Oops, make the connected/disconnected 33 degrees connected and 57 + 90 =
147 degrees disconnected during the half cycle.
73/K5GW
In a message dated 4/12/2012 10:43:16 P.M. Central Daylight Time,
TexasRF@aol.com writes:
Roger, the transformer only delivers current during the time the
transformer voltage is greater than the plate voltage. The rectifiers
disconnect the
transformer when it's voltage falls below the plate voltage. After the
peak, the filter C is fully charged and the rectifiers disconnect the
transformer.
So, if the plate voltage reads 3000v and there is 100v peak to peak ripple
on it, the actual voltage falls between 2950 and 3050v which makes the
3000v average.
The transformer then charges the filter C only during the interval the
voltage is above 2950v.. There are I X R voltage drops in the transformer
primary and secondary as well as in your house wiring and the commercial
power
feed. To reach the above mentioned 3050v, the transformer has to over come
all I X R voltage drops plus 3050v peak. The transformer may well need to
have a 3500v peak (2475 rms) rating or more to make the needed voltage.
Once the filter C is fully charged, at 3050 volts at the peak of the
transformer AC cycle, the rectifiers disconnect the transformer. So, you
can see
that the energy to run the plate supply is delivered by the transformer
only on the up swing from 2950v to 3500v. That means the transformer is
disconnected for 57 degrees and connected for 33 degrees of a 180 degree
half
cycle. 180 to 33 is a 5.45 ratio so the transformer current while
charging the
filter C has to be 5.45 X dc current in this case. And all of the other
currents in the primary circuit have to be 5.45 X as well so no wonder
power
supply regulation suffers!
I have not seen the 1v, 1A, 1 second, 1 microfarad theory before. 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.
73,
Gerald K5GW
This is the cause of power supply regulation not being perfect. The more R
involved, the more I X R drop and the worse the regulation.
If the transformer is conducting only when the instantaneous voltage is
greater than 2950v,
In a message dated 4/12/2012 10:07:48 P.M. Central Daylight Time,
k8ri@rogerhalstead.com writes:
On 4/12/2012 10:13 PM, Manfred Mornhinweg wrote:
> Jim,
>
>>> How do you conclude the proper value (both the voltage and
capacitance)
>>> for a filter capacitor for a 1500 watt linear amplifier?
> I will go into a little more detail than Carl did! ;-)
>
> You need to decide how much ripple is acceptable, and then simply go by
> the definition of the units. One farad is the amount of capacitance that
> will gain or loose one volt in one second, if a current of one ampere
> charges or discharges it. As simple as that.
>
> Let me calculate it with an example: Suppose you have an amp that takes
> 3000V and 1A, and that you decide that 100V ripple is OK. Suppose too
> that you live in a country that has 60Hz line frequency, and that the
> power supply uses a bridge rectifier and a simple capacitive filter,
> without a choke.
>
> In that case, a half cycle lasts for 0.0083 seconds, and the filter cap
> will have to power the amp for ALMOST that entire half cycle, before
> being charged by the next AC peak. We can simplify that "almost" and say
> that your filter cap needs to power the amp for 0.008 seconds only, and
> in that time it's allowed to drop by 100V, when delivering 1A. So this
> turns to simple math:
>
> 1A * 0.008s / 100V = 0.00008F, or 80uF.
>
> Note that this assumes that the transformer and the power line can
> actually deliver the intense charge pulse,
Doesn't the transformer have to make up the "difference" when the cycle
is drawing from the cap and finish recharging the cap during the rest of
the cycle? It's the total charging current that over works the
transformer or AC line with the big capacitor banks.
Of course you can go whole hog and end up with a 100 or more mikes, BUT
at that point you now have a small bomb under the hood. Hence the long
handled "chicken sticks" for the "big stuff" as in a few hundred KW.
Those we didn't worry about blowing up those big oil filled caps, but
these small caps we typically use in amps, (not the oil filled ones)
literally can explode with considerable force. One of the reasons I
don't think we'll ever see superconducting batteries in electric cars.
I've been about 3 feet from a 50,000 Joul GMOV when it blew/disappeared
and left two #6 leads sticking straight out. My ears rung for over an
hour. Now imagine a superconducting battery with basically zero internal
resistance and a few million Jouls of charge that was broken or lost its
superconducting capacity. Theoretically you could hold such a thing in
the palm of your hand, if they ever figure out how to make one that
doesn't have to be cooled.
73
Roger (K8RI)
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|