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Newbie fader led circuit

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A logic level one like Alec posted would be best. It would keep the time from turn on to some light the shortest.
 
I bought most of the stuff from ebay to give it a try. I couldn't find a 600k trimmer resistor but I ordered a 500K and a 1 meg. Which would you think I need to use? I also bought 470uf capacitors instead of the 330uf. That way I bought everything from the same vendor with free shipping.
I do have another question. I am going to try to retrofit my existing lights with these. As you know, they have only 3.g AAA battery packs. I am going to put in a voltage booster to get the 6 volts you need for this circuit. How much extra drain will this all put on the battery pack. The light as of now only has 30 minutes of "on time". If I can't use the battery pack, I can just use the 6 volt external battery pack. I will go ahead and bring in the 3.6 volts out of the voltage regulator from the external battery pack, the boost the voltage after the pir circuit so I am sure I won't fry the original light circuit. Will this work? I know it is redundant--cutting the voltage then boosting it, but I think this will protect the integrity of the circuitry.
 
Alec
Can you please try to write for me the flow of the circuit. I am trying to follow it but being a newbie, I don't understand everything. Here is what I think I know, but I am probably wrong. When the PIR circuit hits the base of Q1, it lets the 6 volts go to ground through R2 and to the trigger of the 555. This lets the vcc go to the out and through R10 and fills C1 and to the gate of Mi. This will let the 6 volts going to the LEDs to go to ground through R13. I think that when the PIR turns off, and Q1 quits conducting, that m1 will discharge through R11 and D15. I don't know what the other connections do. I would really like to try to understand what all of the parts do. I don't understand what the Reset, Discharge and Threshold connection do from the 555. I don't understand what Us andD1 or R12 and C2 do. Is there a U1 that I do not see? How much voltage will be put to the LEDS? I will need to size the resistors to the LED board.
Thanks again
Dennis
 
Well, I did more research on the 555 and see that the trigger is activated when it goes low. So that is the reason that Q1 is connected to ground. When the PIR circuit triggers Q1, the voltage to the trigger goes low and allows the 555 to conduct. I have so much to learn. Thank you so much for taking the time to help me
 
I couldn't find a 600k trimmer resistor but I ordered a 500K and a 1 meg.
The trimmer in my circuit is 500k, not 600k. (Sorry if the image wasn't clear). The 470uF will be fine (giving slightly longer LED fade-in/out times than the original 330uF)
I wouldn't recommend trying to boost the AAA voltage to run the LED array; battery life would be very short. My circuit is designed so that the AAA cells power only the PIR (hence have a long life) and a separate external 6V source powers the circuit and the LED array. The voltage needed to trigger my circuit is anything from ~1V to >50V and the current drawn is negligible (so 3.6V from the PIR is fine).
Circuit operation is as follows:-
A positive-going pulse from the PIR turns on Q1, pulling the Trig input of U1 (the 555 IC) low. This starts the 555 timer running and U1_Out goes high (6V). C2 charges up via R12. When the C2 voltage reaches ~4V (as sensed by the U1_Thrs input) the timer stops, U1_Out goes low (0V) and C2 discharges quickly via U1_Dis. Thus the timer is reset.
While U1_Out is high C1 charges up via R10. The rising C1 voltage drives the gate of the FET (M1). When the gate-to-source voltage Vgs exceeds the turn-on voltage Vt (~2V depending on FET type) the FET begins to conduct and current passes through the LEDs and R13. The voltage developed across R13 by the current raises the FET source terminal voltage. The result is that with increasing C1 voltage Vgs is held at Vt and the FET (and LED) current increases too, thus giving fade-in. When U1_Out goes low C1 discharges via R10, R11, D15, giving fade-out. Trimmer U2 and D16 allow a bias voltage of slightly less than Vt to be applied to the FET gate and prevent the C1 voltage dropping below Vt (except when the circuit is initially powered up). This ensures the fade-in starts as soon as the PIR triggers the circuit. The trimmer should be set so that the LEDs just turn off fully.
How much voltage will be put to the LEDS? I will need to size the resistors to the LED board
You don't drive LEDs with voltage; you drive them with current. As shown there are 7 strings of LEDs, each string having 2 LEDs. The current (~22mA peak) in each string is the FET current (~150mA peak) divided by 7. R3-R9 are present to allow for manufacturing variations in the forward voltages Vf of the individual LEDs and help to balance the currents in the strings.
 
