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If we have a inductor and capacitor in parallel, then from what I can tell, the equation for the current is just the simple harmonic motion equation: i = Acos(w0t) + Bsin(w0t).

However I am struggling to solve it for the initial conditions of the circuit suddenly being attached to a power source. I have read that the immediate voltage across the capacitor when the circuit is connected is 0 and that the immediate current through the inductor is also 0. Hence I've assumed that the voltage across the inductor must be 0 too due to KVL and that the current through the capacitor must also be 0 cos of ohms law (v=IR, I = 0 therefore V = 0).

Hence when t is 0 then A = 0

0 = Acos(w0t) + Bsin(w0t)

Bsin(0) = 0

(Acos(w0t) = Acos(0) = A

0 = A

However I am finding it really hard to find a value for B, taking the equation di/dt = v/L and solving for i to get the integral of v/L = ((vt)/ L) + C, but how can I find a value for V that doesnt just make i = 0. taking ((vt)/ L) + C = w0Bcos(w0t) doesnt seem to be getting much closer to a solution either, as I am not sure what to put as the value for v.

Any help would be much appreciated! Im sure there is something Ive missed like an extra initial condition or something in the calculating of the maths.

For some context, I have put together a driving circuit for a ZVS and I want to work out what values for the resonating inductor and capacitor I need to get a 20khz natural frequency and a resonant current of an amp or less (and from that appropriate values for chokes to protect the driver circuit) so I can test the circuit without worrying about damaging any of the measuring instruments.

Bren
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  • The current through the capacitor at t=0 is infinity hence, you can't solve it with this situation prevailing. – Andy aka Jul 03 '23 at 19:40

1 Answers1

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For some context, I have put together a driving circuit for a ZVS and I want to work out what values for the resonating inductor and capacitor I need to get a 20khz natural frequency and a resonant current of an amp or less (and from that appropriate values for chokes to protect the driver circuit) so I can test the circuit without worrying about damaging any of the measuring instruments.

You don't need to worry about initial conditions and suchlike for this.

If you want a specific frequency, you need to choose an LC product given by

$$f = \frac{1}{2\pi\sqrt{LC}}$$

Choose an L or a C, maybe you only have a few Ls, and compute the C.

If you want a more systematic method to choose both component values, then your resonant impedance is given by

$$Z = \sqrt{\frac{L}{C}}$$

Z is the ratio of the peak voltage across the capacitor, to the peak current in the inductor.

If you want a resonant current of 1 amp, and you know what sort of voltage you want across the capacitor at resonance, maybe 10 V, you can choose an appropriate impedance, in this case 10 Ω for 10 V at 1 amp.

Tim Williams
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Neil_UK
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  • How can I get the capacitor to have 10 V across it? doesn't it just become a 2nd order differential equation again to work that out? Cos without hardsetting the voltage across the capacitor doesn't it just became any linearly dependent solution of Z = root(L/C)? – Bren Jul 04 '23 at 14:37
  • @Bren To get 10 V across the capacitor, use 10 V DC power supply, and connect it to the capacitor, having first opened a switch to the inductor. Remove the power supply, close the switch to the inductor, the pair will then ring at 20 kHz, if you've chosen the values by plugging 20 kHz into that first equation. When the cap voltage is zero, you'll have 1 amp flowing through the inductor, if you've plugged 10 ohms into the second equation. Note that the energy stored in the L at 1 amp is the same as that stored in the C at 10 V, another condition for resonance. – Neil_UK Jul 04 '23 at 14:46