The AM Forum

Hi to all folks,

The HV supply consists a transformer with sec at 3850vac/2A. Fully bridged, C input with smoothing cap 30uf/6kv (13 x 390uf/450v) and some ceramic .01uf/10kv. No load 5500v.

Problem is :
If I have 0.3A idling current the voltage drops 10% to 4950Vdc and if I have 1.1A key down the voltage drops another 20% to 4000Vdc.

The mains 230v drop to 220V which is 5% decrease and so is obvious that the rest -25% comes from the HV supply issues.
Transformer's secondary has low R around 7 ohms and the wire is more then sufficient for 2A. It also weighs around 60lbs - 25kg

Another strange thing is that during the modulation peaks - speech - the mains ac current in the primary of the transformer moves to lower from 20A to 16A as it also happens to the dc current from 1.1A to 0.9A.

Any ideas about improving the HV supply and avoid that 25%-30% voltage drop in full load before I add more turns in the secondary of the transformer in order to get the desired 5000Vdc at 1.1A?

Adding capacitance seems too simple to be helpful ...

Hi!
Just to verify facts,  may I suggest that you check transformer secondary voltage.
Load vs no load voltage.

SM6OID,

I can check it only by vdc panel meter. I can only see 5500v no load, 5000v with 0.3A and 4000v with 1.1A. I can't check the ac sec voltage with no and full load.
It is difficult to find ac meter for that hv sec (3750vac).

Make a voltage divider out of a series of metal film resistors.  That will allow you to see secondary voltage.  Or, spend 30 to 50 bucks on eBay and get an HV probe.

Also, measure primary V Drop.  My last 3cx3000 project was dropping 10 pct on the primary.  The secondary had the same amount of V drop. Because of that, not much could be done on the amp side to make sure the plate and fil voltage didn't sag, except increase the current capability on the input side.

If your primary isn't sagging 20 pct under load, then look at the resistance of the secondary.  If it's high, not a lot can be done to fix that, either.  More filter C, or CLC filter maybe.....?

Need a bit more info.

--Shane
KD6VXI

KD6VXI,

Secondary R is only 7 ohm, primary drops 5% on full load so no big problem.
Secondary voltage dropping - which I will try to measure somehow - or capacitance increase could be a matter of thinking.
I see few topics and find some HV supplies with as much as 150uf of smoothing capacitance but this could mostly influence on ripple percentage and not on supply's load regulation.
I should also mention that the amp's service is as AB1 linear. The voltage's drop however is noticeable before modulation occurs.
Fortunately I have 3 phases mains and filaments plus other supplies are supplied from a different phase in order to be stable.
 

Regulation with cap input filtering is always poor.

You're not going to see 5KV @ 1amp with that xfmr.

Fred

Fred,

I agree with you. I was wondering though if I should firstly invest on capacitance or on adding more ac voltage. I will definitely go first for an extra 1000 vac in order to achieve the 5kv dc and I'll see later if I need more smoothing capacitance. I had few problems in the past with L input hv supplies and so it is out of my plans.

Stefano

It's not a great insight, but ur drawing 1500watts from the wall and you are seeing a % drop.
The typical "rule of thumb" for CCS operation is a 10% drop.
That's for the supply as a whole, and also for the iron without rectification/filtering.

I'm a bit concerned about the voltage drop on the AC mains with a 1500w draw.

HV regulation has always been a challenge for me.  I ran into the same issue where the 240 primary dropped only 5% but the DC HV output dropped  MUCH more and ruined the overall regulation.

You certainly want to get to the bottom of this because a linear amp running class AB/B will have its IMD go to pot as a result. The idle current helps, as well as the carrier on AM to provide a "bleeder current" but if the regulation is really bad, it will simply drop further.

I've found that the solution is to first, use a HV transformer that is over-rated for the job by at least X2. There is nothing wrong using a big MO-FO transformer to supply the filter its needed pulses.

Next, a choke input filter will help regulation a lot. This means planning ahead since you will see about 0.9 X your AC HV.  

Next, think about this: What if you used a near-INFINITE sized filter capacitor? Let's say a tera-farad.  :-)   Once this capacitor was charged, under full load there would be virtually no DC HV drop at all for a LONG time. So, certainly, using as big a filter capacitor value as possible will help your regulation.

I like the suggestion of using a resistive divider or getting a HV probe to get in there and determine exactly where most of the existing drop is now coming from.

But generally, if the 240 VAC mains is at least #6 (and preferably bigger), the trouble is either caused by IR drop in the transformer wire, insufficient core size, too little C, rectifier drop when using hollow state or not using an L-C filter.  Watch the drop across the inductor too.

For my 4KV HV supply, I use a pair of paralleled 5 KVA pole pig transformers, a 200 uF oil-filled capacitor bank, a 30H large choke that is switchable in and out and 6A solid state diodes. It took a long time of dicking around, but my overall HV DC regulation is now down to under 8% from no load to full load.  

Once my rig is keyed and the bleeders, idle current or AM carrier puts on the initial load, there is barely any drop under modulation on both ssb or AM. That is the critical issue: How much does it drop under modulation? How much does the audio cyclic rate affect regulation? The less drop, the cleaner the signal will be. (Better linearity, less compression and less flat topping effects).

T

60lbs for a xfmr trying to deliver over 5kva might be a little light.  Could be the iron is lacking.  I have xfmrs that are only 1.5kva that weight more than 60ibs.  If you load down the supply for any length of time, how warm does the core get.

