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I'm having a problem with a circuit commanded by a PIR.

I managed to ring a chinese doorbell when someone pass in front of the PIR but the main problem is the two transistors become very hot and I don't understand why. The doorbell is a chinese model like this:

chinese wireless doorbell

This is the circuit:

fritzing circuit

When the button is pushed it connect the positive from an LR23A battery to the circuit and send the radio signal to the receiver so I use the PIR and the two transistors to close the circuit when the PIR detect a presence. The battery can provide 55mAh so the circuit has a very low power consumption compared to the TIP122 and TIP127 capacity.

What's wrong with this circuit?

Thanks in advance for your attention.

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    \$\begingroup\$ You need a resistor between the collector of your TIP122 and the base of your TIP127. \$\endgroup\$
    – brhans
    Oct 24, 2018 at 16:00
  • \$\begingroup\$ He still has to add a current limit resistor for the LED. \$\endgroup\$
    – user201365
    Oct 24, 2018 at 16:33

1 Answer 1

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The resistor below TIP127 is miswired, instead of acting as a current limiter from 12V to the emitter of TIP127, it's a a (non-functional) shunt from 12V to the base of TIP127, as a result both transistors short the 12V load.

You need to place it in series from 12V to the emitter terminal of the TIP127.

With 2200 ohms and 12V you have about 5mA of current through your indicator LED.

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    \$\begingroup\$ If possible send schematics of what you 'think' the circuit is, tracing out a layout is painful, a quick schematic cross-check would find this very quickly. \$\endgroup\$
    – user201365
    Oct 24, 2018 at 16:26
  • \$\begingroup\$ Thank you for the answer. I put that resistor as a pullup resistor for when the PIR is not giving any input. Without the resistor the TIP127 was always saturated and the "LED" always on, causing an overheating of the two transistors and after 5 seconds they were smoking! (The LED is there to simulate the load, but there isn't a LED in the real application, there is the doorbell remote that should not drain more than 55mAh, as the original power source, the LR23A battery, can not provide more current) \$\endgroup\$
    – Fabio Sasso
    Oct 25, 2018 at 10:14
  • \$\begingroup\$ At the end I removed the resistor between the +12v and the collector of the tip122 and add a resistor between the ground and the emitter of the tip122 \$\endgroup\$
    – Fabio Sasso
    Nov 22, 2018 at 22:22

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