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Math Help - Calculating Inductor discharge time into a charging Capacitor.

  1. #1
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    Question Calculating Inductor discharge time into a charging Capacitor.

    This is an electronics engineering related question but it is mostly mathematical so I thought I should post it here. Hopefully you know something about electronics to help me.

    What I cannot figure out is how to calculate the time it takes an Inductor to fully discharge into a Capacitor that it is charging up. Assume that the capacitor does not discharge back into the inductor like a simple LC Resonant Tank circuit.

    Inductors:
    Inductor - Wikipedia, the free encyclopedia

    Capacitors:
    Capacitor - Wikipedia, the free encyclopedia

    LC Resonant Tank circuit:
    LC circuit - Wikipedia, the free encyclopedia

    Other info:
    L - denotes the inductance of an inductor.
    C - denotes the capacitance of a capacitor.

    Hopefully I have posted enough info to get the ball rolling.

    Your help is greatly appreciated.
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  2. #2
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    So let me see if I've got this straight: you've got some current stored in an inductor, and then you switch in a capacitor (and, let's say, a diode for the uni-directional assumption you've laid out). The current is a flow of charge that causes a buildup of voltage on the capacitor. You want to know how fast the inductor completely discharges into the capacitor. Is this correct?
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  3. #3
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    Well sort of. I want to be very strict on my wording here to prevent confusion.

    You said:
    " You want to know how fast the inductor completely discharges into the capacitor. Is this correct? "

    I say:
    I don't want the rate of change, I want the time it takes for a pre-charged inductor to completely discharge into a capacitor. I think that's what you were saying anyway but just to be sure.

    Your assumption of a diode was correct. Saying that current gets stored in the inductor is not correct. For an Inductor they store energy in the form of a magnetic field which is proportional to the current (charge/second) flowing through it.

    Thanks.
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  4. #4
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    I understand you meant the time of discharge, not the rate at which it discharges. Also, I do know that the energy is stored in the magnetic field; however, for the purposes of this problem, what's of interest is the current that comes out of the inductor.

    There are several things we know: the current through the inductor cannot change have a jump discontinuity. To have a jump discontinuity is not physical. Also, the voltage across the capacitor, being proportional to the charge on the capacitor, and also being equal to the integral of the current, cannot have jump discontinuities. Taken together, nothing is going to be able to change instantaneously.

    In addition, the current through the inductor and the capacitor will be the same, as will the voltage across the inductor and capacitor. That is, we have

    i=-C\,\dfrac{dV}{dt}=CL\,\dfrac{d^{2}i}{dt^{2}}. Hence,

    \dfrac{d^{2}i}{dt^{2}}=-\dfrac{1}{LC}\,i.

    Solutions are i=A\sin(t/\sqrt{LC})+B\cos(t/\sqrt{LC}).

    Without the diode, of course, you'll get this sort of oscillation. With the diode, you'll only get a portion of the time series here.

    Let's say you start out with an initial current. Then I say the sin term goes away, and you're left with the cosine term:

    i(t)=i_{0}\cos(t/\sqrt{LC}). The current will go out of the inductor until there is no more current to go out of the inductor, which will happen when i=0 for the first time. That occurs when

    \dfrac{t}{\sqrt{LC}}=\dfrac{\pi}{2}, or

    t=\dfrac{\pi\sqrt{LC}}{2}.

    That's my analysis. There could be some very important factors I'm not considering, so please take this with a grain of salt.
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  5. #5
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    You may want to refer to this:
    Help for Boost Converter
    and this,
    http://en.wikipedia.org/wiki/Boost_converter

    Boost converter mode of operation: Discontinuous

    There are some important factors missing in your analysis such as Diode losses and series resistances of the inductor and capacitor. I would guess that the higher the resistance of the series resistances, the less time it takes for the inductor to fully discharge.

    Also we must not forget that the time it takes for the Inductor to fully discharge depends on the voltage of the capacitor as SPICE simulations and practical experiments show. So therefore T_discharge is a function of L, C, RL, RC, Vf (forward voltage drop of diode) and Vc (the current capacitor voltage). How on earth do I solve this problem?

