Ok, here's my next thought. (Hopefully better than the last) 
According to the application guide on Cornell Dublier's site, it sounds like if the capacitors are connected back to back, i.e., the two minuses together or the two pluses together, bypass diodes shouldn't be required as long as the ripple current rating of the capacitors is higher than the line fuse.
According to the application guide on Cornell Dublier's site, it sounds like if the capacitors are connected back to back, i.e., the two minuses together or the two pluses together, bypass diodes shouldn't be required as long as the ripple current rating of the capacitors is higher than the line fuse.
Hi,
the parallel diode/diodes limit the capacitor voltage to 0.7v or 1.4v (for 1 or 2 in series).
In start up or fault condition, the diodes will pass the bulk of the short term excess current. eg. till magnetic flux builds in a toroid or a fuse blows.
Thus allowing fairly low voltage capacitors to be used e.g. 16Vdc instead of 250Vac
the parallel diode/diodes limit the capacitor voltage to 0.7v or 1.4v (for 1 or 2 in series).
In start up or fault condition, the diodes will pass the bulk of the short term excess current. eg. till magnetic flux builds in a toroid or a fuse blows.
Thus allowing fairly low voltage capacitors to be used e.g. 16Vdc instead of 250Vac
Hi,
you missed my point. I'll try again.
the AC voltage will not just pass through the series caps. A voltage will be lost across the caps. As the current rises the voltage rises in proportion.
At high peak current the voltage loss could be significant. The diodes are there to ensure that this voltage never exceeds your preset (diode number) limit.
BTW. I think you are right about the level of DC likely to appear on the line. So again a high voltage cap is not required.
Another thought, since the caps are operating at line voltage both terminals and the case could be at line voltage. I suggested in another thread that it might be better to locate the DC block on the neutral line for safety. No one came back to refute the idea.
you missed my point. I'll try again.
the AC voltage will not just pass through the series caps. A voltage will be lost across the caps. As the current rises the voltage rises in proportion.
At high peak current the voltage loss could be significant. The diodes are there to ensure that this voltage never exceeds your preset (diode number) limit.
BTW. I think you are right about the level of DC likely to appear on the line. So again a high voltage cap is not required.
Another thought, since the caps are operating at line voltage both terminals and the case could be at line voltage. I suggested in another thread that it might be better to locate the DC block on the neutral line for safety. No one came back to refute the idea.
AndrewT said:
You just have to make the capacitance large enough to keep the series impedance down to a reasonable level.
The problem is you need 4 times the total capacitance when you put two capacitors in series versus a single or paralleled capacitors. After looking at the numbers, a couple of diodes are much cheaper than the capacitors necessary to achieve this, so while I like the simple elegance of the capacitor only solution, I'll be using the protection diodes.
Maybe most people agreed with you about installing the circuit in the neutral line. I certainly do.
Hi,
yes, keep increasing the uF to achieve your desired amperage capacity. multiple // caps, say 3 * 3.3mF + 3 * 3.3mF all at 16Vdc are pretty cheap and give an effective 5000uF with a current capacity of about 1.1Apk @ 0.7Vpk or 2.2Apk (400w) @1.4Vpk.
I'm appreciative of your confirmation on Neutral line blocking.
Now that I have your ear let's tell you why some may say NEVER interfere with the Neutral return line. If due to a fault in the caps and/or diodes that causes the DC block to fail open circuit, the whole mains side of the system becomes live. Your amp seems to be dead but everything around the transformer, fuses, switch, DCblock, softstart is now at 240Vac just waiting to KILL any fool that pokes a finger in!!!!
Comments please.
yes, keep increasing the uF to achieve your desired amperage capacity. multiple // caps, say 3 * 3.3mF + 3 * 3.3mF all at 16Vdc are pretty cheap and give an effective 5000uF with a current capacity of about 1.1Apk @ 0.7Vpk or 2.2Apk (400w) @1.4Vpk.
I'm appreciative of your confirmation on Neutral line blocking.
Now that I have your ear let's tell you why some may say NEVER interfere with the Neutral return line. If due to a fault in the caps and/or diodes that causes the DC block to fail open circuit, the whole mains side of the system becomes live. Your amp seems to be dead but everything around the transformer, fuses, switch, DCblock, softstart is now at 240Vac just waiting to KILL any fool that pokes a finger in!!!!
Comments please.
Resurecting a thread
This thread has been inactive for a while, but suddenly I feel compelled to ask a question. Probably a dumb one.
An inductor introduces a 90 deg current lag. Adding a series cap offsets the lag so that current and voltage are once again sychronized. (Resistive elements make this approximate.) So my question is: how much, if any, of the a DC blocker's action with respect to reducing hum is actually due to reducing current lag? Particularly if there are solid state relays (triacs?) being used for soft start.
This thread has been inactive for a while, but suddenly I feel compelled to ask a question. Probably a dumb one.
An inductor introduces a 90 deg current lag. Adding a series cap offsets the lag so that current and voltage are once again sychronized. (Resistive elements make this approximate.) So my question is: how much, if any, of the a DC blocker's action with respect to reducing hum is actually due to reducing current lag? Particularly if there are solid state relays (triacs?) being used for soft start.
