Hi Gerald, Dwayne,
> Hi Manfred, I totally agree with the Farad definition but we were
> discussing ripple voltage and that is related not only to the Farad but time
> constants as well.
It can be calculated in several ways, of course. I was just volunteering
the method I prefer, because it's the simplest and most practical I have
found.
When I calculate the capacitance I need, I like to consider the charge
pulse as being infinitely short, so that the filter cap has to deliver
the full load current for teh full semicycle. Since the truth is that
teh charge pulse is definitely not of zero length, that gives me some
headroom: The actual ripple will be somewhat lower than the one I calculate.
> The charging time constant resistance includes transformer secondary and
> primary resistance (as reflected into the secondary) and house wiring.
And in some cases the resistance of the power grid outside the house
can't be neglected! Think of rural homes, for example, which might have
long runs of power wire.
> The
> discharging time constant is determined by the plate load resistance plus any
> bleeder current.
Not at all! The plate load resistance of a tube (if you mean that!) is
the value of load resistance that should be applied for optimal
operation, best power transfer, or whatever. It's NOT that the tube
behaves like a resistor of this value!
A triode, driven with a certain grid voltage swing, and loaded at a
certain level, will behave as a roughly constant resistor to the power
supply. But a pentode or tetrode will not! It will instead tend to
consume a pretty constant current, even if the supply voltage changes.
That's assuming that the screen voltage stays constant.
> One time constant is the interval for the voltage charge to 65% of the peak
> value or discharge from the peak value. In the stated case of 3050v peak,
> the voltage change would be 3050 X .65 = 1982.5 v.
But how useful is that, in this case? Sure, the time constant given by
the capacitance value and the total equivalent resistance of the
transformer and line, will tell you how much time it takes for the
capacitor to charge to 65% of the final voltage, after being plugged in.
But that's not so very practical for calculating ripple!
> With an 80 uFd filter, the 65% charge time
> would be 2.4 milliseconds. Since there are almost 3 such time constants
> between the 8.3 millisecond charge cycles, the filter charge reaches about
> 96% of
> the peak charging voltage.
Let me translate this: When you switch on the amp precisely at a zero
corssing of the line voltage, then after the first half cycle the filter
cap will be charged to 96% of its final voltage. And you will hear a
mighty THUMP as that happens! ;-)
> The discharge time constant is plate load resistance plus bleeder current
> times the filter Farads. For 3000v at 1A plate current and 3 mA bleeder
> current, the total resistance os 3000 / 1.003 = 2991 ohms.
OK, assuming it's a triode, and defining "plate load resistance" as "the
equivalent resistance presented by the tube to the power supply", which
isn't the usual definition!
> The time constant then is 2991 X 80 e-6 = .239 seconds.
OK.
> The voltage decay of the charged filter
> C then is 1982.5 v in 239 milliseconds.
Yes, but it's not a linear curve!
> in our 8.3 millisecond charge
> cycle the voltage drop would be 8.3 / 239 X 1982.5 = 68.8 volts.
No. It would be slightly more than that, due to the nonlinearity of the
discharge curve.
Only if you draw a constant current, instead of the voltage-proportional
current drawn by a resistor, would you get a linear discharge curve, and
only in that case would your simple proportion hold true.
> I will leave it to others to comment on 68.8 volts ripple being ok or not
> ok. It is only about 2.25% of the plate voltage so it seems like it would be
> adequate.
I think that this amount of ripple would be fine in most or all ham
applications. Specially for SSB voice it's quite OK, and won't cause
audible hum modulation of the signal. But other modes and applications
call for far lower ripple.
Many ham amps get by with much larger ripple, and the hams using them
are still happy.
> I have been going over the same calculations for a 10GHz twt amplifier
> power supply. In that case, the tube displays .008 dB power change per volt
> on
> the cathode. To keep the carrier hum low enough, I was shooting hum at the
> -20 dB level which is 1% of the power. So, the hum would be -.0436 dB /
> .008 dB = 5.4 volts ripple.
>
> That is going to require lots of microfarads or a really good electronic
> filter or both.
In fact, the most practical way to satisfy that requirement is using a
switching power supply. It allows getting a very low ripple (much better
than 5V), tight regulation, coupled with small size, low weight, and low
cost.
Dwayne,
> I recently changed out the electrolitic caps in my 3800 volt power supply.
> from 10 210 Mfd at 450 volts to 10 530Mfd at 450 volts per cap. I did
> this
> thinking that I would lessen the decay of the voltage when the amp is
> keyed..
It won't, or not very much. It will only slow down the decay just after
keying. But it will reduce the ripple.
> But I'am thinking perhaps all I did was provide to much capacitance for
> the capability of the transformer.. Perhaps this is more stress on the
> trnasformer.???
Yes, it is more stress for the transformer, the power line (including
fuses and circuit breakers), and to a lesser extent also for the
rectifier diodes. Whether that will cause any trouble, depends on the
safety margins built into everything, and how demanding your operating
style is.
> All I have noticed is a longer decay time for the voltage
> to return to zero when shutting down the amp.(makes sence,more
> capacitance)
Yes.
The ripple will be lower. But due to the highe rcapacitance, the peak
voltage to which the caps will charge during TX will be lower than the
one to which the old caps charged. And the valley voltage will be
higher, and so the average DC voltage, the one you see on the meter,
will be mostly the same as before. Slightly higher, but not much. How
much higher, depends on how "stiff" the transformer and the power line
are (how low equivalent series impedance they have).
Manfred
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