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Security Light & Switch with PIR Sensor Update

This is a simple update to Mr. Hareendran’s PIR Sensor Security Light circuit. It has a shortcoming limits the relay voltage to approximately 3.3V. While this may function with some 5V relays, it will not function with all. The nature of an emitter follower Darlington forces the output voltage to be two junction drops below the base drive voltage plus a small voltage drop across R1.

In keeping with the spirit of his design, I retained the Darlington arrangement, and the transistor type number – however, I changed the transistors from common collector to common emitter configuration, and connected the relay between the collector output and the positive rail. The relay now utilizes a 12V coil – to use an existing 5V relay, simply add a resistor in series with the coil whose value is scaled up from the coil resistance. Also, I connected the LED across the relay coil rather than powering it via the 2nd relay contact – this way, only one form “C” relay contact is required thus making the relay less expensive – of course it can be done either way. I eliminated some additional, unneeded components.

Security Switch & Light Circuit Schematic

Datasheet links

SB0061 PIR Sensor (complete PIR module):
epled-2008.en.made-in-china.com/
BISS0001 PIR Sensor (interface IC that goes inside many PIR modules):
http://www.ladyada.net/media/sensors/BISS0001.pdf

How it works

When motion is detected, the output (pin 2) goes to 5V and current flows through R2 to the base of Q1. The emitter current of Q1 becomes the base current of Q2 – the current gain of a Darlington is the hFE squared or about 10,000 to 50,000 (very high). This easily provides adequate drive current for the relay. R1 provides a means of conducting any potential leakage in Q1 to common and also helps Q2 to turn off faster. D1 is the back diode that circulates the relay coil current when Q1 & Q2 turns off thus preventing a damaging voltage transient.

Testing

No, I did not build and test this one – I just happen to know from experience that this circuit will work OK – if the circuit was new to me, I would definitely Protoboard and test because Murphy may be lurking somewhere. Murphy’s Law states that anything that can go wrong will go wrong… Such is true as I have proven it numerous times.

For the Future

PIR crystal amplifier circuit – yes, it is not difficult.

14 Comments

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  • Prakash

    Jim,

    Thank you for the clarification.

  • Jim Keith

    Supply voltage is roughly 5 to 20V as indicated by both the SB0061 module specifications and the BISS0001 IC application example. I believe that it is typical for such modules to include a 7805L voltage regulator.

    The only confusion factor is how it works at 5V due to the drop-out voltage of the voltage regulator. This would reduce the “regulated” voltage to perhaps 3.5V or so. As a result, I would recommend simply running it at 9 to 12V. Also, there is no standard 5V battery.

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