HV supply load regulation issues

[WBear2GCR]

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.
<snip>

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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.  

T

Wholly Moley Batman!

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

                                 _-_-

[SM6OID]

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

[AB2EZ]

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

[K1JJ]

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.  

T

Wholly Moley Batman!

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

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

[AB2EZ]

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

[KA2DZT]

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.

[AB2EZ]

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

[KA2DZT]

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

[AB2EZ]

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

[AB2EZ]

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

[Opcom]

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

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."

[KA2DZT]

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

[Opcom]

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.

[AMLOVER]

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...
  

[W3GMS]

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   

[AMLOVER]

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...

[SM6OID]

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

[AB2EZ]

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

[KD6VXI]

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

[WD5JKO]

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. 

Jim
Wd5JKO

[AMLOVER]

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.

[AB2EZ]

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

[AMLOVER]

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.
 
   

[Opcom]

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?

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.

[KA2DZT]

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.