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

In summary, the conversation is about designing a system that can source DC power while transmitting a high impedance sourced AC signal. The main concern is how to separate the two signals without distorting the AC signal or excessively dissipating the DC power. Possible solutions are using audio output transformers, a bias tee, or an inductor in series with the power load. The conversation also discusses the need to consider the impedance and frequency content of the AC signal when using a twisted pair. The ultimate goal is to create a low profile and low cost design.
  • #1
TheAnalogKid83
174
0
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.
 
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  • #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.
 
  • #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.
 
  • #4
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
 
  • #5
chroot said:
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

Yes indeed. You inject DC with inductors, and couple AC with capacitors.
 
  • #6
waht said:
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.

chroot said:
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

waht said:
Yes indeed. You inject DC with inductors, and couple AC with capacitors.

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.
 
  • #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.
 
  • #8
TheAnalogKid83 said:
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.

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.
 
  • #9
Bob S said:
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.

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!
 
  • #10
berkeman said:
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.

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.
 
  • #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.
 
  • #12
Here is what I was thinking of in my post #2. See attached pdf
 

Attachments

  • AC_modulator.pdf
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  • #13
Bob S said:
Here is what I was thinking of in my post #2. See attached pdf

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.
 
  • #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
 

Attachments

  • blockidagram.pdf
    30.8 KB · Views: 692
  • #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.
 
  • #16
It will work without the op-amp. Do you need the op-amp to just amplify the AC?
 
  • #17
waht said:
It will work without the op-amp. Do you need the op-amp to just amplify the AC?

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?
 
  • #18
TheAnalogKid83 said:
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?

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:
  • #19
waht said:
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]

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:
  • #20
TheAnalogKid83 said:
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.
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?
 
  • #21
Bob S said:
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?

DC current in that setup would only need to be 20mA at 1.5V at the most, I will probably need to increase the voltage at the other end to drive the LED. Audio signal will be probably ~60Hz to 10kHz, although the higher range I can get, the better. The amplitude will be about 1 to 2 Vpp at most, and of course is weakly driven by the source. The line length is variable, and will be anywhere from a few feet to say 20 feet, and it will be shielded with low impedance.
 
  • #22
You don't need to use a summing circuit at all. All you need to do is ac couple your signal to your transmission line with a suitable capacitor. (Suitable meaning that the ac impedance seen by your signal source is suitably low.)

- Warren
 
  • #23
TheAnalogKid83 said:
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

You show a single-ended circuit. Is there no cable?

You show a power source short-circuited to the output of the signal source. That would be be bad, no?
 
  • #24
Going with the AC coupling cap and passive circuit ideas:

I built up a circuit and was able to get the AC + DC signal onto the wire with an AC coupling cap. Now, my problem came, as I suspected, when I loaded the combined signal down with a DC load, it completely knocked out the AC signal from reaching the source. I need to find a way to separate the DC from the AC after they have been coupled.

I may just need a much bigger inductor on the DC load as I was only able to scrounge up a ferrite bead used for EMC that is rated in the MHz and this is obviously way too high of a frequency for an audio signal. Still not completely convinced the inductor will keep the audio signal from being loaded down since the AC load is very high impedance and any inductor won't provide enough impedance to be isolated. I'll find out tomorrow with simulation and another circuit if I can find a good inductor.
 
  • #25
On the receiving end also need the same setup; an inductor tap DC, and a cap to tap AC.

You can model this with inductor reactance,

[tex] X_L = 2 \pi f L [/tex]


The higher the inductance, the greater its resistance to AC.

So a 1 H coil @ 1 KHz would exhibit 6.2 kohms. Which should be more than ample. But at 60 Hz you would get about 360 ohms, which isn't that bad either considering you are working with low DC bias.
 
  • #26
waht said:
On the receiving end also need the same setup; an inductor tap DC, and a cap to tap AC.

You can model this with inductor reactance,

[tex] X_L = 2 \pi f L [/tex]


The higher the inductance, the greater its resistance to AC.

So a 1 H coil @ 1 KHz would exhibit 6.2 kohms. Which should be more than ample. But at 60 Hz you would get about 360 ohms, which isn't that bad either considering you are working with low DC bias.

Ya, using a ferrite bead was just stupid, but I was desperate for a part, and pulled it off a board. I'm going to need to get some big inductor values, and I'm also going to make a preamp with a better drive ability so my AC signal has some balls behind it.
 

1. What is the purpose of superimposing an AC small signal on a DC power signal?

The purpose of superimposing an AC small signal on a DC power signal is to add a varying component to the steady DC signal. This allows for modulation and amplification of the original signal, making it useful for applications such as audio and radio transmission.

2. What is the difference between AC and DC signals?

AC (alternating current) signals constantly change in magnitude and direction, while DC (direct current) signals maintain a constant magnitude and direction. AC signals are used for transmitting power over long distances, while DC signals are typically used for powering electronic devices.

3. What are some common methods for superimposing an AC small signal on a DC power signal?

One common method is to use a transformer, which can isolate and amplify the AC signal before combining it with the DC signal. Another method is to use a capacitor, which can block the DC component of the signal and allow the AC component to pass through.

4. How is the frequency of the small AC signal determined when superimposing it on a DC signal?

The frequency of the small AC signal can be determined by the specific application or desired output. In some cases, the frequency may be determined by the source of the AC signal, such as a microphone or radio transmitter. In other cases, it may be determined by the components used in the superimposition process.

5. What are some potential challenges when superimposing an AC small signal on a DC power signal?

Some potential challenges include interference from other signals, difficulty in isolating and amplifying the AC signal, and ensuring proper synchronization between the AC and DC components. It is important to carefully design and test the circuit to minimize these challenges and achieve the desired output.

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