[Amps] Filter Capacitors

[TexasRF at aol.com]

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 at 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)




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