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Transformer and inductive load

  1. Jan 18, 2010 #1
    I have a ferrite 12V to 36V transformer driving a purely inductive load near the maximum load capacity of the transformer. The inductance of the load is much less then the transformer secondary inductance.

    Twelve volts is placed across the primary for a period of time T1, and the magnetization current ramps to somewhere near the saturation knee.

    Next, the primary is open-circuited into a clamp to keep the primary difference voltage equal to, or below +/-12 volts.

    How long does it take to reset the transformer flux to zero?
    Or what voltages and currents are found in the primary and secondary?
     
  2. jcsd
  3. Jan 19, 2010 #2

    uart

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    Assuming that you're talking about clamping the voltage across the transformer primary (and not the switch) at |Vp|=12 volts then it will take approx T1 to reset the flux. (During which time the switch voltage will be 24 volts BTW).

    If you clamp the secondary at |Vs|=12 volts then it will take approx 3T1 to reset.
     
  4. Jan 19, 2010 #3
    This sounds very much like the persistent currents in superconducting magnets (like in MRI machines), except your transformer winding is not superconducting. In superconducting MRI magnets, the decay time constant is L/R, where L is the magnet inductance, and R is the coil resistance, which is nano-ohms or less. So in your case, my initial guess would be the time constant is L1/R1 where L1 is the primary inductance and R1 is the coil-plus-current-clamp resistance. So 5 time constants or more to reach a nominal zero. The primary current decays with this time constant, the primary voltage is clamped so

    V1 = I1R1 + L1 dI1/dt = 0 (Note: dI/dt negative),

    and the secondary voltage V2 is 3 L1 dI/dt.
    Bob S
     
  5. Jan 19, 2010 #4
    Yeah, I should have said that the primary has the clamp. But will the primary voltage rise high enough (or go low enough) to clamp?
     
  6. Jan 19, 2010 #5
    The governing equation here is

    V1 = L1 dI1/dt

    If you interrupt I1, you will get a very high V1. If you clamp V1, then dI1/dt is limited, and the time constant is long.

    This circuit is vaguely similar to the old automobile ignition circuits (coil, capacitor, points). Your 12 volt primary is the same as the 12 volt primary in the ignition coil. The ignition coil was a very high "Q" transformer with very low eddy current losses. During the charging part of the cycle, the primary charged up to about 2 amps. When the coil primary current was interrupted by the "points" (actually in parallel with about a 0.02 uF capacitor) on the breaker plate, the voltage on the PRIMARY shot up to about 300 volts. This would happen even if there were no capacitor.

    If the voltage were clamped at +/- 12 volts, the coil primary current would persist. Very roughly, the discharge time constant is the primary inductance over the coil primary resistance, L1/R1.

    This is true only if you have a transformer with a low-loss core. A standard 60-Hz laminated-iron transformer won't work very well, but you could still get a jolt.

    Bob S
     
  7. Jan 19, 2010 #6
    Well, the problem Bob, is that when the primary opens the transformer secondary wants to saturate the core from the current drawn by the inductor. So L is not always constant. L is proportional to mu by the way. L = N^2 mu K, where K =Ae/le, from the effective core area and effective magnetic path length. But before saturation, L of the transformer should be considered much larger than L of the inductive load (I think).

    Maybe I'm considering too many varied parameters and changes at once, and it's driving me stupid.

    Considering what you've said as the small signal solution, where things don't saturate: The large and sudden current change should cause the voltage to rise on both primary and secondary. The primary initially clamps at 12 volts, right? Then what happens?
     
  8. Jan 20, 2010 #7

    uart

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    Nonlinearity of the "iron" circuit doesn't matter. All you need to know is that,

    [tex]\frac{d\Phi_B}{dt} = \frac{-V}{N} [/tex]

    This tells you that whatever time integral of voltage you applied to set up the flux then you require the same (time integral of voltage per turn) but opposite polarity to reset the flux to zero. This of course is neglecting resistance. The effect of resistance is to somewhat reduce the required time.
     
  9. Jan 20, 2010 #8
    Here is the transformer model suggested by the OP, shown in the attachment. A 12-volt supply charges the primary of the transformer for 10 ms. The transformer has a primary inductance of 1 Henry, and a secondary inductance of 9 henrys. The mutual coupling coefficient is 100%. The secondary is attached to a 0.1 Henry inductance load. The transformer primary is clamped both to 12 volts by diode D2, and to ground by diode D1.

    The waveform for the voltages and currents are shown in the attachment. The primary (L1) voltage is shown in green (left scale). The primary (L1) current is shown in red (right scale). The secondary (L2) voltage is shown in black (left scale). The secondary (L2) current is shown in blue (right scale).

    The primary and secondary charging voltages are 12 volts and 36 volts respectively. The peak primary (L1) and secondary (L2) currents are 11 amps and 3.7 amps (scale inverted) respectively. The effective primary inductance, due to L3, is L3/9 = 0.011 Henrys. With a 10-ms charging pulse, the peak current in L1 should be (12/0.011) x 0.01 s = 10.9 amps.

    The 1/e discharge time constant seen in the attachment is about 120 ms, roughly 12 times the charging time (10 ms). Using V = L dI/dt, V = -0.4 volts (Schottky diode) and dI1/dt = ~30 amps/sec from the plot, the effective primary L would be 0.4/30 = ~0.013 Henrys. The secondary load inductance L3 is 0.1 Henry, reflected through to the primary of the transformer is 0.1/9 Henry = 0.011 Henry in parallel with the primary open-circuit inductance of 1 Henry, so the discharge time constant is determined by the inductance L3 reflected through to the transformer primary.

    Bob S
     

    Attached Files:

  10. Jan 20, 2010 #9
    That's the very constraint that's giving me grief.

    As long as the primary or secondary has some kind of clamp that doesn't place much more than +/-12V on the transformer primary, it seems I don't have to worry about core saturation.
     
    Last edited: Jan 20, 2010
  11. Jan 20, 2010 #10
    Bob, that's very nicely done. Wonderful. I wish I had a spice program. Which one do you use?

    I made some oversimplifications and an error in the explaining. Do you think you could try another variation in spice?
     
    Last edited: Jan 20, 2010
  12. Jan 20, 2010 #11

    uart

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    Why should Bob try variations instead of you showing us exactly what circuit you are considering. Show me the circuit and I will tell you how it works.

    BTW. In a "forward" DC-DC converter there is a diode to prevent the magnetizing current from freewheeling through the secondary. Bob's simulation has a freewheeling diode so it's not clamped (which is why the magnetizing current resets so slowly in that simulation). You cant have a freewheeling path on any of your windings (prim sec tert etc) if you want flux reset at a clamped voltage. This is a basic consideration of any pulsed-DC transformer circuit design.
     
    Last edited: Jan 21, 2010
  13. Jan 20, 2010 #12
    I am using LTSPICE IV, free from www.linear.com
    Bob S
     
  14. Jan 20, 2010 #13
    It looks like what I need. The last time I downloaded the free version of LTspice it wasn't very good.
     
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