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it's my first time working with piezos and I'd like to use one as an input to my iPhone via the microphone port to allow my app to detect vibrations.
in order to do this, I was thinking that I'd use a splitter to separate the headphone and microphone channels for my iPhone, then plug my piezo directly into the microphone port.
first of all: would that work? second of all: is there any danger that I'd harm the microphone-in of my iPhone?
thanks!
You can separate the mic audio from the earphone audio fairly easily. But I don't think that a piezo pickup is going to work into your phone very well (or at all) without an active buffer.
The phone is expecting the mic to have a source impedance down around 1k and your piezo pickup wants to work into something with an input impedance in the megohms range (or higher).
The electret mic that is part of your existing headset has a buffer transistor inside that does this for you. This buffer is a special FET that has extremely-high input impedance and is powered by the bias supply that is present on the mic audio line coming from your phone.
The easiest way for you to interface your piezo pickup is to grab an old earphone / mic cable from somewhere and take the mic capsule apart. You will see the electret element at the front of the mic capsule and a tiny circuit board at the rear (where the wires are soldered). Simply take that circuit board out of the old mic. There is a single wire that attaches to the electret element - that is the hot side of the input. The ground side of the input is the common lead on the back of the circuit board - the trace around the perimeter of the PCB where the aluminum can was crimped.
Be very careful about ESD when working with that input lead into the buffer transistor - it is extremely sensitive to over-voltage and it's easy to destroy the buffer and not even know it.
The piezo pickup will connect directly to that buffer without needing any extra components.
The only problem that you may run into is that your piezo pickup may have too much signal amplitude and may need to be attenuated. Just post another question if that is the case.