    You might have noticed in the link I provided that we can derive the T_discharge from the second last formula. I'm currently using that in my model however as you can see it doesn't take into account all of the variables in my last paragraph.

    Thanks.
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  6. #6
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    Ok, so just to be crystal clear, you're wanting to find the time of discharge of a boost converter operating in discontinuous mode. Do you accept the circuit diagram in the wiki article as an accurate model of your situation? Is there anything more you want to include? (I already know that you're not wanting to leave out the voltage drops across the diode, the inductor, and the capacitor.)

    I'm looking at Figure 4 in the wiki article, and I'm guessing that your target variable is \delta T. Is that correct?
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  7. #7
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    "Do you accept the circuit diagram in the wiki article as an accurate model of your situation? Is there anything more you want to include?"

    Yes I need to include the series resistance of the Inductor and Capacitor. So imagine a resistor R_L in series with the Inductor, and another resistor R_C in series with the Capacitor.

    "I'm looking at Figure 4 in the wiki article, and I'm guessing that your target variable is . Is that correct?"

    You are correct.

    Thanks.
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  8. #8
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    I haven't forgotten you, but the problem is difficult. Post more later.
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  9. #9
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    Quote Originally Posted by Ackbeet View Post
    I haven't forgotten you, but the problem is difficult. Post more later.
    Yes I know. That's why I came here.

    Thanks for your help so far and get back to me when you can. There is no rush, I've got exams to study for anyway.
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  10. #10
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    Ok, here's my shot at a circuit diagram. I've put in labels for everything, including the mesh currents for a mesh current analysis.

    Calculating Inductor discharge time into a charging Capacitor.-inductor-problem.jpg

    Note: meshes 1 and 2 will be joined together when the switch is open. We'll call that mesh 1 for that analysis, and we'll keep mesh 3 as mesh 3.

    Question: how accurately do you need to model the diode's VI diagram? What type of diode are you putting in there?
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  11. #11
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    Your diagram is correct.

    The exact diode I will be using is undecided, however I can tell you that it will probably be a Schottky Silicon Carbide type (SiC) and that it will most likely have a V_f of 1.7V.

    A very accurate model of the diode would be nice but since I don't know the exact one I will be using a simple model of the diode will be good enough.

    Thanks.
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  12. #12
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    Ok. With the switch open, since there are resistors in both of the far right paths, and no resistance through the switch (ideally), we'll say that all the current goes through the switch, and none through the two resistor arms. This is, then, a series RL circuit. The steady-state condition is obviously going to be that the inductor looks like a short, and you have the current i_{ss}=V_{i}/R_{L}. You're not concerned, apparently, with how long it takes to charge up the inductor to this state. Is that correct?

    Well, then. We now open switch S. Let's assume that the VI characteristic of the diode is a constant V_{f} for whatever forward current you have. Then using mesh current analysis around loop 1 (consisting of the voltage source, the inductor, resistor R_{L}, the diode, the capacitor, and the resistor R_{C}, we have the following equation:

    \displaystyle V_{i}=L\,\frac{di_{1}}{dt}+R_{L}i_{1}+V_{f}+\frac{  1}{C}\int_{0}^{t}(i_{1}(\tau)-i_{3}(\tau))\,d\tau+v_{C}(0)+R_{C}(i_{1}-i_{3}).

    For mesh 3, we get

    \displaystyle -R_{C}(i_{1}-i_{3})-\frac{1}{C}\int_{0}^{t}(i_{1}(\tau)-i_{3}(\tau))\,d\tau-v_{C}(0)+R_{\text{load}}i_{3}=0.

    We need some initial conditions here. We have i_{1}(0)=i_{ss}=V_{i}/R_{L}. We'll say that v_{C}(0)=0. I think all the other initial conditions are zero.

