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PCB Fabrication
19 December 2010
Author: Giorgos Lazaridis
LED Fade-In Fade-Out Dimmer

Some time ago, i made an LED fade out dimmer for my Men Of War game console. The circuit was ridiculously easy, using only one transistor and a couple more passive components. Now, i tried to make a similar circuit, but with a Fade-In Fade-Out cycle (the previous had only fade-out). So, here are some very interesting observations that i made while i was designing the circuit.








A very simple Circuit with a slight flaw

Here is the first circuit schematic that i designed, and i tried to keep it as simple as possible:





Similar to the LED fade out dimmer, multiple LEDs can be connected in parallel with one resister per node. Again, you need to take into account the maximum ratings of the transistor before connecting a million LEDs.


The circuit on a breadboard for test

The operation is as follows. When the pushbutton is pressed, the capacitor starts to charge through R1. For now, we will not take into account R2, which is at least 7 times larger than R1. While the voltage across the capacitor raises, the same happens to the base of the transistor as well. Gradually, the voltage across the LEDs is raised. When this value reaches the lower LED forward voltage, the LEDs will turn on very dim. But the capacitor's voltage still increases, and the same happens to the voltage across the LEDs, and so the fade in slowly.

The opposite will happen when he button is released. The capacitor will now start to discharge through the parallel resistor R2. The LEDs will fade out slowly, until they are completely turned off.

This circuit is ideal for applications that simplicity is needed. Yet there are some things that you need to take into consideration. First of all, the R1 and R2 resistors perform a voltage divider. If you try to reduce the fade-in time by increasing the R1 value, this will have a negative impact on the total voltage that can be delivered to the capacitor. A reduced capacitor voltage means reduced base voltage on the transistor, and in turn this means reduced LED voltage. Same will happen if you try to reduce the discharge time by reducing R2.

What this means, is that you can only make small changes to one of the R1-R2 resistors at a time. It is wise to keep R2 at lest 7 times larger than R1. You can experiment with the fade-in-out times also by increasing or decreasing the capacitor's value. In general, this circuit is not very easy to change the times.

But the major disadvantage of this, is the turn-on time. If the capacitor is completely discharged and you give power to the circuit, it will take some time before the LEDs start to fade in. This time depends on the C1 and R1 values. Greater RC, means more time. The values that i have chosen makes a delay of approximately 3 seconds. Which means that, if i power the circuit, the LEDs will start to fade in after 3 seconds. Why is that? It has to do with the LED minimum forward voltage. They will not light until this voltage is reached. If this is not a problem to you, then this is the proper circuit for your application.

Like many other circuits, it can operate in different voltages, but you will have to experiment yourself to find the proper R1 R2 and C values.






Bill Of Materials
Resistors
R1Resistor 4.7 KOhm 1/4 Watt 5% Carbon Film 
R2Resistor 33 KOhm 1/4 Watt 5% Carbon Film 
R3Resistor 47 Ohm 1/4 Watt 5% Carbon Film 
Capacitors
C1Electrolytic Capacitor 470 uF 25 Volts
Transistors
Q1BC548 Switching and Applications NPN Epitaxial Transistor 
Semiconductors
LED1-3LED 3mm white high brightness



To solve the problems - Circuit #2

To solve the above "problems", i introduced a PNP discharging transistor to the circuit and a potentiometer to provide a constant base stand-by voltage to the first transistor:





This circuit is a little bit more complicated. Lets see how it works. And first of all, what this R2 potentiometer does. This small potentiometer is connected as a voltage divider, the output of which is driven (through a diode) to the base of the transistor. By adjusting this potentiometer, you actually change the bias of the transistor, or should i say the stand-by voltage of the base (and of course the capacitor that is connected). The fact that the potentiometer is small (500+100 (R1) max resistance) means that the RC time constant is very small. The maximum time that the capacitor will take to be charged at the stand-by voltage, is theoretically R x C = 600 x 470 x 10-6 = 0.282 seconds = 282 mSec, a quoter of a second. The usual value of the potentiometer will be somewhere in the middle, so the usual RC time constant will be about 165 mSec. So, when you power the circuit, the capacitor will be charged to this voltage within a sixth of a second.