I think you have too little capacitance to maintain good regulation.  Do you have any bleeder current and if, how much??

Only real way to get good regulation is to use a choke input filter and a xfmr that has a much higher secondary voltage.  With choke input you need close to 5800 vac secondary voltage to end up with 5KV DC after filtering.

Fred

Another thing,  3850vac @1amp puts a nearly 20amp current in the primary.  What size wire is in the primary??

If your iron is found lacking, get a three phase one as you say you have it available.  Much better regulation, lower ripple, higher efficiency.

They are usually cheaper and easier to find than single-phase iron of comparable power.  Look for one from old induction heating equipment.

73DG

Not all capacitors are the same. If a cap input supply is what you want choose capacitors with very low ESR. 1 or 2 capacitors with good ESR ratings and sufficient voltage capability is far better then the common 8, 10 or 12 stack.

WBear2GCR,

In the case of idling current of 0.3A and 1500w consumption I get 5% drop in the mains which is acceptable but I get a total of 10% drop in the dc output of the supply. This is what I should cure...

Tom,

It's true that supplies must be twice stronger then our needs. I think that the super cap, 200-300uf, would be helpful in the modulation peaks like it does in the B class audio amps that I also use in the high level mod transmitter. Things become of course worse with modulation peaks and instead of going a bit positive the dc ammeter goes 200-300ma back.

Fred,

60lbs would be enough at least for the idling 300ma but even there there is significant drop. I will add anyway some 50-100uf, I use 13 x 47k/10w resistors one to each cap for discharging and nothing else as bleeders. Choke input needs huge ac voltage and when there is no load the voltage goes very high. it is dangerous as far I always keep the hv on. The pri wire is 2.7mm and the sec is 1.1mm. The core becomes very warm after 15-20min of continuous transmitting with 100% music modulation.

W7TFO,

Many folks here are using a 3 phase transformer with 3 rectifiers and one cap around 10-20uf - successfully. They have almost no hv dropping and very little ripple for such a small cap.
Some others though, needing higher voltages, use 3 phase mains but with 3 individual one phase transformers. They make 3 different supplies, one for each phase, and all the 3 supplies in series. In this case they also suffer from serious voltage dropping. It seems that the 3 phase transformer is the easier way to make a very stable hv supply. I had spare this one and gave it a try but ...

WD8BIL,

I could test your indication before I order the 3 phase transformer.
I will try 2-3 oil caps in series with 30-40uf total capacitance and see the difference. I thought that uf are uf wherever they come from. Is it true that the oil caps are superior of the stacked electrolytics?  I only use the stacked electrolytics because they are newer and possibly less tired...

With no real load (13x47Kohm very little bleed) the caps are going to charge up the the peak AC voltage which is about 5500 volts.  That peak voltage will drop considerable with any amount of real load.  300ma will easily cause a 10% drop from the peak voltage. I'm surprised it is only about 10%.

That xfmr will never make 5KV@1amp with any amount of real world capacitance.  You're expecting an only 10% drop from the no load peak voltage to a full 1amp load voltage (5500V to 5000V) and that is never going to happen with the set-up you're using.

When looking at regulation you have to start with the DC voltage at the minimum continuous load to full load voltage.  Looking at regulation starting with the peak no load voltage is meaningless, especially true with cap input filters.

Fred.

3CX3000 can consume 4800V@2A which is 9600W. A 20KVA transformer is 2x the max load. A choke type filter is the easiest way to get really good regulation. The tests on this supply bear out the simulations for regulation. Regulation is about 5% with the resonant choke setup which also has the lowest transients. Adding a large choke and cap after it gives less ripple, slightly more transient, and about the same regulation. Using just a big 30H choke gives the lowest ripple, about 10% regulation, and quiote higher transients.
Note here that the load goes from 50mA to 2A and is 'snapped' on and off with a (indestructible transistor) switch. This would not reflect real use with voice so the transients would in practice be less. The point is that the choke input or resonant filter with an oversized transformer is what you may want if you want good regulation.

No worries if you use a step start. Mine uses a T/R keyed 10 ohm, 200 watt resistor that is in the 240 primary for 1/2 second and then shorted out. It does a few things.... first the surge is greatly reduced especially with my 200 uF cap bank and SS rectifiers. Though the choke, when in circuit, cushions this.

Second, the voltage never soars. It starts out at about 3300V for 1/2 second and then rapidly rises to 4KV. It never exceeds my target voltage.

Third, is safety. I just hate hearing that HV hum when it's on. The step start keeps it off 95% of the time, or however much I have the rig unkeyed.

Fourth, it saves on power - eddy currents and bleeder heat.  I don't feel as guilty when the big rig is on and I'm just listening.... :-)

A HV step start is a really nice item to employ in the shack.

T

Series resistance in the capacitors will limit how fast a capacitor charges with each voltage peak. Limiting this will improve your regulation. Some of the 450uf 450VDC caps I've seen coming out of wherever have extremely high series resistances. Stringing them in an 8 or 10 cap string just multiplies this resistance.

I'll second BIL comments on the caps.

I bought a cheapie ESR meter.

Wow.  Like the bird Watt meter, it's a heart breaker!

I would say the caps are 'bargain' electrolytics.  Or you have some REALLY puny wire interconnecting.

Looking back at your numbers, your DC voltage is sagging to within 150 volts of your transformers AC secondary output......?  3850 AC 4000 DC.  That's horrible.  Sure you don't have a crappy diode?

