>
>>Well, so far the only significant reason that has been offered, by Steve K.,
>>is that for fast transients, the rectifier string "appears to be a string of
>>small capacitors." True enough, and indeed small they are. Down in the 10pf
>>range.
>>
>>Now. lets consider this in context. These rectifiers will typically appear on
>>the secondary side of a large transformer that is not exactly a shining
>>example of high-frequency transformer design. Although I have never gone
>>looking for it, I would not expect to see much high frequency energy on the
>>secondary side. Especially peaks in the range of many kilovolts.
>>
>Why not?
>
>It is quite common to see mains spikes of 1kV and more on the *primary*
>side of the transformer. According to ANSI/IEEE statistics, "medium
>exposure" sites in the USA can expect several hundred 1kV-peak mains
>spikes per year, and even 5kV-peak spikes at about 1 per year.
>
? Only with no load on the transformer. "Why"? - - With a full-wave
rectifier and filter C there is a low-ESR load - as he explained.
>Plate modulation transformers and output transformers for tube audio
>amplifiers will transmit energy up to at least several tens of kHz, and
>there is nothing very different about mains transformers of comparable
>size. Therefore there is every reason to expect very large
>high-frequency spikes reaching the diode strings, from time to time.
>
? Only if the filter-C is removed.
>
>>In order to get anywhere with this line of reasoning, we need to adopt a (or
>>some) generic models of transient pulses and look at the possible results I
>>think it is futile, however, because of another practical factor. In a
>>typical full-wave rectifier, one leg of the rectifier is always in
>>conduction.
>
>This is not true for the capacitor-input supplies that we use, so the
>rest of the argument doesn't hold up.
>
? Ian -- Do those whom you call "we" put in inductor in series with
the filter-C to prevent it from seeing the spike in voltage,?
>In a typical capacitor-input supply, both strings of diodes may be
>non-conducting for almost all of the time! In the first quarter-cycle,
>the "forward" diodes will not start to conduct until the applied AC
>voltage swings up above the DC voltage on the capacitor. The diodes
>conduct and charge the capacitor until the AC voltage reaches its peak,
>but just over the peak the diodes become reverse-biased again and stop
>conducting.
? He is talking about a spike condition which undoubtedly causes the
diodes to conduct.
>
>Therefore the diode current is a short, sharp pulse. The duty cycle of
>this pulse is very low in the RX condition, because very little DC
>current is being drawn. When transmitting, the duty cycle increases at
>zero-signal anode current, and increases further as you draw more DC
>current. But in linear amplifiers the maximum current you can draw is
>limited by the sag in the DC voltage, so the on-time of the "forward"
>diodes is very unlikely to be greater than about 15% of the full mains
>cycle.
>
>All the rest of the time, no diodes are conducting at all, so they
>provide no damping to the circuit.
>
? This is true only without a spike/anomaly.
>>
>>The basic technical issue at work here is still hanging wide open, and I
>>maintain my challenge to anyone to offer something we can chew on. That is,
>>how do you get all the rectifiers in the string into simultaneous conduction?
>>Bear in mind that when reverse conduction happens, it does so at the nominal
>>PIV+. The diodes do not go into an irreversible SCR-type avalanche where the
>>voltage drop essentially goes to 0. So even if a transient happened to kick
>>the string into conduction, as soon as the transient dies away, we have
>>sub-PIV conditions again.
>>
>That's correct. If the avalanche current is limited, all modern diodes
>will avalanche gracefully, and recover when the voltage falls below the
>PIV (approximately).
>
>And that is precisely the reason for the "no shunt components"
>philosophy! If one or more diodes do go into reverse-bias avalanche, the
>current is limited by the rest of the string which has *not* avalanched.
>Shunt Rs and Cs may do more harm than good because they allow more
>reverse current to flow.
>
? Agreed, which is essentially what the 1995 Handbook says....
- R. L. Measures, a.k.a. Rich..., 805.386.3734,AG6K,
www.vcnet.com/measures.
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