The second circuit on a breadboard for test

Now suppose that you press the button. The 12 volts will be delivered to the capacitor through R3-D2, and it will keep on charging with a new time constant, that is R3 x C. But if you notice now, there is no discharging resistor, at least not as long as the button is pressed. Actually, the discharging resistor is the R4, but as long as a positive voltage is supplied to the PNP Q2 transistor, this resistor takes absolutely no part in the circuit. So, the charging of the capacitor (and the LED fade-in) depends only on the capacitor and R3.

Now, suppose that you release the button. The Q2 gets no more positive voltage on its base and is pulled down through the R5. R4 now acts as discharging resistor. The capacitor will be discharged through R4 and Q2. So, the discharging cycle depend only on the capacitor (of course) and R4.

What have we achieved so far:

First of all, the charging and discharging times are independent. You can adjust different fade-in and fade-out time by simply changing the R3 or R4 value. One limitation: Try to keep R3 as low as possible, maximum 1.8K. This way you will ensure that enough current will flow through the LEDs. If you need to further increase the fade-in time, you can increase the capacitor as well. So, the setup of the circuit is a piece of cake.

Most important though is what the potentiometer does. Suppose that the capacitor is fully discharged and you give power to the circuit (the pushbutton is already closed). The fade-in cycle of the LEDs, REGARDLESS the R3-C time constant, will start within less than 125 mSec, because the capacitor will be charged through the potentiometer up to the stand-by voltage! No more delay.

To adjust the potentiometer, you power the circuit WITH the pushbutton OPEN. Then, you adjust the potentiometer so that the LEDs are barely off. That is the idea. This is the stand-by voltage. You can of course do something more artistic: You can adjust the potentiometer so that the LEDs are very dim but NOT off. So, when you power the circuit, the LEDs will be very dim, when you press the button the LED will fade in to maximum brightness, and when you release it they will fade out to the dimmed level they were.






Bill Of Materials
Resistors
R1Resistor 100 Ohm 1/4 Watt 5% Carbon Film 
R2Potentiometer 500 Ohm Linear Rotary 1/2W
R3Resistor 1.5 KOhm 1/4 Watt 5% Carbon Film 
R4Resistor 4.7 KOhm 1/4 Watt 5% Carbon Film 
R5Resistor 220 KOhm 1/4 Watt 5% Carbon Film 
R6Resistor 330 Ohm 1/4 Watt 5% Carbon Film 
R7Resistor 47 Ohm 1/4 Watt 5% Carbon Film 
Capacitors
C1Electrolytic Capacitor 1000 uF 25 Volts
Transistors
Q1BC548 Switching and Applications NPN Epitaxial Transistor 
Q2BC327 Switching and Amplifier Applications PNP Epitaxial Silicon Transistor 
Semiconductors
D1-21N4148 Switching Diode 
LED1-3LED 3mm white high brightness



Need more current?

A power transistor driving 36 high brightness LEDs

If you intend to power many LEDs, then you will notice that this BC548 transistor is just not powerful enough to deliver the current you need. So you need to add another amplifying stage with a power transistor. For example, you can use the BD243 like i did. Here is a schematic how to connect this transistor. This schematic can be applied to both circuits shown above, directly at the emitter of the BC548:





Q1 is the BC548 transistor from the previous circuits. T1 is the BD243 power transistor. The above schematic has 15 LEDs (5 lines of 3 LEDs each), but you can add more lines. The limiting resistors i used are 10 Ohms, but this has to do only with the type of LEDs you will use! This resistor may be completely improper for use with other LEDs. Either calculate the resistor you need if you are sure about the specifications of the LEDs, or use different resistors and measure the current drown from each LED line (which should NOT be more than the nominal current of one LED - usually 25 to 30 mA).








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Comments

  • At 29 April 2012, 2:12:57 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
      • @Ruben Better check these out:
      http://www.pcbheaven.com/userpages/Flexible_555_LED_Pulsing_Circuit
      http://www.pcbheaven.com/circuitpages/555_Breathing_Pulsing_LED

  • At 28 April 2012, 18:09:56 user Ruben wrote:   [reply @ Ruben]
      • Cant the first circuit be use at the output of a 555 timer to make a simple led flasher fade in and out instead of just blinking

  • At 28 April 2012, 7:50:19 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
      • @Darren 30-12 = 18 volts at 320ma=5.76 watts of dissipation on the transistor. It is absolutely normal that it will get too hot. anything above 2 watts is already enough to heat up the transistor fast. What you can do is reduce the input voltage to nearly the voltage required by the LED - for example you need 12 volts on the led? use 14 volts supply.

