Author Topic: Negative rail generator, why the "polarized" cap doesn't explode or ruined ?  (Read 4173 times)

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Offline BravoV

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I've been seeing a lot at this kind of trick in generating the low current negative rail like below circuit. Like supplying op-amp's negative rail.

Why the polarized capacitor (the one labeled C in the circuit) never get toasted ? I mean at every half cycle, that cap experienced reverse polarity, right ?  ???


EDIT : Made a mistake quoting the circuit, see post #6 below for the actual circuit I'm asking.

« Last Edit: August 13, 2013, 02:58:08 pm by BravoV »
 

Offline IanB

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The diode prevents that. There is no way for current to flow the wrong way through the capacitor on that circuit diagram unless you apply an external voltage to the load terminals.

[Edit: oh, you mean the one labelled "C" -- ignore my above comment while I think about it some more]
« Last Edit: August 12, 2013, 02:41:28 pm by IanB »
I'm not an EE--what am I doing here?
 

Offline IanB

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You could simulate the circuit and plot the voltage and current waveforms.

I think the relative sizes of the capacitors will matter.
I'm not an EE--what am I doing here?
 

Offline IonizedGears

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The diode protects C just as Ian said
I am an EE student with interests in Embedded, RF, Control Systems, and Nanotech.
 

Offline IonizedGears

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Actually both those diodes protect C from experiencing reverse voltage no matter if it is coming from the coupling diode straight to the ac line or the diode going to the negative terminal.
I am an EE student with interests in Embedded, RF, Control Systems, and Nanotech.
 

Offline IonizedGears

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So even if you put a voltage across the terminals, C will be protected by either diode either case
I am an EE student with interests in Embedded, RF, Control Systems, and Nanotech.
 

Offline BravoV

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Ok, thanks for the replies, actually I made a mistake, the 1st one is one of the alternative, actually this is the one I'm asking.

Btw, this trick is also used a lot and it works btw.





@IonizedGears, dude, there is a "Modify" button to edit your post instead of spamming doing multiple posts like that.  :palm:



.
« Last Edit: August 13, 2013, 02:58:35 pm by BravoV »
 

Offline amyk

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I think the situation is similar to a voltage doubler, where the capacitor will be subject to reverse polarity for 1/2 a cycle at first turn-on, but in the other half it gets charged up and thereby is not subject to reverse polarity thereafter.

Also the circuit can be simplified greatly to only 4 diodes and 2 capacitors, as seen below. This is similar to PC PSU's input stage.
 

Offline Hero999

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That circuit only uses two diodes in the bridge to form two half wave rectifiers.

The disadvantage is you need larger capacitors for C1 and C2 and if the circuit draws most of the current from the positive rail, there will be a lot of DC in the transformer's core.

The separate negative doubler circuit is better, if you draw most of the power from the +V and only need a low current negative supply.
 

Offline BravoV

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The separate negative doubler circuit is better, if you draw most of the power from the +V and only need a low current negative supply.

Separate neg doubler circuit from different winding ?
 

Offline Hero999

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That's a valid way of doing it but a single winding transformer could be more convenient, especially if it's inside a wall plug supply, rather than the equipment.
 

Offline jimmc

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Referring to the second circuit:
1)  The current drain from the positive output must be more than that from the negative output for the circuit to work correctly, if not the negative rail will reduce and the positive rail increase.
2) The voltage across the capacitor does not reverse, ideally it would remain fixed and equal to the peak AC voltage (less diode drops).
3) This circuit does not cause DC in the transformer winding, a capacitor cannot pass DC!

Jim
 

Offline Stove

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Is it possible this problem seems counter intuitive just because you assume GND is the same as the lower tap of the transformer? It's not. This makes me feel like building the circuit just to scope it. Maybe I'll deliver some scope measurements if I can find the parts.
 

Offline BravoV

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Referring to the second circuit:
1)  The current drain from the positive output must be more than that from the negative output for the circuit to work correctly, if not the negative rail will reduce and the positive rail increase.

Thanks, I'm not aware of this, but I think its safe to assume that normally the pos output will be loaded a lot more than the neg output. As I mentioned, the neg output is just for low current usage, like biasing neg rail of the op-amp for example.


2) The voltage across the capacitor does not reverse, ideally it would remain fixed and equal to the peak AC voltage (less diode drops).

Though sounds so simple, this is it that I want to hear, makes perfect sense, thank you.  :-+


3) This circuit does not cause DC in the transformer winding, a capacitor cannot pass DC!

I'm assuming this to counter the post made by Hero999, lets hear from him.


Is it possible this problem seems counter intuitive just because you assume GND is the same as the lower tap of the transformer? It's not. This makes me feel like building the circuit just to scope it. Maybe I'll deliver some scope measurements if I can find the parts.

Share it please once you built and tested it.
« Last Edit: August 20, 2013, 05:34:53 pm by BravoV »
 

Offline Hero999

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3) This circuit does not cause DC in the transformer winding, a capacitor cannot pass DC!

I'm assuming this to counter the post made by Hero999, lets hear from him.
No, he's right, that's what I was saying.

The capacitively coupled negative supply draws power on both halves of the cycle so there's no DC in the transformer's core.

Two half wave rectifiers will result in some DC flux, if the current from both the positive and negative supply isn't matched.
 


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