    Before I go on, let me ask if you agree with this model. Think it's ok? Accurate enough?
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  13. #13
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    " The steady-state condition is obviously going to be that the inductor looks like a short, and you have the current You're not concerned, apparently, with how long it takes to charge up the inductor to this state. Is that correct? "

    Yes that's is why I didn't mention R_D_S(on), the switch on state resistance. I already have the charging cycle covered. I'm only interested in the discharging cycle, so this will hopefully cut our work in half.

    You're initial conditions are mostly correct. Don't assume the initial inductor current at the beginning of the discharge cycle to be V_i/R_L, just give it a single variable name. The reason is because I_L at the end of the charging cycle will never reach V_i/R_L.

    You're diode model is good enough, I don't need anything fancy.

    Thanks.
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  14. #14
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    Ok, that's fine. We'll use I_{L} for the initial condition.

    I don't know about you, but I'm inclined to use the Laplace Transform to solve this problem. How about you? Let's do that. On the first equation, we'll get (after canceling out the v_{C}(0)'s)

    \displaystyle\frac{V_{i}-V_{f}}{s}=L[sI_{1}(s)-I_{L}]+R_{L}I_{1}(s)+\frac{I_{1}(s)}{Cs}-\frac{I_{3}(s)}{Cs}+R_{C}(I_{1}(s)-I_{3}(s)),

    and the second equation will yield

    \displaystyle -R_{C}(I_{1}(s)-I_{3}(s))-\frac{I_{1}(s)}{Cs}+\frac{I_{3}(s)}{Cs}+R_{\text{L  oad}}I_{3}(s)=0.

    Notice that we've included all relevant initial conditions here.

    We now need to solve for I_{1}(s) and I_{3}(s). This is a system of linear equations - two equations in two unknowns. I will drop the s-dependence of I_{1} and I_{3} to save typing. We have

    \displaystyle I_{1}\left[Ls+R_{L}+\frac{1}{Cs}+R_{C}\right]+I_{3}\left[-\frac{1}{Cs}-R_{C}\right]=\frac{V_{i}-V_{f}}{s}+LI_{L}

    \displaystyle I_{1}\left[-R_{C}-\frac{1}{Cs}\right]+I_{3}\left[R_{C}+\frac{1}{Cs}+R_{\text{Load}}\right]=0.

    I won't bore you with the technical details of solving the system. What we're really interested in is I_{1}-I_{3}, because that's the current through the capacitor, right?

    The expression I get is the following:

    I_{1}-I_{3}=\dfrac{C R_{\text{Load}}(I_{L}L s-V_{f}+V_{i})}{Ls(1+C R_{C} s)+R_{L}(1+C(R_{C}+R_{\text{Load}})s)+R_{\text{Loa  d}}(1+C s(R_{C}+Ls))}.

    Now the denominator factors as

    \displaystyle \left(s+\frac{a+b}{c}\right)\left(s+\frac{a-b}{c}\right), where

    a=L+CR_{C}R_{L}+C(R_{C}+R_{L})R_{\text{Load}},

    b=\sqrt{(L+CR_{C}R_{L}+C(R_{C}+R_{L})R_{\text{Load  }})^{2}-4CL(R_{C}+R_{\text{Load}})(R_{L}+R_{\text{Load}})}  , and

    c=2CL(R_{C}+R_{\text{Load}}).

    I think that, before this post gets too much longer, I'll break it up here. However, I will point out that the general strategy from here is to use partial fractions to break up the expression into two fractions that are of the form A/(s-B). Then the inverse Laplace Transform will just be Ae^{Bt}. So the solution should look like

    i_{1}-i_{3}=Ae^{Bt}+Ce^{Dt}.

    Solving that for when i_{1}-i_{3}=0 will be tricky, I think.
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  15. #15
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    Sorry for leaving you in the dark for these past few days, I have been very busy and will continue to be for quite a few more weeks. You might have sparked some ideas already to help me out but for now thanks for your very generous help, I greatly appreciate it. If I need further help on this topic I will definitely come back for more assistance.

    Later (:

    Regards,
    Johnny
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