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Old 23rd July 2008, 10:07 PM   (permalink)
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Quote:
Originally Posted by blckscab View Post
Hmmmm.... I can always buy more components if i need to?
Would that other TL072 chip be able to fix this issue? Since ill be buying a few more resistors from there i would like to save on postage
I would not expect that the TL072 would fix it, no. . .it was just because that's what I simmed with and I was wondering about it. However, sims don't get everything right and can't really be trusted to tell you when things will work or not.

Quote:
and what voltage LEDs would be desirable? Although the ones i have didnt frazzle out when they where on the 12v... I could also email the website i got them from to see if they can tell me what there voltage is?
You can actually run LEDs at pretty much any reasonable voltage, as long as you make sure the current running through them doesn't get high enough to burn them out. The simplest way to do that is to use a current-limiting resistor, such as the 100Ω resistor used in that schematic. You can calculate the needed resistor if you know the current demand and forward voltage of the LEDs but we're having to guess at what your LEDs' values are since there's no datasheet.

If you're ordering more parts, I'd recommend getting a value pack or grab bag of resistors so you will have various values to play with. Schematics often show what worked for the author; you will often need to tweak the values to get things to work for you depending on what your actual circuit ends up looking like.


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Old 23rd July 2008, 11:39 PM   (permalink)
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Just bought myself a pack of assorted resisters so i can play with the values a little to see what happens

is there no way of just reducing the 12v (after rectification and filtering) into around 9V to stop the problem all together?
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Old 24th July 2008, 12:56 AM   (permalink)
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You can use an LM7809 voltage regulator IC plus two capacitors to reduce the 12V down to 9V.

But my calculations show that when the supply is only 12V and two LEDs are in series and the timing capacitor is two back-to-back then it should work.
Try connecting three LEDs in series and in series with the 100 ohm resistor. Then try four LEDs.
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Old 24th July 2008, 04:16 AM   (permalink)
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Quote:
Originally Posted by audioguru View Post
You can use an LM7809 voltage regulator IC plus two capacitors to reduce the 12V down to 9V.

But my calculations show that when the supply is only 12V and two LEDs are in series and the timing capacitor is two back-to-back then it should work.
Try connecting three LEDs in series and in series with the 100 ohm resistor. Then try four LEDs.
Yup, worked for me, although when I built it a couple of hours ago I did have some differences in my circuit: I don't have a 1458 around and all my 3904s are in use. I used a TL082 and a 4401 instead. Also, I just used a 22uF non-polarized cap instead of the back-to-back 47uFs.

Worked fine anywhere from around 7 volts clear up to 18 (well, just under 18; I didn't feel like cooking my op amp). The fades got quicker and brighter as I boosted the voltage but it was perfectly happy at 12V. I threw in a third LED just to keep the current relatively safe (didn't bother with the resistor because I only had the thing running for less than a minute).

I don't have any ideas right now as to why blckscab is getting that result at 12V, but that will have to wait for after dinner. I never have any good ideas right before dinner.


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Last edited by Torben; 24th July 2008 at 04:17 AM.
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Old 24th July 2008, 04:23 AM   (permalink)
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If the collector and emitter of the transistor are wired backwards then the transistor will conduct like a zener diode when it is supposed to be turned off.
The collector and emitter of BCxxx transistors are reversed from 2Nxxxx transistors.
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Old 24th July 2008, 04:37 AM   (permalink)
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That could be it. Blckscab, have you tried flipping the transistor around and seeing what happens?


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Old 24th July 2008, 04:44 AM   (permalink)
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What if he has a huge AC component on the adapter? Just a thought.
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Old 24th July 2008, 05:22 AM   (permalink)
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I went back out to the shop and tried flipping the transistor with my power supply set to 12VDC. The LEDs stayed off. However, when I boosted the voltage to about 13-14V, the LEDs came on and stayed on. So that could be the problem: the 12VDC adaptor is probably not loaded enough by the circuit to get down to 12V, which would make a backwards transistor conduct like audioguru mentioned.

I just realized while writing the above paragraph that this didn't explain to me why the thing would have worked on the 9V battery, so I ran back out and tried it with the transistor backwards at 8, 9, 10, 11, 12, 13, and 14 volts. The results were as follows:

8V: No action from the LEDs (they just stayed off).
9V: Weak fading on the LEDs.
~9.5-~10.5V: Strong fading on the LEDs; working nicely.
11V: LEDs stayed off.
12V: LEDs stayed off.
~12.5V-14V: LEDs stay on.

With the above results, that might explain nicely why blckscab saw the thing working on the 9V battery, but the LEDs stayed on with the 12V adaptor.

Blckscab, can you try flipping the transistor? Given my above results it seems likely that your results could be explained by the transistor being backwards in the circuit.


