LED fade in/ fade out

Wendy

Joined Mar 24, 2008
23,415
My personal convention is Vcc is the + side of the power supply, ground is the negative.

Most 555 chips also use the Vcc convention.

As to the experiments, try using 3 LEDs in series along with a 8.2Ω or 10Ω ¼W resistor, then see how much current is used with the 12VDC power supply. It should be in the ball part. Use the resistor to measure the current, and use the number to adjust the resistors you will need. Just in case keep the duration short as possible.

With 3 LEDs per chain there will be 7 chains, this will work out to approximately 1.05A, the power supply will barely work for this mode. If you were using 2 LEDs per chain the resistor would be around 33Ω 1W, and it would draw 1.5A, overloading the power supply.

The problem using 3 LEDs per chain and a small resistor is stability. It isn't very good, and predictability is another issue. Basically you need to try it out on several sets of LEDs (without the dimmer) and see how close it comes for you.
 
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Wendy

Joined Mar 24, 2008
23,415
First, try the experiments, with a 33Ω 1W resistor and two LEDs and with a 8Ω ¼W resistor 3 LEDs. We need to see how well this works. I also need to look up the specs on your LEDs (just getting around to it). Just in case, you may want to double those resistance values. The idea is to check out what you have. The spec 3.0-3.8V bothers me, lets see what it really is. Do you know what Vf is? That is what we're researching.

We could build a current regulator per chain, but I think that would be unnecessarily complex. If you want to go that route we could, but my opinion is once we find a resistor that works it will be good enough.

These LEDs are going to get warm (.4-.6W), you will need to either give them a heat sink or make sure the finished project has good air flow around the LEDs.
 

Wendy

Joined Mar 24, 2008
23,415
Don't trust calculators, they are very often wrong, and have no common sense. I'm after real world values here. This is why I would like you to connect them up for a test.
 

Thread Starter

ucitelot

Joined Mar 14, 2011
29
i think ive done schematic of the pcb, when you have time if you can look at it to know is it ok


and what value are Q1 Q2 Q3

about 7555, what better to use 7555 or 555 (will ask at shop what they have)
 

Wendy

Joined Mar 24, 2008
23,415
You still left U3A off. Pins 1 - 4 of U3 are grounded.

C1 is going to be a bit larger than you have shown, you also need to mark the + lead since it is a electrolytic.

C3 is missing, it is also a large electrolytic, so the + side will need marked.

I'll do a more precision look over later. I like your work to date, may I incorporate your PCB in the article?

/

Here is the reason I would like you to try the circuit out as chains. The specs (which I suspect are a bit bogus) say Vf of 3.0 to 3.8.

Work the numbers, if you use a 10Ω resistor and a 12VDC power supply.

If Vf = 3.0V drop, total 9.0V, The resistor drops 3.0V, I = 300ma. Not good, LEDs smoke.

If Vf = 3.3V drop, total 9.9V, the resistor drops 2.1V, I = 210ma. Still not good.

If Vf = 3.6V drop, total 10.8V, the resistor drops 1.2V, I = 120ma. Much better, very acceptable.

If Vf = 3.8V drop, total 11.4V, the resistor drops 0.6V, I = 60ma. Pretty low.

This is why I think you need to try it with a 20Ω resistor first, measure the Vf of the chain, then adjust the resistor accordingly.
 

Thread Starter

ucitelot

Joined Mar 14, 2011
29
about c3 and c6
c6(0.1mF) to connect across power suply (on LM393 its pin8)
and also c3 (220mF)to ground also on pin8

and q1/2 will put 2N2222Ai got them at home, what about q3
 

Wendy

Joined Mar 24, 2008
23,415
Q1 - 2N2222A NPN
Q2 - 2N2907A PNP

Q3 - nMOSFET, such as IRF510 (Not a very good part) IRF520,IRF530 (better). Any decent MOSFET will work.

C3 is shown across U3, but it isn't critical where it is. The small caps C4, C5, and C6 should be close to the chips to supress any digital noise.
 

Thread Starter

ucitelot

Joined Mar 14, 2011
29
no i mean on LM393 (U3)
on pin 8 to connect c3 and ground, and also on pin 8 to connect C6 and power
or C6 its on other pin
C4,5,6 i put as close as can, i think thats good
 

Wendy

Joined Mar 24, 2008
23,415
Again, it doesn't matter where C3 is, as long as it is across the power supply. It was drawn that way because it was easy to draw that way. C3 can be anywhere. I mentioned C4-C6 in a note because those were important where they were. C3 was not part of that note.
 

Thread Starter

ucitelot

Joined Mar 14, 2011
29
here is with changes, take a look, so dont make any mistake


i order the components i dont have, till then will try to simylate with Multisim
 

Wendy

Joined Mar 24, 2008
23,415
Couple of minor errors, both U1 and U2 have pin 8 not connected to Vcc. Connecting them will cause problems for C4, C5.

Pin 8 of U3 also not connected to Vcc. Connecting it will cause problems for C6. C3 not connected to Vcc

Still looking.
 

Wendy

Joined Mar 24, 2008
23,415
C4, C5, and C6 are shorted, those are the problems I mentioned.

I have to say, you are one of the more motivated people I've helped. It is refreshing.

I was thinking about your current limiting resistors, I redrew the schematic a little.



The idea is you can use a resistor value guaranteed not to damage your LEDs, say 18Ω ¼W, then adjust the value with the second resistor to bring it up to full brightness. The LEDs survive, and you get what you want.

If you want to eliminate D21 the two resistors will help adjust for that too.

.
 

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ucitelot

Joined Mar 14, 2011
29
thanks man, when i start something i wanna done well, hope this will end well,
and you are most helpfull person, thanks a lot

here is the change, i put closed pins for C4,5,6


about the resistors to leds the one whats best 18Ω and other to adjust, corect
 

Wendy

Joined Mar 24, 2008
23,415
Yes. C4 - C6 are still shorted. They need to connect to pins 1 and 8 of U1 - U3. Move the capacitors to the end of the ICs.
 

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