[deadpoet]

Hello,

I'm trying to power up a Led strip which came with a battery holder. I connected a USB cable and tried different chargers who basically did the job. The only problem is that the led are way too bright compared to when they are powererd by batteries: the effect is not nice as they are slightly blinding.

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I tried three different chargers:

1. Output: 5V 1A (the one in the picture)
2. Output: 5V 0.7A
3. Output: 3,7V 335mA

The fact is that I can't see any difference between the three. The leds light up with quite the same brightness (too much), whereas with the batteries the light is less bright. This doesn't make any sense to me: the batteries provide 4,5V, but the charger outputting at 3,7V makes them brighter!

What charger should I use?

 

[RAL238]

LED brightness is determined by current, not voltage. I don't know of any USB chargers that will let you manually adjust current. Some "smart" chargers will change current output based on resistors that are wired across the wires in the cable, using them as a code from the device to tell the charger what the device can handle. The lowest is usually 0.5A, which might still be too bright for your case.



I look for a different LED strip and a better LED driver that allows brightness control, unless you want to hack into the wiring to the LEDs.

 

[deadpoet]

Well, the output of the three chargers that I tried ranged from 1A to 0,335A (I guess the latter is less than the current output of a AAA battery, or am I wrong?), yet the brightness remained the same... how come?

 

[RAL238]

Since the LED strip probably has no resistors on the signal lines to set a current level, all three chargers probably defaulted to the lowest level. So they are all delivering about the same current, even though they could deliver higher levels if connected to a proper device.



It's hard to say how much current the batteries are providing. There appears to be a circuit board in the battery pack, and that may contain some circuitry that limits the current from the batteries to something lower than the AC adapter provides.

 

[deadpoet]

So, basically I should try to find an AC adapter that outputs less than 0,3A ?

 

[RAL238]

It's hard to say what the right current level is for the brightness you want. One problem is almost all USB AC chargers put out a minimum of 0.5A.

 

[deadpoet]

Thanks for your explanations!

May it be helpful if I take a picture of the small circuit board in the battery pack and post it here?

 

[deadpoet]

Here it is the small circuit board. Hope it helps!

The switch has three positions:
[ALWAYS ON---FLASHING---OFF]
In this picture is in OFF position.

The wires:
RED: AC Hot
BLACK: AC Neutral
CLEAR TOP: To LED Strip Hot
CLEAR MIDDLE: To LED Strip Neutral
CLEAR BOTTOM: From Battery Pack Neutral

 

[FrodoOne]

To get back to basics, if you have three 1.5 V "Zinc Carbon" cells in series to form a "Battery of Cells", the maximum voltage available will be 4.5 V.
If you "apply" this voltage to a "string" of LEDs (however configured) you will get a certain current flow and a consequent "briteness" in the LEDs concerned.

If you now apply about 5V to the LED string (without introducing any in series dropping resistor to reduce the current flow and, hence the brightness of the LEDs), why would you not expect that the LEDs would glow more brightly?

So, if "the effect is not nice as they are slightly blinding." it is up to you to measure the current when the brightness does not so offend you,
measure the current when it does offend you
and then,
figure out how to introduce sufficient resistance into the circuit concerned to make it pleasant for you in the latter case.

One does hope that you do understand "Ohm's" law, in order to perform these calculations.
If Ohm's law is beyond your understanding, please post the measured current in each of the Voltage situations concerned and I, or others, will calculate the required resistance for you, no doubt giving you advice concerning the theory of "Preferred" resistance values in the process.

Good Luck!

 

[deadpoet]

Done.

Pleasant bightness (battery pack) is about 0,175 / 0,2 A.

My current AC adapter output is 1 A (offending brightness).

Mighty Ohm's Law is beyond my understanding.

 

[RAL238]

The HR706 component appears to be a special purpose LED voltage regulator/blinker device. (Do your LEDs blink?) There's not much you can do to control its function.

The resistor seems to be in series with the power source, and is probably acting to limit current. If you were to remove it and replace it with a larger value resistance, it might reduce the brightness. But that might also affect the proper operation of the HR706.

 

[deadpoet]

Yeah, the switch has three positions: always on, blinking and off, but I don't care about blinking. Which resistance value should I look for the replacement resistor?

 

[RAL238]

I can't be certain of the colors of the bands on the resistor. Looks like they might be brown, black, black, gold? Or maybe red instead of brown?

 

[deadpoet]

They are brown, black, black, gold.

 

[RAL238]

This is where you get into the experimental phase. That's a 10 ohm resistor in there now. It's hard to predict what value would give you the right brightness. You could start with an 15 ohm, 1/2 Watt resistor (brown, green, black, gold) and work up or down from there.

The problem I see is that as the resistance is increased, it will also lower the voltage to the LEDs and they will not work at all if the voltage drops too low. This is a very primitive circuit and there is not much flexibility to alter its operation.

How are your soldering skills?

 

[dmxtothemax]

Lets see, output of charger is five volts'
Output of batteries is 3.6v
There's your answer
if you want to run it off the charger ?
Then you have lose 1.4 volts.
If you can measure the current you can use maths to find resistor required.

 

[FrodoOne]

If the current drawn is 0.2A, as indicated in Post #10, and you need to drop 1.4 V, the resistor required is 1.4/0.2 = 7 Ohms which is nor a preferred value.

The nearest preferred value resistors are 6.8 or 8.2 Ohms.
(Then they go 10, 12, 15, 18, 22, 27, 33, 47, 56, 68 etc for 5% tolerance resistors.)
Heat dissipation will be over 1/4 Watt, so you should use a 1/2 watt resistor.

You do not need to replace any existing resistor.
Just solder the new dropping resister in one of the leads supplying the lights.
It would be a good idea to slip a piece of heat-shrink tubing over the wire concerned so it can be slipped down to cover the resistor when you finally determine that you have installed the correct value.

 

[deadpoet]

You do not need to replace any existing resistor.
Just solder the new dropping resister in one of the leads supplying the lights.

So should I solder the new resistor here?

(The dotted wires are the one supllying the lights)

 

[FrodoOne]

You could, if you have room.

However, you could just as easily solder it in series with the Red wire, since the pattern on the circuit board shows that the terminal concerned is directly connected tho the LED supply wire other than the one which you indicated.

In any case, you really need to redo those soldered connections.
As it is, the wires are just "pasted" on and are not held firmly.
It looks as though either
the iron was not hot enough or
the iron was not in contact for long enough.

The existing and new solder should melt together in a common pool so that, when solid, it forms a shiny solid "globule", like the original ones.

 

[deadpoet]

Thank you, I will. And thanks to everybody for your answers