Power Conditioning / Surge Protection

We agree to disagree.

It is not a matter of disagreeing of agreeing. This is objective engineering, not subjective listening.  Everything in what I wrote is easily verified through simulations or building and testing.  I will absolutely agree that added resistance whether real resistance or AC impedance will limit /reduce the maximum rail voltage and hence the peak volume, but I cannot agree as it simply is not true, that the effect and operation is as you described and to the to the level you described as that is simply not the case, no more than the real world performance of the Furman will be as they describe. I am not making that statement because "I believe" it to be true, I am making that statement as I know it to be true both from a fundamental engineering standpoint, and a practical, having done it, implemented it, tested it. measured it standpoint.


There may be something perceived as lack of dynamics with some filters and some amplifiers, but any real world loss of peak power using real music is going to be fairly small, and unless you are driving into clipping, loss of dynamics should not be an issue with any competently design amplifier. Instead of just taking the easy answer, which means the problem is never solved, it is better to find out what is really behind the perceived difference.

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.