# LC Circuit with Diode

Discussion in 'Homework Help' started by shespuzzling, Oct 5, 2010.

1. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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Can someone please help me understand what happens in a simple, DC driven circuit with a diode, inductor and capacitor?

If there was no source, does the diode prevent resonance if you assume ideal components and no resistance in the circuit? I think that it does but a problem I came across in my book implies that if you assume no resistance, an LC circuit with a diode will be sinusoidal. I also think that the diode would be damaged in such a circuit because the built up magnetic field in the inductor would need a place to go once the current = 0.

I attached an image of the type of circuit I had in mind. Thanks in advance for your help.

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2. ### Georacer Moderator

Nov 25, 2009
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In a DC circuit there is no reasonance. In this particular circuit, if we assume that the switch was open and then closes, the following will happen: The current will flow through the diode since it is forward biased. It will initially find great resistance on the inductor which will gradually diminish. It will initially find no resistance on the capacitor which will gradually reach infinity and an open circuit. How will those two effects overlay depends on the size of the two elements (in Henrys and Farads respectively).

If the switch was initially open the above state will be reached. The capacitor will be open circuited and no current will flow through the circuit. Opening the swith will not alter this, as the capacitor cannot be discharged anyway.

3. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
1
Thanks for the clarification. It seems to make a bit more sense, but what happens then when the current in the circuit is 0? At this point, the voltage across the inductor would have to be some non-zero number and that magnetic field, unlike the electric field in the capacitor, has to discharge somehow, right? I would guess that would have to be through the diode in the reverse bias direction, but I'm unclear on that because I think it would damage the diode.

Are you saying that in this particular circuit the diode has no effect?

4. ### Georacer Moderator

Nov 25, 2009
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Fist of all I don't see why we should consider the coil's magnetic field more "capable" than the capacitor's electric field.

Other than that, the capacitor will force the current to drop to 0. Consequently the coil's current will rise at the beginning and fall to 0 again at some point. Intuitively only, someone please confirm or reject, I think that the coil will resist as much in the current's increase as in it's decrease, resulting to a final 0 volts over it.

The diode prevents current from going to the opposite direction by any means. So either it's the capacitor pushing it back or the coil, none of them would accomplish anything.

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I've attached 3 plots showing what happens in an LC circuit like this where a constant voltage is suddenly applied.

The first plot shows what happens if there is some resistance in the circuit, which there always is in the real world (in the wires, the voltage source, etc.), and if the diode is temporarily removed. The inductor current is shown in red, and the voltage across the capacitor is shown in blue. The inductor and capacitor resonate, and some energy is repeatedly exchanged between the inductor and capacitor in the form of an alternating current and voltage, but this component gradually dies out due to losses in the resistances of the circuit. Finally, the current in the inductor is zero and the capacitor is charged up to the source voltage.

The second plot shows what would happen if there were no diode, and no resistance anywhere in the circuit, and no loss by radiation or any other mechanism. The energy in the circuit would "slosh" back and forth between the inductor and the capacitor forever.

The third plot shows what would happen when the diode is in place and the resistance in the circuit is zero (or very small). The inductor current builds up to a maximum then begins to decrease; the current follows a half sine wave, resonating with the capacitor. The voltage in the capacitor begins to increase (also following a half sine wave, 90° out of phase) and reaches a value of twice the voltage source just when the inductor current reaches zero. At this time the voltage across the capacitor attempts to cause the current in the inductor to reverse (which happened in the no diode cases), but the diode prevents this. The diode is reverse biased, and the final situation is that the current in the inductor remains zero, and the voltage across the capacitor remains a constant equal to twice the source voltage.

This behavior is used in so-called resonant mode switching power supplies.

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6. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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My understanding was that while a capacitor can maintain its electric field even when disconnected from a circuit, once current stops flowing in an indcutor, it loses its magnetic field, and that energy has to go somewhere (i.e. and arc)

Maybe this is where I'm getting messed up: Assuming an ideal circuit with 0 resistance that consists of a DC voltage source, an inductor and a capacitor; will you get sinusoidal voltages and currents without any damping? If no, why not?

