# Building circuit concern and question

Hi all, I’m looking to build the attached circuit to analyze audio for a LED strip project. I was wondering what the 47nF ceramic capacitor was used for. Also, what is significant between using the 3.3v supply and the 5v supply for the amplifier+dc bias portion of the circuit? If I’m following this circuit, how am I supposed to supply the LEDs if the 5v supply is being used for the audio analyzing portion of the circuit?

The 47nf and 100k resistor look like a short at 33 kHz. I can't see a valid reason for the capacitor.
The Arduino is a 3.3V processor. 5V is dropped to 3.3v with an on-board regulator. The 5V has absolutely nothing to do with the amplifier. The two resistors provide a 2.5V zero point for the analog input because the amplifier output will be between +/- the battery voltage. The Arduino analog input only reads 0 to +3V3.

What LEDs?

The 5V power can only supply, barely, 1 Amp. But the voltage divider draws practically no current. About 30 nanoAmps.

That makes no sense at all to me, but I'm not that experienced with analog circuits. Have you considered using an MSGEQ7 chip?

SteveMann:
The 5V has absolutely nothing to do with the amplifier. The two resistors provide a 2.5V zero point for the analog input because the amplifier output will be between +/- the battery voltage. The Arduino analog input only reads 0 to +3V3.

In the circuit, why is the analog circuit connected to 5v instead of 3V3 and AREF pins?

SteveMann:
What LEDs?

How can I power the LEDs from the Arduino if the 5v pin is taken up by the analog circuit?

You can connect more than one thing to 5V but you can’t power the LED strip through the Arduino because it requires too much current. The LED strip needs it’s own connection to the 5V power supply.

The bias should be half of the Arduino’s power supply voltage which is 5V for a standard Arduino, and that’s also the default ADC reference. So the ADC should read 512 with silence (on a standard Arduino with a 10-bit ADC). That means you connect the voltage divider (2 equal-value resistors) to +5V.