Power Conditioning / Surge Protection

I'm just tired of arguing, especially most of people don't follow the subject.  I also don't understand some of your statements.  Capacitor is charged only from the bottom of the ripple to next peak.  Bigger ripple means longer charging time - ALWAYS!, but you disagree with this.

Ripple is a function of capacitance and load current.  At constant load current when capacitors get larger (big capacitance) voltage ripple gets smaller hence current charges will be narrower (and usually of higher amplitude because of lower ESR).  Yes, additional impedance will make charging pulses smaller, because they depend on source voltage divided by impedance in the charging circuit - but charging time will not get longer!  It will be charged exactly the same amount of time - from the bottom of the ripple to next peak. Additional impedance in the charging circuit will only result in the voltage drop and the lower voltage on capacitors. It does not affect charging time within each cycle.

Yes, Furman's output capacitor appears to be connected in parallel to electrolytic caps, but is not.  It is on the the other side of the rectifier on the AC side.  Voltage on this capacitor follows AC voltage cycle but any loss of charge caused by PS charging pulse is replenished from energy stored in the inductor thru the whole cycle.  In addition this (huge!) capacitor has very low ESR.  That way during narrow charging pulses current comes from this capacitor and not from the inductor.  Removing this capacitor would have huge effect on the amp.  Just connect big inductor in series with mains and you will see results.  There are power supplies that charge/discharge capacitors during whole cycle because they have big choke in series but then capacitors' voltage is an average and not the peak of the waveform.

Perhaps we're hijacking this thread?

Bigger ripple means longer charging time - ALWAYS!, but you disagree with this.

No it does not and until you understand this you will never be able to move forward. It only means more ripple IF the charge current is the same, and the charge current will be much different with added resistance.

Ripple is a function of capacitance and load current.

No, it is a function of capacitance, load current, and the way the capacitor is supplied with power.

Yes, additional impedance will make charging pulses smaller, because they depend on source voltage divided by impedance in the charging circuit - but charging time will not get longer!

Of course it will get longer. How do you think an LC based PFC works? It works by essentially extending the conduction angle .... i.e. the time the capacitor is being charged.

Yes, Furman’s output capacitor appears to be connected in parallel to electrolytic caps, but is not. It is on the the other side of the rectifier on the AC side. Voltage on this capacitor follows AC voltage cycle but any loss of charge caused by PS charging pulse is replenished from energy stored in the inductor thru the whole cycle. In addition this (huge!) capacitor has very low ESR

Again totally wrong interpretation of what happens. The diode starts conducting when the AC capacitor that is not at all huge in terms of capacitance compared to the amplifier even adjusted for voltage (it will be a film capacitor versus electrolytics in the amp ....way less storage per volume). And no, the capacitor is not "replenished" the whole cycle. Some of it is being discharged, some it is not. It’s voltage follows the AC line with a lag angle determined by its capacitance and the inductors inductance. Once the diodes start conducting, then the lag angle is a factor of the inductor, the AC capacitor and the dominance DC capacitors due to their much higher reflected capacitance. When the diodes are conducting, the "PFC" capacitor is essentially in parallel with the capacitors on the amp (with the exception of the diode drop). The ESR of the film capacitor is almost meaningless as it does not charge the capacitors of the amplifier.


We are not "arguing". No offence, you appear to have some technical acumen, but you view of how this all works is quite wrong.

Charging pulse starts at the bottom of the ripple and ends at the peak of the waveform.  When ripple is larger this time gets longer.  When you comprehend this we can go forward.

Amount of ripple depends on capacitance and load current and does not depend how capacitor is charged since ripple is an effect of voltage drop during capacitor discharge cycle hence has nothing to do with capacitor charging.  The way capacitor is charged will affect voltage on capacitor but ripple will be always the same percentage of this voltage.  When you comprehend this we can go forward.

The ESR of the film capacitor is almost meaningless as it does not charge the capacitors of the amplifier.

ESR is very important since any ESR in the charging circuit will limit maksimum voltage on capacitor.  When you comprehend this we can go forward.

Of course it will get longer. How do you think an LC based PFC works? It works by essentially extending the conduction angle .... i.e. the time the capacitor is being charged.

