Paul,
Surely one half cycle won't fully charge the filter capacitor(s). How
many half cycles would required to fully charge them?
In theory, infinite half cycles! The charging current gets ever lower,
but never arrives at zero.
Why is the diode surge rating for a single half cycle?
I think, mostly because they are esy to test that way.
If it can handle that surge once, why not more than once?
Because the diode chip ends up hot, and first needs to get rid of that
heat, before it can take another such current surge.
Does heat become the limiting factor after only one or two half cycle
surges?
Yes. The diode chip is small, so it has only a very small thermal
capacity. The heat has to flow away from the silicon chip into the
terminals and the board and ultimately the air, and that can happen only
at a certain rate.
What would the diode surge handling capability be on the second...
third... etc. half cycle? What would be the point at which this
repetitive but decaying surge would harm the diode?
This is often rated in the data sheets. For example, at this moment I'm
watching a Vishay datasheet for the 1N540x series. Figure 2 is a graph
showing how much current those diodes can take, depending on the number
of cycles at 60Hz. Some point values are: 200A for a single cycle, 130A
for ten cycles, 60A for 100 cycles.
When you have a typical capacitor charge pulse, of course teh current
starts very strong and then decays, so the above ratings are not the
most useful. The I^2*t rating given in some datasheets can be more
useful in this case. It tells the product of current squared by time,
that will fuse the diode. The Vishay datasheet for the 1N400x series
gives a value of 3.7A^2*s. Taking the capacitance value, final voltage
on the cap, and the total circuit's resistances, and doing the proper
math, you can know how far the diode will be from fusing, in the worst
case. Of course in any practical design you want to be WELL clear of
that fusing level!
Doing some simple math: Let's say that your amp has a filter capacitance
of 50uF that charges to 2500V. Charge in coulombs is simply volts
multiplied by farads. So you need to charge .125C into your capacitor.
And one coulomb is nothing other than one As (ampere second). So, 0.125A
will take one second to charge that cap, or 1A will take one eighth of a
second, and so on.
If the total effective, equivalent resistance (transformer secondary
resistance, ESR of diodes and capacitors, any series resistance, and the
transformed values of primary resistance, line impedance, and inrush
limiter impedance) is for example 1000 ohm, then your capacitor would
start charging at 2.5A, which then goes down asymptotically to zero,
with a time constant (resistance * capacitance) of 50 milliseconds.
That means that 50ms after switching on, the capacitor will be up to
about two thirds of its final voltage, and the current will be down to
about 0.8A, which is already within the continuous current capability of
an 1N4007. So the overcurrent (above 1A) would last only only about 45
milliseconds, with an average value of about 1.8A. The average of the
squared currents would be roughly 2.2A. That makes a surge load of
about 0.24 A^2*s, about one fifteenth of what a 1N4007 survives. This is
yet another proof that 1N4007 diodes are plenty big enough for a typical
ham amplifier, as long as it uses some sort of inrush limiting - even if
some people feel by their guts that using 1N4007's is cheap and bad and
ugly and merits the culprit's degradation to novice class!
Note that in this example, 1N5408 diodes can handle the inrush current
within their CONTINUOUS current rating, without having to enter any
surge rating at all!!!
BTW, my 4000 volt supply has a step start that slowly ramps the
voltage to about 3000 volts over 4 to 5 seconds before the step start
relay actuates. There still must be a considerable surge to finish
charging the 32 uF oil cap, because when that step start relay pulls
in the house lights flicker!
Either the charging is too slow, so that at the end of those 4 to 5
seconds there is still a lot of charge missing, or else the bleeders
take so much current, that the voltage tapers off to a value too low. 4
to 5 seconds is far longer than necessary to smoothly bring up such a
supply. If you want to limit the initial surge to no more than the full
normal current, that would be about 2kW or 3kW input, you can use a time
constant of about 0.1 second between the step start resistor and the
capacitor. Or make it 0.2s for even lower startup current. In that case,
after your 4-5 seconds time, the capacitor will be charged to a voltage
very close to the final one, and the pulse when teh relay actuates
should be minimal.
Note: my step start is controlled by a ramped DC voltage. The relay
coil has no voltage applied until the DC control voltage ramps up to
a pre-determined point. When that rising DC voltage reaches the
pre-determined trip point, the control circuit momentarily applies
150% of rated coil voltage to the relay. This causes it to pull in
quickly and cleanly, which may not be said of some step start
circuits I have seen. This may (or may not) have some influence on
the surge taking the cap from 3000 to 4000 volts.
It's fine just as you are doing that. Just your limiting resistor in
parallel with the relay contacts seems to be too high. If the
transformer's magnetizing current is low, a resistor of about 68 ohm
would be fine for a 240V input, with 18 Ohm being about right for 120V
input. If your transformer draws a high magnetizing current, you will
need lower resistance values.
How much resistance do you have there?
After the relay has pulled in its coil voltage is reduced to 50% of
rated, which is more than enough to keep it solidly closed while
limiting heating of the somewhat large coil.
That's fine.
Manfred
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