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.
The charging time constant resistance includes transformer secondary and
primary resistance (as reflected into the secondary) and house wiring. The
discharging time constant is determined by the plate load resistance plus any
bleeder current.
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.
The charging time constant will be quite short since R is hopefully small.
A typical plate transformer might have a total of 30 ohms secondary plus
stepped up primary resistance. 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.
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. The time constant
then is 2991 X 80 e-6 = .239 seconds. The voltage decay of the charged filter
C then is 1982.5 v in 239 milliseconds. in our 8.3 millisecond charge
cycle the voltage drop would be 8.3 / 239 X 1982.5 = 68.8 volts.
Therefore the peak to peak ripple is also 68.8 volts.
Note there is charging and discharging current for the filter C but it not
come into the math above.
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 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.
73,
Gerald K5GW
In a message dated 4/13/2012 1:38:13 P.M. Central Daylight Time,
manfred@ludens.cl writes:
Gerald,
> I have not seen the 1v, 1A, 1 second, 1 microfarad theory before.
It's not a theory. It's the very definition of the unit "farad"!
But it's "farad", not "microfarad"! A capacitor of one microfarad
instead would gain one volt in one second if charged by one microampere.
Or if charged by one ampere, it would gain one volt in every microsecond.
If you get confused by where to add "micro" and where not, better use
the base units only, without any prefix, even if then a typical Pi tank
tuning capacitor might end up having a capacitance of 0.0000000002 farad.
If you don't believe me about the definition, look it up here
http://searchcio-midmarket.techtarget.com/definition/farad
or here
http://en.wikipedia.org/wiki/Farad
or google for additional sources!
> I will
> need to look into that as I was under the impression that voltage drop
was
> based on time constants and they are related to 65% voltage charge or
> discharge.
Those time constants happen when you charge a capacitor through a
resistor, resulting in a non-constant charging current. In many
situations it's good to use those time constants, for example when you
use a 555 timer chip. But in a power supply filter calculation it's more
practical to consider the load current as being constant, and thus apply
the definition of the unit.
In other cases, like RF work, you would typically use neither the linear
volt ampere second farad relationship, nor the RC time constant, but
instead you would use the LC resonant frequency, Q factor, and the
reactance of your capacitor at a given frequency.
For a power supply filter, a calculation using the reactance of the
filter capacitor isn't really applicable, because the waveform delivered
by the rectifier contains an infinite series of harmonics, and so a
given capacitor has an infinite number of different reactances to consider!
So we have to use the most suitable method for each particular situation.
Manfred.
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