My last 3cx3000 has a 12kva Pete Dahl ccs rated xformer in it.  It will do 10kw out the snout, pep, for 8 minutes (longest song I had).  HV sags from 6kv to about 5500 under modulation.

This is in a high level xmitter?  If so, those voltage drops explain the carrier shift downward.

I'm afraid adding windings to increase the loaded B+ isn't going to help.....  Your still going to have horrible regulation.


Also, another thing to check.  All screws tight, from the wall socket to the plate choke? I2R means you'll have bad regulation if there is a loose connection.  And that regulation will get worse as the load increases (ask me how I know (stupid computer electrolytics with screw terminals!)). 

--Shane
KD6VXI

The voltage drops to 4KV with a 1.1amp load.  That's a lot of current to ask the 450 volt caps to supply between peaks.  At that load I wonder how much ripple is riding on the DC.  He needs to make voltage divider to lower the DC voltage to a lower level to use a DVM.  Measure some level of DC (say 500VDC).  Then switch the meter to AC and measure the level of the AC component.  Or, just use a scope.

At the 1.1amp load if the DC is still clean without any ripple, the problem is not the caps.  But, highly unlikely the DC under load is clean.

Fred

Hi!
3-phase is the path to walk, I have 400 V 3-phase.
One of my Harris RF-130 has got the 3-phase power supply, voltage drop is low, even if beeing pushed "above limits".

Once I tried primary regulation in a project, approx 1 kW level, worked like a charm. Design suggestion can be found in mid 90's handbook.

And, another thing to consider, with huge capacitors the stored energy will become as impressive as destructive.
Unless, the design can handle hard faults.

Last but not least, make sure that you have proper equipment to measure what needs to be checked.
Be warned, voltage is high and the stored energy can be phenomenal.

My set-up is the same like in the schematic below except the 2400uf/450v caps.
The results though are different :(
I had few 'bargain' 390uf/450v and thought like all of you that with less capacitance the possible negative will be a worse ripple % but definitely the real negative is the extra 20% dropping under load.
The schematic comes from w2dtc.com who supplies 3cx3000s and he has a very acceptable 10% drop under full load with the same transformer secondary ac as mine and the same cap input regulation system.
I use 24 x 10A10 diodes (1kv/10A) 6 per leg of a full bridge.
The supply in the schematic uses 150uf instead of mine 30uf. Can't believe that this is the problem of my extra 20% drop.
Even if Opcom's 3rd option seems simple and manageable, choke input needs a new transformer with at least 5500vac plus the 2.5H choke and care on controlling transients even with the resonant filter cap. In such a case the 3 phase transformer with smaller cap and no choke seems preferable.
There is soft start but for my set-up is more convenient to keep HV always on the plate.
The amp is AB1 linear with carrier indications 1.1A plate, 0ma grid and 30ma screen (bleeder's current). On modulation peaks, grid shows 10ua, screen 100ma and plate a horrible 900ma.
I will check the net for "primary regulation" it sounds interesting.

Stefano

Hi!
Stefano, when I get back home, I will look in the archive.
AT least I can find the original article. My "own" design documentation got lost years ago when I changed employer. Perhaps it was in QST I found the original drawings. DC regulation less than 1% is easily done.

SM6OID,

It will be appreciated, thank you.

I have attached my set-up and few meter indications.
The strange thing as I now realize is that when the dcA go down with mod peaks the acA in the primary go down, too and in the same time the dcV in the supply's output go a bit up as also happens to the primary acV.
In mod peaks the amperes dc and ac go down meanwhile the dc and ac volts go up...

Put your dvm across the bottom capacitor or two.

The caps / bleed / equalizing R will act as a voltage divider and allow your dvm to make real measurements.

Scale readings up by the amount of caps in series.

Do both the AC and DC test described by another post.  Reply back with results.

--Shane
KD6VXI

Move the ground on the cap string to chassis ground,  you have it going through the meter.  Redo your measurements.

Fred

Wholly Moley Batman!

Wocka wocka! Did ya build that stout enough or what??

                                 _-_-

Hi!
I'm home...
Have a look at QST 1987, May.
Pg. 23-25

Gentlemen

Looking at this from another perspective:

Each 60Hz cycle, the energy flowing out of the output filter capacitor has to be equal to the energy flowing into the output filter capacitor (conservation of energy).

If the DC (average) voltage is 4000V and the DC current is 1A... then (neglecting the effects of residual AC output ripple) the energy flowing out of the output filtering capacitor in each 60Hz cycle is 4000V x 1A x (1/60) seconds = 66.7 Joules. The size of the capacitor does not affect this (again, neglecting residual AC output ripple).

For the case of capacitor-fed power supply... power (energy per unit time) can only flow into the capacitor when the time-varying voltage being produced by the transformer (with respect to ground) on one side of one of the diodes is larger than the DC output voltage on the other side of that diode. I.e. one side of the bridge or one side of the full wave rectifier must be forward biased.

As, a result, the DC output voltage must be below the peak AC voltage being produced by the transformer by a large enough amount to allow current to flow from the transformer into the capacitor for a sufficient portion of each cycle to inject 66.7 Joules of energy.

The current that flows into the capacitor from the transformer during each cycle is time varying... and has a peak value limited by the series resistance of the transformer, the AC power source, and the diode(s).

