AndrewT said:
Happy to join you in that very cause. Working on a sim right now. I think Nixie woke up with a dirty diaper though, didn't much like the tone myself. Back to the sim. Btw I am an advocate of putting it in the live line myself. You can't guarantee it will never fail, so err on the safe side. You can easily insulate the circuit.
Tweeker said:
We're already talking about playing with mains so whatever wattage it is becomes irrelevant as to what can kill a man. I'm talking about a practical DC blocker that doesn't require more capacitance than the supply itself, and stands a good chance of lasting awhile.
I know it clips the peak but odds are it's far from ideal to begin with, just add a pi filter after it. I want the diodes doing the conducting, that's why I used smaller caps for a higher impedance.
Just call it a proof of concept it's not something I'd wire as shown.
If you want it to have less capacitance than the amps supply itself, then build a tube amp.
In terms of energy storage needed in the capacitors all these setups are much less than most amps, 16V parts are plenty. Have existing DC blocking supplies of normal design using big elcos been dying left and right?
In terms of energy storage needed in the capacitors all these setups are much less than most amps, 16V parts are plenty. Have existing DC blocking supplies of normal design using big elcos been dying left and right?
Hi,
the DC block works if the capacitor blocks the DC.
If only the diodes are put in place then the DC bias still passes through, although I think it was EVA that published some tests showing diodes alone can work IF the load is fully reactive (we never agreed an outcome to this argument).
I believe the diodes are there to prevent excessive voltage across the caps during periods of high current. This being during start up, a fault situation and possibly if very high power outputs are being sustained in a ClassAB amp. I would not recommend undersizing the DC blocker in a PA amp nor in a ClassA amp. The diodes will get hot if they are repeatedly asked to pass current, particularly if the amp also happens to be drawing high current at the same time.
If the low current is passed by the cap and the higher, but normal current, bypasses the cap then I think the DC bias still appears across the transformer and the resultant saturation is still there.
I have arrived at this by considering the area of the curve above a DC bias that is partially cut off by the diode and the area under the other peak curve. The two areas will not be the same due the bias and the effect of unequal areas is DC bias across the transformer.
Some clever person will know how to simulate this or even measure it.
the DC block works if the capacitor blocks the DC.
If only the diodes are put in place then the DC bias still passes through, although I think it was EVA that published some tests showing diodes alone can work IF the load is fully reactive (we never agreed an outcome to this argument).
I believe the diodes are there to prevent excessive voltage across the caps during periods of high current. This being during start up, a fault situation and possibly if very high power outputs are being sustained in a ClassAB amp. I would not recommend undersizing the DC blocker in a PA amp nor in a ClassA amp. The diodes will get hot if they are repeatedly asked to pass current, particularly if the amp also happens to be drawing high current at the same time.
If the low current is passed by the cap and the higher, but normal current, bypasses the cap then I think the DC bias still appears across the transformer and the resultant saturation is still there.
I have arrived at this by considering the area of the curve above a DC bias that is partially cut off by the diode and the area under the other peak curve. The two areas will not be the same due the bias and the effect of unequal areas is DC bias across the transformer.
Some clever person will know how to simulate this or even measure it.
The use of two diodes is series is probably not a good idea as the forward voltage drop of a silicon power rectifer will be closer to 1V (or higher) rather than the notional 0.7V that seems to be assumed in the discussion so far. Certainly in power amplifier applications - especially at turn-on.
Two diodes in series will then have a combined voltage drop of >2V which likely exceed the reverse voltage rating of the electrolytic capacitors.
Two diodes in series will then have a combined voltage drop of >2V which likely exceed the reverse voltage rating of the electrolytic capacitors.
Hi,
the disadvantage of a single polarised capacitor is that it will be reverse biased every other half cycle.
That's why we use the back to back format. So that one cap in the series string is correctly oriented to resist the voltage.
The disadvantage of a single diode is that you ONLY have a peak voltage across the AC conducting capacitors of about 600mV. This demands very large capacitance to keep the AC (rms) voltage drop down to about 400mVac (conveniently ignoring the high current charging peaks that Nixie correctly referred to)
The advantage of double diodes is that one can halve the required capacitance. There is no added risk to the protected transformer, nor to the user, nor to the back to back capacitors.