No it won't.  Voltage droop on capacitor after peak is related only to capacitance and load current.  Amount of this voltage drop (amplitude of ripple) defines width of charging pulse, since capacitor is charged only from the bottom point to next peak.  This bottom point was defined by discharge cycle thus charging has nothing to do with it.  PFC does not work by by extending conduction angle but rather eliminating (shifting) phase between voltage and current to present resistive load to mains.  Averaging (filtering) current pulses coming from PS to draw current from mains during whole cycle is exactly what Furman does.  They call it (improperly) Power Factor correction but it does not change any conduction angle.  It only averages current pulses over whole period.  When you comprehend this we can go forward.

Your understanding of electronics is poor, IMHO and I find you keep arguing just for the sake of it.  Sorry, for the "when you comprehend..."  but that's the unpleasant language you use and another reason I don't want to continue discussing this and perhaps anything else in the future. 

Charging pulse starts at the bottom of the ripple and ends at the peak of the waveform. When ripple is larger this time gets longer. When you comprehend this we can go forward

.
It is impossible for you to discuss this as you don't understand what is happening. It is unfortunately obvious to me you have never had something like this on the bench, never had something like this in a simulator. You are just trying to run it through your head and are doing it wrong. You repeatedly assume the charging current is the same in all circumstances. If the circuit changes, i.e. you add resistance or inductance to the charging circuit, then everything changes. The waveform of the charging current changes, the total conduction angle changes, etc.

Amount of ripple depends on capacitance and load current and does not depend how capacitor is charged since ripple is an effect of voltage drop during capacitor discharge cycle hence has nothing to do with capacitor charging.

You are painfully showing here that you have no idea what is actually happening in the circuit. Sorry can't call it any way but that. Unless you learn this is not the case, you are unable and uneducated enough to discuss this. That is simple reality.  It is not my job to teach you about the basic concept of choke regulation in a power supply, but it is your job to learn it if you want to call other people "low in knowledge" and not have egg on your face.

Amount of this voltage drop (amplitude of ripple) defines width of charging pulse, since capacitor is charged only from the bottom point to next peak. This bottom point was defined by discharge cycle thus charging has nothing to do with it. 

Again, you are showing that you simply don't understand what is happening in enough detail to form any coherent view on what is happening. Amount of voltage drop does not define the width of the charging pulse. The amount of time the input voltage (rectified) is above the output voltage (rectified) defines the time of the charging pulse, literally by definition and by any coherent view of how the circuit operates. We have already determined that the average rectified rail voltage will be lower. Guess what, that means that the amount of time the rectified input waveform will be higher that the output increases.

PFC does not work by by extending conduction angle but rather eliminating (shifting) phase between voltage and current to present resistive load to mains. 

I said the LC circuit in the Furman works by extending conduction angle, I did not say that was what PFC in general was. I can't post pictures here, but perhaps this will help you understand it a bit better as this is exactly what an inductor will do in a linear power supply.  The C in the Furman as pointed out earlier, when combined with most audio amplifiers, will be near useless in impacting PF or THD for that matter.

https://www.allaboutcircuits.com/technical-articles/power-factor-thd-why-linear-power-supplies-fail-...
The fact this article shows the inductor in the output circuit is meaningless. It would work exactly the same in the input circuit, obviously with the inductor value adjusted based on the transformer turns ratio.  The L in the Furman absolutely increases power factor in a linear amplifier. It absolutely is power factor correction for a linear amplifier, however, its effectiveness will be highly dependent on amplifier capacitor bank size and odds are it will rarely result in a high power factor.


Based on what you have written here, by understanding of electronics and especially power supplies and power electronics is obviously way more extensive and accurate compared to yours.


Here, maybe you can increase your knowledge so we can have a proper discussion:

https://sound-au.com/lamps/pfc-passive.html#acc

Oh, and P.S., even if your capacitor bank is larger once you hit a practical size, the peak current and the charging waveform starts to look the same.   I will let you figure out why, but this may give you some hints, see section 5.3.

https://sound-au.com/power-supplies.htm

spatialking285 posts11-07-2020 6:07pm"....The "inductor in parallel with the line" term must mean in series with the line, since it is counter productive to put it in across the line, and in parallel means the line is shorting it out.  If they mean something else, then it is not in parallel with the line."

The SoundStageNetwork and Stereophile reviewers (see links above) both state that the inductor is wired in PARALLEL with the AC line. An interesting departure from more modern designs in which series-mode surge suppressors use an inductor wired in series with one leg of the AC line.