  • At 27 April 2012, 4:03:19 user Darren wrote:   [reply @ Darren]
      • @Giorgos Lazaridis @Giorgos Lazaridis

      Hi, i'am using the circuit to power-up a LED downlight,the source from the LED power driver is about DC 30 volts (320mA), the BD243 is extremely hot(with heatsink). What should i do? Thanks

  • At 10 February 2012, 0:37:58 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
      • @Annonomous you will need certainly to add a power transistor (as explained in the document) and a heatsink. I cannot tell you more. You have to do some tests.

  • At 9 February 2012, 18:04:18 user Annonomous wrote:   [reply @ Annonomous]
      • Hello I am in middle school and i want to put some led light strips in my locker that fade in when i open the door. I was wondering if i need any different components for a led light strip I found here.

      http://ledwholesalers.com/store/index.php?act=viewProd&productId=474

      Please tell me if this will work or what kin of components I need

  • At 9 February 2012, 10:57:00 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
      • @Jon strange indeed. sign up in the forum and send me the manual as attachment

  • At 9 February 2012, 10:46:25 user Jon wrote:   [reply @ Jon]
      • @Giorgos, I got very strange result. First, when the PIR is INACTIVE (live but not sensing motion difference), the OUTPUT reading shows 10.36V. However, when the PIR is active (i waved my hand), the OUTPUT reading shows 10.45V.

      Based on the PIR datasheet (how can I send it to you?), it is active LOW but yet I don't see this behaviour in the output. The LED does shows brighter and it is INACTIVE and dimmer when the PIR is ACTIVE (contradicts with the Voltmeter reading). How can I test if the PIR is working or faulty?

      On the other hand, when the pull down resistor is connected, the OUTPUT reading shows between 0V and 0.22V.

      Thanks and appreciate your help !

  • At 7 February 2012, 10:10:37 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
      • @Jon Oh, i see what is the problem here. The output does not go lower than 2.5 volts. 2 things you should do and reply: First, measure the voltage with multimeter for the 2 different states of the PIR, and second, instead of a pull-up use a pull-down and tell me the results. We will sort it out.

  • At 7 February 2012, 5:46:44 user Jon wrote:   [reply @ Jon]
      • Hi Giorgos, thanks for the advice. My previous connection is quite similar to your proposed connection. I had connected the LED strip on the collector instead of the emitter. By doing so, the LED strip will have the full 12V across it instead of the 0.7V drop on the emitter. Is this advisable? I have, however, wired up the proposed connection and found the result are the same regardless if the LED strip is connected to the Emitter or the connector. The LED strip remains lit (same level of intensity) regardless of the PIR state.

      Thus, I fall back to the basic connection which is just a pull up resistor and a single LED (not LED strip) to the PIR output. The LED does shows visual difference. When the PIR is inactive, the LED is bright. When it senses motion from my hand, it becomes dimmer. However, I do not find this differences in the LED strip. Could you advise what should I do?

  • At 7 February 2012, 2:26:41 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]
      • @Jon You need an amplifier, probably an emitter follower to get maximum current. Connect the base of the transistor to your output (with a pull-up resistor around 470 Ohms or 1K). The collector is connected directly to the power supply of the strip and the emitter directly to the strip positive. This will get you enough current for the LEDs, only that the output will be about 0.7V less than the input.

  • At 7 February 2012, 1:29:45 user Jon wrote:   [reply @ Jon]
      • Hi, you are amazing. Could you help me on this design?

      I'm intending to use a PIR to control the LED strip. When there is someone in proximity, the LED strip will fade in and after a couple of secs when no one is there, the LED strip will fade out.

      I have a PIR with a open collector output and I have connected a pull up resistor to the output lead. I have realized the OUTPUT is too weak to power up the LED strip. If a normal LED is connect to the OUTPUT lead, i could see the light but not when LED strip is connected to it. Could you please advise how should it be done? Thanks.

  • At 8 November 2011, 3:33:42 user Kammenos wrote:   [reply @ Kammenos]
      • @vimal it is not for 1 LED, it is for 3 in series with a resistor. The output voltage shifts up to (around) 1v bellow the supply voltage (which is 12V). If you read the complete page, i add a BD243 to increase up to 6A the ouput. You may use a bigger transistor to increase further more. As for the lower voltage, prototype the circuit on a breadboard and you will find out how to use the potentiometer.