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Thank you, that makes more sense. If I want to shorten the fade out time, can I make R11 a smaller value? The way my lights are set up, the external power source connects to the same port that the original solar panel connected to. Without using the external power source to charge the batteries, There is no way to keep the AAA battery pack charged. I may be able to bring the 6 volt source in and move my voltage regulator to inside the light housing and tap off to maintain the batteries. If I put my voltage booster inside the light, it would only be drawing off of the 6 volt battery and the external voltage regulator. I will have to try different ideas.
Thanks again
 
I also have a light that has 20 leds. Will this circuit work with it also or is the current through the fet too small?
 
If I want to shorten the fade out time, can I make R11 a smaller value?
Yes. And if you need to increase/decrease the fade-in time you can increase/decrease R10.
I also have a light that has 20 leds.
With my circuit you could add extra strings of 2 LEDs plus resistor to enlarge the LED array to 20 or more. R13 would need to be reduced proportionally, e.g. to 6.8 Ohms for a 20-LED array, if you wanted the same 22mA peak current through each LED.
 
Another question. The lights I have now have the LED boards already in place and are in parallel. I cut the board in my 15 led lights and insert a 4.7 ohm resistor so that the green LEDs even come on. The voltage going to the board when hooked up is 3.0 volts going to the green LEDs and the resistor keeps the voltage to the red LEDs at 2.0 volts. R13 will keep the current low so I do not blow the LEDs, correct? If I do not have the capability of splitting each set of LEDs and putting an adjusting resistor like R1 into the board, will I have to increase the value of R13? The 20 LED light has individual resistors for each LED and I size them to 20ma. They are the .5 watt 10mm LEDs capable of 100 ma. I just keep the current low to conserve battery life and I cannot tell that much difference in brightness.
 
The lights I have now have the LED boards already in place and are in parallel. I cut the board in my 15 led lights and insert a 4.7 ohm resistor so that the green LEDs even come on. The voltage going to the board when hooked up is 3.0 volts going to the green LEDs and the resistor keeps the voltage to the red LEDs at 2.0 volts.
Sorry, I don't understand that. Can you post a schematic/sketch of the LED interconnections you now have? In my circuit each string has a red and green LED in series to make best use of a 6V supply. Any other configuration would require a re-design of the circuit.
R13 will keep the current low so I do not blow the LEDs, correct?
That depends how you've got the LEDs wired. In my circuit the max FET current is 150mA, which exceeds the 100mA rating for an individual LED.
The 20 LED light has individual resistors for each LED and I size them to 20ma.
Not understood. What value is each resistor? What is the forward voltage rating of each LED?
 
Okay, I do not know how to draw a circuit so I will try to explain how the board is made.
On the 15 LED board, all LEDs are in parallel. The board is a rectangle. Originally both of these lights were security lights with white LEDs. I changed out the LEDs for the colored ones. There are 8 green LEDs that are on one side of the board and that is where the power hooks up. Then I cut the board and installed a 4.7 ohm resistor between the two sides of the board. The other half of the board has 7 red LEDs. Without the resistor, only the Red LEDs would come on. The forward voltage of the green ones is 3.2 to 3.4 volts and the forward voltage of the red ones is 1.9 to 2.0.
On the 20 LED board. Again all of the LEDs are in parallel with a resistor on each LED. 18 ohms on the red ones and 32 ohms on the green ones.
On both lights, the voltage coming from the PIR circuit is 3.6 volts open-ended. When hooked up, the voltage is less because of the draw of the LEDs.
 
I'm still unclear as to your setup. Are you saying the 15-LED array shares only one resistor, or does each red LED have its own resistor? Do the red LEDs form one parallel array and the green LEDs another parallel array?
18 ohms on the red ones and 32 ohms on the green ones.
I'm surprised it's not the other way round?
 
I am surprised too, I had these boards made for me and evidently they put them together wrong. I will fix them and reverse the resistors.
The LED board had no resistor originally. There is still no resistor between the green leds a d the pir circuit. The only resistor is the 4.7 I installed between the green end of the board and the red end of the board. Just one resistor between the 2 halves.
 
The reason for the low resistor on the red LEDs on the 20 light is to make the red lights as bright as the green ones. Since these LEDs can run 100 ma, we were able to make the red LEDs run more current than the green ones . Usually the red ones aren't as bright as the green ones
 
OK. That explains the resistor values. I'm still not sure of your board wiring. From your description the boards seem to be like this:-
LED-boards.gif
Can you confirm that?
If not, can you post a clear .gif or .png image of the boards?
 
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