Injecting AC to DC power supply

Discussion in 'General Electronics Chat' started by Ataleph, Apr 20, 2009.

  1. Ataleph

    Thread Starter Active Member

    Apr 20, 2009
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    Hi All,
    How can I combine AC source of known frequency and amplitude and DC power supply?
    Thanks.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    If by that, can you let some AC ride on your DC output, yes. Helps to know details, of course.
     
  3. Ataleph

    Thread Starter Active Member

    Apr 20, 2009
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    I just want to setup a test in which I will need a known amount of sine wave riding on DC power supply (lets say 20mVp-p 10MHz sine riding on 5V DC). I do have such sine wave generator and DC power source. How I can put them together to get the "noised" DC supply at output?
    tnx
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    A small capacitor should let you inject the signal. Have you got an o'scope? If so, start with around 1000 pF and see if you get the desired level.
     
  5. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
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    That would be difficult because the usual 10MHz signal generator output most likely cannot provide sufficient power(more than a few mA) to drive the external loading.

    If your load takes less than a few mA, then it is possible to "combine" a power supply and a signal generator to get what you want. You just put the output of the signal generator via a high frequency transformer in series with the output of the power supply.

    Wind 10 turns each on a ferrite toroid to form a primary and secondary winding. Connect the output of the signal generator to the primary and the toroid secondary winding in series with the +5V output.

    You can also try the same if you have got a powdered iron toroid instead.
     
  6. Ataleph

    Thread Starter Active Member

    Apr 20, 2009
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    Thank you Beenthere and eblc1388,
    1) In my case the load takes aproximately 0.5 A, but if it was only a few mA, why should I need the transformer? Why just not a signal generator output in series with the output of the power supply?
    2) The combined source should give power to a load (chip) with several stages of bypass caps in parallel (all together a large capacitance), therefore a small capacitor would not help, since almost entire AC signal voltage will fall on it and not on the load as required.
    3) Indeed, the signal generator I have can only supply a few mA and its output resistance is 50 ohm.
    4) I don't have any power amplifier, so will need a passive circuit solution.
    TNX
     
  7. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
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    You could damage the signal generator by injecting current across its output.

    The signal generator output socket is most likely connected to a variable resistor used to varies the RF signal output.
     
  8. peranders

    Well-Known Member

    May 21, 2007
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    Ataleph, just curious but what are you up to? What is your load?
     
  9. Ataleph

    Thread Starter Active Member

    Apr 20, 2009
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    The load is a chip containing SerDes having its own PLL, what I want to do is to characterize PLL's capability to tolerate noise on the chip's power supply. For this purpose I need to inject a known amount of sine noise to the power supply of the chip. That's it.
    Any help is appreciated ( attached schemes are preferable ,of course)....
    TNX.
     
  10. David Bridgen

    Senior Member

    Feb 10, 2005
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    A power supply, i.e. a voltage source, has a source impedance approaching zero, and would therefore be a short to the signal source.

    A transformer in series sounds like a possibility.
     
  11. AdrianN

    Active Member

    Apr 27, 2009
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    1
    Last edited: Dec 6, 2009
  12. Ataleph

    Thread Starter Active Member

    Apr 20, 2009
    31
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    Thank you very much, AdrianN!
    It really seems it would work. I will try to build such a setup and update you about the progress.
    I have already think about using capacitor and inductance, but didn't manage to calculate the values. In your blog it is very clearly explained. Thank you again!
    P.S. There are a few misprints like "XL=10 RL=100" instead of "XL=100 RL=10".
     
  13. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    Indeed the multiplication dot was missing. It should read
    XL = 10 times RL = 100 ohms

    I made a slight change to be clear. Let me know if you see any other typos.
     
  14. Ataleph

    Thread Starter Active Member

    Apr 20, 2009
    31
    0
    AdrianN,
    -To make your article more practical, I think you can count in the sinewave generator output impedance, which is 50ohm in most modern signal sources.
    -More interesting case is when RL is actually a power supply system on a board consisting of traces, bypass caps, parasitic inductances and finally chip's VDD and VSS inputs. In this case there is a problem with the big value of the inductor (micro Henry) , because it will affect the power supply chain. Am I right and what is possible solution for this?
    TNX.
     
  15. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    Ataleph, once you have a method to calculate the components of your circuit, you can complicate it as much as you want. Do you need to take into consideration the output impedance of the generator? Very well. Just replace, in the calculations I posted, XC with RG+XC, where RG is the generator output impedance.

    If this 20mVpkpk, 10 MHz signal needs to appear on a power plane, where you have a lot of 0.1uF caps, then I don't think what you are trying to do is practical. Let's ignore the stray inductances for the moment. If you have twenty 0.1uF decoupling caps on the power plane (as an example), their total capacitance makes 2uF. At 10MHz, it presents an impedance of 8 milliohms. So, at 10MHz, the caps will short out the DC load of 10 ohms. It is normal. It is their job as decoupling caps. At 10MHz your 10 ohm DC load is huge compared with the decoupling caps equivalent impedance. The generator energy will go into these caps. In order to have 20mVpkpk AC drop on these 20 caps, your generator would have to source 20mV/8milliohms = 2.5Apkpk. That is why I said it is impractical.

    I don't know your schematic, but I am sure you can recalculate your RL based on whatever is in series/parallel with it and the result you can call ZL. Then replace RL in the same calculations, with ZL and see what you get in terms of generator output current and voltage. I am sure they will be huge. Your 50 ohm generator output impedance and the decoupling caps will make sure of it.
     
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