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RGB Potentiometer help! Answered

Hey guys and girls, I've got a question! I've got 3 voltage regulators set up in parallel, wired to output 3.27V @ an amperage that I don't know (for some reason I can't get a reading on my e-meter) Anyway, The outputs from those 3 regulators are going to 3 pots, which in turn go to a set of 3 LEDs, 9 in total (3R 3G and 3B). My problem is that those pots have a max resistance of 2k ohms, and that the LEDs aren't shutting off when the resistance is maxed out. Can you guys find a way to make the LEDs shut off, yet still go on at full blast when the Pots are open? Thanks, -Josh

Discussions

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lacrimax

10 years ago

For a dimming a led you need work with PWM pulse width modulation, non for voltage variation, in a few days a i gonna make an instructable and get you a pcb for a dimming control using an le555, now i need to find a mosfet that works fine, other thing that you need yo think if you gonna make you one is you gona need a npn transistor for sitching the mosfet that work about 190 mhz for fast and good switching dimming. the led is not a inductance light , the led works with ions . don´t waste your money, varing the voltage you can the led in 10 percent 30 percent and 100 percent , read about pwm. or wait my pcb is very cheap.

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NachoMahma

11 years ago

. Not sure I understand, but try using a "larger" pot. I'll guess that 47K would be easy to find and work well enough to tell if that's the problem. If 47K is too sensitive, try something a little smaller until it feels right.

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T3h_MuffinatorNachoMahma

Reply 11 years ago

The thing is, I want the ratio between the resistance and the power output of the LED to be almost direct, not exponential. I'm going to draw up a schematic, but even with the 2k pot, the LED just suddenly clamps within the last 10 degrees of the pot's rotation. This isn't what I want. I think I might have the wrong set-up. I think my pots are in the wrong place, according to Dan's schematics for his RGB high-powered lamp, that is.

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NachoMahmaT3h_Muffinator

Reply 11 years ago

. If you are using a linear taper pot, get a log (audio) taper. And vice versa.

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NachoMahmaT3h_Muffinator

Reply 11 years ago

. Maybe this will explain a little better. Assume a 100 ohm pot with a dial calibrated 0-100. . With a linear pot, 25% on the dial will give 25 ohms, 50/50, 75/75, etc. . . Don't have a real good explanation for logrithmic. If the pot gives 10 ohms at 10% on the dial, it will give 100 ohms at 20% and 1000 ohms at 30%. That may not be exactly right, but it gives you the idea. . Ever notice how on a audio frequency response chart the low frequencies are spread out and the high frequencies are jammed together? Same thing.

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T3h_MuffinatorNachoMahma

Reply 11 years ago

ahhh, got it now. The linear pots have a direct relationship, they follow a "linear curve", whereas the taper pots follow a logarithmic curve.

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NachoMahmaT3h_Muffinator

Reply 11 years ago

. Doh! I misjudged your level of knowledge/experience. That's it exactly.

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LasVegas

11 years ago

I have a circuit that will allow adjusting the brightness of LEDs by adjusting the current rather than the voltage. While it's quite a bit more complex, it allows driving many more LEDs and a more linear brightness change. If you'd like, I'll share the circuit with you in private.

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NachoMahma

Reply 11 years ago

. It might be more efficient to use the feedback loop to adjust the regulator voltage, instead of dropping it across your 2K pots. Not sure.

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T3h_MuffinatorNachoMahma

Reply 11 years ago

What's the feedback loop? (I'm sorry, all I know about the Vreg is what I picked up from a brief skimming of the spec sheet)

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T3h_MuffinatorNachoMahma

Reply 11 years ago

It actually is kind of broken, the relationship between the pot's resistance and the intensity of the LEDs is certainly not direct. Hmmmmmmmmm ' I'll try that, definitely!

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NachoMahmaT3h_Muffinator

Reply 11 years ago

. The loop from the regulator output that "feeds back" to the adj terminal (the 1K and 1.4K resistors). This is what sets the output of the reg.

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T3h_Muffinator

11 years ago

Eureka! I've got it, but I don't know if it's efficient Man, I'm tired. I wasted my whole day on this! (and ACTs)

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lemonie

11 years ago

I don't think that I understand your set up. I seems like you are throwing 3.27V at an LED through a resistor (0-2K Ohm), but putting 2K in front isn't shutting the LED off? - circuit diagram? L

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T3h_Muffinatorlemonie

Reply 11 years ago

You've got it right, but I'll draw up a schematic anyway.