  • At 8 November 2011, 3:22:14 user Kammenos wrote:   [reply @ Kammenos]
      • @Markus184 you may wanna increase R3 and C1

  • At 6 November 2011, 22:16:25 user Markus184 wrote:   [reply @ Markus184]
      • Hello. I was wondering if you could help me. Is there a way to make the fade-in and fade-out times 3 minutes or higher without reducing the current. I am planing on using the first circuit with 12v psu running 2 LEDs.

  • At 6 November 2011, 1:50:02 user vimal wrote:   [reply @ vimal]
      • wow.........thats 95% what i was looking for.great work man.

      your circuit is for one led only.in fact i want such circuit because i wanted to put led lights beneath my car,i will use led strips for that.

      (1) but the led strips uses around 6amps at 12v.how should i modify the circuit so as it can power the led lights at 6amps.

      (2) your circuit uses 5v,is there a way to use 12v to power your circuit?

      (3) i dont want the lights to fade out completely,how can i use a potentiometer to do that

      i know i am asking a lot but i wanted such circuit since 2 years.please help

      thanks

  • At 5 November 2011, 20:51:42 user Kammenos wrote:   [reply @ Kammenos]
      • @vimal try this one instead:
      http://pcbheaven.com/circuitpages/555_Breathing_Pulsing_LED

  • At 5 November 2011, 15:42:31 user vimal wrote:   [reply @ vimal]
      • hello friend,nice experiment.

      i would like to ask you something.is there a way if i want to make this circuit to work automatically with a 12v car battery.meaning,adding a potentiometer to adjust fade in and fade out at equal time intervals without the press button.

      simply, what i want to make is a circuit like the last one in your video with the power transistor and with leds with 2 changes.
      (1) eliminate the press button so that when i connect the curcuit to the battery,it starts fading(i think it should work with a timer)
      (2)a potentiometer to adjust speed of fading in and out

      is it possible to make?please help me.

  • At 19 October 2011, 9:13:16 user Steve wrote:   [reply @ Steve]
      • Hi, i was wondering if you could help me, ive set up a series of lights under my fishtank top hood and have red green and blue in sequence on a light bar, white leds on a light bar and on a seperate circuit four dichoric 12v warm white led downlights.
      So ive basically got three individual lights groups all 12v and wanted to know if you have ever heard of any way to fade randomly between the three groups slowly, to give it a different fading/changing colour effect?
      Cheers
      Steve

  • At 10 October 2011, 11:37:11 user Kammenos wrote:   [reply @ Kammenos]
      • @Mike there will be no problem if you power with 15 volts. a relay will not work at all, you need to have this power transistor.

  • At 8 October 2011, 16:46:59 user Mike wrote:   [reply @ Mike]
      • Kammenos,
      Oops one more question... I went ahead and build this circuit on a breadboard and tested it. It seems to work fine. One thing I noticed that the value of R3 directly effects the brightness of my LEDs... the lower the resistance the brighter they are. If I connect this circuit to a 12v power source (batteries), then the LEDs only get about 10v. With the 220ohm resistor you recommended and the 2k trimpot adjusted to give a decent amount of brightness at the expense of a shorter fade-in time the value of R3 is about 330ohms. I am not sure if there is a better way to get more power to the LED Array but I am thinking about putting more than 12v in the power source. I went ahead and added another 1.5v battery bringing the power source to 13.5v and the LEDs now get about 11v and are much brighter. They are still not to their full potential since they are designed to work on 12v. If I add another battery (1.5v) and bring the power supply to 15v it should increase the voltage to the LED array to about 12v. However, I do not know how this would effect the longevity of the circuit components. Is this a plausible course of action or is their a better way to get 12v to the LED array? Maybe a relay would work better than the power transistor. Let me know what you would do. Thanks.

  • At 8 October 2011, 12:09:44 user Kammenos wrote:   [reply @ Kammenos]
      • @Mike i cannot tell you because i do not know these transistors, never used them before.

