Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Changing a Ground while a circuit is live

  1. May 2, 2012 #1
    Let's say I had a very large circle of wire, and I introduced a charge to it (with the positive terminal on one end of the circle directly opposite the ground). What would happen if before the charge reached the ground I moved it to the other side with a relay, and kept doing this for an extended period of time?
  2. jcsd
  3. May 2, 2012 #2
    I just joint in only, don't necessary have the answer.
    This is in most condition, is a transmission line propagation issue. Assuming the wire loop is very long, and the wire loop is on top of the ground ( earth, plane etc. ) where the other end ( called B) of the wire loop is first connected to ground. When you introduce a charge at the end of the wire ( called A), it launch an EM wave that propagate down the wire towards B. The result is you see a voltage and a current propagate down the wire.

    Then you switch the B from ground to open before the wave front reach B. So B is open when the wave front reach B. Because it is open, the voltage and current wave bounced back and travel back towards A,

    If you switch again, AND if the wire is a perfect conductor that has no resistance, this theoretically will keep bouncing forever. But in real life, there is no perfect conductor, so the amplitude of the voltage and current decrease as loss when going through the resistance. So it eventually die down.
  4. May 2, 2012 #3
    That's what I thought, are there any instances of people doing this?

    Furthermore, can a DIYer do this in a garage? Or is the speed of electricity simply too great for timed relays?
  5. May 2, 2012 #4
    No I don't think this is even practical. The loss of the wire will make it die down quickly.

    When come to traveling wave like this, you get bouncing even if you keep the ground at B, the polarity just inverted every time.

    I thought this is just a theoretical question, you can never use a relay to do this unless you have a wire that is hundreds of miles long and the loss will be so great you might not get a signal from the other end.
  6. May 4, 2012 #5


    User Avatar
    Gold Member

    yungman is exactly correct...you could never use a mechanical relay on your setup in the garage! You yourself said the reason above: The speed of the electrical signal is FAR too great for a mechanical relay to act...by thousands of times faster!
  7. May 4, 2012 #6
    I know I couldn't use a mechanical relay, but how about a SSR or some other switch? I have seen some switches quoted in the 100 ns response time range. Combined with insulated wire which slows the signal down to 60% the speed of light it might be possible?
  8. May 4, 2012 #7


    User Avatar
    Gold Member

    Hey, good idea, ANarwhal! There are FETs that switch on and off within a few nanoseconds, so why not experiment? You may be on to some new discovery...and your experimental apparatus appears to be inexpensive. Go for it!
  9. May 8, 2012 #8
    The question is ......What are you trying to achieve? You have to design a circuit that time the signal and switch.

    You know if it done right, the signal will stop quite fast due to power loss in the wire. You are not going to get a perpetual bouncing if that's what you are driving at. If you observe signal for an extended period of time, it is highly likely the switching circuit actually introducing the glitch every time it switches and that has nothing to do with your initial pulse.

    You can use transmission line equation to calculate how long before it dies down. Even if you don't switch, when the signal hit the end of the wire which can be either a short or open, wave will reflect back to the source, and if the source is shorted or open, it will reflect back again without you switching back and fore. And also it will die down fairly quick.
  10. May 8, 2012 #9


    User Avatar

  11. May 8, 2012 #10
    I want to put a lot of voltage into a wire (around 30kV) and see how long it can ionise air for while going in a loop. Usually the power would just run from a to b, but perhaps I could get a bit more out of it if I redirected the signal back around in a loop.

    If I use this formula: http://www.paigewire.com/volt_loss_formulas.htm [Broken]
    With a current of 5mA and 10 gauge wire the voltage loss is 0.004995V per 1000 ft, so if I make the switching system small enough I think I could ionise air on 1 charge for an extended period of time.
    Last edited by a moderator: May 6, 2017
  12. May 8, 2012 #11


    User Avatar
    Gold Member

    ANarwhal, Thank you for finally explaining what you want to do! If you had made that clear earlier less time would have been used guessing and speculating on your project.

    To ionise air molecules (mostly nitrogen) it requires the expenditure of energy. You don't get any "free lunch". If you apply 30 kV to a wire loop power does not "run from a to b or around in a loop". No current flows, only the HV "sits" on that wire. It may or may not ionise surrounding air.

