0
\$\begingroup\$

This question already has an answer here:

I'm thinking of building a project that would only include PIR sensors and LEDs.

Could I use a PIR sensor like this one, https://www.adafruit.com/products/189, to power 8 LEDs or would those 8 LEDs draw too much power for the onboard circuit of the PIR sensor? How would I wire it?

If this is possible how much voltage would I need to provide the PIR Sensor?

If it's not possible as is, is there any way I could still make this work without having to introduce a microcontroller like an arduino?

\$\endgroup\$

marked as duplicate by Community Mar 8 '17 at 19:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ No you do not need a micro to do that. Look at the "Testing a PIR" page (pg.18) of their spec sheet. It's a simple battery led circuit. If you want to drive 8 LEDs, you may need to boost the output current using a transistor circuit. The specs are really incomplete on details. READ THEM ANYWAY.... \$\endgroup\$ – Trevor_G Mar 8 '17 at 19:10
  • \$\begingroup\$ Voltage, power , current, schematics. \$\endgroup\$ – Marko Buršič Mar 8 '17 at 19:10
  • \$\begingroup\$ @Trevor Thanks for your input. I did read through the specs and that still left me wondering if it would work or not. From looking at this datasheet, it seems like the signal out only pushes 10 mA. What would be a simple implementation of a transistor circuit that you mention? \$\endgroup\$ – Paul Masek Mar 8 '17 at 19:25
  • \$\begingroup\$ @skvery Just read your comment and answer. That looks to be what I'm seeking! Thanks! \$\endgroup\$ – Paul Masek Mar 8 '17 at 19:27
-1
\$\begingroup\$

I quickly rigged up a circuit which would do the trick. Hope it helps!

As mentioned in the link provided by you, the PIR sensor can work on 5v-12v. For this application, 5v would be sufficient.

The BC547A BJT transistor acts as a switch here. It acts like a closed switch when suitable voltage is applied to it's base terminal. It acts like an open switch otherwise.

When the PIR sensor senses an object, it's output would go high. This high output voltage would make the transistor act as closed switch. As a result, all the LEDs are turned on whenever an obstacle comes within the sensing range of PIR. When there is no obstacle, the output of PIR will be low and transistor acts as open switch. Hence, all LEDs will be turned off.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Perfect! This precisely address my question. I appreciate the time you took to answer this and diagram it out for me! I'm still just very much dabbling with electronics and this is extremely helpful. \$\endgroup\$ – Paul Masek Mar 8 '17 at 19:37
  • \$\begingroup\$ Only issue with that circuit is the data sheets say some models are open drain output @EBasu \$\endgroup\$ – Trevor_G Mar 8 '17 at 19:37
  • \$\begingroup\$ Bad idea. You have LEDs in parallel attached to your series resistor. At best, the LEDs won't light up evenly. At worst, one will hog all the current and burn out. Then another snarfs up all the juice and burns out, and so on until all the LEDs are gone. \$\endgroup\$ – JRE Mar 8 '17 at 19:40
  • \$\begingroup\$ It is highly recommended that you use one current-limiting resistor for each LED when you are using LEDs in parallel. \$\endgroup\$ – Peter Bennett Mar 8 '17 at 19:45
  • 1
    \$\begingroup\$ Rather put them in series. This will increase the power efficiency of the circuit as the resistor will be smaller for the same current flow. \$\endgroup\$ – skvery Mar 9 '17 at 5:44
0
\$\begingroup\$

The 12 V will be enough to power the LEDs. The 3.3 V out will be far too little.

You will need to switch the 12 V from the 3.3 V .

(Oops, this is a duplicate of "How to switch 12 V from 3.3 V. :-)

\$\endgroup\$

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