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I've connected a Raspberry Pi to my Friedland "Ding Dong" doorbell. The doorbell has two (apparently unused) terminals, over which there seems to be a stable voltage of 1 mV when the button is not pressed, and a stable voltage of 4.8 V when the button is pressed.

To get this signal into the Raspberry Pi, I've come up with this simple design:

enter image description here

Between the switch and the resistor is 10 m of UTP cable. The resistor is the largest one I had, and it can still trigger the transistor. I'm running the GPIO input high (pull-up resistor) so a signal on the base pulls it down to ground.

My problem is that every once in a while, a signal will be detected when there is none. I know noone rang my doorbell, but the Raspberry Pi detects a signal (I'm using Python3 GPIO.wait_for_edge, like this:)

 GPIO.wait_for_edge(pin,GPIO.FALLING)

So I suppose some noise can trigger the transistor? But what can I do about it?

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  • \$\begingroup\$ Keep in mind that many doorbells use AC current to actuate. They have a simple transformer connected to mains which outputs a few volts AC. Are you sure your doorbell uses a DC supply? \$\endgroup\$
    – Ryan Griggs
    Jan 1, 2016 at 21:54
  • \$\begingroup\$ Good point! But it does have four 1.5 V cells, and no mains connection :-) \$\endgroup\$
    – OZ1SEJ
    Jan 1, 2016 at 23:28

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Your voltage-detection circuit has a very high impedance. I would go for a (much) lower impedance, for instance an 10k resistor, with an additional 10k resistor between the base and emitter of the transistor. This makes it much less likely that a stray voltage (moisture?) triggers your circuit.

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  • \$\begingroup\$ Thank you for your suggestion! But I'm just a bit curious (I don't have very much electronics experience): Why would reducing the base resistor to 10k reduce noise? And why would adding a resistor between base and emitter reduce noise? \$\endgroup\$
    – OZ1SEJ
    Jan 1, 2016 at 21:57
  • \$\begingroup\$ Together, the two resistors from a voltage divider, which raises the voltage that will trigger the transistor to ~ 1 Volt, and the lower impedance means that ~ 0.05 mA is required to get 1V. Together these requirements make it less plausible that a high-impedance source (statics, moisture-leakage) will trigger the circuit. \$\endgroup\$
    – Wouter van Ooijen
    Jan 2, 2016 at 9:39
  • \$\begingroup\$ Hm, I do see your point. Two things; first: If the voltage is 4.8 V and both resistors are the same, won't that raise the required voltage to 2.4 V (not that that's a problem)? And second: 4.8 V / 10 k = 0.48 mA - is that not too much for the base-emitter junction? Is this sketch correct? i.imgur.com/rak6OYg.png \$\endgroup\$
    – OZ1SEJ
    Jan 3, 2016 at 20:21
  • \$\begingroup\$ With both resistors the same the required voltage is a little over 2 x 0.6V = 1.2V. The maximum base current can be found in the datasheet, look it up. \$\endgroup\$
    – Wouter van Ooijen
    Jan 3, 2016 at 20:32
  • \$\begingroup\$ Okay, so, now I've made a voltage divider like in this diagram: i.imgur.com/rak6OYg.png. But the GPIO pin still triggers every once in a while even when the button is not pushed - typically when the kids are jumping around or I'm pulling the wires, like it's a loose connection. I just don't see how that can trigger the circuit! Especially now with the two 10 kΩ resistors... \$\endgroup\$
    – OZ1SEJ
    Jan 5, 2016 at 21:43

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