    High voltage is used in commercial air ionisers all the time. They use sharp electrodes and consume power while working, as we expect. I suggest you study how air is ionised by others, and then consider your alternate method, based on what you learned from the others. Your idea may have some merit, don't give up!

    "Ions cannot be produced without an energy source. An "energy source" means, heat or flame, radioactivity, frictional rubbing, electricity, evaporation (which is a heat process), etc. Minerals that are not radioactive do not spontaneously emit ions. Normal fair-weather ion concentrations are 200 to 800 negative and 250 to 1500 positive ions per cubic centimeter. Indoor levels are usually lower. Several hours before a storm, + ion concentration will increase dramatically, sometimes exceeding 5000 ions per cubic centimeter (cm3). During a storm, - ions increase to several thousand while + ions decrease, often to below 500.
    Ions can also be produced by high-energy events, such as an open flame or a glowing hot object. Hot objects usually emit equal numbers of + and - ions. High DC voltage (over 1000 Volts), especially when connected to pointed metal edges or needles, will produce ions of the same polarity as the voltage source. This is the basis of electric ionizers."
  13. May 9, 2012 #12
    I don't expect to get a "free lunch", just a reduced cost. Perhaps I'm not understanding the concept of HV correctly, but I assumed that it needed a ground from which to draw a current from? What is the physical mechanism of voltage in this case? (I was made to understand that it was an EM wave)

    Please forgive my naivety on this topic, I'm still very much learning when it comes to electrical engineering.
  14. May 9, 2012 #13
    I can't speak for ionized discharge as this is physics. You are dealing with Microwave frequency, you cannot use relay as it is too slow. There is no active device that can take this kind of voltage and to open and short the terminals. I designed a lot of high speed pulsing circuit in my days, you cannot get perfect open and ground.......I could not get perfect open or ground. switchs are usually done by MOSFET. It has Rds, and output capacitance that suck up power.

    For 30KV, you need to stack many of the MOSFET in series. I did a 6KV switch by stacking 8 MOSFETs before, the switching time are in nS range, it gets slower and slower because isolation circuits are needed to drive the FETs in series. Good luck in doing 30KV.

    As I said before, you can get the pulse going back and fore just by leaving the end open as it is traveling wave. You don't need to switch anything. You just have to induce the pulse and let it ring, and regenerate again and again. But again, what I said above still stands. The devices you used to pulse the line is going to load the line. With imperfect open or short, you loss energy and the pulse die and you pull too much power into the circuit.
  15. May 9, 2012 #14


    User Avatar
    Gold Member

    Oh, ANarwhal, building a cheaper air ionizer is in the true tradition of our capitalist system! As you know, there are lots of claims of health benefits and of air purification that ionizers can provide.

    If you are to be successful you need to get yourself familiar with some basic electronics and also some basic physics. Briefly, High Voltage (HV) can "tear off" electrons from neutral atoms or molecules, making them into charged ions. So you would need to make or buy a HV power supply. I would suggest “buy” for you now, and consider making one later. There are various types but, since they all can be DANGEROUS, here on PF we are prohibited from giving information to folks for making dangerous things like rail guns, stun guns, and yes, HV generators. I can say, however, there is always Google...

    Once you have a HV source, you'd need to attach that HV to sharp points (not a loop)where the HV gets "concentrated". If you study some HV chapter in a Physics textbook you’d find that a sphere distributes the charge evenly all around the sphere, whereas a needle point concentrates it...this is the technique ionizers use. If you bought a cheap ($30.00) ionizer and could find a schematic diagram of it you would find that some tiny current (a few microamperes) flows from the HV power supply during operation. This represents the power consumption. Yes, you are exactly right, for current to flow there must be a complete (closed) circuit. So, if you were using HV, there must be a "return path" for current to flow. I leave it to you to discover that path.

    Here is some material you may find useful:
    This topic was already discussed here on Physics Forums. See: “Photon Absorption wavelength(s) to aid ionization?”


    For a technical description of the physics of ionization see: “Electrical Discharges, Introduction”
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Changing a Ground while a circuit is live
  1. Circuit ground (Replies: 3)