Bridge-Totem Pole-Balance

[trem]

OK...First, let me address my failings...not the WHOLE list, that would take too many pages...just what is applicable to my question here.
1. I could not find it (not saying it isn't there) in my copy of Merlin's Power Supply book.
2. I cannot find a diagram of a bridge rectifier into a series reservoir cap.

So finally to my scenario. (just theoretical and round numbers)
A 500 Volt PT secondary into a bridge rectifier.
That rectifier into a 200-500/200-500 totem pole reservoir cap.

Assuming there is a 220k "balance resistor" across each cap...how do you estimate the wattage required for the resistors.?
Is it as simple as 500/440k = 0.0012A x 500 Volts = 0.6 Watts for each resistor.?
So a 1 Watt resistor across each cap should be OK.?
Thank You

[Enzo]

REsistors do not know or care where the current through them comes from. The only law they know is Ohm's Law. If you put 250v across one 220k resistor, I get about 0.28 watt. Your 0.6W is for two resistors, each dissipates half that. I'd use 1W for them even so. Or even 2w.


You just want an example? Stacked caps are everywhere. Here is a Fender:
http://bmamps.com/Schematics/fender/hotrod_deville.pdf

[trem]

Thanks Enzo _
I see the mistake I made.
With 2 caps in series...there could NOT be 500 Volts across each cap...AND 500 Volts across both caps.
There would need to be 1kV across both for there to be 500 across each.....assuming all else is equal.
Thanks Again

[Dave H]

...A 500 Volt PT secondary into a bridge rectifier.
That rectifier into a 200-500/200-500 totem pole reservoir cap.

Is that a 500V AC (RMS) PT secondary? The calculations will be different if it is.

[trem]

Hey Dave -
It is theoretical...but yes. That 500 VAC would be the number you would see if you put a meter across the PT Secondary.
I am not even sure if my meter could take that...I will have to check.
Sense the question has come up...if a meter says it can read 600 VAC...do you need to worry about peak to peak voltage...or just the RMS that the meter shows.?
Thanks

[g1]

The meter (or probes) would be rated for RMS unless otherwise specified. I believe the point DaveH was making was that for calculating the power or current in the bleeder or balance resistors, you need to use the rectified DC voltage, not the AC voltage of the PT secondary.

[Dave H]

It is theoretical...but yes. That 500 VAC would be the number you would see if you put a meter across the PT Secondary.

In that case the capacitors could charge up tp the peak voltage of 500V AC which is about 700V or 350V per capacitor so the power in the 220k resistors would be 350^2/220k = 0.57W

[trem]

The meter (or probes) would be rated for RMS unless otherwise specified. I believe the point DaveH was making was that for calculating the power or current in the bleeder or balance resistors, you need to use the rectified DC voltage, not the AC voltage of the PT secondary.

Got it, and.....

In that case the capacitors could charge up tp the peak voltage of 500V AC which is about 700V or 350V per capacitor so the power in the 220k resistors would be 350^2/220k = 0.57W

...got it.
Thank You Both.
I Appreciate It

[R.G.]

While we're at it

I usually assign balance and bleed resistors by deciding how much - OK, really how little - power I will put up with them eating. I decide that I could stand for a total of ... pick one... about 1/2 or 1W to be lost in the bleeders. From that, I divide the power by the number of caps being balanced (so this works for more than two) and get the power in one resistor. I know the voltage and the power, so the resistor value is then the voltage squared divided by the power. Pick the nearest standard value.

Pick a standard power that's twice the power you have alloted to the resistor for derating, and then check the voltage rating of your resistors. Most resistors have max voltage ratings too, often 200 or 300V, which is right on the edge of the voltage you'll usually be wanting in a tube amp.

If the voltage rating on the resistors you want to use is not enough, divide the resistance in half again and make each balance resistor be two of half the resistance in series. This doubles the voltage rating of the composite resistor, and coincidentally also doubles the wattage it can stand, as the power is now divided among two resistors.

You can play the same game with power. If you have, say, 1/4W resistors and need a rating of 1W in each balance resistor, make up each balance resistor out of four 1/4W resistors in series, adjusting the resistor value.

[trem]

Thanks RG -
I never even considered the max Voltage of the resistor(s).
I will have to look at the data sheet for a 1 Watt carbon film resistor.
Thanks again for the equations.
btw.....it is Duct Tape. The actual "Duck" brand came later.