Adding a PIR sensor to a string of LEDs - Page 1
Forum section: electronics / projects

author: sentry7
author_rank: Regular Contributor
timestamp: July 20, 2016, 03:07:26 pm
content: Hey guys,I have a string of decorative LEDs that's powered by 3 AA batteries. There seems to be a series resistor as well (looks like 5.1ohm, 2 watts?...I can't see it very well). I need the lights to turn on when people come into close proximity, so I want to add a pir sensor. The sensor is 3.3v, but I'm not sure about the current draw.So...something like this?

author: georges80
author_rank: Frequent Contributor
timestamp: July 20, 2016, 04:10:06 pm
content: You say PIR sensor, but is that a raw sensor device or a module.Raw sensor needs amplification and lens etc...If a module then it must have some basic specs.cheers,george.

author: sentry7
author_rank: Regular Contributor
timestamp: July 20, 2016, 04:19:13 pm
content: It is a module. Sorry, not at the bench ATM, will get back with some specs.

author: SeanB
author_rank: Super Contributor
timestamp: July 20, 2016, 05:01:02 pm
content: If it is a 3V3 module simply put a 3v tap in the battery pack, so it will get a 3V supply. Many though are 5V modules so will run direct of 4V5. A 1000uF 16V capacitor across the power leads to the PIR will help it as the batteries age. These typically are a low power draw though, so should last quite a while on a set of AA cells. You can increase the series resistor for the leds to 10R without much light reduction, but with a good increase in battery life.

author: ajb
author_rank: Super Contributor
timestamp: July 20, 2016, 06:19:47 pm
content: If the LEDs are blue, green, or white, then current draw is ~300mA. I = (Vsupply-Vled)/R = (4.5-3)/5.1 = 0.294, or if they're red, I = (4.5-2)/5.1 = 490mA.You haven't shown the LEDs in your diagram. Assuming R1 is the current limiting resistor for the LEDs, then the LEDs should be in the same spot. So Vcc->Resistor->LEDs->collector of the transistor. This will work as long as the PIR produces a positive pulse when triggered; if it's inverted then you'll need to flip everything around and use a PNP transistor (or add a second transistor as an inverter).

author: edavid
author_rank: Super Contributor
timestamp: July 20, 2016, 07:52:05 pm
content: Quote from: ajb on July 20, 2016, 06:19:47 pmIt's extremely unlikely that a battery powered LED string would draw that much current.They could be lower voltage LEDs, with 2 in series.Or there could be resistors in series with the individual LEDs in the string.Or OP misread the value (OP, do you have a multimeter?).

author: sentry7
author_rank: Regular Contributor
timestamp: July 22, 2016, 02:00:49 am
content: Sorry for the delay. So I did a meter check, and the resistor is indeed 5.1 ohms. I'm not sure about the power rating, I have a pair of calipers and can check the diameter.

author: edavid
author_rank: Super Contributor
timestamp: July 22, 2016, 04:08:37 am
content: Did you try measuring the current draw of the LED string when you connect it to the batteries?

author: sentry7
author_rank: Regular Contributor
timestamp: July 22, 2016, 10:24:01 pm
content: Quote from: edavid on July 22, 2016, 04:08:37 amOkay, I didn't measure the current directly, because I have a non-operational ammeter, but I got a voltage of 1.35v across the series resistor, so I = 1.35v ÷ 5.10ohms ~ 260mA. I'll find some specs for the sensor.