Can someone from the Arduino team or anyone who have the Make your UNO kit or Arduino experts confirm the diode connection in the Make your uno kit documentation [Using the Audio in-line section] (https://makeyouruno.arduino.cc/makeyouruno/module/projects/lesson/04-make-your-synth-shield)? The image in the above url shows a diode connected to Audio IN 's AUX pin. They are supposed to mix audio from a smartphone or external audio output with the audio generated by the Synth Shield. Should not be the diode in the reverse direction as shown in the image to let the audio come in to the Shield and should block any current going out to smartphone? I am attaching the image here for a quick view.
Here is the schematics of the relevant portion of the connection that shows Arduino Uno D9 output pin (which generate audio) is connected to the AUX IN.
it might be important to remember that In a typical mono audio signal, the waveform is an alternating current that oscillates between positive and negative values. So, while the connector has a positive (Tip) and a ground (Sleeve), the audio signal itself oscillates between positive and negative values.
I'd assume the diode is there to cut half of the signal
Thanks @J-M-L for the hints! I totally forgotten that the signal coming in is an AC. But still I do not understand the direction of the diode. Will it let in the positive part of the AC and stop negative part or vice versa? Also, it would loose the waveform and distort the original audio.
sure - cutting off the negative part of the signal removes half of the waveform, which can lead to distortion
I'm unclear from the circuit you posited where the diode goes (and I'm more of a SW engineer than a hardware engineer - so I might not have all the answers)
The mono audio from the jack enters into the Arduino Synth Shield via the diode to the AUX IN pin that is connected to an audio amplifier and eventually is played on a speaker. The Synth Shield allows to mix synthesized audio programmatically to this mono audio.
I guess I would need to bias the AC signal to shift it above 0V.
can you draw exactly where it fits on the circuit?
I see AUX-IN as a pad to the left of your drawing and then the name is reused along the D9 blue wire. so I'm not sure I get what the real circuit is.
I would have thought that a resistor in the position occupied by the diode would have been a better option to mix the external signal with the output from D9.
It's a shame that the PCB layout didn't have a position to fit the component, there is plenty of space available.
what's the purpose of the diode ? It does not prevent a signal coming from D9 to be injected into the phone and cuts half the signal (half-wave rectification?)
if D9 is an output and LOW and you have HIGH on the other side on the jack, don't you create a short on D9?
I think a capacitor + resistor would be better in the place of the diode to bias the AC signal to shift it into DC voltages. Also, I think D9 will never output the current to the external device since it is connected to ground through resistors (is my assumption correct?). But, of course, we cannot take risk on the expensive smartphone so I am planning to connect the audio output of a cheap FM receiver.