Back to the original problem, the equations I found are below. I plotted each function, and saw that when the current is equal to 0, the voltage across the capacitor is 2*Vs (where Vs=source voltage) and the voltage across the inductor would have to be Vs but of opposite polarity.

Here's the equation I used for current:
i(t)=Vs*$\sqrt{C/L}$*sin(t/$\sqrt{C*L}$)

For capacitor voltage:
v(t)=Vs*(1-cos(t/$\sqrt{C*L}$))

So at one point, you have zero current that is decreasing, a voltage across the inductor and twice the source voltage across the capacitor (but of opposite polarities). So:

1. Does the diode turn off when i=0 like it should? If so, you are left with an inductor with a voltage across it, which must be dissipated because there is no longer any current flowing. Which way does current flow? What happens to the voltage across the capacitor after i=0?

7. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
1
Thanks! That answered a lot of my questions (I posted my previous reply right before I saw yours).

Going back to that third graph, what if you added a plot of the voltage across the inductor? I got a negative value when the current is 0, as expected. I still am not sure I understand how the voltage across the inductor at that time suddenly goes to 0 without any arcing or any damage to the diode. Isn't that why they have freewheeling diodes; to create a path for current to go when an inductor has a large voltage across it? I guess I'm not sure why the diode trumps the magnetic field in this case, and decides the ultimate fate of the circuit, while in other cases we're concerned about arcs with inductors when a circuit is suddenly opened or closed.

Also, so it is true that a voltage source can charge a capacitor to twice its own value? That seems like it shouldn't make sense but I'll buy it!

8. ### Georacer Moderator

Nov 25, 2009
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Assume that current flows through the circuit at the direction of the diode. At some point the voltage of the inductor and the capacitor combined will reach the voltage of the source. That is the peak point of the third graph. At this point no more voltage is applied on the inductor and the capacitor by the source, because of the diode. However, the diode resists current going reverse and will not inhibit current going through it forward, no matter the voltage on it. In other words, if something makes current flow through the forward direction, the diode is ok with it.

This something is the inductor. The "inertia" that the current has in it makes it to continue to flow through the diode. The inductor "spends" all its stored voltage and magnetic field to force current through the diode. This describes the declining slope of the current on the third graph. When all the inductors energy is depleted, all of it will be stored on the capacitor (hence the double voltage) and no more current will flow through the diode in any direction.

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9. ### Georacer Moderator

Nov 25, 2009
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This seems impossible but think about it: The source charges simultaneously both the inductor and the capacitor storing energy in both of them and expending energy for both of them. Then the source stops giving anymore energy and this time the inductor gives all its energy to the capacitor. The capacitor ends up with double the energy, but in the end all the energy came from the source.

10. ### Georacer Moderator

Nov 25, 2009
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@Electrician:

I was trying to simulate the circuit as you did without the diode and I 'm having some trouble. The capacitor starts right away with the voltage of the source and the inductor starts with 0 voltage. In short, the graphs are straigt. Any idea where I 'm going wrong?

I used Mulitisim 11 with 10 and 1 Volts at the source, 1m and 1 Herny at the inductor and 1mF at the capacitor.

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I'm not familiar with Multisim, but the problem might have something to do with the way some simulators initialize state variables at time t=0..

If you actually have a switch in the circuit, try having it close at a time later than t=0.

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12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Freewheeling diodes are there to provide a path for the current when the inductor has a current in it. The issue of whether the inductor has a voltage across it has, at any instant of time, nothing to do with the current in it; it's the voltage history--its integral--that determines the current in an inductor.

Imagine an inductor that is carrying a current. If you suddenly place a short across the inductor, the current will remain unchanged at that instant of time. The current only changes as the integral of the voltage applied across the inductor, so for an instant of time when you apply a short to an inductor, the current in the inductor doesn't change. Ideally, if the inductor was wound with superconducting wire, and a superconducting short were placed across it while it was conducting a current, that current would persist forever. However, if the short is a wire with finite resistance, or the forward voltage of a diode, the current in the inductor will decay slowly, at a rate determined by the small, but non-zero, voltage provided by the short.

