How to wire a potentiometer for use dimming multiple LEDs?

djsfantasi

Joined Apr 11, 2010
9,156
The light panel did function for a while, and the LEDs are white.
...Which implies that after a “while”, the panel stopped functioning and no longer functions.

Is this correct? Have you gotten any light from the panel since? Have you considered that the LEDs may have been destroyed?
Have you been able to test the strip since?

Secondly, your posted picture looks like each LED has a current limiting resistor. So it should operate with a 5V supply. Assuming the 3W rating, all LEDs are drawing 0.6A. You could divide by the total number of LEDs to obtain the operating current of one LED. Then you could calculate the inline resistor value. But you don’t need to do that.

Your pot must be rated for more than 3W or 0.6A. At 100% (wiper to panel connection = 0 Ohms), the panel should be bright. Then, turn the pot down to the lowest level of light you need. The wiper to connection resistance is the resistance you need for your pot (which is rated at 3W.) You don’t need a second pot nor additional resistors, unless you also want to set a maximum brightness.
 

Bernard

Joined Aug 7, 2008
5,784
Pot dissipation is less than 1 W. With an emitter follower there is about a 1 V loss which is noticeable with a 5 V supply. There is an example on page 6 , Chat, " Attaching an EL Wire to a switch---".
 

Thread Starter

birckcmi

Joined Jan 1, 2018
210
...Which implies that after a “while”, the panel stopped functioning and no longer functions.

Is this correct? Have you gotten any light from the panel since? Have you considered that the LEDs may have been destroyed?
Have you been able to test the strip since?

Secondly, your posted picture looks like each LED has a current limiting resistor. So it should operate with a 5V supply. Assuming the 3W rating, all LEDs are drawing 0.6A. You could divide by the total number of LEDs to obtain the operating current of one LED. Then you could calculate the inline resistor value. But you don’t need to do that.

Your pot must be rated for more than 3W or 0.6A. At 100% (wiper to panel connection = 0 Ohms), the panel should be bright. Then, turn the pot down to the lowest level of light you need. The wiper to connection resistance is the resistance you need for your pot (which is rated at 3W.) You don’t need a second pot nor additional resistors, unless you also want to set a maximum brightness.
The problems with this panel are the crappy power connection and the switch. I have three panels. The other two are different sizes and voltages, but those both have a power cable coming out of them with either a 3.5mm female barrel connector on the end, or a 5.5mm connector. Both of them work just fine. I follow the instructions, punching the membrane switch on the front of the panel, and the light comes on or goes off, and goes through its levels of brightness. On this one, the female connector is a mini-USB socket molded into the frame of the panel, designed to accept power(5VDC)from a USB outlet . In other words, a POS. It sort of worked when I got the panel, then it worked sometimes, then it didn't work at all. So I routed the whole switch and socket assembly out of the panel, winding up with one black, one red wire in the arrangement in my photo of the LEDs above. I'm pretty sure the LEDs are OK. Thanks for the suggestions regarding components and voltage. And the 25-ohm pot is rated at 5 Watts.
 
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ElectricSpidey

Joined Dec 2, 2017
2,758
I wish you would have asked in public, so people don't think I'm being presumptuous.

Here is the schematic.

A_Emitter_Follower.jpg

1/2 watt pot
High gain TO220 Transistor
Small heatsink

The 10Ω resistor is also optional.
 

Thread Starter

birckcmi

Joined Jan 1, 2018
210
Thanks for the details and the schematic. If I use the 10-ohm resistor, does it need to be one of those 2" long cement resistors, or will a standard color-banded brown-black-black resistor do the job?
I retried post # 6 with a 3W LED. 100 ohm pot connected from bottom of 25 ohm pot to common gave full cutoff of light when set to 20 ohms. 25 ohm pot then gives full range of dimming. Results are slightly skewed from the large number of clip leads. First light @ 2.3 V, 3.2 V at 500 ma, losses too great to reach full 600 mA.
Bernard, here is my front-view version of your schematic. I'm not sure about it, so I'm sending it for approval or...not. Mostly it makes sense, but I'm not certain about the connections on the 25-ohm pot. Negative from the 5V source(?) goes to the negative side of the LEDs (sounds good), then to the 100-ohm wiper, and from there to the #3 connector on the 100-ohm pot, and back out to the #1 connector on the 25-ohm pot?Bernard circuit.jpgBernard circuit.jpgBernard circuit.jpg
 

Tonyr1084

Joined Sep 24, 2015
7,852
First, regarding what size resistor to use - that depends on wattage. Multiply current times voltage to get your wattage. Then go to the next size up if your wattage is close to the resistor's rating.

