2
\$\begingroup\$

I have multiple generators that are producing a variable DC voltage between 0 and 18volts each that are being sent to an inverter for an AC output used to power various devices.

I really need to be able to transmit an audio signal back to each of the generators, and would really like to avoid having to add additional wires. Would it be possible to piggyback an audio signal of a usable quality over these variable DC lines?

I've searched around and suspect this answer might be appropriate, but they're mostly talking about communication signals, and I'm not sure if the required fidelity for an audio signal would be possible. I'm also not sure the effect a highly variable DC voltage would cause.

Thanks

\$\endgroup\$
6
\$\begingroup\$

You would need to use a modulated signal, not the baseband audio because the low frequencies will be heavily attenuated by capacitance on the d.c. voltage, and because of the rate of change of your d.c. voltage could be at low audio frequencies (rumble). I'm sure you could get reasonable audio quality if you use a suitable modulation that could be seperated out from the d.c./low frequency signal.

\$\endgroup\$
  • \$\begingroup\$ Hey Martin, I have to say I'm not particularly hot on this topic. But from your answer I guess you mean I could use Frequency Modulation on the audio signal, merge that onto the DC lines, and then decode the FM signal at the other end in order to get the audio out again? \$\endgroup\$ – Ben Delarre Mar 31 '11 at 10:04
  • \$\begingroup\$ Yes, you could use FM or AM (analogue modulation) with a carrier that wasn't necessarily RF although needs to be higher than the audio. Or there are digital modulation techniques, or combinations of the two. \$\endgroup\$ – Martin Mar 31 '11 at 10:12
  • 1
    \$\begingroup\$ Martin has a point about the DC supply voltage variation possibly leaking into your audio signal. However, I disagree with the reasoning that lower frequencies will be attenuated more by the shunt capacitance on the DC supply line. Capacitive reactance (i.e. impedance) is inversely proportional to frequency. Therefore as frequency increases, the impedance decreases and the capacitance will effectively short the high frequency signal to ground. That is to say, it will attenuate high frequencies and not attenuate low frequencies. \$\endgroup\$ – Adam P Mar 31 '11 at 19:08
  • \$\begingroup\$ Adam, you are right you would lose the high frequencies more than the low. I was thinking of the high ESL of large electrolytics but that wouldn't come in at audio frequencies. \$\endgroup\$ – Martin Mar 31 '11 at 19:27
  • \$\begingroup\$ OK that makes sense. Upon further discussion with a friend who is better at this than me, I have more concerns now about reflection due to impedance mismatches between the generator and the inverter, and the noise coming from the brush-based generators. Guess I need to test this out, and then see if these issues can be overcome. \$\endgroup\$ – Ben Delarre Apr 1 '11 at 8:58

Your Answer

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

Not the answer you're looking for?Browse other questions tagged or ask your own question.