There is an advantage to your pocket (cash flow), the extra pair of diodes are a lot cheaper than the bigger capacitor.
I do not advocate a single polarised capacitor for this DC blocking duty. I cannot adequately test a production run of selected capacitors nor statistically identify if the risk to the user is adequately low. Yes, I could test the prototype until failure happens, but one failure whether after a week or a month or 10years is a failure and I cannot forsee (and not prepared to accept the risk of) all the consequences of that shortsighted penny pinching attitude.
I have even seen three diodes in series advocated. If that is the designer's decision then I will not argue against. But it has to be an informed decision, not guesswork.
the disadvantage of a single polarised capacitor is that it will be reverse biased every other half cycle.
That's why we use the back to back format. So that one cap in the series string is correctly oriented to resist the voltage.
The disadvantage of a single diode is that you ONLY have a peak voltage across the AC conducting capacitors of about 600mV. This demands very large capacitance to keep the AC (rms) voltage drop down to about 400mVac (conveniently ignoring the high current charging peaks that Nixie correctly referred to)
The advantage of double diodes is that one can halve the required capacitance. There is no added risk to the protected transformer, nor to the user, nor to the back to back capacitors.
There is an advantage to your pocket (cash flow), the extra pair of diodes are a lot cheaper than the bigger capacitor.
I do not advocate a single polarised capacitor for this DC blocking duty. I cannot adequately test a production run of selected capacitors nor statistically identify if the risk to the user is adequately low. Yes, I could test the prototype until failure happens, but one failure whether after a week or a month or 10years is a failure and I cannot forsee (and not prepared to accept the risk of) all the consequences of that shortsighted penny pinching attitude.
I have even seen three diodes in series advocated. If that is the designer's decision then I will not argue against. But it has to be an informed decision, not guesswork.
classd4sure said:
AndrewT said:
Reading the last pages, I am getting confused . . .
most probably because I am reading late night just after 6-hour driving from Seoul to Busan . . .
As far as I know, ac current through the caps of the blocker is ignorable. Why . . . ? The ac voltage drop across the diodes would depend on their bulk resistance which is low, say 0.25ohms or lower. Then, even if 4A peak ac current flows through the diodes of 0.25ohm bulk resistance, the peak ac voltage drop across the blocker would be vac peak = rs X iac = 0.25 X 4 = 1v. If so, the maximum ac current through the caps could not be big according to i = C x dv/dt. It means we could assume that all ac current flows through the diodes.
When we consider caps of the blocker only with respect to ac, we could be satisfied with small sizes.
Am I wrong . . . ?
Hi,
I think it is easier to think of the DC blocker as two mechanisms at play.
The AC signal passing through the cap (or pair of polarised caps).
and
The AC signal passing the diodes.
Just try to identify the conditions when each operates and then see if the conditions overlap and can to an extent be summed.
Firstly,
the cap will pass AC to the transformer.
The voltage drop across the cap is related to the cap impedance at the mains supply frequency.
As before you can predict the effective impedance and use this to determine the peak current that produces a limiting voltage drop. This can happen without any assistance from the diodes.
The complication as pointed out by Nixie is that my manual example only applies to sinusoidal waveforms and the charging currents through the transformer and rectifier to the smoothing caps is decidely not sinusoidal. This is a capacitor input PSU, with all it's inherent pitfalls.
Secondly,
the voltage drop across the diodes is nearly constant and once the voltage of the mains has moved away from 0v, the mains voltage passes to the transformer with just the loss of a diode junction or two. This will produce a waveform reminisant of a ClassB amplifier with no bias.
I contend that diodes on their own will not remove the DC element of the mains supply and that this will show as unbalanced areas (voltage vs time) above and below the 0v axis.
I further contend that the diodes should not bypass the capacitors if one wants the caps to remain as effective DC blockers for current flowing to the transformer.
That leaves just case one as the normal operational mode.
