Electrolytic polarity

[jan.didden]

The mains DC is from an asymmetrical waveform due to incorrect power distrobution, loads or interference on the lines. Because of that, the circuit is, IMO, useless.

Funny circuit. I would think it actually CREATES offset (low at that) what with the diodes which are probably not matched. The caps across the diodes are pretty useless, no? the waveform across the diodes looks very much a square wave with 1.2V amplitude or so. That means that the voltage across the transformer, which is the mains minus the voltage across the diodes, looks like sine with the middle part missing, that's disturbingly similar to good old xover distortion waveform. Why would that stop a toroid from buzzing?
Strange.

Jan Didden

 

[Prune]

I ran this in LTSpice. If you simulate it, the sine wave goes through just fine. Without the capacitor that does not happen. The cap lets through part of the waveform until the diodes begin conducting, omit it and the waves get distorted.

 

[Sch3mat1c]

As long as the reactance of each is well under the resistance of the load (so it produces less than a volt AC drop), it won't matter. As a plus, you can also use 6.3V caps, odd as it may seem on a 120V line.

The L+C is a filter, even if it did oscillate (read: ring due to an external source) it would be at some random frequency, depending on leakage inductance.

Tim

 

[peranders]

Hmm, about the mainsdcblocker.gif , i might be paranoid, but doesn't
the combination of L1/L2 + C2/C3 form a nice oscillator ? Aren't
resistors in series with C2/3 a must ?

Mike

If you have a LC-filter you can always get a resonance, but not an oscillator! You must have an active element for this. But nevertheless the filter designer must know how much Q the circuits has in order to determine the filter characteristics.

 

[peranders]

Funny circuit. I would think it actually CREATES offset (low at that) what with the diodes which are probably not matched.

The diodes are for inrush current transients. If the DC level is more than 0.6 V you must either change the caps to a bigger value or add more diodes in series or use Zeners.

The cap values is determined by the max steady state current. If one diode is needed the max voltage drop over the cap is 0.5 - 0.6 V.

4 A, 50 Hz, 0.6 V max = 21221 uF

 

[MikeB]

If you have a LC-filter you can always get a resonance, but not an oscillator! You must have an active element for this. But nevertheless the filter designer must know how much Q the circuits has in order to determine the filter characteristics.

Yes, that's what i've meant. Anytime a spike comes in, the LC is
pushed, and could create a HF-ringing. I am missing a lot of english
words for these things. In german the combination of L & C is called
"Schwingkreis", this is normally used for AM-receiving.
My problem with the filter is, that these Spikes appear often in the mains.

Mike

 

[Sch3mat1c]

Any tank can be considered an oscillator (an "oscillating system", pendulum, spring and weight, inductor and capacitor, etc.), but us electronics-heads take it to mean an active device.

Tim

 

[Prune]

Peranders, what's the formula you used for that calculation?

 

[dnsey]

My question as to how to match capacitors remains unanswered

Put them in opposite arms of an AC bridge (easily thrown together if you don't have one).

 

[Prune]

Peranders, did you solve for the -3 dB corner, as that's what it looks like? Don't you think that the -3 Db is a rather arbitrary choice? Why not -1 dB or -5 dB instead?

I wonder if the distortion added by insufficiently large caps may be worse than a little DC.

 

[Prune]

And what about this one:
http://www.diyaudio.com/forums/attachment.php?s=&postid=328821
I don't understand why the capacitors are in parallel here...

And this one:
http://www.psaudio.com/products/images/306_1.jpg
How can it work with such tiny caps?

 

[Sch3mat1c]

And what about this one:
http://www.diyaudio.com/forums/attachment.php?s=&postid=328821
I don't understand why the capacitors are in parallel here...

I doubt the designer does, either. Note that any cap I'm aware of can withstand 3V or less in either direction... it easily reforms under such conditions. There's so little voltage, faliure isn't going to happen either (don't quote me on that, everything dies (to quote Type O Negative) eventually).

The formula to find capacitance has nothing to do with decibels, just ohm's law, capacitive reactance, load current and the maximum tolerated voltage drop.

And this one:
http://www.psaudio.com/products/images/306_1.jpg
How can it work with such tiny caps?

Probably in parallel, to shunt rectifier hash.

Tim

 

[Prune]

Diodes do not block DC by themselves, you need large caps. In a thread where the photo (commercial product) was mentioned, the maker said the capacitors couple the AC waveform near the zero-crossing where the diodes do not conduct. Again, I don't see how small caps can do that. Maybe I'll build a transformer model in the simulation instead of using a resistive load and see if it changes anything. As for the first circuit, it's also from a commercial product, an amplifier that's supposed to be very reliable. But I'm wondering if someone messed up the schematic.