DC blocking filter for toroid Xfmr
My apologies in resurrecting a long dead thread, but I have a question I'm hoping to get an intelligent answer for.
I'm calculating much larger values for capacitors than the examples listed in the thread show, and I'd like to understand how my assumptions are different (or wrong).
I have a 2kVA toroid which will be connected as an isolation transformer. Max average current for 240 VAC in will be 8.33 amps.
My understanding is that the impedance at 60Hz of the capacitor must be such that the voltage drop will be less than the forward voltage drop across the diodes.
If I assume 2.8 Vdc (4 diodes) to be blocked (high, but this relaxes requirements on the cap size), then the target Z is .336 ohms (2.8/8.33). This impedance requires ~8000uF @ 60Hz. Which seems really large relative to the other examples in the thread.
Are my assumptions incorrect, or perhaps overly conservative?
Thanks and regards,
Rob W.
My apologies in resurrecting a long dead thread, but I have a question I'm hoping to get an intelligent answer for.
I'm calculating much larger values for capacitors than the examples listed in the thread show, and I'd like to understand how my assumptions are different (or wrong).
I have a 2kVA toroid which will be connected as an isolation transformer. Max average current for 240 VAC in will be 8.33 amps.
My understanding is that the impedance at 60Hz of the capacitor must be such that the voltage drop will be less than the forward voltage drop across the diodes.
If I assume 2.8 Vdc (4 diodes) to be blocked (high, but this relaxes requirements on the cap size), then the target Z is .336 ohms (2.8/8.33). This impedance requires ~8000uF @ 60Hz. Which seems really large relative to the other examples in the thread.
Are my assumptions incorrect, or perhaps overly conservative?
Thanks and regards,
Rob W.
Hi Weinstro,
go back two posts to 108 and you'll find a suggestion for 10mF +10mF at a lot less current than you plan on drawing.
So your answer is YES you need lots of capacitance to keep the reactance low at your planned current for 50Hz or 60Hz operation.
I think the diodes are there for short term peak current bypass purposes. You seem to be suggesting that you can tolerate a higher reactance and let the transformer see a lower drive voltage and then using the diodes at just a little bit higher voltage so they don't conduct in normal operation.
Just a thought but should you work with peak sine current to find peak volts drop across the caps and then select the diodes to non conduct ( my complex maths cannot rise to this - Help.)?
And finally you will need to ensure that your chosen caps can cope with the ripple current - lots of low quality electros in parallel or a very high quality, high capacitance, expensive electrolytic.
go back two posts to 108 and you'll find a suggestion for 10mF +10mF at a lot less current than you plan on drawing.
So your answer is YES you need lots of capacitance to keep the reactance low at your planned current for 50Hz or 60Hz operation.
I think the diodes are there for short term peak current bypass purposes. You seem to be suggesting that you can tolerate a higher reactance and let the transformer see a lower drive voltage and then using the diodes at just a little bit higher voltage so they don't conduct in normal operation.
Just a thought but should you work with peak sine current to find peak volts drop across the caps and then select the diodes to non conduct ( my complex maths cannot rise to this - Help.)?
And finally you will need to ensure that your chosen caps can cope with the ripple current - lots of low quality electros in parallel or a very high quality, high capacitance, expensive electrolytic.
Peranders,
Residential service in the US is 240V phase-to-phase, with a common neutral to form two 120V line-to-neutral legs that are 180 degrees out of phase. I simply ran a 240V, 20A circuit to behind the stereo cabinet.
I have some equipment I purchased in Europe while living abroad. The 2kVA transformer, from Plitron, is a medical isolation transformer. The original plan was to power this from a standard 120V, 20 amp circuit, but the size of the transformer pretty much dictated that a new circuit would have to be down in either case.
Thanks and regards,
Rob
Residential service in the US is 240V phase-to-phase, with a common neutral to form two 120V line-to-neutral legs that are 180 degrees out of phase. I simply ran a 240V, 20A circuit to behind the stereo cabinet.
I have some equipment I purchased in Europe while living abroad. The 2kVA transformer, from Plitron, is a medical isolation transformer. The original plan was to power this from a standard 120V, 20 amp circuit, but the size of the transformer pretty much dictated that a new circuit would have to be down in either case.
Thanks and regards,
Rob
AndrewT,
OK, thanks.
Is my method of sizing the parts OK?
I'd like to maximize the voltage delivered to the transformer, just became concerned at the size of the capacitors suggested.
That's a good point. The current, 8.33amps, is an RMS figure.
Here's another consideration: all of the loads that will be connected have full-wave bridge rectifiers in them, which pull current in short spikes (not sine waves). The peak currents here can be several times higher than average current draw. This factor is known as "crest factor". So, if instantenous current actually hits ~25 amps for an RMS of 8.33 at 2000 VA, shouldn't the diode be able to handle this current? And the capacitive reactance result in a drop lower than my desired threshold?
