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LRC Circut with a polarized cap

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Myke

Fri Jan 02 2009, 04:58AM

Registered Member #540 Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969

Once the voltage starts to drop, the coil will use the magnetic energy in the core (be it iron or air) to try to keep the current flowing. This current would charge the cap in the reverse direction. The diode shorts this out and prevents the reverse charging.

Imagine the coil as a small battery that turns on when the voltage starts to drop. This battery has the positive connected to ground and the negative connected to the cap. This would charge the cap backwards so that's why you need the diode to prevent that.

Marauder709

Fri Jan 02 2009, 06:31AM

Registered Member #1895 Joined: Thu Jan 01 2009, 03:12AM
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Posts: 22

I'm in total agreement with you Myke except for your final statement. My question is not what is happening, my question is how do I prevent this from happening and why it works. So far I have the how, a diode in anti-parallel. I do not have a why. I don't doubt that this arrangement works. However, I cannot conceptually rationalize why it works. The whole idea somehow does not connect in my mind.

This is the way I see it, as plain as I can make it.

There are two branches to this circuit, the diode branch and the inductor branch. The diode branch however, at least in my mind, does not prevent the cap from charging with a reverse voltage. This is my rational:
As the capacitor discharges the diode opposes the current in its branch of the circuit so that all the current flows through the inductor branch. The voltage of the capacitor decreases in exponential decay. The current however, does not. This continuation of current is caused by the inductor. The current therefore continues to persist even after the voltage across the capacitor is zero. Because the current is still flowing, the plates of the capacior begin to acquire an opposite charge, reversing the Electric feild and creating reverse polarity. An electrolytic capacitor cannot acquire a reverse polarity and therein lies the problem. Is any of this not true?

From what I can glean from the various attempts to explain the diode in anti-parallel, the inductor, instead of discharging to ground, will discharge through the diode branch and so negate reverse in polarity of the capacitors. I see no reason however, why the inductor would not also discharge to ground. At the very least, the diode branch and ground would be in parallel and while the effect of the reverse polarity would be less severe, it would nevertheless persist. Is this not also true?

blackgrunge

Fri Jan 02 2009, 06:39PM

Registered Member #1889 Joined: Mon Dec 29 2008, 07:36AM
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If its in parallel with the inductor its in parallel with the capacitors. This diode your using is also called a quenching diode. Not only does it prevent reverse voltage across the capacitor its also helps to essentially short the current that an inductor stores after the capacitors finishing discharging. For multistage guns this can be especially useful because it allows coils to be switched off much faster. Be wary though the current the inductor tries to push through the diode is quite substantial. Sometimes it can be as much as 75% of what the capacitor pushes through coil to begin with (that's if you have ungodly inductance). As far as current during discharge is concerned. Current doesn't remain constant through the discharge phase at all. Its a dampened oscillating current meaning it has a relationship that resembles a logarithmic decay (exponential). The voltage in the inductor isn't zero either because if the voltage is zero there wouldn't be any current flowing and there wouldn't be any energy stored (E=P/t; P=IV) Here's an example of what the current may look like in a dampened circuit.

RLC Series Imag

Try using Barry's RLC circuit sim. It might give you a better idea of what's going on. The reason the diode works is that when current is flowing in the 'correct' direction it blocks the current from flowing through the diode and shorting but as soon as the current is flowing in the 'wrong' direction aka reverse voltage, it shorts through the diode and because that diode has a significantly lower resistance than the capacitors do, most of the current is channeled into it.

badastronaut

Fri Jan 02 2009, 09:04PM

Registered Member #222 Joined: Mon Feb 20 2006, 05:49PM
Location:
Posts: 96

Ok, how bout this: you said it yourself that current takes the path of least resistance.

It has two choices to go, either reverse charge the cap or through the diode.

Which will the current want to go through? An object with an impedance of 1/jwC or a short circuit?

The answer is always the short circuit.

A real diode can be approximated as a short in this case. However, if you are really anal about it, there is a small voltage drop across that diode as well as a resistance. These two cause the voltage across the diode to not be zero, and the cap will charge to whatever voltage that may be and in this case it is a negative voltage.

Why don't you do what everyone else does and analyze the circuit with equations. Use the node voltage method or the mesh current method, whatever you prefer.

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