Second, parallel LED's is not good practice. It's doable, but you have to make sure the forward voltage of each LED is nearly identical. I have a video on that.

 

Bernard

Joined Aug 7, 2008
5,784
birckcmi, your schematic looks good & correct. Review post # 19. If you do not have a 100 ohm pot just try 18, 20 or 22 ohm Rs, 1/2 W.
 

Thread Starter

birckcmi

Joined Jan 1, 2018
210
I didn't calculate wattage for the 10ohm resistor, but I'm sure a 1/4 watt will do.
ElectricSpidey, let me restate the consequences of this circuit, to see if I have this right: Resistance inversely affects current, and current is what needs to be varied to increase or reduce brightness of the LEDs, right? To do that, the 10K pot can be set between zero and 10K ohms resistance, but at the zero end the power supply makes considerably more current available than the 3 watt LED strip requires. To reduce the current accessible by the LEDs to a quantity closer to 3 Watts, i.e., .6 Amps, the power transistor is added between the pot and the LED strip such that the actual power to the LED strip is provided by the transistor, which in turn is controlled from the pot.
I can understand that the 10-ohm resistor between the pot wiper and the transistor Base reduces the control range of the transistor to .5 amps max, which is just under the LED strip wattage (.6A), but that doesn't tell me what current is passing from the Transistor's Collector to the Emitter. What is the relationship between current to the Base and current out from Emitter? I'm sure there's a stupid question buried in there somewhere, but that won't surprise anyone on the forum.Lightbox-transistor.jpgLightbox-transistor.jpg
 

ElectricSpidey

Joined Dec 2, 2017
2,758
The 10 Ω resistor will not reduce the current output from the circuit, it is only there to reduce possible oscillations, at a very small reduction of output voltage.

Yea, basically the follower circuit does exactly what the name implies, that is it follows the input voltage presented to the base from the pot, set up as a divider.

(so if the base sees 2 volts the load sees 2 volts minus the loss across the transistor)

The emitter follower is a unity voltage gain circuit that has only current gain.

But, after thinking about this you are better off just using a pot, because for some reason I was thinking you were using 12 volts and not 5.

Sorry, I should have said something earlier.
 

djsfantasi

Joined Apr 11, 2010
9,156
I agree with @ElectricSpidey

You just need the pot. Your led strip is designed to operate in 5V. It has built in current protection, so initially you don’t need to worry about the current available EXCEPT it must be greater than the current needed.

When the pot is at “zero”, it’s just as if it isn’t in the circuit at all. As you move the pot, it’s effectively reducing the current to the LEDs and thus they dim
 

Thread Starter

birckcmi

Joined Jan 1, 2018
210
I agree with @ElectricSpidey

You just need the pot. Your led strip is designed to operate in 5V. It has built in current protection, so initially you don’t need to worry about the current available EXCEPT it must be greater than the current needed.

When the pot is at “zero”, it’s just as if it isn’t in the circuit at all. As you move the pot, it’s effectively reducing the current to the LEDs and thus they dim
So I don't need the transistor at all? Just the pot? So that with the pot at zero, all current is going to the LEDs, thus bright LEDs, and as I turn the pot toward greater resistance, the LEDs dim? But the available current must be more than .6A, or is .6A enough? Would this be the new schematic?
Lightbox no transistor.jpg
 

djsfantasi

Joined Apr 11, 2010
9,156
So I don't need the transistor at all? Just the pot? So that with the pot at zero, all current is going to the LEDs, thus bright LEDs, and as I turn the pot toward greater resistance, the LEDs dim? But the available current must be more than .6A, or is .6A enough? Would this be the new schematic?
View attachment 220497
If the switch is open, how is the LED strip going to get power? The switch should go before the junction with the pot wiper.

If the power supply can provide 0.6A or more, that is the only current consideration that you need.
 

Irving

Joined Jan 30, 2016
3,843
All of the above is true, but LED brightness is not directly proportional to current. If you want 'proper' control of light output from 0 to 100% you need to use a different technique, Pulse Width Modulation, which has the benefit of not wasting energy as heat in resistors, works by turning the LEDs on full for a short time repeatedly around 50 times a second or more. Your eyes don't see the flashes, they see an average brightness proportional to the time they are switched on for. There are lots of ways to do this, I can provide a simple circuit if you want.
 
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