I have a problem and some of the components I think I need to solve the problem but I do not have the knowledge or understanding to put it all together and what I read online is just not connecting the dots for me.

Ultimately what I need to have happen is a PIR sensor detects movement and triggers a string of LEDs. Which seems like it should be simple enough. I can get the PIR sensor to light an individual LED but anything beyond that and the PIR is to weak.

The LEDs I am trying to light are in a ribbon format and I want them to run off of bare minimum 12v but preferably 18vdc as this is when they are at their absolute brightest. At 18vdc they draw about 2 amps of current.

The PIR is powered off of 6Vdc and outputs 3.3VDC with around 300uA.

I want to use a relay that I have which the coil can be pulled in at a little under 3 volts but needs around 100mA of current and the contacts could handle the 18vdc and 2 amps I want to run my lights.

My problem is how can I get more current from the output of my PIR to trigger the relay, being that I can't increase the voltage coming out of the sensor?

I have some transistors to work with but I don't have enough understanding to connect a proper circuit up and I believe I have been burning out some components needlessly just playing around with stuff.

My PIR is an HC-Sr501, my Relay is a JQC-3F (T73), the LEDs are a 12V RGB ribbon style, and I have available some standard NPN and PNP transistors BC547, BC557, 2N2222, and 2N3904. I also have a variety of common resistors and other standard components. Either a circuit somebody has already created or just some specific instructions of how the connections should be made in order to accomplish my goal would be appreciated.

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I don't have a data sheet for the LEDs. This is what they are and all the specs I could get on them.
SUPERNIGHT (TM) 16.4FT 5M SMD 5050 Waterproof 300LEDs RGB Color Changing Flexible LED Strip Light
Voltage 12 volts Wattage 60 watts
I don't have a problem running them at 12v if I have to and I understand it would be a better idea. here is a data sheet to my PIR sensor http://www.datasheet-pdf.info/entry/HC-SR501-Datasheet-PDF . Although it isn't marked as an HC-sr501 when I purchased it from a chinese supplier this is what they claimed it to be.

I know you need real information from data sheets but I don't have much of that stuff.

I know what I can read from my DMM. The sensor puts out 3.3 volts as expected when triggered but the current is only 300 micro amps. my relay coil will react at about 2.8 volts but draws about 100 milliamps in order to do so. The LEDs are designed to operate on 12 volts but when I crank it up to 18 volts they are bright and although they will eventually burn out I have run them at 18volts for more than 24 continuous hours without fail, that doesn't matter so much to me anyways, even if I only supplied them with 12 volts that would be ok. I just can't get my PIR to trigger my relay to turn my leds on in the first place. There isn't enough current coming out of my PIR output to activate the coil.

  • 1
    \$\begingroup\$ Running your 12 V LED off 18 V is a recipe for disaster. Don't do it. If your PIR output is that weak, you can always buffer it, but it seems strange that would be the problem. Please draw a diagram if how you connected everything. \$\endgroup\$
    – winny
    Dec 1, 2016 at 22:14
  • \$\begingroup\$ Link to the datasheet of the PIR sensor and the LED string. It is possible that the PIR sensor needs a pullup on its output. If the LED string is really 12 V, then run it from 12 V. Anything out of spec is a bad idea. What you want to do should be doable with a transistor or two instead of a relay, but we need real specs. \$\endgroup\$ Dec 1, 2016 at 22:31
  • \$\begingroup\$ You want to run the 12 V LEDs at 18 V, using a PIR running off of 6 V, and discussing the use of a 3 V version of a relay (which does come in a 12 V version?) I'd probably go for 12 V all the way, if possible. \$\endgroup\$
    – jonk
    Dec 1, 2016 at 22:32
  • 1
    \$\begingroup\$ A logic level FET is preferred for simple low current drain , but Vmax is 14.2V applied to led-strip, so choose a better power supply. \$\endgroup\$ Dec 1, 2016 at 22:59
  • \$\begingroup\$ That's not a datasheet, just a product splash page. All it says is that the output is "TTL", 0 to 3.3V. It says nothing about current source or sink capability. \$\endgroup\$ Dec 1, 2016 at 23:42

1 Answer 1


The below schematic demonstrates a circuit that will do this for you. When the PIR signal is low, the NPN transistor will be off, and no current will flow through the relay. When the PIR signal turns goes high, this causes the transistor to turn on, drawing current through the relay coil turning the relay on.


simulate this circuit – Schematic created using CircuitLab

Winny in the comments is correct, you should not use 18Vdc on the 12Vdc rated LEDs. While they may be brighter in the short term, this can lead to a few issues:

The LEDs will dim much faster over time than if you used 12Vdc.
There is a chance of overheating, and by extension, fire.

Now, you also say you have an RGB LED strip, not a simple one color strip. You will need to supply power to all three colors, not just one, if you want a white(ish) color out of them. (This is also assuming this strip is a common cathode model, datasheet/links haven't been provided)

To address @user2913869's concern that the sensor may not be able to drive a single NPN transistor, according to this website, and this datasheet, the actual output of that chip is approximately 10mA @ 5V supply voltage.

D6 is what is called a flyback diode. This is a necessary component when using relays (and other inductive loads). Inductive loads don't allow current to change immediately in a circuit. This diode allows current to flow back to the relay, rather than trying to push that current through the transistor, which can break the transistor. To put it simply, it allows the energy stored in the coil of the relay to dissipate without breaking the transistor.

  • \$\begingroup\$ If his spec of 300 uA output current is current, I'd probably recommend a darling pair configuration for the transistors, just to ensure enough current gain. \$\endgroup\$ Dec 1, 2016 at 22:46
  • \$\begingroup\$ Like I said I cannot say what the PIR sensor is for certain as nothing is printed on it. I can only say the manufacturer claimed it was the HC-Sr501 but I don't know for sure what the specs are. I only see 300 microamps when the sensor is triggered and putting out its 3+ volts on the output pin. I don't see 10mA and I am feeding it 6 volts dc. I am going to try and connect this circuit you provided and test it out. As far as the RBD LED goes, I only want to light one of the colors right now. \$\endgroup\$
    – Jmay
    Dec 1, 2016 at 23:16
  • \$\begingroup\$ Because of \$V=IR\$ you wont necessarily see all 10 mA coming from the output of the sensor, you should just see the 3.3 V output, and it should be able to remain at 3.3 V up to the 10 mA rating. \$\endgroup\$ Dec 1, 2016 at 23:24
  • \$\begingroup\$ Thanks, I just connected this circuit and my sensor is pulling in the relay now when triggered. So I can understand this circuit more can you give a brief explanation for what the diode D6 is accomplishing? \$\endgroup\$
    – Jmay
    Dec 1, 2016 at 23:30
  • \$\begingroup\$ @Jmay added to the answer. \$\endgroup\$ Dec 1, 2016 at 23:36

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