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Solenoid current profile

  1. Jul 21, 2008 #1
    Why is the steady state current through a solenoid greater than than the current when the plunger is first fully pulled in? I would expect it to be identical to the low peak in the current profile when the plunger hits the stop when fully engaged.

    Thank you,
  2. jcsd
  3. Jul 21, 2008 #2


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    Do you have a pointer to a solenoid datasheet that shows what you are asking about? Also, what is the inductance before and after the plunger is pulled in? Finally, if this is a homework/coursework question, I can move the thread to the Homework Help forums.
  4. Jul 21, 2008 #3
    Thanks for the quick response.
    No, it is not a homework question, but something I ran into as I was trying to measure solenoid performance (response time). The current profile shows an increase in current when it reaches steady state. I tried posting the profile here, but I could not figure out how to paste an image. However, I found a link that shows a similar response:


    Check item 24, response time.
    ( I think I just managed to attach a jpeg of the response time to this posting).

    Thanks again.

    Attached Files:

  5. Jul 21, 2008 #4


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    I'll hypothesize that you are seeing the usual exponential rise in current through an L-R network in phase I. When the field builds up far enough to overcome friction and the force of the return spring, the core begins to move into the coil (phase II), but this causes the effective coil inductance to increase dramatically. Accordingly, the L-R time constant also increases and the rise in current slows. Once the plunger stops, the current can finally pop up to the DC value determined by the winding resistance.
  6. Jul 21, 2008 #5


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    I'd have to agree with marcusl.

  7. Jul 22, 2008 #6
    Thanks again for the responses.
    I understand everything leading up to the final statement. That was my theory as well, but I was having trouble explaining the "sudden pop" in current. Although I realized it was the current resulting from the DC resistance of the coil only, I could not explain it. I guess the fact that there is no more change in flux because the plunger is no longer moving is the reason for the inductance going to 0. That makes sense.
    However, I still can't reconcile this with the notion that the hold current for a solenoid is less than the pull in current. What am I missing?

  8. Jul 22, 2008 #7


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    You’re a bit confused about inductance. The inductance doesn’t go to zero as the core moves in, it goes to its maximum value (the iron core increases the coil inductance). The sequence is this: a) connect coil to voltage source and the current rises as
    Here L_min is the inductance of the empty coil. The circuit time constant is L/R.
    b) force becomes large enough to move core into coil. L now rises rapidly, so the rise in current slows and, in the graph you provided, even drops
    c) core is fully inserted so inductance reaches its max value L_max. Current is some value I_0. Current again rises exponentially

    The final value of current is i=v/R. It is always greater than the initial pull in current, which is always 0 for an L-R circuit.
  9. Jul 22, 2008 #8
    That curve is not what I would have suspected, at all. And the interesting part is phase 2 where the current drops from a peak value. This isn't going to happen with a first order RL equation. So what's going on?

    If it's really an accurate graph of what goes on, it looks like boh the mass of the slug and the spring enter into the calculation as well. Where the plunger limits can be ignored, it becomes second order, where apparently, phase 2 represents a DC offset, dampened, sinusoidal current.

    If all this really makes sense, the rapid rise in current after phase two could only be explained as the result of the plunger hitting its insertion limit.
    Last edited: Jul 22, 2008
  10. Jul 23, 2008 #9
    Thanks for the explanation. Although I did say inductance, I meant to say the "impedance (Xl) of the inductor goes to zero", leaving only the resistance of the coil. However, from your explantion, I think that was still not correct either. As I understand your explanation, the effective inductance is changing (increasing) as the plunger is pulled in, and the rate of the rise in current is changing during this time as the effective inductance is changing (since it is increasing, the rate is slowing down). Once it stops changing (plunger at the stop), the current now rises exponentially once again, with Lmax/R time constant. At steady state, the current max's out at V/R.

    With regard to pull in current being greater than holding current, maybe this is a misinterpretation on my part. I'm wondering now if what they are referring to is the magnitude of the current required to maintain the magnetic field necessary to hold the plunger inside the bobbin, and not the DC value of the current at steady state.

    Thanks again!
  11. Jul 23, 2008 #10
    This isn't enough to explain so much structure in the curve. The rapid rise in current after phase 2 is opposite of what would occur if inductance were simply increasing, as tau=L/R informs us. It's not a simple RL circuit with R monotonically increasing. The core when in motion, induces a counter EMF on the coil. It's can't regauded as a passive circuit element. Core saturation could be a component, as well.

    Usually the pull-in and hold voltages are quoted, with the hold voltage less than the pull-in voltage.
    Last edited: Jul 23, 2008
  12. Jul 23, 2008 #11
    I asked a person about this today and he said that holding coils have higher current than pull-in, otherwise just the vibration of the machine might release the coil. Maybe that is what is happening?
    I found this on the Internet:
    A typical solenoid requires 10% of the normal current to remain energized. To accomplish this, use one of the following:
    Mechanical hold in resistor
    Capacitor discharge and hold in resistor
    Transistorized hold in circuit
    Pulse-width modulation
    Pick and Hold
    Dual voltage
    Multiple coils
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