Pulsing coil using High Voltage

Discussion in 'General Electronics Chat' started by wes, Dec 25, 2010.

  1. wes

    Thread Starter Active Member

    Aug 24, 2007
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    So my question is about pulsing a inductor. Here are the specs from the NL lab Multisim software I have been using.

    Voltage = 100KV
    inductor = 10uh
    Resistance = 1 ohm


    According to the software, the inductor will reach 1 amp in about 150 picoseconds, I double checked it with a calculator and that seems about right for a 10 uh inductor driven by a 100kv source.

    BUT here is the question. I know that at that level the power pull would be about 100KW if you calculated it by

    100KV * 1 amp = 100KW

    but is that the proper way to calculate a pulse of that sort?
    You see according to Multisim, the voltage at that point in time ( 150 picoseconds) the voltage has only reached about 1 volt.
    That makes since to because 1 volt / 1 ohm = 1 amp

    So if I calculate the power pull using what I get from Multisim, then the power pull is only 1 watt at 150ps.

    1 Volt * 1 amp = 1 watt

    So would this be the correct way to calculate it. I know as time goes on the power pull would increase. for instance at 300 ps (2x )
    the Voltage = 2 volts
    Amperage = 2 amps
    Power = 4 Watts
    again this makes since even after double checking.

    So is this correct way to measure the pulse power. I guess I mean to measure power usage in a pulse, Should I use the Voltage and the current at that time in the pulse? Because at these short times, the pulse has not reached the source Voltage yet.
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Multisim is working with ideal components.

    What is the series resistance of the inductor?

    Impedance of voltage source?

    Tesla Coil, or other pulsed power project?

    It goes into piecewise calculations, an integral of the exponential response of an inductor over time, there isn't an easy way to have a perfect answer, but the approximations sound about right if using some sort of low impedance source.
     
  3. wes

    Thread Starter Active Member

    Aug 24, 2007
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    Well I was just working with ideal componets to make it easier to see how it works. I didn't bother with putting in resistance for the wires and the inductor.

    I just wanted to know which way would calculate the power usage.

    I ran it again with some differn't settings and it seems like the way you would calculate it is by using the voltage where it is at that point in time instead of using the overall source voltage.

    Think about it. If you use the source voltage then your power usage would be extremely high for fast pulses.

    Example: You pulse a inductor to 10 amps

    Inductor = 100 uh
    Source voltage = 1000V
    the inductor would reach one amp in 100 ns

    if you calculate the power by using the source then this thing would use close to 10,000 watts if left on 100% of the time. that is insane for such a small device?
     
  4. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Pulsed Power at Extreme Levels

    The wiki for Pulsed Power has some good info too.

    The reason it seems "easy" in multisim is getting a 100kV supply with more than a few milliamps isn't a trivial task. It is usually created with banks of charged capacitors, think rail gun / coil gun applications.

    The inductor simply dissipates that energy as RF and heat.

    To be sure we are talking about the same concept, can you post a screenshot of your circuit?

    Your calculation for power is correct. Slice up all discrete dv/dt and di/dt (where dt is the same discrete segment), and add them (Integration).
     
  5. wes

    Thread Starter Active Member

    Aug 24, 2007
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    Here is a Screen-shot of the circuit I have been messing with. (it's an attachment)

    So just so I understand, taking the voltage and current at that same instance in time is the correct way to calculate the power?

    in the circuit, you can see the rise time is 1 ns. the pulse width is 100 ns, the voltage drop is 10 volts and if continued, the period is 10 us.


    So Would I just take the voltage drop x the current at that point?
    or would I take the applied voltage x the current?

    1. 10V * 10 Amps = 100 Watts

    2. 9,926V (org 9,936V without voltage drop) * 10 = 99,260 Watts

    Doesn't two seem way off?
     
  6. marshallf3

    Well-Known Member

    Jul 26, 2010
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    The problem with a voltage that high is the tendency to arc over or through the windings of a coil unless it's air wound and spaced properly. I've got some neon sign transformers and several made to ignite oil burners but haven't done a thing with them since I bought them years ago. The intention was to build a large copper ion laser but I never got around to it or the ND:YAG flash tube pumped one I was going to build. Got the proper tubing for the copper one and a nice used rod for the ND:YAG but the time involved far exceed my actual need for such beasts.

    One of my last projects when I was working for the Physics department involved the effects of high intensity light on a quartz crystal oscillator circuit held around 1* Kelvin and yes, we had the ability to get things down that low in temperature with a liquid helium refrigeration system. I got sidetracked on sensors made from electrodeposition of various materials on various insulators and ended up with some rather interesting discoveries in that area.
     
  7. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    With 100A through a 1Ω, 10kW of power is dissipated.

    The supply is only 10kV, so this all works out.
     
  8. wes

    Thread Starter Active Member

    Aug 24, 2007
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    So then the total power pull would be 1MillionWatts

    10KV x 100Amps = 1 MW

    ??
     
  9. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The figure is from the RMS. If you mix the peak-peak values with RMS values, then it does go higher which only "looks like" it violates conservation of energy.

    Put a probe on the resistor like you do on the resistor to be sure, enable the options for showing power on both probes, the resistor only shows current.

    i.e. Since you have a squarewave drive, using the scope measurements aren't needed.
     
  10. wes

    Thread Starter Active Member

    Aug 24, 2007
    242
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    ok, I replaced the first probe before the resistor and this time it shows everything. (pic included)

    I also included probes all over showing everything, lol, just to make sure.
    I also included two just so you can see that voltage and current does stay the same even as they change.

    I don't know how to get a probe to show power though, :(
    I am using the Mulitsim 10 edition, I am still learning to use it.
     
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