optimum equalizing resistor value for electrolytic cap stack.

[N3DRB The Derb]

OK. I got all my parts in today to wire the new HV line in the Gonset.

Is there a quick way to determine an B- to B+ optimum value rx string?

There are 5 caps. Gonset used 100uf @ 450 with 100K 2 watt carboniums. 5 of them in series string. So they had 20 uf.

Later on when they went to using 572B's, they added another 100K @ 2 watt in parallel so they had 50K @ 4 watts across each cap.

I have 30 of each size, 100K, 200K, 220K, 270K. All 1% metal oxide 2 watts. I'd like to know some underlying theory about what size should be picked and why. I just dont wanna copy Gonset, I want to learn something.

The new caps are made in USA TCI's 180 uf @ 500 vdc for a total of 36 uf choke inpoot supply. B+ is around 1600.

In all cases, selection should be based on achieving munky swingage and slopbukit p0wnage.   TIA.

[AB2EZ]

Derb

In order for the equalizing resistors to do the job... it is important that the current that flows in each equalizing resistor be at least 10x larger than the leakage current of any of the individual capacitors.

I.e.,  if you think of the leakage current of a capacitor as equivalent to a resistor in parallel with it, you want the balancing resistors to be at least 10x smaller than those "leakage resistors"... because, otherwise, the differences in the leakages of the capacitors will defeat the purpose of using equalizing resistors.

The capacitors you are using now are possibly better (lower leakage current at a given voltage) than the capacitors that Gonset was able to obtain years ago... so you may be able to use somewhat larger balancing resistors.

Keep in mind that the higher leakage current in the "worst" capacitor will cause the other capacitors to have a higher fraction of the total voltage across them. Also, as you know, the leakage current of one or more of the capacitors might increase with aging.

I suggest that you see if you can find a manufacturer's specification sheet, for the type of capacitor you are using, which tells you the leakage current at its rated voltage (e.g., 450 volts). If the leakage current specification is given over a range of temperatures, pick the highest leakage current. Then convert that to an equivalent leakage resistance (e.g. 450 volts / the worst case leakage current in the specification). Then pick your balancing resistors to be no larger than 10th of that value.

In the end, you should probably (very carefully) check the actual voltages across each capacitor, while in operation, to verify that the measured voltages are all reasonably close to each other. If there is no load on the supply, then even if the capacitors in the chain have different actual values of capacitance... the equalizing resistors should cause the voltages to all be the same (assuming the balancing resistors all have the same value). Therefore, with no load on the supply, if any capacitor has a significantly lower voltage across it... that implies that the capacitor has much too much leakage current... and it should be replaced. With full load on the (choke input) supply, the voltages measured across each capacitor should still be the same... even if the individual values of the actual capacitors in the chain are not the same. [This last statement would not be true for a capacitor-input supply][/font] Post script:The statement I highlighted in Red is not correct. See my other posts below... Stu

Separate from the above is the issue of how fast you want to bleed down the total stack of capacitors. If you want to bleed down the total 36uF with a 5 second time constant... then the total resistance should be around 140,000 ohms. Therefore... if the balancing resistors are also serving as your bleeder resistor... each balancing resistor should be no larger than 28,000 ohms.

Last of all, there is the issue of the regulation of the output voltage of a choke input supply. In order for a choke input supply to achieve its best voltage regulation performance... there is a minimum value of load that should be maintained. Sometimes this minimum load is provided by the bleeder resistor... which, in turn, could be the balancing resistor chain.

Best regards
Stu

[The Slab Bacon]

IIRC, the gonshit also has a pretty stout bleeder resistor across the whole stack as well. those resistore are most likely, just equalizing resistors, in an attempt to keep all of the voltages accross each cap somewhat uniform..

100k across each cap is probably not enough to keep capacitors from that day (they were all somewhat leaky back then) uniform in voltage. The current draw from the resistors (5x100k) should only be around 3.5 mils. Caps from that day could have enough leakage resistance to draw that on their own. (total draw from the whole resistor stack would be just over 5 watts @ 1600vdc.