If you want the DC output to be close to the peak of the AC voltage (i.e. close to 1.414 x the rms voltage) you need a very low series resistance... because the current will only flow from the transformer into the capacitor for a small portion of each cycle.

Stu

Ya know what the disheartening thing is about this for us Old Schoolers?  With today's new SDR technology and clever programming, someone can build a piece of crap badly sagging HV supply, add a dirty final tube beat to hell with low emission - then use Pure Signal and come out with superior -55 dB 3rd IMD RF results.

How's that for progress?

T

Continuing from my earlier post today:

The effective series resistance of the transformer is probably much larger than the measured 7 ohm secondary resistance of 7 ohms.

The effective resistance is: R(secondary) + N x N x [R(primary) + R(line)]

where

R(secondary = the resistance of the secondary = 7 ohms

N = the turns ratio of the transformer = 5000VAC/220VAC = 22.7

R(primary)= the resistance of the primary winding

R(line) = the resistance associated with the 220VAC line

If one assumes that [R(primary) + R(line)] = 0.25 ohms, then the total effective series resistance of the transformer's AC output is: 7 ohms + 22.7 x 22.7 x 0.25 ohms = 136 ohms

If the DC voltage is expected to be reasonably closerto 1.414 x the AC no-load secondary RMS voltage (I usually assume something like 1.25 x the AC no load secondary RMS voltage for a decent supply)... then current will flow from the transformer into the output capacitor (through 1 or more diodes) only around 20% of each cycle (or less). I.e. 1/5th of the duration of each cycle. The rest of each cycle, all of the diodes will be reverse biased.

If the average current being drawn from the capacitor(s) by the load is 1A, then the peak current flowing from the transformer, through one or more diodes, during the 1/5th of each cycle that current flows, must be more than 5 x 1A = 5A. This must be the case if we expect the charge flowing into the capacitor during each cycle to balance the charge being drawn from the capacitor by the load during each cycle. I.e. the energy being delivered to the load by the output filter capacitor during each cycle is equal to the energy being provided by the transformer to the output filter capacitor during each cycle.

If the peak current being supplied by the transformer is more than 5A... then the peak voltage drop associated with the transformer's effective series resistance will be more than 5A x 136 ohms = 680V

This is the reason that a low effective transformer output resistance is so important.

All of the above can be easily (and accurately) simulated using the program called LTspiceIV... which is available for free from Linear Technologies... and is easy to learn and use.

It is interesting and instructive to see how large the peak current being drawn from the transformer is... compared to the average current being provided to the load... as a function of the effective series resistance of the transformer.

Note also that, under load, the voltage across the transformer secondary is not a sine wave... and attempts to measure this voltage using a resistive divider and an ordinary voltmeter may lead to misleading results.

Stu

Stu

I think the turns ratio of this xfmr is around 16.7  (3850v/230v).  I'm not understanding why the turns ratio plays a role in the calculation of the series resistance.

What about the ESR of the capacitors,  seems that would have a great affect on the loaded output voltage.

Also, take a look at the schematic the OP posted on page 1.  I have an issue with the capacitor ground being returned to the amp-meter instead of directly to chassis ground.

Fred

Thinking about it,  I guess the turns ratio squared would be the impedance step-up primary to secondary.  Still not sure why that figures in the equation.

Fred

Okay regarding the correct turns ratio.

From the standard transformer relationships, when calculating the Thevenen equivalent circuit of a "black box" containing a transformer, whose primary side is connected to a voltage source with series impedance Z, and whose secondary side provides the output of the "black box"... one can move that series impedance to the secondary side by multiplying it by the square of the turns ratio.

I looked at the schematic. I think it is the high impedance voltage divider that returns to ground through the ammeter ... not the output capacitor.

Stu

OK, I now understand the turns ratio role in the equation.

The schematic is the last one shown on page 1, third post up from the bottom.  The cap (13x390ufd) is grounded to the amp-meter.  Also don't like the two diodes across the meter.  Thinking about what affect these two things have on the accuracy of the amp-meter readings.

Fred

Fred

The 3A ammeter will have a very low resistance shunt... that will provide the equivalent of a direct connection to ground... in the sense that the presence of the ammeter will not affect the behavior of the supply. The diodes across the ammeter will provide some degree of protection for the ammeter from current surges that pass through the load... but they should not affect the ammeter reading under normal operation.

The resistance of anything in the path of the current that flows from the transformer through the capacitors... including the effective parasitic series resistance of the capacitors, and the rectifier diodes... will add to the problem of low regulation. But their effects will be small if their total effective series resistance is much less than the effective series resistance of the transformer (which, itself, is dominated by N x N x the AC line resistance).


Stu

Fred
et al.

Note, again (mentioned in one of my earlier posts):

The current flowing in the primary (and the secondary) of the transformer will consist of two large spikes in each AC cycle [probably having a peak value of 5 (or more) x N x the average DC output current] that correspond to the two brief portions of each AC cycle in which the diodes in one half of the bridge or the other half of the bridge are forward biased... and injecting current into the output capacitor(s).

These 2 large current spikes in each cycle will each produce a large voltage drop on the input side (and the output side) of the transformer... due to the AC line resistance. However, the AC voltmeter on the input side of the transformer will not provide an accurate reading of the effect of those voltage drops... because it is designed and calibrated to measure the RMS value of a sine wave.

The peak amplitudes of these voltage drops (on the output side of the transformer) determines the regulation ...i.e. the drop in DC output voltage under load.

