Electronic – How to convert the current from a fan to the microphone headset input on an iphone

Apparently you have a audio signal with 3.3 Volt peak to peak amplitude and want to couple that into a "microphone" input of some other audio device. Microphone inputs are meant to take the very small signals produced by microphones. These are often 1 mV or less, with peaks maybe a few mV in normal operation, although this depends a lot on the microphone.

For starters, you probably want to attenuate your input signal by at least 1000 in voltage. That may still be a bit high, but most microphone inputs can probably handle that. You will have to experiment. Maybe you need a attenuation of 3000 or 5000 so that you set the volume control near the middle of its range when listening. A simple start would be:

That will attenuate by about 2000 in voltage. Note the capacitor to AC couple the signal, which removes the DC bias of the input signal.

The top trace is a perfect output for a differentiated square-wave:

The \$C\$ is your coupling capacitor. This is a high-pass filter with

\$ f_c = \dfrac{1}{2 \pi RC}\$

So the higher RC the lower the cutoff frequency. If \$f_c\$ is lower than the signal frequency the signal will pass without much distortion, that's the rightmost waveform. As RC becomes lower the cutoff frequency will move to the right, and ultimately only the signal's highest frequency components will pass the filter. That's the leftmost waveform, only the edges will pass.

So to get a good reconstruction of your signal you'll have to increase RC. I'm not quite sure where exactly you measured this signal, so I don't know the components' values, but you may increase \$R\$ by a factor 10 to start with.