Decreasing the resistor value to 50k will double the draw to somewhere around 6.5 mils. This increase in current will equalize the voltage accross each cap mich better as it will help overcome any inherent leakage current the individual caps might draw.
At 1600v the total draw from the whole stack will still only be around 10 watts. which is NOT enough to be much of a bleeder for total regulation.

My opinion is to use 50k accross each cap at that lower voltage (1600v). Since you have them, put a pair of 100k in parallel accross each cap (just like gonshit did) and you'll be OK-Fine!!

                                                      The Slab Bacon

[N3DRB The Derb]

yeah, it's got a 50K 100 watt across everything. My main concern was making sure of the balance across the whole string. When Aerotron took over Gonset was when the Mark IV came out and it was expressly changed to support the 572B tubes. That change in the string of EQ resistors is one of the few changes that were made.

Stu, thank you for the lesson. I got it!  Slab, theres not much left to do in it before blowsmoke testing into the termaline. I only have one more connection to make in the matchbox, put the cover and all 47 screws back on and thats it. The Yaesu is workin perfectly.

All of the underlying guts that make it go are ready....gotta install the measures mods, the tank coil, and hardwire the loading padder I need to make it draw current into a 50 ohm load.

Its amazing what you can get done when you are home 4 days a week instead of gone 5 days a week being babysat on a seizure watch.

[K1DEU]

I have used 120K  voltage dividing resistors successfully since 1957 across many different 450 Volt electrolytic condensers.

In my SB-220 and SB-200 and DX-100 (where I use a different modulator SG2 supply) this keeps the heat to a minimum!  Electrolytics prefer cooler operating temperatures.

These SB heath kits have voltage doubler power supplies with only C filtering. I use higher temperature higher one cycle current, Digi Key Panasonic 470 Ufd at 450 Working volt caps. Six in the 200, eight in series the SB-220 3,000 volt supplies and two in the full wave w/choke in 800 volt DX-100 plate supply. With this amount of C there is no trace of ripple on my AM or CW linear carrier and excellent regulation. I use my standard transformer primary clamp  http://hamelectronics.com/k1deu/pages/ham/general/pages/surge_suppression.htm and common relay/restive step-start.

For resistors I use 120K Ohm Matshuta/Panasonic 3 Watt flame proof ceramic type. Very interesting resistors that maintain their resistance when testing with a red color overload !  73  John

PS  Bacon, WA3WDR is home successfully from a little Hospital Stay, Yea !!!

[w8khk]

With full load on the (choke input) supply, the voltages measured across each capacitor should still be the same... even if the individual values of the actual capacitors in the chain are not the same. [This last statement would not be true for a capacitor-input supply]

Hi Stu,

Thanks for the details on calculating the equalizing resistor values, it is timely for me because today I am building the capacitor banks for my Valiant.  Could you explain the reason why the statement above does not apply to capacitor input supplies?  I knew the voltage out of a cap input supply is typically higher than choke input, but I cannot explain why the voltage across each cap would not be the same.  Thanks!

73,
Rick

[AB2EZ]

Rick

Sure...

The effect is probably not too great if the capacitors in the stack are approximately equal in value... but here is what I was referring to...

1. In a capacitor input supply... the load draws current constantly... but the diodes only recharge the capacitors during a brief period each half cycle (for a full wave or bridge rectifier design).

2. The current drawn from the capacitor string by the load is the same for all capacitors in the string, and thus the charge withdrawn in each half cycle is the same for all capacitors in the string.

3. If the capactors have different values (for example, in an extreme case: 100 uF +20% and 100 uF -20%) then the voltage change during the time when there is no recharging current supplied by the diodes will be 56% larger for the capacitor whose value is 80uF than it will be for the capacitor whose value is 120 uF...

i.e. change in voltage (dV) = change in charge (dQ)/value of capacitance (C)

Change in voltage across the 80uF capacitor = dQ / 80 uF
Change in voltage across the 125 uF = dQ / 125 uF

1/80 = .0125 
1/125 = .008

.0125/.008 = 1.56

Meanwhile, the balancing resistors will be drawing currents that are proportional to the voltages across their, respective, capacitors (which won't be the same, because the voltages across the capacitors are not the same for the reason given above)

Finally, the diodes, when they conduct, will supply an amount of charge equal to Q... i.e. the same as the charge that the load removes each half cycle.