Stu

It's very common paradigm. In a different life style the old school harley biker builds his individualized hog from parts and his own craftsmanship, but the motorcycle-loving lawyer goes to the HD dealership and whips out the plastic for the best machine in the house and a lot of doo-dads to go on it. Either one goes down the road in style.

Neither is right or wrong  but each derives the pleasure from a different aspect of the activity. For my part, "Never use a 1U-sized solid state piece of equipment when a 6 FT rack full of tubes will do the same thing."

Thanks Stu,

Yes, I know the amp-meter has a low resistance shunt.  Just seems the caps should be grounded to the chassis, which is how I would do it.  It looks like the current that flows through the caps is by-passing the meter.  Wonder if that would cause the meter to possible read low.  As a result, the OP may be loading the supply to a greater current than the meter is indicating worsening his regulation problem.

OK on the diodes, I know they're there for protection of the meter.  I guess it would take a high current to create enough voltage drop across the meter to cause the diodes to conduct, he does have two in series.

Fred

This may be a clue to the issue. OK these are for 1.8A, so your current would be lower. put that value in in (). this is just for broadly looking at the situation. referring to the pictures:

with a 2 Ohm ESR on the 300uF cap the peak current is 252A (154A). 5109VDC, 59.6V ripple.

with a 2 Ohm ESR on the 30uF cap the peak current is 246A (150A). 5018VDC, 415V ripple.

with a 0.5H choke, 2 Ohm ESR on the 30uF cap the peak current is 81.5A (49.8A). 4232VDC, 249V ripple.

Bigger cap=good but from here it looks like the first two cases which are C input, and with those the mains is called upon to deliver a lot of peak amps and it has to be done without a lot of voltage drop.

Can your mains wiring deliver to the primary of the transformer a peak current of 150A without sagging during that peak?  

Can you use a current transformer or similar device to look at the primary current and at the same time observe the primary voltage? It may sag at the peaks. It will look like clipping or peak flattening on an audio waveform. A small sag is all that is needed to spoil the performance because the peak is when the charging takes place.

just my 2 cents.

Fred,

I thought that is common practice to ground the -dc of the supply due the ammeter in order to avoid the +dc hv side. The diodes are glitch protectors for the meter like it is published from Rick Measures in the attached page below. It works fine till now and never had a problem to any of the supplies till now. I use this design to all my supplies except the one for the bias of the AB1 amp because there I have to control microamperes and the bleeders ruin the precise indication.
ESR is a serious matter but the w2dtc in his set-up uses 450v common capacitors and he claims very good load regulation...

Stu,

The transformer's turns ratio is the factor that mostly increases the effective resistance but I can't avoid that as far I need 4kvac. In my set-up under your calculations the effective resistance is 77ohm, very high though.
In the schematic of w2dtc there is the same ratio like mine (16.7), let's suppose a much better secondary resistance because of Dahl 3ohm and a much better primary+mains resistance 0.15ohm again because of Dahl. His effective resistance then is approximately 3+16.7x16.7x0.15 = 45ohm. Could those extra 32ohms be the reason of my excessive drop? I am wondering if I use a 1500vac transformer with 5 ohms sec and ratio 6.8 under the same load my dropping % will be better than the one I have now? hmmm...

SM6OID,

I unfortunately don't have the QSTs but I will search for the source you recommended to me in the net.

Shane,

I have done that and get the same results but with 1/13 ac and dc indication on the meter. It is not accurate though because of many factors out of bleeders etc...
  

I always enjoys Stu's great analytical examination. 

If you have the liberty to find another transformer, then find a single transformer solution with a center tap and go with a conventional two diode scheme and utilize the CT as a return.  If doing that is an option, you would also have the liberty to find a higher voltage transformer so you could use a choke input filter with enough bleeder current to take care of the minimum amount of current through the filter reactor so it does not look like a cap input filter.

Joe-GMS   

Can you use a current transformer or similar device to look at the primary current and at the same time observe the primary voltage? It may sag at the peaks. It will look like clipping or peak flattening on an audio waveform. A small sag is all that is needed to spoil the performance because the peak is when the charging takes place.

Opcom,

Yes, I have a current transformer in the primary and I really see what you are describing. Every time I say 'oooooola' the pri ac amperes go from 20A to plus minus 16A with all the bad results you have referred.
Adding capacitance from 30uf to 300uf will fix the ripple but not the high demanding current 150A.
Adding a 0.5H choke and an even larger capacitance let say 60-70uf seems to solve the problem of ripple and drop, right? In this case I will however need a 5600Vac sec transformer in order to get 5000Vdc. Is that right or not?

W3GMS,

I want to avoid a new single phase transformer but I would do anything else in order to get acceptable ripple and drop before I order a 3phase transformer.
I can't however understand how w2dtc and many others with almost the same set-up have got less than 10% regulation and me around 30%. Do they have mains wiring that can deliver 150A...? By the way I have it but using all the 3 phases. My mains are 3x230V/50A or 380V/50A. As it is now I can only have 50A far away from 150A...

Stefano, I can scan the article and email you it as a pdf.
However, it happen in mid of next week.

Further clarifications:

A. With respect to the series resistance of the 240VAC circuit leading from the main circuit breaker box to the wall outlet where the equipment is "plugged in":

In the US, the National Electrical Code mandates that the voltage drop at every outlet must be no larger than 5%, when the full rated load is drawn from the outlet. The voltage drop is primarily caused by the round-trip resistance of the copper wire cable being used... and, for a given length of copper wire cable, and a given maximum RMS current rating, one can calculate the minimum gauge wire that must be used in the copper wire cable.

