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Newbie fader led circuit

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Drcarnine

New Member
Hello
I have a pir driven security light that I want to fade in over the time of 30 seconds to a minute when the pir triggers it. Right now the unit operates on 3 AAA batteries or 3.6 volts. That is the voltage going to the 15LED board. the LEDs each use 20ma. I can remove the battery pack and install the circuit I am requesting help on. I can then drive the pir circuit externally with either 6 or 12 volts. I am using a 6 volt 4.5ah lead acid battery that I can put a voltage booster on if I need 12 volts. I would rather stay with the 6 volts if possible. The fader circuit will just have an in and out from the pir to the LED board. I want the circuit to fade on and then remain at full 3.6 volt brightness until the pir removes the power. It can instantly drop to zero or fade down. I am not too good with the parts and how they work so I will need some guidance. I have been researching the 7555 and the 555 timers but all I have found are oscillators that do not stay powered up. Is there someone out there that would be willing to take the time to help me. I would greatly appreciate it. Thanks Dennis
 

alec_t

Well-Known Member
Most Helpful Member
Welcome to the forum.
Are you trying to run 15 LEDs, each having a Vf of ~3.6V, in parallel from 3 AAA cells? Battery life will be very short :(. If you have a 6V 4.5Ah battery why not run them from that?
 

Drcarnine

New Member
Thank you for your reply
The original security light comes with a 3AAA battery pack internal. When the pir is triggered, it comes on and goes off about 10 seconds after the motion that triggered it leaves. There is about 30 minutes of light in the original design. It is solar powered so the batteries recharge each day. The light was originally 15 white LEDs. I remove the white LEDs and replace them with 7 red and 8 green LEDs. I put a 4.7ohm resistor between the red and green LEDs so that the greens will come on. I wanted to make it last longer so I developed the 6volt external battery pack. It has a 6 volt 4.5ah Lead acid battery with a voltage regulator in a weatherproof box. I plug this into the solar panel jack on the light. The voltage regulator is set to 4.0 volts so it charges the internal batteries.
I thought that if I could use the 6 volt battery and remove the regulator, and the internal battery, that I could put a fader type circuit in the space where the original batteries were inside the light. I also purchased a voltage booster so that I could bring 12 volts from the battery box to the light. I tried to make a fader from some of the designs I saw online. I used 12 volts with a 2300uf capacitor and a 10K resistor connected to the base of a PN2222a transistor, but when connected to the light board, the maximum voltage I can get is 2.8 volts. the green LEDs do not brighten up. The transistor gets really hot too so I think I am overloading the transistor. I didn't know if I put 2 transistors in parallel if that would divide the current and allow more to get to the LEDs. When I try to use the 6 volt battery, I only get 2.4 volts and the green LEDs do not light up.
When the original light triggers, with its 3.6 volt battery pack, the voltage across the light board is 3.2 volts. That is what I am trying to achieve. Is it possible without too many complex components? Thanks again
 

alec_t

Well-Known Member
Most Helpful Member
After the fade-in how long should the LEDs stay on? What will turn them off?
When the PIR triggers, what is its output voltage? What current can the PIR output provide (or what is the minimum load resistance it can drive)?
 

Drcarnine

New Member
I will try my best to answer your questions. I also have a similar light that has 20 LEDs. If I can get this one to work, then I can use the same setup for the little 15 LED light. The pir circuit will turn on the circuit and I want it to fade in over a 30 second period at that time. The pir circuit has a built in timer that shuts off after 10 seconds when motion is not detected anymore. If motion remains under the light, the pir circuit does not shut the light off. I want the lights to remain on until the pir circuit shuts it off. I measured the current between the pir circuit and the light board and it is 350ma. The output voltage of the pir circuit is 4 volts if I use the battery pack and voltage regulator open-ended. The voltage is 3.5 volts when the light board is connected. I can then put my buck voltage booster after the pir circuit. With the adjustable pot on my buck booster, I can increase the voltage to 36 volts if necessary from the 4 volts coming from the pir circuit. So either the 4 volts from the pir circuit or the voltage from the voltage booster can be the trigger. I then have up to 36 volts to work with to make the fader circuit work. I do not know how much this will drain my 6 volt battery but I would like to keep it to a minimum if possible. Right now, the 6 volt battery pack will allow the light to run for at least 15 hours between solar charges so it can remain on all night if movement is detected.
Thank you again
Dennis
 

alec_t

Well-Known Member
Most Helpful Member
The pir circuit will turn on the circuit and I want it to fade in over a 30 second period at that time. The pir circuit has a built in timer that shuts off after 10 seconds when motion is not detected anymore.
I see a possible conflict there. What do you want to happen if, say, motion is detected but stops after 1 sec? Should the LEDs turn off at 11sec, or at 31sec?

