Heater resistor values to hit target filament voltage drop - iterative method

[Deke609]

I recently added EML 300B tubes to my Beepre and needed to adjust the heater resistor values to get the 5V +/- 4% filament voltage specified by EML.

After installing the EMLs, I measured 9.77V dropped across the filament and the heater resistors to ground, 4.65V across just filament, and 5.12V across the heater resistors (2 X 8R, for an effective R-vale of 4 Ohms). Knowing V and R, I calculated I (current) using Ohm’s law: I = V / R = 5.12V across 4 Ohms = 1.28A Knowing I and voltage dropped across the filament I also calculated the filament resistance: Filament DCR (FIL-r) = 4.65V / 1.28A = 3.63R

Since I wanted 5V of the 9.77V dropped across the filament, that meant I needed 4.77V dropped across the heater resistors (H-r): 4.77V / 1.28A = 3.727R. Using a parallel resistor calculator, I calculated that I needed to add approx. 54R in parallel with the existing heater resistors.

BUT, installing a 47R 50R (which I had on hand) in parallel only got me to approx. 4.86V dropped across the filament! Within spec of +/- 4%, but not that close to 5V (or slightly over as my math would have predicted). Through a process of trial-and-error (without math), I ultimately landed on adding both 47R and 150R 50R in parallel, which gave me 5.055V dropped across the filament.  Using the parallel resistor calculator again, this means my actual H-r value was 3.598R  3.565R

This confused me, until I realized that in reducing H-r, current must have increased and that increase could be calculated: 9.77V across revised R-total (FIL-r + H-r) =   9.77/(3.63 + 3.727) = 1.328A

Re-running the math for calculating H-r using 1.328A gets me very close to the R-value that ultimately worked: 4.77V/1.328A = 3.592R

To be even more precise, I figure you could do another iteration by recalculating current using the new H-r value (because in further reducing H-r a bit, current will have crept up a bit).

The above assumes that the 9.77V presented to R-total (Filament R + Heater R) remains constant. I figured that since filament voltage is shunt regulated in the Beepre, this assumption was fairly safe. But I think a similar iterative calculation could be done even if voltage would vary based on R-total – it would just involve more steps.

Does the above make sense, or did I just get lucky with a wrong method?

If the above iterative calculation method is theoretically and mathematically sound, it would make arriving at the final R-value easier, save time/money when ordering resistors, and hopefully cut down on the number times one needs to add/remove resistors.

cheers and thanks,

Derek

[edited to correct some misstated values; I misread my math scribblings]

[Paul Birkeland]

Did you double check the actual resistance value of the two paralleled 8 ohm resistors?  I wouldn't just assume that they are exactly 4 ohms in parallel.

I would suggest bringing the voltage just barely to within 4%, as more current will heat up the regulated filament supply.

[Deke609]

Did you double check the actual resistance value of the two paralleled 8 ohm resistors?

No, I just assumed they'd be within the spec'ed tolerance (2% I think) and used their stated values.

I would suggest bringing the voltage just barely to within 4%, as more current will heat up the regulated filament supply.

I want to get as close to 5 as possible b/c I think Jac has suggested that 4% is truly the outer limit and that for better performance/life, one should try to get as close as possible to the actual target.  So I'd prefer if possible to go only 2% low. 


Is there a way of modding regulator circuit to handle the extra heat? Higher wattage rated resistors, bigger heat sinks etc?


Many thanks,


Derek


ALSO: regardless of whether the regulator can be modded to handle more heat, does the previously posted "iterative method" make sense? I.e., dropping resistance, increases current, so you need to re-run the calculations to take the higher current into account?

[Paul Birkeland]

Realistically you would need to know the input voltage, output voltage, and actual desired current to re-balance everything.  When I say "everything", it could also mean adjusting the operating point of the 300B.  The regulator thermal issues are just part of the equation, as dropout voltage can also become problematic.  If you can increase the available input voltage doing into the regulator (by using this year's most efficient Schottky diode), then you may be able to nudge up the regulated voltage coming out of the filament supply, and this will increase the 300B filament voltage as well.

[Deke609]

Thanks PB. I'll look into the new diodes. And in the meantime, I'll just cut out the 150R 47R to bring things back down to approx. 4.86V.

[Deke609]

  If you can increase the available input voltage doing into the regulator (by using this year's most efficient Schottky diode), then you may be able to nudge up the regulated voltage coming out of the filament supply, and this will increase the 300B filament voltage as well.

@PB: can you please help me in identifying the new Schotkky diodes? There are a bizillion versions and I don't know what I'm looking for. 


Many thanks,


Derek

[Paul Birkeland]

Look for the lowest possible voltage drop at 1.5A of current passing through the device.  You'll have to dig into the datasheets.

[Deke609]

Will do. Thanks PB.

[Deke609]

@PB - I've found 2 types of Schottky diodes with very low forward voltage, but I'm unsure whether their other properties make them suitable/unsuitable.  Both are TO-220 AC, but I can make that work if they are suitable. I've not found an axial version with anywhere near as low forward voltage.


My first choice would be the attached "STPS20L15" -- it has very low voltage drop @1.5A even when cold -- going by the graph, it looks like < 250mV @1.5A @25C and approx. 175mV @75C -- for comparison, the 80SQ045N axial type that comes stock with the BP has a forward voltage > 350mV @1.5A @25C and 300mV @75C.

My second choice would be the attached "IXYS" diode.

Would either of these be suitable?

Many thanks again,

Derek

[Paul Birkeland]

Your selection of 15V diodes is making me a little nervous.  I wouldn't go below 20V.

When you find a promising one, order up a pair, then put them in place of the the two rectifier diodes in one channel, then use a variac to see which channel drops out (it will hum) first when you turn down a variac and start dropping line voltage.  Rinse and repeat with whichever diodes you find that seem promising until you have the winner, then you can look at what kind of DC voltage is sitting at POS_IN, then the POS_OUT can be adjusted by replacing Rset on each filament regulator board to make a little more voltage available for your 300Bs.

[Deke609]

Awesome! Thanks PB. This is turning into a real adventure.

[Deke609]

@PB: I will do the diode experiment this wknd and I have a few follow-up questions.


Diodes - do I need a heatsink?

Once you go 20V and above, it's hard find diodes with lower "on-paper" voltage drop. I found a few dual anode/common cathode ones, but only one single diode -- another IXYS in a TO-220 AC package (datasheet attached). The IXYS is 10A 45V rated, so a good replacement match for the 8A 45V 80SQ045N. On paper, the max forward voltage drop of the IXYS looks to be about 100mV lower than the 80SQ045N (0.2V versus 0.3V) @1.5A @125C.  Unfortunately, IXYS doesn't provide "typical" voltage drop figures - so the 80SQ045N might still turn out to be better.


Getting to my question #1: since (a)  voltage drop decreases with increasing temperature, and (b) the IXYS diodes will be used well below their rated current and voltage handling, I assume that they are unlikely to overheat w/o heatsinks and that I want to install them w/o heatsinks to get them warm and operating more efficiently.  Is this right?


Testing Procedure


Question #2: I am to use the variac to slowly lower the line voltage to the amp as a whole, right? I don't need to disconnect the heater circuit from the 6.3V secondary and then variac only the heater circuit.


Question #3: Do I need to send a signal through the amp, or will a dropped out channel hum even without signal? I am assuming that I don't need signal, but just want to check.


many thanks again,


Derek

[Paul Birkeland]

I wouldn't use a heatsink.  Yes, you want to lower the line voltage to the entire amp.  When the regulator drops out from low input voltage, the channel will hum on its own.

[Deke609]

Perfect. many thanks PB.