For example, if a 240VAC outlet is rated to deliver a maximum of 50A, then the RMS voltage must not drop by more than 12VAC when 50A is being drawn by a resistive load. 12VAC/50A = 0.24 ohms of resistance in the wiring between the circuit breaker box and the outlet.

This is why I used, in my example, 0.25 ohms as an estimate of the sum of the AC wiring resistance and the resistance of the primary winding of the transformer. The actual resistance may be higher.

The Thevenin equivalent series resistance on the secondary side of the transformer is obtained by multiplying the primary side series resistance (AC wiring resistance + primary winding resistance) by N x N... where N is the ratio of the transformer's output voltage to the transformer's input voltage. For a given input voltage (e.g. 240VAC) and a given output voltage (e.g. 4500VAC) it doesn't matter whether you use one transformer or two transformers in tandem. If the input voltage and the output voltage are specified, then N is specified. The only impact of using a different transformer (or two transformers in tandem) would be on the resistance of the primary winding and the resistance of the secondary winding... but the effect of those is probably dominated by the resistance of the wiring of the AC circuit that is powering the equipment.

B. With respect to modulation

If the amplifier is behaving as a linear amplifier... and if the mode of operation is AM...then the application of modulation will have no effect on the DC output voltage of the power supply. The filtering provided to remove 60Hz ripple, will smooth out any instantaneous changes in the power supply's DC output voltage associated with instantaneous changes in the current being drawn from the power supply (because of the modulation).

However, if the amplifier is not behaving as a linear amplifier (or if the signal driving the amplifier is not a proper AM signal with a constant short-term average value)... then the average (not instantaneous) current being drawn from the power supply could change with the application of modulation... and that would cause the DC output voltage to change with the application of modulation.

For example, if the modulation is causing the amplifier to 'flat top" on positive modulation swings, then the average current being drawn by the amplifier will decrease when modulation is applied. If the modulation is causing the amplifier to "bottom out" on negative modulation swings... then the average current being drawn by the amplifier will increase when modulation is applied.

In any event, if the output filter capacitor(s) is large enough to sufficiently filter out the 60Hz ripple... making the capacitor larger will have no effect on the linearity problems that are causing the average output current drawn by the amplifier to change when modulation is applied.

Stu

AMLOVER :

From http://amfone.net/Amforum/index.php?action=dlattach;topic=42633.0;attach=54997;image

That's through #6 wire.

I have too much VDrop, but it's what I had and it fits in the pipe, which I also had.

My 8877, with a 4kva ccs supply is fed with #8.

The 4-1000, with a 3kva ccs choke input supply is fed with 10 gauge.

What size is your wire main to the panel?

--Shane
KD6VXI

This is a little off topic, but....

I repair 3KW SS RF amplifiers as part of my job. These run off 3 phase 208vac, and use very little filtering off the line. The power factor at full load is just below unity (1). There is no need to handle massive AC line current surges since there is almost no energy storage in the shoe box sized amplifier. The Lumpy DC (about 5% ripple) made is followed by a precision H-Bridge 100Khz switcher that makes up to 200v at 25 amps (5KW).

My point here is that a unity power factor switcher to make 5KV at 1-2 amps is getting more and more practical when compared to needing pole pigs, and massive capacitor banks to essentially do the same thing, but with no active voltage regulation.

Back to the normal broadcasting.  :D

Jim
Wd5JKO

Stu,

Here we have 230v mains and we have 8% drop limit for full load.
In my case even the wires in the cable are awg5 or #16 the distance from the public electricity connection on the edge of the land to the radio is 170ft or 50m
If the drop during mod peaks was only in the sec side, as it is from 1.1A to 0.9, would be as you have mentioned a matter of non linearity or driver's lucking capabilities but in such a case the primary should go up as far energy should be kept equal in both sides of the supply.
    No mod : sec = 4000v/1.1A   pri = 218v/20A
mod peaks : sec = 4300v/0.9A  pri = 218v/16A
Beside the poor load regulation with dc output 30% drop, it is strange that on mod peaks the pri ac amperes drop to 16A.

Shane,

The mains wiring is 5 x AWG5 or 5 x #16, 3 for the phases, 1 for neutral and 1 for ground.

SM6OID,

Thank you for the pdf. It is very clear but it is a regulator that takes higher voltage 4kv to regulate at the 3kv level with 50v ripple from ma to amperes. I need a regulator that keeps the supply voltage regulated with not more than 10% drop.

Repeating what I have already said in earlier posts:

The measurement results  you are reporting for AC voltages  and currents are not useful/meaningful because the time varying voltage across the primary of the transformer and the time varying current flowing through the primary of the transformer are not sinusoidal waveforms.

For 50m of awg 5 wire, the round trip loss is 0.103 ohms. I can't tell from your post if the entire run is awg 5.


The effective secondary series resistance  of this run (not including the primary winding resistance of the transformer) is: 16.7 x 16.7 x .103 ohms = 28.7 ohms.

If the peak (not rms) secondary current in each 60Hz AC cycle is (for example) 5A, the peak voltage drop associated with this equivalent secondary resistance would be 144V. This is just a few percent of 4400V... so the effect of this resistance on the regulation would be just a few percent of extra drop at full load v. zero resistance.