  • At 8 October 2011, 10:20:06 user Mike wrote:   [reply @ Mike]
      • Kammenos,
      Thanks for the fast reply and the advice on using the small resistors in series. I have one more question for you. They didn't have the transistors you specified at my local shop but they did have some substitutes. I wanted to run them by you before I turn the circuit on and see what you think. Instead of BC548 they gave me NTE123AP. They substituted BC327 with NTE159. Also, they didn't have the BD243 power transistor, their computer actually recommended NTE377 but they didn't have that either. The computer also listed NTE331 as a substitute for BD243A, BD243B, and BD243C. So we went with NTE331. So, hopefully my last question to you: Is it acceptable to use NTE123AP, NTE159, and NTE331? Thanks

  • At 8 October 2011, 2:09:36 user Kammenos wrote:   [reply @ Kammenos]
      • @Mike yes you can substitute all 3 components as you said. But for r3/r4, if you do not put a limiting resistor and the trimpot is set to very low number, the transistor will fry. For R3 use a 2.2K potentiometer with a 220 ohms resistor in series and for r4 use a 5K potentiometer (or 10K) with a 1K resistor in series.

  • At 7 October 2011, 22:41:45 user Mike wrote:   [reply @ Mike]
      • Kammenos,
      Is it possible to substitute a 500ohm trimpot for the 500ohm potentiometer? I am not sure if it matters or not but I will be using the T1 power transistor connected to an LED array (G4 LED disc w/integrated resistors with a total of 12 5050 SMD LEDs). It looks like it draws about 2.4w or about 200mA. Also, is it ok to use trimpots in place of R3 and R4 to make the fade in and fade out times adjustable?
      By the way this is a great tutorial and a lot of very good information. Thank you for sharing.

  • At 7 October 2011, 16:58:55 user Kammenos wrote:   [reply @ Kammenos]
      • @Roger the schematic you sent looks fine. i wonder why this does not work... If you connect the potentiometer like that, what happens?

  • At 6 October 2011, 0:49:45 user Roger wrote:   [reply @ Roger]
      • Ya I'm not sure what I did I must have accidentally bridged to the (+) rail. I still cant get it working though, is this how im supposed to be hooking it up ?
      http://i6.photobucket.com/albums/y225/LegendGS/dimmer2.png

  • At 5 October 2011, 14:23:19 user Kammenos wrote:   [reply @ Kammenos]
      • @Roger it cold not kill the transistor because it could only deliver less or equal current as before. The way you describe it, i think that you had one pin of the potentiometer connected at power supply.

  • At 5 October 2011, 13:01:36 user Roger wrote:   [reply @ Roger]
      • Tried it last not but no luck. It wouldbt work as a voltage divider only as a variable resistor but the leds would flare up in brightness if you turned the pot too fast. Then at some point it must of killed the transistor because it would only get like 80% brightness. Replaced the transistor and it back to full power. Not sure if I hooked it up wrong Im pretty sure everything was correct? Any ideas?

  • At 4 October 2011, 5:21:03 user Kammenos wrote:   [reply @ Kammenos]
      • @Roger no, T1 operates are emitter follower. You wanna add a potentiometer connected as a voltage divider to the input of the T1. This is how:
      Cut the line that goes from D2 to the base of T1. Then connect this line to one end of the potentiometer. The other end of the pot will go to the ground. The middle connector of the potentiometer now goes to the base of T1. This way, you can control the voltage on the leds with little current and almost no power loss. As for the pot size, begin with 1Kohm. I bet this will be ok, but you may need to go up to 10K.

  • At 2 October 2011, 13:53:48 user Roger wrote:   [reply @ Roger]
      • @ Kammenos

      Hoping you can help me out here. In desperation for a suitable dome light circuit I took your awesome circuit here and added a few components. Now I have it setup with a 12V reg for auto transient protection & added another PNP transistor allowing for low side switching from my car. I hooked it up and it seems to work great! The only thing and its not a big issue but my OEM ground wire is actually PWM ground controlled so even with the POT (Which I know mine is low) There is a slight delay before the ground triggers the PNP transistor to even activate the fade/in. I could probably just use a door triggered ground but it would mean running extra wiring and like I said its really not an issue.

      Ok to the point! When the circuit is active and the LEDs are fully on I would like to add a dimming feature. Would you suggest a POT maybe in addition to the R7 resistor for dimming?

      Here is my Schematic
      http://i6.photobucket.com/albums/y225/LegendGS/CarDomeLight.png

  • At 1 October 2011, 1:38:27 user Kammenos wrote:   [reply @ Kammenos]
      • @Roger i'm sorry, i rarely make PCBs for circuits from the "Circuit" section. I usually design them either for educational or for experimental reasons.

