# Noise in Audio (DFPlayer) when adding LED strip

I'm having a problem with a weird audible noise in my sound player circuit.. It's due to an interference with a LED strip.

I've used this circuit (Arduino + DFplayer + Amplified Mini Speaker) many times but since I decided to add a LED strip to the circuit, a very loud noise (which sounds like high frequency strings) appeared..

I've tried to use a 104 capacitor with 270ohm resistor to create a filter, but it didn't work.

Maybe a diode with a capacitor somewhere??

Could anyone give me some suggestions to fix this?

simulate this circuit – Schematic created using CircuitLab

• How is the speaker amplifier powered? I think that wire is missing from your schematic... – Vladimir Cravero Feb 24 '17 at 14:43
• Most likely your LED strip uses PWM, so it draws pulsed current. This could cause either lots of ripple in the supply, or ground noise. Please show a picture of the setup, or at least a detailed schematic of how the grounds are connected. – peufeu Feb 24 '17 at 15:42
• The amplifier is powered by USB (5v + G)..the wire is there (usb in). – Chu Feb 24 '17 at 16:53
• I have uploaded pictures now. – Chu Feb 24 '17 at 17:19
• The USB lead powering the loudspeaker makes a huge ground loop via the hub. Test with a separate power supply for your loidspeaker. – peufeu Feb 24 '17 at 17:25

The source of the noise is the PWM controlling of the WS2812 LEDs within the stripe. Their PWM frequency is approx. $1\,kHz$ (some say $400\,Hz$), what is audible for humans.
The first fix is to add a electrolyte capacitor close to the terminal of the stripe. Chose a $\ge470\,\mu F$ cap with a voltage rating from at least $\ge6\,V$. You even can use several caps in parallel.
The next step is to filter noise on the player supply lines. Therefore use a resistor of $\approx 100 \Omega$ and again a cap $\ge100\,\mu F$ with again $\ge6\,V$ voltage rating. If you have inductors available replace the resistor with a solenoid. The suitable equation for determining the values is $f_{Filter} = \frac{1\,kHz}{10} = \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}$