However, if the wire size is not awg 5 for the entire run, or if there is significant additional series resistance (circuit breaker? fuse? screw terminal connection?), that would explain the poor regulation.

Try plugging in a purely resistive load that draws a known rms current... and measure the voltage drop (e.g. a toaster or a space heater)

Stu

Stu,

My mains situation is in the attachment. There are few interruptions with switches and inlets/outlets and for sure much more R in the mains.
It seems that the easier road me to go at this moment is to add one more supply (1400vdc) in series with the existed one in order to reach the desirable 5kvdc plus some more capacitance for a bit better ripple level.
Future improvement will be the 3-phase transformer and a second tube in parallel.

WD5JKO,

A switcher 5kv/1-2A is my dream but at the moment there is no production or even schematics around.

Thank you all for the help and the extra knowledge you have shared with me.
 
   

No, because the small value of the choke is just enough to broaden the charging pulses so that the peak current is lower. To have 5000V@1.1A, the transformer result with the 60uF cap is 4190V and 80V ripple. The 3850V transformer is pretty close.

A real-life experiment could possibly be done using a 'variac' as instead of 240VAC you could try 261VAC to see if it works. Providing the transformer could handle the 9% increase.

Only if using a traditional regulation-oriented LC filter are the much higher voltages like 5600V needed from the transformer to get 5KV out.

I purposely simulated the small value 0.5H in order to help only with the peak current issue. (300uF was tried in calculations out of curiosity only to see how it affected the rest of the parameters. I guess someone uses that sort of value but it's usually excessive.)

A low-inductance choke of some sort, rated for the current you want, would be easier to find and cheaper than some large value. 3 Phase supplies with choke input filters use such low values of 0.25 to 3 Henries but many people overlook them when searching through surplus, because they want the 'regulation' benefit of choke input.

The larger filter cap 60-70uF after a small choke gets about the same ripple result as the more usual large choke and small capacitor arrangement would. The only pitfall is that the smaller choke will not regulate as well at low currents as the larger inductance - but it is still better regulation than the straight capacitor input.

If the insulation of the existing HV parts is adequate, there is nothing wrong with putting a DC supply in series with the (-) end of the one you have now (my opinion only). It should have its own rectifier, small choke, and capacitor. It should be made so it draws only small peaks from the AC line and because it will be small like 500VA it could have a decent choke and not eat too much space.

I think you may need to add only 500V or so, once the mains peaks are shaved down to size. This could be full of errors, anyone can point them out.

You can stack supplies by simple connecting the diode output of the lower voltage xfmr directly to the negative terminal of the FWB diodes of the higher voltage supply.  No chokes, no caps.  Just the diode output directly to the other diode bridge.  If the higher voltage xfmr is a FW CT, connect the diode output to the CT.  The secondary of the second xfmr will be at a higher voltage above ground by the amount of the first xfmr DC output.

The filter caps off the second xfmr will have to have a higher voltage rating.  The filter caps have to be grounded to the chassis.  The OP has his filter caps grounded through the amp-meter which IMHO is a mistake,  ground you filters caps to the CHASSIS.

If you try the two xfmr set-up, your negative ground connection and amp-meter are moved to the first xfmr, either the CT or the negative terminal of the diode bridge.

Fred

After giving the cap grounding more thought,  I now thing the way it's grounded to the amp-meter is correct.  The power supply can be thought of as a two terminal unit, positive and negative.  In the positive the amp-meter would be between the cap (+) and the load.  Likewise, in the negative the amp-meter would be between the cap (-) and the load.

They can be stacked without filters but the filters isolate the power transformers behavior from each other. There is no real difference. It's a choice.

The xfmrs are isolated by the diode rectifiers.  I have done this with a lot of different xfmr combinations and never had any problems.

Opcom,

I also prefer to connect separate bridges and filter caps to individual supply and then connect them in series.
As far I don't have access to any supply simulator, could you please simulate for me the attached 3-phase supply with individual single phase transformers in order to get the rates for chokes 1,2 and 3 for a reasonable ripple 3-5%...?
There are 2 conditions, one is the 1.8A steady carrier and the other is 3.6A mod peaks for 50% duty cycle.
Thank you in advance.

There is of course this set-up which needs only one choke but I have never used it before and I don't know if it is efficient.

Yes, this is the correct way to do it.  I've done this many times using many different xfmr combinations and it works perfect.  The PIV rating of each FWB diodes only have to be rated for the xfmr they're connected to.  You can lower the voltage by simple turning off any of the xfmrs leaving everything else connected  You can step-start by turning on one xfmr at a time.  Doesn't matter which xfmr is turned on first.  Only thing is the voltage insulation of the secondary winding has to able to handle the higher voltage above ground.  This is for the xfmrs higher up the series.  All the xfmrs have to be able to handle the current load.

The secondary voltages of the xfmrs do not have to be the same,  any voltage combination will work. The highest voltage xfmr should be at the top of the series.  Use as much input inductance you have to improve regulation.

Do not put any chokes or capacitors between the FWB rectifiers,  just connect the FWB diode output directly to the next higher FWB negative terminal.

Fred

What might be the effect of one transformer having a higher resistance and it being in series through the diodes with a lower resistance transformer? The currents must be equal in all parts of the circuit, right? 50 Ohms secondary resistance is not out of line for a 500V 1A transformer. I concede that 10 Ohms shown is very low for a HV unit but I think these are worth considering if no filters between them. Not so much a problem as a performance detail.