Otherwise, the diodes will conduct every cycle, which I thought was bad. Or, is it that this is better than having DC offset?
Thanks and regards,
Rob W
OK, thanks.
Is my method of sizing the parts OK?
I'd like to maximize the voltage delivered to the transformer, just became concerned at the size of the capacitors suggested.
That's a good point. The current, 8.33amps, is an RMS figure.
Here's another consideration: all of the loads that will be connected have full-wave bridge rectifiers in them, which pull current in short spikes (not sine waves). The peak currents here can be several times higher than average current draw. This factor is known as "crest factor". So, if instantenous current actually hits ~25 amps for an RMS of 8.33 at 2000 VA, shouldn't the diode be able to handle this current? And the capacitive reactance result in a drop lower than my desired threshold?
Otherwise, the diodes will conduct every cycle, which I thought was bad. Or, is it that this is better than having DC offset?
Thanks and regards,
Rob W
Hi,
I don't remember anyone mentioning crest factor in earlier discussion. I think you are right and then all the more reason to choose even bigger caps.
I also agree that the diodes should not be designed to conduct every cycle but maybe allowed to conduct on intermitant high peaks.
Then it comes down to average load and consideration of peak load on your 2kVA isolator.
I don't remember anyone mentioning crest factor in earlier discussion. I think you are right and then all the more reason to choose even bigger caps.
I also agree that the diodes should not be designed to conduct every cycle but maybe allowed to conduct on intermitant high peaks.
Then it comes down to average load and consideration of peak load on your 2kVA isolator.
This is nothing you have to worry about because it's about dB's (in output power) PLUS natural variation on the mains.weinstro said:
I'll guess you are talking about MUSIC, not continous sinus wave?weinstro said:
2 kVA transformer is one thing but how much power is possible to get out of the amplifier with real load?
When you calculate the parts you must have real world signal in mind and then worst case.
Eva's oscillographs show that the current spike caused by DC offset in the AC power occurs at the zero crossing of the AC signal. I believe this is the concept behind PS Audio's Humbuster, which uses diodes to block current flow within 2.4 volts of the zero crossing, rather than using a series capacitor to completely block the DC component.
PS Audio's Humbuster appears identical to the circuit shown in the lower left quadrant of mlloyd1's schematic, except that:
(1) PS Audio's capacitors are merely small RF bypass capacitors so that almost all the AC current flows through the diodes rather than the capacitors; and
(2) PS Audio uses two bridge rectifiers in series (one on each leg of the AC power line) to raise the zero-crossing threshold from 1.2 volt to 2.4 volt.
The operating principle of the Humbuster is that 4 back-to-back pairs of high power diodes in series (ie, 2 bridge rectifiers; see photo) prevent current flow within 2.4 volts of the zero-crossing of the AC signal.
High power diodes for PS Audio's circuit are cheaper and much more compact than huge capacitors, and they should provide less impedance to high current surges during amplifier output peaks.
PS Audio's Humbuster appears identical to the circuit shown in the lower left quadrant of mlloyd1's schematic, except that:
(1) PS Audio's capacitors are merely small RF bypass capacitors so that almost all the AC current flows through the diodes rather than the capacitors; and
(2) PS Audio uses two bridge rectifiers in series (one on each leg of the AC power line) to raise the zero-crossing threshold from 1.2 volt to 2.4 volt.
The operating principle of the Humbuster is that 4 back-to-back pairs of high power diodes in series (ie, 2 bridge rectifiers; see photo) prevent current flow within 2.4 volts of the zero-crossing of the AC signal.
High power diodes for PS Audio's circuit are cheaper and much more compact than huge capacitors, and they should provide less impedance to high current surges during amplifier output peaks.
Hi,
I'm shooting from the hip here so I could be wrong.
The DC that creeps into our mains supply is not a pure DC instead of an AC signal.
It is a DC superimposed on an AC waveform with a lot of distortion thrown in. In fact that distortion is the clue.
The AC waveform is distorted to such an extent that the tops of the waveforms can be SEEN to have flattened tops due to other loads taking non linear current. It is the asymetrical flattening that gives an effective DC component to the AC waveform. The negatives do not balance out the positives equals net DC.
Stopping a voltage/current as it passes the zero crossing will not remove the distorted waveform. The flattened waveform will still exist and still show it's effect as distortion.
Could there be another effect that the diodes are using that reduces the hum?
Any counter argument?
I'm shooting from the hip here so I could be wrong.
The DC that creeps into our mains supply is not a pure DC instead of an AC signal.
It is a DC superimposed on an AC waveform with a lot of distortion thrown in. In fact that distortion is the clue.
The AC waveform is distorted to such an extent that the tops of the waveforms can be SEEN to have flattened tops due to other loads taking non linear current. It is the asymetrical flattening that gives an effective DC component to the AC waveform. The negatives do not balance out the positives equals net DC.
Stopping a voltage/current as it passes the zero crossing will not remove the distorted waveform. The flattened waveform will still exist and still show it's effect as distortion.
Could there be another effect that the diodes are using that reduces the hum?
Any counter argument?