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Best way to superimpose an AC small signal on a DC power signal

  1. May 4, 2009 #1
    I'm trying to design a system that will source DC power while transmitting a high impedance sourced AC signal superimposed onto this power signal. I'm sure this has been done in many applications, but I'm new to it.

    Currently I'm trying to use a basic adder circuit with an opamp to superimpose the two signals, but my concerns are that the DC load (a simple resistor + LED) off of the output of the opamp will load down my AC signal on the receiving circuit. I need a way to separate the two signals transparently ie not distorting the AC signal AND not excessively dissipating the DC power where its not needed.

    First is there a simpler or more ideal way to superimpose the two signals? Second, is there any suggestions on how to extract the AC signal off of the DC power signal without wasting power or distorting my AC signal?

    EDIT: I'll draw up a simple block diagram in the near future.
     
  2. jcsd
  3. May 4, 2009 #2
    We used to use audio output transformers in vacuum tube radios that had superimposed DC plate current and audio signal (input impedance several kohms), with an 8-ohm secondary. The major problem with many transformers is that they are not designed to carry a dc current. I would try the Radio Shack Audio Output Transformer
    Model: EI-19 | Catalog #: 273-1380. 1000 ohm pri, 8 ohm sec. For the DC sourcing, run the dc through the 8-ohm output winding. On the DC receiving end, run the signal through the 8-ohm winding of an identical transformer. Now apply the AC source on the 1 k-ohm winding, and on the receiver, pick the ac signal off the 1-kohm winding. To get 1 volt ac superimposed on the dc, you will need sqrt(1000/8) = 11 volts. I will guess you could put 100 ma current through the 8-ohm windings w/o saturating the iron core.
     
  4. May 4, 2009 #3
    This is done all the time and the solution is extremely simple. If you have a satellite, then the power is supplied to the dish LNB through same coax.

    You know that a capacitor will only allow AC to go through and block DC. On the other hand, an inductor will tend to block AC but will let DC to go through. Put those together and you have a bias Tee.
     
  5. May 4, 2009 #4

    chroot

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    To insert the signal, just capacitively couple it with a suitable capacitor. To receive it at the other end, use a high-pass filter.

    - Warren
     
  6. May 4, 2009 #5
    Yes indeed. You inject DC with inductors, and couple AC with capacitors.
     
  7. May 4, 2009 #6
    My problem is that I need to tap into this signal for the DC supply with a resistive load. If I load it down, I will be loading down the AC and it simply won't pass through the capacitor . . Unless you suggest I put an inductor in series before my DC load? I just don't want to dissipate the AC signal across any of my DC powered components.
     
  8. May 4, 2009 #7
    I just reformatted my PC and don't have any tools on it yet, but I'll install office and make something so you can tell me if I'm on the right track or if I'm making a problem out of nothing.
     
  9. May 4, 2009 #8

    berkeman

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    Yes, as mentioned earlier, you put an inductor in series with the power load (and any power source that has a low output impedance), and you capacitively couple your AC signal onto and off of the main twisted pair. Be sure that your inductor has high enough impedance at signal frequencies, and does not saturate when the load current passes through it.

    Also, depending on the length of your twisted pair, and the frequency content of your AC signal, you may need to consider forward terminating the twisted pair.
     
  10. May 4, 2009 #9
    Hmm this is interesting, and I'd investigate it further, but the design I'm doing has to be low profile and low cost, I'm worried two transformers would take up more space than I can afford. I'll keep this in mind though, thanks!
     
  11. May 4, 2009 #10
    Alright,

    Luckily the LED current won't saturate a cheap inductor.

    Unfortunately, I am dealing with very low frequencies (audio) so my inductive reactance will be low without a large inductor value, and that will take up space and cost.

    This is leading up to another important question I have that I hope you guys can help me with, but I'll wait until I get more feedback and the block diagram. Thanks again to everyone.
     
  12. May 4, 2009 #11
    Also, does everyone think that the OpAmp summer circuit is suitable for forwarding DC power? I didn't say it explicitly, but I can't imagine needing more than 100mA of current at about 1.4V out to the receiving circuit depending on what I do on that end.
     
  13. May 4, 2009 #12
    Here is what I was thinking of in my post #2. See attached pdf
     

    Attached Files:

  14. May 4, 2009 #13
    Do you think I could make something like that in the size of say a flash thumb drive for each end? The part you suggested looks huge, but this looks like it could be promising.
     
  15. May 4, 2009 #14
    Alright, attached is sort of what I've been planning to do, and I've simulated it with real opamps, except for the DC power distribution scheme, which is what worries me the most.

    blockdiagram.jpg
     

    Attached Files:

  16. May 4, 2009 #15
    On second thought, I'm wondering if I can just delete out that audio signal conditioning block all together if I can modify my summing circuit to give me what I want by itself. That way I don't have to worry about the DC load changing because of additional active circuitry.
     
  17. May 4, 2009 #16
    It will work without the op-amp. Do you need the op-amp to just amplify the AC?
     
  18. May 4, 2009 #17
    Well, ideally, I'd like to touch the AC signal as little as possible, so no opamp would be great for me. I just was always under the impression that its bad to tie two voltage sources together, and that they wouldn't necessarily add together just by shorting them together. I'll run through the circuit analysis real quick.

    Edit: How exactly is the AC supposed to arm wrestle with a DC supply? I guess the only way for that to happen is if I took the battery's series resistance into account?
     
  19. May 4, 2009 #18
    By running DC first through an inductor, then couple AC with a capacitor. The inductor will block out any AC trying to get on your DC source

    http://img89.imageshack.us/img89/2735/0610mwj32fig01s.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
  20. May 4, 2009 #19
    Hmm, I did this for a GPS active antenna circuit, but the ind and cap values were very small since the frequency was 1.575 gigahertz and at low power. I will probably need large values for audio signals without distorting them.

    Thanks for the help, I'll give this a hard look.
     
    Last edited by a moderator: May 4, 2017
  21. May 4, 2009 #20
    How much dc current do you need to send over the line? What are the frequencies and amplitudes of the ac signal? How long is the line?
     
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