp »

Look into the Microchip CL220. It is a 2 terminal 20ma current regulator with an input of 5V-220V designed specifically for LED's. You series the CL220 with as many led's desired, and as long as the input voltage is adequate this IC regulates the current to 20ma. They can be paralleled for more current output. About $1.50 at Mouser, through hole and surface mount is available.
Len

“To invent, you need a good imagination and a big pile of junk.” (T. Edison)
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cpprioli
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Re: how do you keep I at V

Post by cpprioli »

One thing that nobody has mentioned yet is the concept of current limiting resistors used with LED's. In order to preserve the life of the LED(s), it is necessary to insert a resistor in series with the LED(s). The proper resistance is calculated using the formula RLIMIT = (VSOURCE - (VFORWARD * LED count)) / IFORWARD.

SO... if you were to be placing three LED's in series, with each LED rated at 2.1V for the forward voltage and 20mA for the forward current (you would get these values from the LED datasheet), and if you were using a 9V supply, it would work out like this: R = (9 - (2.1 * 3)) / .020 , or R = (9-6.3) / 0.020.... or R = 2.7 / 0.020, which is equal to 135Ω calculated. The closest E24 resistor would be 130Ω, which would then dissipate 54mW.

To determine what limiting resistance you need for your project, simply plug the correct values into the formula.

BTW -- if your LED's are specified as being 5V devices, they may already have limiting resistors built into the LED unit...
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