To see how this actually plays out... attached is a simulation (SWCADIII) of a full wave, capacitor-fed supply with 2 capacitors in the stack: 80uF and 125uF; a 100,000 ohm balancing resistor across each of the capacitors, a 5000 ohm load, an ac supply of 1200 volts (rms), center tapped (850 volts peak on each side of center tap), and a series resistance in each half winding of 100 ohms.

As can be seen, the peak voltage across the 80uF capacitor (blue curve) is 465 volts
The peak voltage across output (red curve) is 775 volts
Thus the peak voltage across the 125uF capacitor is 775 - 465 volts = 310 volts

The last attachment shows the same circuit, with the capacitors changed to 100uF each.

As you would expect, with equal capacitors, the peak voltages across the two capacitor are equal: 390 volts (total 780 volts)

 

 

[w8khk]

Cool explanation!  So it is not only necessary to use matched equalizing resistors, it is also prudent to evaluate leakage currents and actual capacitance values of all the caps in the string. 

I am planning to use full wave choke input for the final, and capacitor input for the modulator of the Valiant, similar to the Turbo Ranger mod article by Keith, WA1HZK.  It appears the matching requirement is more critical for the latter application.   I would assume that the pulses of current drawn on voice peaks by the modulator would also discharge the capacitors unequally, similar to the way the diodes only charge the caps on peaks.  So I should also allow more voltage margin (more caps in series) in the bank for surges and unequal voltage distribution. 

My calculations for + or - 20% tolerance of 100uf caps range from 120uf to 80uf, not 125.  Minor point, I will measure all my caps and use the best matches.  Also more equalizer current than I originally planned seems to be in order, to avoid zorches.

[AB2EZ]

Rick

Just to see... I simulated a choke input supply (schematic attached) with the same values as I used before... except with a addition of a 10H choke.

The simulation (attached) shows:

307 volts (lower red curve) across the 80uF capacitor and 530 volts (upper blue curve) of total power supply output. Thus, the 125uF capacitor has 223 volts across it.

The other red curve is showing the current through the 10H choke... in the direction of the diodes (therefore it is negative in value).

It appears, therefore, that in this example, a choke will not (as I thought before I did this simulation) smooth out the charging current enough to remove the un-equalizing effect of the unequal capacitors.

I even simulated the choke input supply with an unrealistically high choke value of 1000H (see last attachment). The higher choke value obviously passes less 120Hz current to the capacitors ... but this appears to have no impact on removing the un-equalizing effect of the unequal capacitors.

[w8khk]

Thanks for the simulation of choke input supply.  I will match caps carefully for both the choke and cap input supplies.  I think I will borrow some of QIX's clip leads and bench test the thing before I start assembly.  I will also compare the voltages across each cap, in loaded and unloaded configuration.  Should be very enlightening!  With the variac, it will be interesting to see if the variations occur more at lower or higher voltages - depending upon how the caps "form"

[Opcom]

Here are a couple charts that shows an example of a power supply regulation when a LC filter is used versus a CLC filter, including the 'knee' at the Lcrit point. In this case the CLC filter is analogous to a simple C filter so it is a good comparison. The same rectifier and transformer was used for both. The charts are from an article on the effect of power supply regulation on peak RF power under modulation.

[N3DRB The Derb]

glad i asked. i learned a lot out of this thread.

[AB2EZ]

The more I thought about the simulation results for the split in the voltages across the two capacitors (late last night when things were quiet around here) the more the simulation results puzzled me. I know the SwCADIII simulation program is an excellent program... but it just doesn't make any sense (see below) for the voltages across the two capacitors to be different if the balancing resistors are the same .

If that were the case... the results imply that the average currents in the two identical resistors are not the same... which, in turn replies that the current flowing in one or both of the capacitors has an average value that is different from zero. [I.e., at the point where the two capacitors and the two resistors come together, the sum of all of the average currents has to be zero]

I always remind my students not to put blind faith in their simulations. The answers have to make sense, regardless of what the simulation produces.