Yes you can download PSUDII from duncanamps.com. It does not do 3 phase but it is free. Be sure to also download the many additional rectifiers file that users have made. It's a text file called rectifiers.txt. The "Perfect" rectifier I used is something I just rewrote. You will see how to do it. Here is my rectifiers.txt file  including that but it may not have the latest additions by others.

I'm a little puzzled by the fact that AMLOVER is proposing to feed each transformer's primary with a different phase of a 3-phase service. Therefore the various bridge diode paths will not be opening and closing at the same times. I've never simulated something like this, and I can't model it in my head.

Stu

I also gave that some thought but haven't concluded anything about it yet.

Fred

Adding xfmrs in series with just rectifiers works fine provided they're in phase.  Three xfmrs on three different phases probably will not add together to the full peak voltages.  When one xfmr is at its peak another is only at its peak (x sin60) and I think the third is also at its peak (x sin60).

So I think the voltage sum would be the peak voltage + (peak voltage+peak voltage)(sin60)

Stu, maybe you can look at this and figure it out, my brain is smoking.

Fred

Stefano, please send me a PM with your e mail adress.

Stu,

The supply I have attached in my prior post is working 2 years now under this attachment's set-up. It off course sags in mod peaks and after our conversation about how to improve a bit the drop I thought that adding some inductance and larger capacitance would improve it's performance.
The voltage with no load is 10850Vdc rectified from 3 transformers, one in each different phase, with a total of 7700Vac.
I soft start it through 3x10ohm/200w resistors switching on/off one transformer after the other by 50A ac switches. Never something strange happened till now except the sagging that is not that much like the smaller one phase supply for which I opened the topic.
I don't know if that sag comes from the difference of phases by 60 degrees but the DC voltage is there it should be 10850Vdc.
I only thought to save 2 chokes, to series connect only the bridges, to use only one very well isolated choke on the highest voltage point and put all smoothing caps in series on the end of the supply.

Fred,

I have never tried before to series connect only the bridges, even to lower voltage supplies and I am a bit suspicious on how it would behave.
Saving 2 chokes though is enough to give it a try.
I must now find the inductance I need.

Opcom,

Thank you for the simulator link. I will check it this evening.

SM6OID,

I will send you my email by PM as soon I find the way to send PMs here.

The supply attached to your last post allows each transformer and each bridge rectifier to operate completely independently of the other two transformers and the other two bridges.

Each transformer and its associated bridge rectifier can deliver charge to its associated output capacitor, at times and in quantities that are independent of the other two.

Therefore the phase relationships of the three primary AC feeds don't affect the charging process for each of the three output capacitors.

The D.C. output voltages add, and the residual ripple voltages add with relative phase offsets that provide a beneficial reduction (vs. having all three primary supplies in the same phase) in the total residual ripple at the output.

Also, the peak primary current is lower than it would be if all of the primary supplies were in phase (and driven by a single phase of the AC service), and this improves the regulation.

If you force the currents from each bridge rectifier into a single series current flow (by using the alternate design (i.e the single choke design), I think the phase offsets of the primary supplies will cause significant problems.

Stu

Putting xfmrs in series with just the rectifiers works fine provided the xfmrs are all on a single phase.

Three xfmrs on three different phases, I would agree with Stu, seems there would be too many issues.  Using separate filters as you pictured would be needed.

Fred

Opcom,

I have added a rectifier 15kv as you can see in the PSUD2's rectifiers database.
Which way I can add it to my supply simulation?
The program offers a small rectifier's database to choose from.
I couldn't add any rectifier from the user's database in the program choices...
I was worried if these chokes must be gapped or not as far their inductance is small.

Stu and Fred,

I will keep then my set-up (total supplies in series) improving it by 200uf extra capacitance and if possible 3 smoothing chokes with small inductance.

Not disagreeing with you about filter inclusion, by the way, because I know it works fine without filters; but always wanting to learn more I usually try to dig into the smallest details when different circuits are put forth.

I see that you added it online. Now, you have to download the file and put it in place of the rectifiers.txt that comes with the program.

BUT - I notice that you have a huge voltage drop as though it is a vacuum tube rectifier. That is not going to work and is going to mess people up. That unit should be removed from the library of the site. Please use the forum and tell them it was entered in error.

The short answer is that you can use the existing 575A vacuum tube model. It's a 15KV 1.5A mercury rectifier.

Otherwise you can use the file that comes with the program and add a 15KV rectifier to it. Carefully edit, then save, the rectifiers.txt file with windows notepad. You should first copy/back up the original rectifier.txt file as rectifiers-old.txt or something.

For a near-perfect diode rectifier, add this line as the first rectifier:
__perfect, SS, 0.01, 20, 9.353, 100000, 1000, 1000

For a stack of fifteen 1KV 10A diodes, use this line:
_1KV10Ax15, SS, 6, 5, 9.353, 15000, 100, 10
The "__" is not required but I used it to show to myself that it is one of mine whenever I open the file for editing.



===

As to the choke, it does not matter about the gap as long as the inductance is at the rated/desired current of 1.1A or whatever DC current you specify. It should be designed for rectifier filter use with appropriate voltage ratings. If the choke is gapped it may swing the inductance less as the current through it varies and be less likely to saturate. The point is that it should not saturate at the peak current because that is where its effect is wanted.

Opcom,

I have tried hard to find a connection or to erase my wrong data rectifier addition but no result. If you have a communication connection or link to the Duncan's webpage let me know.
Thank you.