Is the PIR output less than 0.7V when the timer has shut off?
 

Drcarnine

New Member
As far as fading out, I do not really care. If is easier to have the circuit fade out at the same rate it fades in then that would be okay. The pir will shut off the power to the fader after about 10 seconds on no motion being detected. At that time, the fader can either drop out or fade down. If it drops out, then it will be reset for the next motion and begin fading up again. If it is slowly fading down and motion is detected again, then the light will begin fading up at that time. Either of these would work. When the pir drops out, the voltage is zero. It completely cuts off the power to the light board which saves battery power. So the circuit would not have any power until the pir circuit was triggered.
 

alec_t

Well-Known Member
Most Helpful Member
Something based on this perhaps?
LED_Fader.gif
 

Attachments

  • LED_Fade-in.asc
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Drcarnine

New Member
Like I said, I do not know much about it. If you say it will work, then I assume it will. I recognize most of the parts except a few. What is U2, and M1. Can the trigger voltage be the original 3.6 volts of the light or does it need to be the 6 volts from the power supply? I can use a buck voltage booster to give me what ever voltage is needed after the pir circuit. I think I can find all of the materials. Will my PN2222a transistor work or do I need something more powerful? Thanks again for taking the time to try to help me.
 

alec_t

Well-Known Member
Most Helpful Member
The circuit works in simulation. An extra component or two would be needed for an actual circuit if built.
U2 is the simulator's reference for a trimmer resistor. M1 is a N-channel FET which would be used instead of your PN2222A. The FET is used because, unlike your transistor, it needs virtually no current to control its load current. This means it does not affect the timing. The FET type shown is for simulation purposes only. In practice, look for a 'logic level' type rated for at least 1A load current. Q1 can be any general purpose low power silicon transistor. D15,D16 are any general purpose low power silicon diode, e.g. 1N4148.
The trigger voltage can be the original 3.6V from the PIR.
 

Drcarnine

New Member
So where would I get pcb's that this circuit could be put on or can I just solder and assemble them myself. How large would it have to be? I am not familiar with most of these components. What would be the identification number you would recommend for Q1? I see numbers for M1 and the resistors and capacitors and the 555 timer. I see you are using an NE555. I have read about 7555, etc. What is the difference between the NE555, a 555, or a 7555? Questions--Questions

Thank you again for your taking the time to help me.
 

alec_t

Well-Known Member
Most Helpful Member
So where would I get pcb's that this circuit could be put on
There is no ready-made pcb. It would be expensive to have one made. A prototype project like this would normally be built on perforated board such as this:-
https://uk.farnell.com/vero/01-0021/veroboard-pcb-121-92mm-x-101-6mm/dp/1536938
Apart from the LED array the components will take up only ~2-3 sq in of board space.
For Q1 you could use your PN222a, or a 2N3904.
What is the difference
555 is the generic code for a well-known timer IC. Commercial variants/brands have those digits somewhere in their part number. The NE555 is one made by Signetics. They're all pretty similar.
C1 and C2 are electrolytics with a working voltage rating of at least 10V. They must be connected with the correct polarity.
All the resistors have a rating of 1/4Watt or 1/8Watt.
All components should be 'through-hole' or 'axial' type (not surface-mount!).
D15,16 are type 1N4148 diodes (or other general purpose silicon).
In addition to the components shown you would need supply decoupling capacitors, i.e. another 100uF electrolytic plus a 0.1uF ceramic capacitor.
 

alec_t

Well-Known Member
Most Helpful Member
It will.
 

Drcarnine

New Member
Where do the other caps go? I have 1uf capacitors. I do not know if they are ceramic. They are not polarized, just orange pancake type. They say 104M on them
 

alec_t

Well-Known Member
Most Helpful Member
The first board will do (though it's not stripboard which I personally find easier to use); definitely not the second.
They say 104M on them
That = 100,000pF = 100nF = 0.1uF and is ok. One should be connected across the + and - supply rails, as close to the power supply pins of the '555 IC as possible. A 100uF should also be connected across the supply rails.
The FET is static-sensitive so needs careful handling. You will need to download a datasheet for your chosen FET to confirm its ratings and pin-out.
 

alec_t

Well-Known Member
Most Helpful Member
The two decoupling caps must both be connected from the top (+) rail to the bottom (-, ground). The 0.1uF needs to go as close to the '555 as possible but the position of the other is less important.
The first lot of FETs will all be fine (albeit heavier duty than strictly necessary).
The second lot are all surface-mount..... best avoided if you're a newbie.
 

Drcarnine

New Member
You said that the mosfets I found were bigger than needed. Can you give me a part number for the one you would recommend? I am getting ready to order the parts and try to make it work. Thanks again
 
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