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With A capacitor initially charged to Vc (assuming ideal diode and ideal switch), I'm curious about determining the max current through the inductor and total discharge time (the switch stays closed).

I know that when the switch closes, the current equation is: IC = IL, where: IC = CdV(t)/dt, and VL = Ldi(t)/dt. This is because the Diode wouldn't have current flow until after inductor stops acting like an open circuit. I was thinking i could solve for inductor current with integration, and plug in the equation for capacitor current. However, I am struggling with solving this problem, because I'm unsure on how the resistor and diode affect the circuit's timing.

Any help on determining the time constant or help solving for the max inductor current would be much appreciated.

enter image description here

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  • \$\begingroup\$ The diode branch can be considered to be absent till the first time capacitor voltage reaches zero from positive value. Once capacitor voltage goes below zero, you have a circuit with a capacitor, resistor and inductor in parallel (ideal diode is a short) until the time the capacitor voltage goes above zero. So you will have two sets of differential equations one for when capacitor voltage is above zero (only L and C in circuit) and one for when capacitor voltage goes below zero (R,L,C in parallel). \$\endgroup\$ – AJN Aug 14 at 3:14
  • \$\begingroup\$ You can imagine it this way : The voltage across the diode resistor combination is forced/set by the capacitor voltage (since they are in parallel). Don't worry about the inductor (or its current) when trying to determine if the (ideal) diode is open or short at any given instant. It is decided by the capacitor voltage for this particular circuit. \$\endgroup\$ – AJN Aug 14 at 3:18
  • \$\begingroup\$ "total discharge time". How is it defined? for LC circuits, the oscillations usually go on for infinite amount of time (at smaller and smaller amplitudes). This circuit may also fall in that category. \$\endgroup\$ – AJN Aug 14 at 3:24
  • \$\begingroup\$ Is the initial charge in the capacitor positive ? Then the answer to peak inductor current may be straight forward. No need to solve differential equations. \$\endgroup\$ – AJN Aug 14 at 3:26
  • \$\begingroup\$ The charge in the cap (for this problem) can range form 100-1000V. \$\endgroup\$ – JohnnyMac Aug 14 at 3:30
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When you close the switch you have energy stored in the Cap and none in the inductor, but you have started an ideal LC resonant circuit for 1/4 cycle until the Cap voltage oscillates negative and conducts the diode heavily with the same current now as the peak current in the inductor which dissipates almost all the energy in the next half-cycle. This is going to be an unrealistic huge current even at 10V, forget about 100 to 1kV since you have ideal reactive parts. (unrealistic)

Analyze it with Conservation of energy with an LC resonant half-wave followed by a negative exponential L/R =T decay towards zero.

\$Ec=½ CV^2=½ LI^2 ~thus~ I^2=\dfrac{CV^2}{L}\$ meaning the initial Cap voltage becomes a peak current wave.
\$f_o=\dfrac{1}{2\pi\sqrt{LC}}=3,577Hz=1/0.28ms\$ or 280us * 1/4 wave =70us (Vpeak to 0)

let's choose Vc=10V for initial conditions, I=0

\$I^2=180uF\cdot10^2/11uH=1,636 \$... I=40.5A

So current ramps up to 40A while voltage decays to 0V asa 1/4 sine wave then the R dissipates the current with an exponent T=L/R = 11u/85m= 129ms decay up towards zero while oscillating without damping on the positive side so that doubles the decay time to 258ms (asymptote.) Then x5 = 1 second for <10% residual.

Now repeat that with practical values.

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  • \$\begingroup\$ Do you need a simulation for proof? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 14 at 4:05
  • \$\begingroup\$ A simulation would help a lot! This was a very helpful way to solve the problem, and is much more intuitive. Howevever, why does the decay time double when not damping? \$\endgroup\$ – JohnnyMac Aug 14 at 4:20
  • \$\begingroup\$ Because it non-damped half the cycle. No decay then .Do you know Falstad’s sim yet? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 14 at 4:49
  • \$\begingroup\$ No, I don't know Falstad's. \$\endgroup\$ – JohnnyMac Aug 14 at 4:54
  • \$\begingroup\$ It has real diodes but will work close enough, but ideal reactive parts so you must model them with ESR, DCR added as a discrete R \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 14 at 4:54

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