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Volt-seconds of magnetic core

  1. Aug 22, 2015 #1
    If I have a voltage magnitude that I can modify, say I'm measureing the RMS of it. And that it is at a set frequency, and I want to know the quantity (that I am only just familiarising myself with) known as volt-seconds, on a coil and magnetic core.
    Say I was looking at how many volt seconds the flux ran up and down a 4 quadrant BH curve. The V.s would be
    I don't really see why...

    Nothing to do with the period of the wave? Also say I was just talking about the top quadrant of the curve, when the flux was at the right most point, I couldn't say that the V.s = V*(quarter time period of voltage) could I?

    Because I imagine at a quater of the time it has run up, at half time it has run back to zero and at another quarter is negative. Given T = 1/frequency

    V.s = Vrms * 1 / (4*frequency)

    I don't see a problem with that, do you?
  2. jcsd
  3. Aug 22, 2015 #2
    I've been plotting some data and I've noticed that the BH curve I have generated is the same as my V.s/N Vs. H curve, by a constant.
    Why is this so?
  4. Aug 22, 2015 #3
    Ah, I see 1/omega is like time period, so I was timesing when I should have been dividing.

    I need to convert AC V.s into DC V.s

    I suppose just say the V.s for rmsAC then divide those V.s by the same DC voltage to get the amount of time for DC V.s
    which will be many orders of magnitude smaller.
    Last edited: Aug 22, 2015
  5. Aug 23, 2015 #4
    I realised that flux is V.s/Turn, not just V.s

    Just to double check, V.s = V/(2*pi*freq) right??
    Why not just V/Hz?
    Because time period = 1/f
    so V/f would be [V.s]
  6. Aug 23, 2015 #5

    jim hardy

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    volt-seconds is the product of volts X seconds , just the area under the v(t) curve
    for sinewave observe it starts at zero, aftr a half cycle reaches some positive value , at end of cycle.returns to zero

    what would be area under first half cycle if Vpeak is 100 volts, frequency is 50 hz?
    I think it'd be 0.636(average of sine?) X 100 volts X 1/100 sec =0.636volt-sec
    Try integrating it. ∫vdt = vt ?

  7. Aug 24, 2015 #6
    Good to hear from you. I've thought on this long and hard.
    Remembering the integral of Sin(ωt) = 2/ω and that T = 1/f:
    Well the area under a sine hump through integration is (the peak magnitude of the Sin) * 2 / ω,
    so the total area of a sine wave for one period is: 2 * [2 * peak magnitude / [2π*frequency] ]
    (That's like 2 * ∫0π/ω dt )
    multiply that by how many periods in an amount of time, for V.s.
    However if it's RMS, it's a straight line a bit lower than the peak of Sine: so the area is: [peak/√2]*Time of period = peak / [√2*frequency]
    which gives a slightly larger area than the total area of a sine wave period.
    How is the average of a 100 volt Sine wave 63.6V? wouldn't it be 100/√2 = 70.7V?
    And for 50Hz, wouldn't the time be 1/50 sec, not 1/100 sec?

    Is it one of these methods:
    Where T = 1/(2*pi*f)
    70.7*T = 70.7/(2*pi*50) = 0.225 V.s
    Or using T = 1/f multiplied by the rms level:
    70.7/(50) = 1.414 V.s
    Or using multiplying the integral area by 2:
    2* 100*2/(2*pi*50) = 1.2732 V.s

    As you can see I am still a bit confused as to the reasoning, graphically the bottom two methods make the most sense to me.

  8. Aug 24, 2015 #7

    jim hardy

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  9. Aug 24, 2015 #8
  10. Aug 24, 2015 #9
    Ok, I see how it's the 'average amplitude of half the period of a Sine wave, I always thought the RMS was the average, as I thought 'mean' was like the average. Oooh boy, you learn something new everyday... so you use RMS for power.
    Could you explain what the difference is between RMS and Average? And when you use average?...I assume you use average in working out the V.s, not the Rms, for a start...

    mathematically I remember it was the averages squared, then square rooted, but that still seems like the same thing...

    Gee wizz...
  11. Aug 24, 2015 #10
    So 100V * 2 / PI * 1/50 = 1.273 V.s
    Which is the same as my:
    "Or using multiplying the integral area by 2:
    2* 100*2/(2*pi*50) = 1.2732 V.s"

    H'mm I'm going to have to check what this does to my V.s/N calculations in excell.....

    See other post for what happend there.

    "Measuring average voltage is cheaper than measuring RMS voltage however, and that's what cheaper DMMs do. They presume the signal is a sine wave, measure the rectified average and multiply the result by 1.11 (0.71/0.64) to get the RMS value. But the factor 1.11 is only valid for sinewaves. For other signals the ratio will be different. That ratio got a name: it's called the signal's form factor. For a 10 % duty cycle PWM signal the form factor will be 1/10−−√, or about 0.316. That's a lot less than the sine's 1.11. DMMs which are not "True RMS" will give large errors for non-sinusoidal waveforms."

    Ok so the difference between average and RMS really is just order of operations to get RMS from average. And we use RMS instead of average because power is propertional to voltage squared...I only sorta get that last bit.
    Well my voltage measurements were taken with a DMM but it was a very sineusoidal supply. (my current was taken with a true RMS meter).
    Last edited: Aug 24, 2015
  12. Aug 24, 2015 #11

    jim hardy

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    when the magamp begins to saturate , does current remain a sinusoid?
    Same question for voltage?
    We must be aware that we're measuring waveshapes with varying form-factors
    slide rule accuracy is as good as we can expect.
  13. Aug 24, 2015 #12
    No the current doesn't, I know the voltage will start to move from being over the ideal TX to be over the resistance of the coil (as the current spikes), so V.s on the core won't be sineusoidal as the core saturates. But you can see from the graphs in post # 198 that the shape of the N.s/N curve is identical to the BH curve, so what does that mean?
    Also, I don't see why B = Vrms/(2*π*f*N*Area) From the EMF of a TX equation (which is essentially Faraday's equ.)
    But if: Φ = V.s/N
    than why V.s/(N*A) = Vrms*√2*2*/(N*π*50*Area) = VPK*2*/(N*π*50*A) isn't equal to B?
    Last edited: Aug 24, 2015
  14. Aug 24, 2015 #13

    jim hardy

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    Sorry for post 11, i shouldn't have mixed the transformer and volt-second threads.

    keep thinking real simple and single-step it.
    Apply this waveform from above to a resistive load so that current will have the same waveform.
    Then calculate power..
    instant by instant, power is volts X amps ,
    and you know amps is volts/R
    so instantaneous power is volts X amps = volts2/R
    and volts2 is always positive.
    So to calculate average power i must use average of the squares of instantaneous voltages, and square root of that is RMS
    to calculate average voltage i use just the average of the not-squared instantaneous voltages , no squares or square roots at all.

    Simplify: How do numbers work?
    Is the square of the average same as average of the squares ?
    Try it
    take an easy set of numbers : 3,4,5
    Their average is 4 and 42 is 16
    their squares are 9,16 and 25 and average of those is 16.667
    so average of squares and square of average are not the same.

    RMS of 3,4,5 is √((9+16+25)/3) = √16.667 = 4.082
    average of 3,4,5 is 4.

    it's an arithmetic process and not just for sinewaves.
  15. Aug 24, 2015 #14
    That's a good way of putting it.

    Aah, fantastic explanation.
    Last edited: Aug 24, 2015
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