The reason for freewheeling diodes is that if an inductor is carrying a current and you suddenly open-circuit it, the voltage across the inductor will suddenly rise because if you do truly open-circuit it, there is no path for the current, and di/dt will become infinite (in the real world, the inductor's distributed capacitances, and other circuit capacitances, limit di/dt to a less than infinite, but large, value) because the current which did have a finite, non-zero value, suddenly becomes zero. The only way that can happen is if di/dt is very large. If di/dt in an inductor is very large, that means that the voltage across the inductor becomes very large. It's that high voltage that can hurt diodes, switches, and transistors/FETs.

On the other hand, if an inductor is carrying current, suppose you open-circuit it when, for example, a transistor that was acting as a switch in series with the inductor is suddenly turned off; if there is no path for the inductor current, the inductor's di/dt will become very large. A freewheeling diode will carry the current in the diode's forward direction, and the current magnitude is the same as it was when the transistor turned off. This amount in the forward direction won't hurt the diode (assuming it has been selected to carry such a current).

In the circuit in post #1, when the inductor current goes to zero, there is no voltage across the inductor; the difference between the voltage source and the double voltage on the capacitor all suddenly appears as reverse bias on the diode. The rate of rise of reverse voltage across the diode will be in the nanosecond range, controlled by various circuit capacitances.

13. ### Georacer Moderator

Nov 25, 2009
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Ok, here are the voltage graphs. In the circuit without the diode the circuit start open, then a switch closes and engages the source, then opens again and closes again. Notice the very large voltage spike on the inductor. In the diode circuit I think the reasonance of the inductor voltage is an imperfection of the simulation, unless someone else says different.

Without Diode

With Diode

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Why are your simulations starting with 10V on the capacitor initially?

The oscillations in the version with the diode are an artifact of the simulator. You can make them go away to a large extent by connecting a 1000Ω resistor in series with a 1 nF capacitor, and placing this series combination in parallel with the diode. You may have to adjust the values of the resistor and capacitor for best results.

15. ### Georacer Moderator

Nov 25, 2009
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I used a 10V source instead of 1V. No reason, just happened, it doesn't change the result's theoretical value.

About the simulation problem I had, you were right on the money. The software automatically started with the steady state reaction of the system. I just had to select zero initial conditions on a list of parameters.

I will increase the time step later to see if the artifact you mentioned dissapears. Off to uni now!

16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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What I mean is, the capacitor should have zero volts when the switch closes. It looks like you have 10 volts on it when the switch closes.

17. ### Georacer Moderator

Nov 25, 2009
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Here's another simulation. I couldn't get rid of those artifacts on the diode circuit, no matter how small time step I chose. It is related on the software's discretion how it looks I guess.

LC with Switch

LC with Diode

I 'll explain again what the first graph shows. The inductor voltage is in red, the capacitor in green. The switch closes at the first time minor tick. As expected the capacitor starts with 0 volts. Electrician, you must have switched the colours, the capacitor does start with 0 volts.

The switch opens again at the thick red ribbon. This red colour is the infinity voltage on the inductor. Notice the green line that stays horizontal throughout this step. It is the capacitor voltage that "freezes" in place as long as the switch is open. When the switch closes again the circuit resumes the oscillation exactly where it left it.

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Besides reducing the time step, did you try the suggestion I made in post #14?

"The oscillations in the version with the diode are an artifact of the simulator. You can make them go away to a large extent by connecting a 1000Ω resistor in series with a 1 nF capacitor, and placing this series combination in parallel with the diode. You may have to adjust the values of the resistor and capacitor for best results."

19. ### Georacer Moderator

Nov 25, 2009
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RC filters ftw!
Red:Capacitor Green:Inductor

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
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347
Good result. I find that when I simulate circuits with diodes, I often get those pesky oscillations if I don't place what amounts to an RC snubber across the diode.