So... this morning, I ran the program again (schematic and simulations attached below) with the unrealistically large choke value of 1000H (to ensure that the current is the choke had a very small 120 Hz component... just to keep things simple).

This time, I looked at both the voltages across the capacitors and the currents flowing through the capacitors  (top flat line is the voltage at the power supply output, bottom flat line is the voltage across the 80uF capacitor)

As I suspected... the simulation is showing about 0.5mA of average negative current in one of the capacitors  and about 0.5mA of average positive current in the other capacitor. [This means that between the two capacitors... there is a total of about 1mA flowing away from the common point.] If you multiply +/- 0.5mA x 100,000 ohms... then you can see where the voltage imbalance across the two capacitors (in the simulation) is coming from. [Why would the average current in one of the capacitors be negative... when the voltage across it is positive?]

So... the simulation results are not correct (i.e., they don't capture what is really happening)

I tried decreasing the time step size by a factor of 100... but that didn't change the simulation results (it just took longer to run)

Therefore... I think the simulator must be having problems with round-off errors... or something like that.

As of this morning .... my thinking is as follows

1. All of the waveforms have to be periodic, with a period of 1/(120 Hz) = .00833... seconds
2. Therefore, every waveform has to be some (linear) combination of: a dc component, a sine wave at 120Hz, a sine wave at 240 Hz, etc.
3. A capacitor cannot pass DC (unless it has leakage current)... because it would charge up forever.

4. Therefore, in reality, and ignoring leakage current, neither capacitor will have any average current flowing through it.... and the average current through each of the balancing resistors must be identical.
5. Finally, since the balancing resistors are assumed to be equal, the voltage across each of the capacitors must be equal, even if the values of the capacitors are different.

This conclusion (item 5) would be the same (again, assuming the leakage currents are small compared to the current flowing through the balancing resistors) for both a choke input supply and a capacitor input supply... even though I thought otherwise until last night.

 

[AB2EZ]

It turns out that we were both right: the SwCADIII simulator and my prior analyses (both yesterday and this morning... even though they came to different conclusions).

I was running the simulation for 3 seconds (and displaying the last 100 ms) in my prior postings (including earlier this morning)... thinking that 3 seconds was enough for everything to reach steady state.

But, it occurred to me on the train ride to Newark this morning that the time constants are much longer than that. I.e. 100uF x 100k ohms => 10 seconds.

So I ran the simulations for the capacitor input supply and the choke input supply again... using a 100 second simulation ... starting from the time the AC supply is turned on. [80uF and 125uF capacitors, and a 10H choke in the case of the choke fed supply]

The results are attached below (top curve is total supply voltage, bottom curve is voltage across the 80uF capacitor).

As you can see... the voltages across the two capacitors are unequal (the 80uF has significantly more voltage across it) for the first 10 seconds or so... and then the voltages converge to an equal split.

This means that every time you turn on the supply (e.g. if you key the supply on and off when you go from transmit to standby)... you put more stress on the smaller value capacitor.

Stu

[w8khk]

Thanks to Kirchoff, we can assume that the current flowing into any point is equal to the current flowing away from that point.  Running the simulation until voltages are static clears up the confusion.

It appears to me that a soft-start circuit would minimize the voltage difference on the unequal capacitances.  But with large capacitors, too large a soft-start resistor in the HV xfmr primary circuit could require too much time before the transmitter could be modulated.   So it is still rather important to select matched capacitors.

The simulation used a steady-state load of 5000 ohms, so after initial charge approximately 150 milliamperes is continuously drawn from the supply.  Voltage changes on the individual caps would probably be minimal during transmitter operation; when the modulator demands large peak current, the duration of these peaks would be rather short, considering the long time constant of the capacitor bank.

[Opcom]

Using Kirchoff's law, a wire size table, and a voltmeter, derive the current in each conductor.

[w8khk]

Ohm's Law and Kirchoff's Law have been repealed in that part of the world.  Electrons know this, and do not behave as expected.  In a maze like in the photographs, electrons get lost forever.  Also, there is no such thing as conservation of energy in those parts.  Ya gotta love it!