6
\$\begingroup\$

I need help with a project. I want to make a circuit to work as a night-time motion detector circuit that will power several LEDs. I bought:

  • 1 LDR
  • some NPN transistors BC547 and 2N3904
  • some resistors 100kOhm
  • boost regulator 3V to 5V
  • PIR sensor

I connected the LDR to 2 NPN BC547 transistors and successfully tested it. Then I connected the (-) from NPN transistor to the (-) of the PIR and the (+) of the PIR to (+) and the output of the PIR to a single led. It worked but the LED had a low output.

If I connect the PIR output to a 2N3904 transistor and after to the LED, the LED is brighter. But when I connect the current to the voltage booster and to all LEDs, the light is very low.

The problem is the 3.3V from the PIR output. I need 5V to work my LEDs. How can I establish this?

Update: This is my PIR Sensor (mod edit: Pretty much a standard BISS0001 Design, no transistor on the output.)

enter image description here

My Power Supply Is 2 Batteries 3.7V 18650

\$\endgroup\$
10
  • \$\begingroup\$ Kudos, @Dave, you cleaned this up rather well. \$\endgroup\$
    – JYelton
    Commented Oct 1, 2013 at 20:32
  • \$\begingroup\$ A schematic of how you have everything hooked up is crucial for us to be able to help you. \$\endgroup\$
    – Passerby
    Commented Oct 1, 2013 at 20:50
  • \$\begingroup\$ How much power can you get from your PIR output? \$\endgroup\$
    – Andy aka
    Commented Oct 2, 2013 at 7:30
  • \$\begingroup\$ PIR = Passive Infrared sensor? \$\endgroup\$
    – AKR
    Commented Oct 2, 2013 at 12:14
  • \$\begingroup\$ Some questions. What are the NPN's doing? (a schematic will probably answer this as Passerby requested). What sort of PIR are you using? (a raw sensor or a module with buffered outputs?). Ive used both and the raw sensor needed a capacitor to connect to the sensing circuit to make it change sensertive and not level sensertive. \$\endgroup\$
    – Spoon
    Commented Oct 2, 2013 at 12:38

1 Answer 1

5
\$\begingroup\$

A boost regulator is not useful here. The problem is that your module, based on the BISS0001 PIR IC, the output pin is VCC (3.3v) 10mA max. Connecting a boost regulator to this output would be really limited.

All you need is a single transistor.

schematic

simulate this circuit – Schematic created using CircuitLab

Problem is that you need a transistor that works on the current provided from the output pin. Normally you see a 1k or 2.2k resistor on the board from Pin 2 of the BISS00001 to the output pin, which means only 2ma or 1.2ma at the output pin.

So you need either 1) a Transistor with a HIGH HFE or 2) a Darlington Pair (Two transistors in a pair).

schematic

simulate this circuit

These numbers are all based on the transistor you choose. A 2n3904 is only 100mA to 200mA max, with a hfe of 30 (So it multiples the base current, 1mA by 30, and that's the maximum current you get at the collector, 30mA).

You need to know how much current your led box needs, and what voltage it can use. I also assumed that your two batteries are in series.

See this page http://www.electrobob.com/fun-with-leds/ for a project that does both LDR and PIR for leds (but not the same way as what I think you want).

Adding the LDR as a night time detector is simple, and also requires a transistor.

schematic

simulate this circuit

Q1 and Q3 can be any weak small signal transistor (2n3904 100mA), Q2 should be a better one depending on your led box current needs (2n2222 1Amp). Adjust R2 for sensitivity.

\$\endgroup\$
16
  • \$\begingroup\$ DEAR FRIEND THANK YOU FOR THE ANSWERS AND HELP.MY BATTERIES ARE PARALLEL. TOTAL VOLTAGE OUTPUT WHEN CHARGED 4.2V NOW THE LAST CIRCUIT THAT YOU MAKE FOR ME IT WILL WORK? IF YES SO I NEED TO BUY TRANSISTORS 2N2222. AM I CORRECT? THANKS \$\endgroup\$ Commented Oct 3, 2013 at 10:02
  • \$\begingroup\$ ALSO THE RESISTORS YOU SHOW IN THE CIRCUIT ARE THESE I HAVE 100K Ohms OR SOMETHING ELSE??? THANKS \$\endgroup\$ Commented Oct 3, 2013 at 15:22
  • \$\begingroup\$ In that case, the batteries in parallel should connect to the Booster circuit you have, and everything should connect to the 5v side of the booster. R2 can be 100k, r1 is inside of the PIR module already, and Rdown is optional, so if you dont have it don't worry. And Yes, a 2n2222 or similar, any transistor that can handle 1 AMP should be fine. @panayiotiskasapis \$\endgroup\$
    – Passerby
    Commented Oct 3, 2013 at 19:02
  • \$\begingroup\$ woooo i did not get it! as i know ( not much) the boost i must connect it to the pir output so the 3.3v make them 5v but it does not because as i understand from you i have low (ma). i try that several times. or if you mean i connect the boost from the beggining of the circuit after the batteries , when there is an output from the pir it is going to be 3.3v again? or more? please drawn it for me if you can i am confused now \$\endgroup\$ Commented Oct 3, 2013 at 19:08
  • \$\begingroup\$ @panayiotiskasapis updated the circuit. The PIR has it's own regulator, and works on 3.3v, so yes, the output is still 3.3v so the transistors are still required. \$\endgroup\$
    – Passerby
    Commented Oct 3, 2013 at 19:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.