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LRC Circut with a polarized cap

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rp181

Thu Jan 01 2009, 06:00PM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529

Heres a text diagram =p

|-----(capacitor anode (+) ->)-----(<-diode cathode())----|
|-------------------------------------------------- -----------------------------|

capacitor anode is the right, diode anode is right.

Barry

Thu Jan 01 2009, 06:06PM

Registered Member #90 Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301

The protection diode across the capacitor terminals is the simplest approach by far. It does have the disadvantage of extending the current pulse in the coil.

Some good alternatives include an H-bridge switching network, and Evgenij Vasiljev's v-switch to dump another capacitor into the coil to kill it.

Cheers, Barry
I live in the state of Washington where they only charge sales tax on 'luxury' items. So why is toilet paper taxed?

Marauder709

Thu Jan 01 2009, 07:19PM

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

I have always thought anti-parallel was this like the diagram I uploaded. However, current takes the path of least resistance. I don't see how adding the below arrangement helps.

I like Barry's V-switch suggestion though it looks expensive.

badastronaut

Thu Jan 01 2009, 09:44PM

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

The diagram is correct, and you are half right on the current path part.

Current will not divert itself completely to one path; it will divided among the different paths related to the relative impedances. If you have two parallel resistors, one 1ohm and the other 100ohms, current will flow through both resistors.

When the diode conducts, it forms a short circuit across the capacitor; it turns into a wire. You can't charge a capacitor when its terminals are shorted. The voltage across a short is always zero. You're also shorting out the inductor when the diode conducts.

Perhaps it will be easier to understand if you divide the circuit into two operating modes. Mode one is when the diode does not conduct, so you replace it with an open circuit. Mode two is when the diode conducts, and you replace it with a short. Also, when the inductor current is falling, the voltage across it will be negative; this forward biases the diode by having a negative voltage on the cathode.

Marauder709

Thu Jan 01 2009, 09:57PM

Registered Member #1895 Joined: Thu Jan 01 2009, 03:12AM
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A diode is not a resistor. I always assumed that (for all intents and purposes) no current passes through the diode because it is in anti - parallel.

A agree with your mode one. You may as well have no diode as its diverting all the current through the inductor. I don't get mode two however. The only way to get that diode to conduct is to reverse the current. The current does not reverse until the capacitor acquires an opposite polarity. This is not possible with an electrolytic and your stuck with the same problem.

Is the diode supposed to break down or something?

big5824

Fri Jan 02 2009, 12:41AM

Registered Member #1687 Joined: Tue Sept 09 2008, 08:47PM
Location: UK, Darlington
Posts: 240

When the capacitor is discharging the current flows against the diode, so it is all forced through the coil. However, as the magnetic field collapses, it generates a reverse voltage in the coil, which would flow in the opposite direction and reverse charge the capacitors. However since the diode is in antiparallel and the current has changed direction, the diode now conducts and shorts out the coil, so the current just flows round the shorted coil until its all disipated as heat (practically instantanous)

rp181

Fri Jan 02 2009, 01:18AM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Think of the coil as a capacitor. When you put a diode across in parrallel, it discharges. When you put it in anti parrallel, nothing happens. If the diode was antiparrallel, and the capacitor was rotated 180 degrees (the back EMF), the capacitor discharges.

Marauder709

Fri Jan 02 2009, 01:39AM

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

There is no reverse voltage provided by the inductor. The EMF applied by the inductor as the feild collapses is in the same direction as the voltage originally applied by the capacitor. The inductor acts to maintain the current in the circuit, why would it then provide an EMF that would oppose the current flow.

rp181, I'm not sure I understand your analogy. "back EMF" is a misleading term. It is not actually a reverse in EMF. On the contrary, the EMF acts to maintain the current which then builds up a back voltage in the capacitor.

There are two ways to charge a capacitor with a reverse voltage. The most obvious is to simply reverse the voltage. The less obvious way is to maintain a current after the capacitor has reached equilibrium. This is the case with any oscillating circuit. The diode in anti-parallel only protects the circuit from the former scenario, not the ladder

badastronaut

Fri Jan 02 2009, 04:10AM

Registered Member #222 Joined: Mon Feb 20 2006, 05:49PM
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The current does not need to reverse for the diode to conduct. Maybe you're confused as how a diode works. Current flows from the anode to the cathode. If you assume that, then there is no where else for the current to go if you model the conducting diode as a dead short.

Why don't you draw the circuit with no ground symbol, just a continous loop.

Marauder709

Fri Jan 02 2009, 04:42AM

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

I drew the two grounds simply for simplicity sake, in the program I use to draw my circuit diagrams its just easier.

Correct me if I'm wrong but in the circuit I have drawn, the diode branch is not conducting any current when the capacitor is discharging. It can be regarded as an open branch or simply deleted. If we are then in agreement that:
a) The direction of the current does not reverse.
b) The diode is oriented so that it does not conduct while the capacitor discharges
Then I fail to see how the diode arrangement will prevent a reverse voltage.

Maybe this is what you all are driving at. The EMF applied by the inductor will be reverted through the diode bypass and form a complete circuit with the inductor as the voltage source and the diode loop as the rest of the circuit. If this is the case then wouldn't the inductor also discharge to ground and thereby creating a reverse voltage in the capacitor?

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