4000V B+ power supply -filter choices

[Opcom]

Most ham-built 3CX3000 amps I've seen have a separate power cabinet, but the commercial 'plasma generator' ones usually are in one cabinet. I'm not sure what to do.. Floor space is low, but it would be nice to have the meters at eye level and controls at mid-chest level, so a 7FT rack isn't too attractive and neither is a pair of 4FT racks. My power components are old and physically larger than the Dahl units in some of those really nice looking amps you find looking up "3CX3000". Same for oil caps - -the oldies are much bigger than new Russian ones, but they are on hand.

I do not yet have a filter choke other than a 3H/3.5A/6.75 Ohm/7.5KV pig from an RCA TX found at the Black Hole.


I sort of insist on a choke input because the ripple waveform @1.5A is too pointy with a capacitor input and to do the same p-p ripple with the transformer on hand a 300uF cap would be needed. ugly!! Believe it or not a 673 or 575A would handle the charging peaks, barely. The solid state rectifiers will be used though. I apologize to the MV tubes for this, but I want to make this project with a more modern power supply because it will be a new thing around here.

Below are three Duncan simulations, all for 4KV output using the 6.7KVA CCS transformer. The same transformer is used for each, but the primary voltage and therefore the HV can be considered to have been set by a perfect variac.

A 10H or better choke would be best, and, a C input filter is unreasonable because of huge peak currents which can't be good for the transformer. Perhaps it is the transformer's high impedance in relation to the DC load that makes this so nasty.
This is no different than any other old transformer with a relatively high impedance compared to one purposely built for a doubler in a so-called modern amp.

By high impedance I mean not so much the DCR measured on the transformer secondary, but one with but one where the impedance (which can be calculated by the Duncan program) is larger compared to the minimum load resistance than others.

explanation:
Transformer designed for FWCT choke input service:
KVA = 6.7 (stated on nameplate)

V_Winding (secondary-one side of CT_ = 4430
 -(whatever this means, it is the part of a secondary winding in use at any half-cycle)

I_Max according to VA/Vwinding is then 6700/4430 = 1.55A maximum current
I_Max=1.55A

R_Load_Min = V_Winding / I_Max = 4430/1.55 = 2858 Ohms
 - (this would be the resistive load that would max out the transformer)

Z_Transformer = 43 Ohms
 (either calculated by Duncan or simply by hand, this is the impedance of half the secondary in a C.T setup or the full secondary in a bridge or a doubler set up)

so, a factor, "F" (for lack of a better name), would be R_Load_Min. / Z_Transformer.

2850/43=67   .. F=67.

This is maybe typical of an old style transformer intended for a full wave center tapped choke input filter.


Next is a transformer apparently designed for doubler service

1760V
DCR 10 Ohms
1.53A (I presume this is meant to be the DC output of the supply)

explanation:
KVA = 2.7 (volts * amps from nameplate)

V_Winding (secondary-one winding = 1760
 -(whatever this means, it is the part of a secondary winding in use at any half-cycle)

I_Max according to VA/Vwinding is then 2700/1760 = 1.53A maximum current. However being a doubler, the current is twice because the winding voltage is half (and is subsequently doubled).
I_Max=3.06A


R_Load_Min = V_Winding / I_Max = 1760/1.53 = 1150 Ohms
 - (this would be the resistive load that would max out the transformer)

Z_Transformer = 18 Ohms
 (either calculated by Duncan or simply by hand, this is the impedance of half the secondary in a C.T setup or the full secondary in a bridge or a doubler set up)

so, a factor, "F" (for lack of a better name), would be R_Load_Min. / Z_Transformer, same as before, BUT since the voltage is doubled, the secondary appears twice over a cycle, and is in series with itself in alternating half cycles. For this reason the Z_transformer should be doubled in this equation and the following figures should be used.
1150/18 (but double the 18)
1150/36=32   .. F=32.

To partly soothe the argument that this string of assumptions exceeds the apparent 2.7KVA rating of the doubler-service transformer by a factor of 2, it is worth it to say that this may be the SSB-service rating. The transformer is not as large as the FWCT example but each transformer delivers the same voltage and current in the appropriate circuit if duty cycle is put aside. Furthermore, if the doubler type unit were rated for CCS duty it would be at least as heavy and large as the 6.7KVA transformer above, probably heavier.

