1
\$\begingroup\$

I'm new here and I'm re-new in electronic hence why I'm seeking for some help. Last time I touched a breadboard with a transistor was 10 years ago.

So, here's my problem and the problem to the solution to try to fix the first one. I have purchased a PIR sensor from ebay. On a breadboard, I connected the sensor to a single LED and I saw that the voltage was low (1.75V circa). Note: I'm powering the circuit with a USB cable, so the input is about 5V.

Now, I thought that the PIR was just giving me that kind of output and I thought of a simple solution: I connected the PIR's output to the base of a 2N3906 transistor. The collector is directly connected to the source (5V) and between the emitter and the ground I have the load which it the LED. Nothing, same issue. The output's pretty low as before.

Any suggestion? Is it because, perhaps, the PIR is drawing too much to itself that I haven't got much left for the LED (through the transistor)?

EDIT: I'm adding the schematic I've followed. Also, the PIR sensor model I'm using is HC-SR501 (Ebay PIR Sensor HC-SR501)

schematic

simulate this circuit – Schematic created using CircuitLab

Cheers :)

\$\endgroup\$
  • \$\begingroup\$ The sensor may not be able to drive the LED. Please update with the PIR sensor datasheet. A schematic would also be beneficial. \$\endgroup\$ – DigitalNinja Apr 19 '16 at 18:06
  • \$\begingroup\$ Could you provide a schematic? If you go back and click the edit button, there is a button for answering that will let you draw one with the tool that is provided, you should try it. \$\endgroup\$ – laptop2d Apr 19 '16 at 18:21
  • \$\begingroup\$ Hi both, thanks for the replies. I've just added the schematic and the sensor model i'm using. PS, the Jumper's on H. \$\endgroup\$ – Simone Apr 19 '16 at 20:22
3
\$\begingroup\$

The circuit makes no sense to me. You should have 5V connected to the PIR's Vcc, the transistor is never going to conduct since its base is always above ground, there's no current limit on the LED and it emitter is above the collector. Don't you want an NPN transistor with its base connected to the output, and the LED and a resistor in series between Vcc and the transistor's collector, with the emitter connected to 0v? Something like that?

\$\endgroup\$
2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ I'm actually having a little problem after further testing.. The LED lights up when motion is NOT detected lol. I checked the tension when the LED lights up and when it doesn't. When it does, I have about 0.30V and when it doesn't I have 3.3V out of the PIR now. I don't get it. As for the transistor design, I might have inverted it. Also note, the LED I'm using has got its own resistor integrated (it's a USB LED Strip that is generally used behind TVs). I don't get why the PIR is now giving me around 0.3V when doing nothing. I'm a bit confused.. \$\endgroup\$ – Simone Apr 19 '16 at 21:08
  • \$\begingroup\$ @Simone the HC-SR501 is active high when it detects motion. So this circuit will work as shown. The Transistor does have to be chosen for the right collector current. The 2n3904 can only handle 100mA to 200mA. \$\endgroup\$ – Passerby Apr 19 '16 at 21:11
  • \$\begingroup\$ I'm an idiot. I missed NPN and PNP around.. PNP will close the circuit when no current reaches the Base. That is why I'm getting the opposite result atm :P - I'm gonna gran the 3904 tomorrow. Cheers \$\endgroup\$ – Simone Apr 19 '16 at 21:16
0
\$\begingroup\$

I got it. Whilst I was looking around for the datasheet of the PIR, I wanted to look at the one of the transistor one more. I noticed that one of the drawings I found online gave me the C and E pins all the way around!! Once I simply rotated the transistor of 180 degrees (so what was connected to the collector is now connected to E and vice-versa), it started working as expected.

Thanks all for the answers. I'll now have to fight with the actual PIR with the sensitivity etc.

Thanks!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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