  • At 30 September 2011, 23:10:58 user Roger wrote:   [reply @ Roger]
      • @ Kammenos Do you happen to have a PCB layout for etching of circuit #2 that you would be willing to share?

  • At 29 September 2011, 19:41:04 user Roger wrote:   [reply @ Roger]
      • Cool sounds like I could easily use a few for like my glove box with smaller LEDs. My dome light however is a different story. My current setup has 4 dome lights 2 front / 2 back. Each spot has (4) Nichie LEDs that im running at 50ma each (60ma max rating) so 200mA for each socket and 600mA total. I would probably have a circuit for each 200mA cluster because they are all independently controllable. I understand about not having time! No worries I have been paying with programming a PICAXE but I cant get it to fade nearly as smooth as your circuit here!

  • At 26 September 2011, 12:33:22 user Kammenos wrote:   [reply @ Kammenos]
      • @Roger with less than 30mA quiescence current (which you can cut in half if you change R2 with 1K resistor, it will take a decade to drain the battery. Regarding the other circuit, i do not have plenty free time, so be patient. I will design one.

  • At 24 September 2011, 21:20:38 user Roger wrote:   [reply @ Roger]
      • Hey Kammenos

      Any thoughts yet on a similar circuit appropriate for a car dome light?? This circuit would work I'm just worried about it constantly draining the battery while keeping the cap slightly charged. Any thoughts?

  • At 14 September 2011, 15:27:19 user Kammenos wrote:   [reply @ Kammenos]
      • @naztee i'm sorry but if you have already tried higher supply voltage for the circuit and did not work, i cannot do much. What happens if you use only the power transistor to switch on the strip?

  • At 14 September 2011, 15:09:43 user naztee wrote:   [reply @ naztee]
      • @Kammenos:
      I'm not using any resistors per se, however, my LED strip has the resistors built in. It's the same as in "Need more current". The strip comprises of some 60 LEDs, of which each 3 LEDs are in series with a resistor, and then all is connected in parallel. It's ready-made and it looks something like this: http://goo.gl/3GN8j
      I have a correction. When connected directly, the strip draws some 500mA@13V
      And I cannot get them to full brightness when using this circuit

  • At 14 September 2011, 7:23:11 user Kammenos wrote:   [reply @ Kammenos]
      • @naztee are you powering the LED strip with a protective RL resistor? (like in my schematics)

  • At 13 September 2011, 3:23:13 user naztee wrote:   [reply @ naztee]
      • hello

      i've been trying to use this circuit #2 for fading in/out a 12V ready-made LED strip, but it seems I cannot get enough voltage/current out of the transistor. sure i've tried the "power" version, but that didn't solve any problems. it fades in/out beautifully, but they never reach full brightness.
      I've been using a 12V and a 16V power supply too. The LED strip draws some 0,03A@13V when connected directly to power source.

  • At 27 August 2011, 19:39:36 user Roger wrote:   [reply @ Roger]
      • Yaaaa! You should fade/on fade/off!!

  • At 16 August 2011, 15:09:05 user Kammenos wrote:   [reply @ Kammenos]
      • @Roger I do not recommend you use any of these circuits for dome light. A friend of mine asked me a dome light LED circuit for his car, and i promised him that i will make one soon. That happened a few months ago, and the \"soon\" has already passed, but i will make one for sure, as many people have asked, you are the fourth from my site and along with my friend that makes 5 people.

  • At 13 August 2011, 19:16:19 user Roger wrote:   [reply @ Roger]
      • Nice circuit! I duplicated the first part so far and the fading is very smooth perfect for what I am trying to achieve in my car. My car had factory fade-in fade-out but when I installed LEDs I lost most of the effect & the dimming is choppy toward the end.

      The only thing that i\\\\\\\'m not sure about is the fade-in and keeping the capacitor charged all the time ready for fade up Routine. Is there any other circuit you might recommend maybe a 555 chip somehow?

      (PS. My dome lights are neg triggered, they have constant (+) so I would need to rearrange anyways )

      I have 4 dome lights and each can be triggered individually independent of each other. So Would probably need for circuits one for each. Leds are (3) 60ma 3.3V Superflux (Nichia).

      Any advice on which way to go would be great!

