I have a project that I'm currently powering with a AAA battery, providing 1.5v to a simple 555 timer circuit. The output of the circuit goes through a voltage divider into the microphone input of a tablet or mobile phone to be measured, i.e. with an oscilloscope app.

Is there a reasonable way to remove the battery from the circuit? When I measure the output of the mic pin with nothing else attached, I get anywhere from 1.5v to 3v, depending on the phone/tablet. Can I draw on this power from the microphone port to power the timer, and still measure the output using the mic?

  • \$\begingroup\$ Does the tablet have a USB port? \$\endgroup\$ – SomeEE Feb 7 '14 at 5:17
  • \$\begingroup\$ The ultimate goal is to use this on any tablet or phone, so assume no USB. \$\endgroup\$ – option8 Feb 7 '14 at 17:09

It's probably possible to power your device from the microphone port itself if your device consumes about 1-2 mA or less. Just connect the "power" rail of the device to the microphone input positive terminal with an RC filter and put a capacitor between the device's output and the microphone input terminal. Choose the resistor's resistance so that the voltage drop is acceptable to you and the capacitor so that the RC is about 5 times longer then 1/f, where f is the lowest frequency present in the signal.

If you need to attenuate your signal, connect a resistor in series with the capacitor between the device's output and the microphone input. Treat it as a resistive divider where the "upper" resistance is this resistor and the "lower" is the input resistance of the microphone amplifier and your RC-filter's resistor connected in parallel.

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  • \$\begingroup\$ What is the purpose of the RC filter between the mic and the timer's supply? \$\endgroup\$ – option8 Feb 7 '14 at 17:13
  • \$\begingroup\$ "if your device consumes about 1-2ma or less." Is this a documented standard or an assumption about the construction of tablet audio input circuits? \$\endgroup\$ – SomeEE Feb 7 '14 at 19:45
  • \$\begingroup\$ After playing around with resistor and capacitor values, i'm finally seeing the waveform on the tablet. Thanks! now i just need to do the math to maximize my output... \$\endgroup\$ – option8 Feb 7 '14 at 19:49
  • \$\begingroup\$ @MathEE, this is my experience in measuring current supplied by typical microphone inputs including celluar phones and PC soundcards. \$\endgroup\$ – motoprogger Feb 8 '14 at 7:05
  • \$\begingroup\$ @option8, the RC filter splits the signal and the power lines. It prevents the signal from the device output from reaching the power line and also prevents the current consumption variability from being recognized as a signal by the microphone amplifier. \$\endgroup\$ – motoprogger Feb 8 '14 at 7:09

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