first of all: noob disclaimer: sorry for the unprofessional sketch and my limited understanding of what I'm doing. I'll try my best to make myself understandable.

I'm trying to make a radio receiver that uses an external audio amplifier, but when I try to use the same power source (the AC-DC adapter), either the DC-DC buck converter or the FM receiver blows out because of over voltage or over current (smoke and sparks effect). Luckily those asian modules are cheap..

But if I use different AC-DC adapters for the FM receiver and the amplifier everything works. Why? How could I use a single power source for powering everything? I suspect the problem is some kind of ground loop etc, but I can't understand how to fix this.

In the below image the AND-gate is supposed to be a 3.5 mm audio jack and not an AND-gate.


simulate this circuit – Schematic created using CircuitLab

  • 3
    \$\begingroup\$ those asian modules are cheap might be the problem ... do you actually have everything connected as per the block diagram? ... measure the voltage between FM receiver GND and audio amp GND \$\endgroup\$ – jsotola Mar 1 at 7:30
  • \$\begingroup\$ … measure that difference before connecting your 3.5mm plug. \$\endgroup\$ – Marcus Müller Mar 1 at 8:27
  • \$\begingroup\$ Yes I have. I get a feeling my problem relates to this [electronics.stackexchange.com/questions/17379/… And per Marcus's suggestion the problems begin when connecting the 3.5 mm plug to the amp. \$\endgroup\$ – ComponentFryer Mar 1 at 8:45
  • \$\begingroup\$ measure the voltage between the sleeve terminals of both 3.5mm jacks before connecting them. (don't connect them, just measure) \$\endgroup\$ – Jasen Mar 1 at 9:14
  • \$\begingroup\$ I was fast enough to discard the blown fm receiver so I don't have it anymore, but I took a measure between the buck converters out (-) and the sleeve of the audio input (-) at the amp and the voltage difference was 2.65 V when the buck converter is set to an output voltage of 5.04 V. \$\endgroup\$ – ComponentFryer Mar 1 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.