So this factor "F" which compares the resistive load required to max out the transformer to the value of the transformer's impedance, should be as low as possible for a C-input filter.

What is written above is illogically composed and I am sure the arithmetic is right and the math is twisted beyond reason but the result makes sense.

"F" for a FWCT style plate transformer is high like 67.
"F" for a doubler service transformer is low like 32.

The math is specious but the simulations give reasonable outputs to run a 3CX3000 with the only real difference being the DC output voltage and that is due to the secondary voltage differences and filter differences.
1750 * 2 = 3500V

(doubler transformer sim. in next post)

[Opcom]

doubler at 4360V/ 1.5A

[KA3EKH]

That’s a lot of work for building a power supply! I just tend to build with what's on hand and go from there. Not smart enough for power supply simulations. I got a huge filter choke from a 3 kW Bauer that I would gladly give you but have no idea on how that can happen being it weighs  around seventy five ponds and in order to ship something like that they usually send them in a wooden crate by truck. We use to have a RPS terminal in town and that was good for shipping such things but that closed a couple years back and the nearest is at least a hour away. Anyway the Bauer ran 3.4 to 3.6 kV around 1.5 amps with a solid state full wave bridge. Don’t know if it would help but can get the DC resistance if you needed it. From what I recall it was a choke input filter with a 12 uf cap. Or if you can figure some method for getting something large, heavy and hard to pack from the east coast to the west let me know.
RF

[KA2DZT]

I agree,  if you use a choke input filter with at least a 8-10hy choke it's hard to screw things up.  I built many HV supplies and never used a computer to build them.

An input choke helps to limit the in-rush current so long as you don't try to use huge capacitors.  For a plate modulated xmtr about 30-40ufd is all you need.  You will still need a step-start.

Stop looking at the simulator and find a proper choke.  You can put a few chokes in series to get more inductance.

If you use a LCLC filter you can run the modulators off the first choke and run the PA through both chokes.  The first cap doesn't have to large, about 15ufd and use the 30-40ufd on the second choke for the PA.

Fred

[W3GMS]

Most chokes will saturate during turn on without any inrush current protection. Only the line impedance along with the DCR of the xmfr and choke will limit the inrush if no other protection is designed in. A good filter cap at turn on when voltage first hits it will look like a very low impedance and then the choke looks like a piece of wire!  The only thing that would save this if all the DCR's when added together are high enough at turn on to limit the current to an acceptable level.  Sometimes one lucks out in that there is enough DCR. This tends to be harder to achieve with high current secondary transformers.    

Variac's in the primary help due to added DCR even if they are preset at the desired voltage.  

Joe, W3GMS    

[KA2DZT]

Joe,

I did say that he will still need a step-start that limits the in-rush current.  A lot has to do with the iron that's being used.  Older HV xfmrs usually have a little more DCR in their windings.  I guess big BC iron may have much lower DCR.  Big BC chokes usually have low DCR also.  It always helps to use the right iron for the job.  Using 2amp xfmrs and chokes in supplies where maybe only 600-700ma may be pulled could be a problem.  Some additional resistance might be needed.

Fred

[W3GMS]

Hi Fred,

My statement was originally directed at you, but I changed it to not address anyone specifically.  Some folks believe that if they use a choke or two that everything is fine from an inrush point of view.  I know you realize differently! 

Hit the turn on at the peak of the AC cycle and the current can be extremely high.  Even the old BA 100W gear, I at least stick a thermistor in for a cheap solution.  I really like the "step start" as you mentioned.

Fun stuff.....

Joe, W3GMS

[KM1H]

If the iron doesnt groan when hit with the full input the "resistance" is sufficient.

As far as determining the transformer impedance the shorted secondary method is simple and quick IF you know the rated secondary current spec.
Place a ampmeter across the secondary and a variac at the input. Inrease primary voltage until the secondary is drawing rated current. Then its Vp/V where VP is the voltage used in the test and V is the primary rated voltage.

If you dont know the current rating then one method is to plot input V and current and pick the point where the linear progression drops by 10%. Ive found this particularly useful for determining the actual rating of various WW2 Mil Spec iron and also to get a feel for how an old BC or other CCS transformer will perform for SSB/CW.

Perhaps others have another method or comments about the seat of the pants accuracy of my method which is directed to those without a lot of fancy test equipment, PC software, or trying to do complex calculations without taking their shoes off.

Carl