  • At 24 July 2011, 10:59:53 user Stefan wrote:   [reply @ Stefan]
      • Wow,how do you desiging these interesting schematics?They are awesome.
      Good job.

  • At 6 June 2011, 11:50:37 user Kammenos wrote:   [reply @ Kammenos]
      • @Mojo the resistor has to do with the current and not the voltage.

  • At 6 June 2011, 2:19:10 user Mojo wrote:   [reply @ Mojo]
      • Hiya, I'm using a 5V output from a pic, and a 12V supply for the collector, what value of R1 should I use to fully switch on the transistor?

      Thank you.

  • At 5 June 2011, 9:38:23 user Kammenos wrote:   [reply @ Kammenos]
      • @Fung depends on the coil of the relay. Usually, 20 to 150mA more

  • At 5 June 2011, 7:39:39 user Fung wrote:   [reply @ Fung]
      • If an SPST switch is used instead of a button for S1, keep it closing, besides the LEDs, how much current would draw? (I am going to find ways to save power even the LEDs are on)

  • At 18 April 2011, 22:21:09 user danbug wrote:   [reply @ danbug]
      • Thank you. This circuit fit my needs exactly.

  • At 17 March 2011, 4:35:10 user Kammenos wrote:   [reply @ Kammenos]
      • There is no typical value for a high brightness LED. Could be 25mA or higher.
      It will emit heat because of power dissipation. If you have 21 LEDs in 7 rows of 3, and each raw draws 30mA, then you have a total of 210 mA @ 12 Volts, that is 2520mWatts. To totally turn them off, the transistor will be required to dissipate 2.5 watts. The one you've chosen can dissipate up to 0.8 Watts, and that will be a problem. BD243 can dissipate 65 Watts

  • At 17 March 2011, 3:22:02 user Fung wrote:   [reply @ Fung]
      • For a normal high-brightness white LED, how much current that it draws?

      In the diagram of powering 36 LEDs, how much current that you have measured? If I want to power 20 LEDs with a BC639 medium power transistor (Ic=1A max), will it emit excess heat?

      Magnus:
      "0V" represents the negative rail (V- or Vss) of the power source while "12V" with arrows represent the positive rail (V+, Vcc, Vdd etc).

      For example such as a battery clamp, there is a red wire and a black wire; the red one is the positive rail and the black is is the negative.

      So, when we connect a power source, the "12V" goes to the red wire while the another wire goes to "0V".

  • At 8 March 2011, 15:10:44 user Magnus wrote:   [reply @ Magnus]
      • What does the OV mean and the arrows that point up to the 12v mean?? please help!! i am making a school project! :D

  • At 30 December 2010, 7:37:41 user Kammenos wrote:   [reply @ Kammenos]
      • the led gets from a NO contact of the button

  • At 30 December 2010, 7:09:41 user Fung wrote:   [reply @ Fung]
      • From the video, a green LED with resistor is added to show that a signal is sent from a push button. Is it just parallel to the button?

  • At 21 December 2010, 10:28:49 user Kammenos wrote:   [reply @ Kammenos]
      • R3 for charging and R4 for discharging.
      You can use lower voltage as well.

  • At 20 December 2010, 12:26:35 user Fung wrote:   [reply @ Fung]
      • Found that R3 affects the time of discharging (may be), except the C1, is there any part that may affect the time? Is R4? Or only R3?

      Can I use a lower voltage such as 9V to power the circuit?

      I have read the article but I cannot test because of the absence of some components (not bought yet), and I still not understood all of them.

      *Sorry that I made mistakes in the last note, I still have one 9013 transistor which is NPN type. And three BC557 transistors.

  • At 20 December 2010, 11:06:11 user Kammenos wrote:   [reply @ Kammenos]
      • i explain the time and the components that affect it in the article. you should test to find the one that fits your need.

  • At 20 December 2010, 10:28:41 user Fung wrote:   [reply @ Fung]
      • So, by using the above values of the components, how long does the fade-in and fade-out time?* To increase the fade-in and fade-out time, which resistor (or more than one of it) should change the value?

      If 470uF is used (or even smaller) for C1, what value of that resistor should be used at least?

      *Since I have used up all the NPN transistors (but 3 PNP's are left...), and I have no 1000uF capacitors, I cannot test the circuit until I go to buy more.

  • At 19 December 2010, 7:28:35 user manosv wrote:   [reply @ manosv]
      • Thank you for your post


     

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