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Old 24th July 2008, 01:19 PM   (permalink)
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Well, isnt this getting interesting lol

im at work at the moment but i will try this out when i get back home and see what happens. I looked at a website to find out what legs of the transistor do what. 1 = collecter, 2 = base, 3 = emitter etc...

So maybe the website was wrong or ive done a silly mistake, but i am using a 2N3904 transistor like the one used in the 'fading eyes' adobe file

Edit:
also got this email from the website i buy my components from...

Hello Alex, the supply voltage for both these LED`s is exactly the same:
1.85v - 2.5v max @30ma. I hope this will help. Good luck with your project.
Kind Regards, Will.

Last edited by blckscab; 24th July 2008 at 03:54 PM.
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Old 24th July 2008, 04:08 PM   (permalink)
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You should look at the photo of the transistor showing which pin is which in its datasheet.

Your LEDs have a wide range of forward voltage. If their voltage is at max then two in series will barely light in the circuit when the supply is 9V.
Attached Images
File Type: jpg 2N3904.JPG (6.3 KB, 2 views)
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Last edited by audioguru; 24th July 2008 at 04:13 PM.
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Old 24th July 2008, 06:22 PM   (permalink)
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The circuit works fine after i turned the transistor round!
Glad that is fixed and i understand the cause (dont understand the transistor acting the way it did, but what i know is good enough, hehe)

I have tried the circuit with 1-3 pairs of 2 LEDs in series and all give a desirable fading effect. Should i still replace the 100ohm resister with a 330 one?

None of the components i recieved came with a datasheet... so i guess that didnt help much, lol. Dead again, i havnt scanned the website for one either. Ill have a look in a second to see if they are on there. That pic of the transister was indeed different to the one i used earlier, but now i know some transistors are opposit and can look out for that

Next step would be the rectification circuit and put the 2 together... (dramatic scary music starts now)
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Old 24th July 2008, 07:02 PM   (permalink)
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Hmmmm, ok good news n bad news

Ive put the whole circuit together, the motor, the rectification and filtering part of the circuit and the fading circuit. (mind you the resistor infront of the LEDs is still 100Ohms and im only using 2 lEDs parellel)

Good news is that it works, LEDs light up, and motor turns

Bad news is that the fade seems to be working oddly. it will fade a little and then turn off and then turn on and restart. After this result i put the 2 LEDs in series just to see the difference, and now the LEDs seem to switch between half on and on with no 'fade' inbetween....

what does that mean?

I also notice that the circuit will carry on for a bit after turning it off. im guessing that is because of the large capasitor used in the filtering circuit?

Edit:
Also when i take the plug out, when it is still goin for a bit it does seem to fade properly.
Is it that the rectification and filtering part of the circuit is messing the fading circuit up?

Last edited by blckscab; 24th July 2008 at 07:06 PM.
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Old 24th July 2008, 07:08 PM   (permalink)
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Hey hey! Good stuff.

OK, your LEDs have a Vfwd range of 1.8 to 2.5 V volts. We can calculate the needed resistor to get a desired current through an LED with Rled = (Vsupply - Vfwd) / Iled

. . .which means we can calculate the current through the LED when using a given resistor: Iled = (Vsupply - Vfwd) / R.

I did up a little table in Calc which shows the current through the LEDs using one and two LEDs, at the top and bottom of the Vfwd range, and using 100Ω and 330 resistors:

Code:
Vsupply	Vled	R	Iled	Notes
12	1.8	100	.102 A	One bottom-of-range LED at 12V exactly.
12	2.5	100	.095 A	One top-of-range LED at 12V exactly.
12	3.6	100	.084 A	Two bottom-of-range LEDs at 12V exactly.
12	5	100	.07 A	Two top-of-range LEDs at 12V exactly.
12	1.8	330	.031 A	One bottom-of-range LED at 12V exactly.
12	2.5	330	.029 A	One top-of-range LED at 12V exactly.
12	3.6	330	.025 A	Two bottom-of-range LEDs at 12V exactly.
12	5	330	.021 A	Two top-of-range LEDs at 12V exactly.
14	1.8	330	.037 A	One bottom-of-range LED at 14V exactly.
14	2.5	330	.035 A	One top-of-range LED at 14V exactly.
14	3.6	330	.032 A	Two bottom-of-range LEDs at 14V exactly.
14	5	330	.027 A	Two top-of-range LEDs at 14V exactly.
According to that, even the 330Ω resistor is pushing it given that your supply voltage is going to be a bit above 12V after rectification and filtering. I'd suggest trying a 470Ω resistor just to be safe.


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Old 24th July 2008, 07:13 PM   (permalink)
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well thats good because im getting a few 470Ω with my assorted resistor pack hehe
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Old 24th July 2008, 07:22 PM   (permalink)
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Quote:
Originally Posted by blckscab View Post
well thats good because im getting a few 470Ω with my assorted resistor pack hehe
Good. The LEDs might appear a little less bright but they'll last longer.

What does the fading circuit do if you simply disconnect one wire from the motor?


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