Balance resistor ratings/size for tube PSU

[middlec]

Put simply, I have filtering caps run in series that need appropriate balance resistors. The set up is this:

• Two sets of two 270uF, 420V(max) caps connected in parallel.
• Those two sets are connected in series
• In effect, I have 270uF rated for 840V
• Using the following formula:
  
  
• and with the values:
  • n = 2
  • Vrated = 420V
  • Vbus = 497V
  • Imax_D_leak = 3 x sqrt(270uF x 420V) = ~0.001A (assumes no leakage in one set and maximum leakage in the other)
• I get an Rbalance of 343K

Here's the thing, I see a lot of schematics that have somewhat similar configs. They typically use either 100K or 220K ohm resistors (Merlin's PSU book recommends 50/C, which gives a conservative 95K). Few give ratings, and they range from 1/4W all the way up to 5W.

IMHO, the nearest common value for optimal balance resistors would be 1/2W, 330K ohm. Unless I calculated wrong, or am missing something?

 

[rayma]

First worry about the voltage rating of the resistors. You may need two in series for each calculated resistor.
Then the power rating will be calculated for each resistor's actual drop.

 

[middlec]

That is a good point, To be clear though, the B+ voltage I'm targeting is Vbus, which is 497VDC. So each balance resistor should only see roughly half that (1% tolerance resistors).

That's still not nothing. My 1/4W resistors are only rated to 200VDC working voltage. They would indeed fail at some point. My 1/2W resistors are rated at 300VDC. The max power dissipation would be about 0.2W.

Perhaps I'm just being overly cautious, since I'm deviating from the usual 220K (or 100K) I see others using.

 

[rayma]

Ok, if the B+ is only 500VDC:

Using 2W resistors, with an actual dissipation of 1W and 250VDC across each,

that would be 1W = 250 x 250 /R ......... or R = 62,500 ohms minimum.

Then I would use 100k at 2W resistors rated at 300VDC or higher.
That would ensure good voltage sharing, and bleed 2.5mA.
Capacitor leakage will be much less than that.

 

[Philbo King]

The power dissipation of a resistor (in watts):
P = V²/R
V is voltage across the part
R is resistor value (Ohms)

The max. power rating of the resistor should be 2x or 3x that number to avoid frying it.

 

[thisusername]

Put simply, I have filtering caps run in series that need appropriate balance resistors. The set up is this:

• Two sets of two 270uF, 420V(max) caps connected in parallel.
• Those two sets are connected in series
• In effect, I have 270uF rated for 840V
• Using the following formula:
  View attachment 1247662
• and with the values:
  • n = 2
  • Vrated = 420V
  • Vbus = 497V
  • Imax_D_leak = 3 x sqrt(270uF x 420V) = ~0.001A (assumes no leakage in one set and maximum leakage in the other)
• I get an Rbalance of 343K

Here's the thing, I see a lot of schematics that have somewhat similar configs. They typically use either 100K or 220K ohm resistors (Merlin's PSU book recommends 50/C, which gives a conservative 95K). Few give ratings, and they range from 1/4W all the way up to 5W.

IMHO, the nearest common value for optimal balance resistors would be 1/2W, 330K ohm. Unless I calculated wrong, or am missing something?

You can always do that or you can use the cosθ derivative of a coulomb charge to define leakage, and derive a bleed balance resistor value.
Even though the best approach afterwards is purchasing the caps at the same time so that they will have the same leakage characteristics.

R = (NVmax−Vb) / (0.0015CVb)

where:
R the value of the balance resistor that will be individually across each capacitor in series (in meg ohms)
N is the number of capacitors in series
Vmax is the rated maximum voltage for any one capacitor
Vb is the B+ applied to the capacitors in series

 

[middlec]

Cool, thanks all!