Electronics > Projects, Designs, and Technical Stuff
Help reusing a PIR sensor
Evangelopoulos Panagiotis:
--- Quote from: georges80 on March 10, 2020, 06:48:51 pm ---There's a lot of 'similar' analog (dumb) PIR sensors out there. The circuit you are looking out is likely 2 amplifiers with MANY X amplification. The signal coming out of an analog PIR can be quite small so lots of amplification is needed. There'll be some filtering in the amp circuit to stop from amplifying everything (RF etc)...
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If you have a template circuit that I could expand upon that would be much appreciated!
--- Quote from: georges80 on March 10, 2020, 06:48:51 pm ---PIR sensors NEED a lens to create the differential IR signal hitting the sensor. Most sensors have 2 elements that both see IR, the lens (Fresnel) is designed to illuminate one element and not the other - to generate the differential signal. Lens design is an art in itself to provide the coverage/range that is required.
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I mistakenly thought that this sensor didn't have a lens in it since the plastic is smooth from the outside and i didn't bother opening it up to check that there actually is a lens there. So thank you for that. I'll do some measurements with the plastic lens to see if I get a bigger voltage swing.
georges80:
^ just google search for terms such as:
pir amplifier circuit
and you'll find various circuits.
The PIR lens has the fresnel pattern on the inside surface and the outside will be smooth. It is also a special plastic that is IR transparent (long IR that is heat versus remote control RF).
cheers,
george.
Evangelopoulos Panagiotis:
Thank you for your help :D
RoGeorge:
In the example from the datasheet, the analog signal you want can be found at pin 7 of the second LM324 op-amp. Everithing on that schematic on the right handside is just to trigger the relay. Chances are you already have that exact schematic already made on the backside on the PCB.
Put a picture of the backside, should be easy to identify the chips, starting from the power and GND connections of each chip.
A PIR analog output can also be found right after C5, just that it will be much smaller.
Zero999:
--- Quote from: EvangelopoulosPanagiotis on March 10, 2020, 05:00:54 pm ---I know there is a circuit there but there is no explanation of it whatsoever and I'm not that experienced to know how this works and if it does what I want. Since it switches a relay then it's final output is binary. Either on or off. I need an output that gives off values between 0 and 5 volts. Since I don't know which stages do what and what to cut off I'll refrain from building it. If though no one else has a better solution to my problem I'll have to build it bit by bit and probe everything till I figure stuff out.
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The circuit isn't that complicated. Just break it down into several parts.
C9 to C14 and R8 to R10 form a power supply filter.
ZD1 regulates the voltage to the PIR and all of the biasing circuitry. U1's power supply isn't shown, but I believe it's connected straight to +12V, as the zener regulator won't provide power for it.
U1-A is a non-inverting amplifier, with some filtering. Its gain is determined by R3 & R4 and the passband by C3 to C4.
U1-B is an inverting amplifier, again with some filtering. Its gain is set by R11 and R5 and the passband by C8 & C5.
U1-C, U-D, D1 & D2 form a window comparator, with the voltage thresholds set by R12 to R13. As mentioned above it's required to activate the relay.
If you just want an analogue signal, then take the output from U1-B. The circuit can be simplified by using the LM358, which is the same as the LM324, but with only two op-amps.
Search for op-amp inverting/non-inverting amplifier and window comparator, for more information.
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