The start up current can and is many times higher than the operational current in the transformer. During the start up the DC is not blocked due the bypassing action of the diodes. For the few tens of mS that this second mode lasts the DC is not blocked and the toroid will suffer from DC induced hum, but it hardly matters. The thump from the transformer (from all the energy it absorbs) as you switch on will completely hide the hum that might be generated for those first few cycles of the mains input. As soon as the transformer flux is established the current returns to a much lower figure as the smoothing caps charge up. This may cause the diodes to keep bypassing but again this is relatively short lived and may just extend to a second or so. The heat in correctly selected diodes should be tolerable.
I think it is easier to think of the DC blocker as two mechanisms at play.
The AC signal passing through the cap (or pair of polarised caps).
and
The AC signal passing the diodes.
Just try to identify the conditions when each operates and then see if the conditions overlap and can to an extent be summed.
Firstly,
the cap will pass AC to the transformer.
The voltage drop across the cap is related to the cap impedance at the mains supply frequency.
As before you can predict the effective impedance and use this to determine the peak current that produces a limiting voltage drop. This can happen without any assistance from the diodes.
The complication as pointed out by Nixie is that my manual example only applies to sinusoidal waveforms and the charging currents through the transformer and rectifier to the smoothing caps is decidely not sinusoidal. This is a capacitor input PSU, with all it's inherent pitfalls.
Secondly,
the voltage drop across the diodes is nearly constant and once the voltage of the mains has moved away from 0v, the mains voltage passes to the transformer with just the loss of a diode junction or two. This will produce a waveform reminisant of a ClassB amplifier with no bias.
I contend that diodes on their own will not remove the DC element of the mains supply and that this will show as unbalanced areas (voltage vs time) above and below the 0v axis.
I further contend that the diodes should not bypass the capacitors if one wants the caps to remain as effective DC blockers for current flowing to the transformer.
That leaves just case one as the normal operational mode.
The start up current can and is many times higher than the operational current in the transformer. During the start up the DC is not blocked due the bypassing action of the diodes. For the few tens of mS that this second mode lasts the DC is not blocked and the toroid will suffer from DC induced hum, but it hardly matters. The thump from the transformer (from all the energy it absorbs) as you switch on will completely hide the hum that might be generated for those first few cycles of the mains input. As soon as the transformer flux is established the current returns to a much lower figure as the smoothing caps charge up. This may cause the diodes to keep bypassing but again this is relatively short lived and may just extend to a second or so. The heat in correctly selected diodes should be tolerable.
EXACTLY,
The caps must acquire a DC voltage in order to "buck" the DC present on the mains.
The voltage change on the cap, during the charging period 120 times per second can be derived simply by equating input power to current, and then Coulombs per half cycle. Charge applied to the capacitance should yield a voltage. This voltage should be kept small, 1/10 (?) of a Volt, so as not to consume the total DC bias, 1 Volt (?), available. This means that greater DC offsets, or, greater loads both will require larger caps.
The caps must acquire a DC voltage in order to "buck" the DC present on the mains.
The voltage change on the cap, during the charging period 120 times per second can be derived simply by equating input power to current, and then Coulombs per half cycle. Charge applied to the capacitance should yield a voltage. This voltage should be kept small, 1/10 (?) of a Volt, so as not to consume the total DC bias, 1 Volt (?), available. This means that greater DC offsets, or, greater loads both will require larger caps.
Hmmm... methinks my mouth is outstripping my mind.
I was thinking purely in terms of bigger voltage = bigger cap... in a volumetric sense. Size = capacity x voltage. In consideration of the fact that these gizmos are limited by max reverse rating of the cap... that does not really apply here... jibberish as such. Were we using something other than electroltyics, it would apply... since we're not, still jibberish.
You're a sharp man Mr. T!
I was thinking purely in terms of bigger voltage = bigger cap... in a volumetric sense. Size = capacity x voltage. In consideration of the fact that these gizmos are limited by max reverse rating of the cap... that does not really apply here... jibberish as such. Were we using something other than electroltyics, it would apply... since we're not, still jibberish.
You're a sharp man Mr. T!
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