Need To Drop Minimum of 4V in Heater Supply

[DWhiteDenim]

Using this PT: https://www.tubesandmore.com/products/t ... -secondary
Heaters will run off secondary three 10.5V @ 2A. Build is a push-pull ECL84 tri-pent with 12ax7 preamp. Heaters datasheet shows total of 1.75A needed. Large watt low ohm resistor is one option. I ask this question on another amp site and one reply suggested to convert the 10.5 vac to vdc with diodes and then use a small step-down dc-to-dc buck transformer w/adjustable voltage output to get to (10%) 6.3v for the heaters. I asked him additional questions but no replies yet in over a couple weeks. My questions:

1) Two rectifier diodes or a bridge (four) diode to convert vac to vdc?
2) Convert vac to vdc and lower via resistor and/or LED light, or zener to 12v to run in series?
3) Convert vac to vdc to run at 6.3v in parallel using one of these to lower and filter voltage https://www.ebay.com/itm/224352155131 or https://www.ebay.com/itm/233395615599
4) If I go route #3 will the PT's secondary three 2A rating be compromised considering the tubes heater requirement + the buck transformer requirement?

[TUBEDUDE]

I'd get an appropriate tranny and ditch the buck transformer, rectifier, regulator, filter etc path

[luix]

You could start rectifing, levelling the filament supply with cap and measure it, it should be around 14V but usually is a bit less.

To be perfectly supplied the filament of ECL84 should be in parallel with 6.3V but in series they will also work with small tolerance and you could give it a try, guitar amplifier the distortion is not to be avoided...

Connect the tubes (ECL84 in series, 12AX7 for 12.6V fil) for short time and measure the voltage drop, since you are near the max current of 2A it could also be possible that you have 2V drop and perfectly match the 12.6V for the heaters, transformers are far than ideal and a load on the high voltage supply will reflect on the filament.

If you don't get 12V than you can use a simple resistor to drop 1V or so, il will dissipate from 2W to 4W so one or two parallel 5W resistors should be OK.

If you want to supply the tubes with 6.3V I'll prefer the SMPS way, it is smaller and with a frequency well over the audio band, let's say 40kHz, it will work properly.
Linear regulator will dissipate a power P = Vdrop*I so for a drop of 7V @2A it will dissipate 14W, SMPS usually have an efficiency of about 80% so with 7V drop and 2A the P will be 20% of the previous, only 2.8W.

[Stevem]

In terms of noisy troublesome magnetic fields I would want no added amount of iron in the amp but for the PT and OT, and a choke if wanted!

[sluckey]

Large watt low ohm resistor is one option.

Simple is better. A 2.3Ω resistor will drop 4V at 1.75A.

[DWhiteDenim]

I'd get an appropriate tranny and ditch the buck transformer, rectifier, regulator, filter etc path

I've built a few of those already and if this doesn't work out I will go that route and buy a proper PT. However, this is a low-dough, lower than normal voltage, experimental type build. I probably should have set my sights lower and went with a simple single ended build to reduce heater amperage requirements but if it doesn't work out that will become this PT's future use.

[Phil_S]

Large watt low ohm resistor is one option.

Simple is better. A 2.3Ω resistor will drop 4V at 1.75A.

FWIW, this would be my first choice, too. I'd get the biggest one that will fit, at least 10W. It will get hot and the higher the wattage rating the better. I'd make sure it is reasonably isolated from other parts or wires due to the heat.

[DWhiteDenim]

You could start rectifing, levelling the filament supply with cap and measure it, it should be around 14V but usually is a bit less.

To be perfectly supplied the filament of ECL84 should be in parallel with 6.3V but in series they will also work with small tolerance and you could give it a try, guitar amplifier the distortion is not to be avoided...

Connect the tubes (ECL84 in series, 12AX7 for 12.6V fil) for short time and measure the voltage drop, since you are near the max current of 2A it could also be possible that you have 2V drop and perfectly match the 12.6V for the heaters, transformers are far than ideal and a load on the high voltage supply will reflect on the filament.

If you don't get 12V than you can use a simple resistor to drop 1V or so, il will dissipate from 2W to 4W so one or two parallel 5W resistors should be OK.

If you want to supply the tubes with 6.3V I'll prefer the SMPS way, it is smaller and with a frequency well over the audio band, let's say 40kHz, it will work properly.
Linear regulator will dissipate a power P = Vdrop*I so for a drop of 7V @2A it will dissipate 14W, SMPS usually have an efficiency of about 80% so with 7V drop and 2A the P will be 20% of the previous, only 2.8W.

Sounds like good advice.
I think I will first take Sluckey's additional advice in the simple is better approach and buy the low ohm/high watt resistor he suggested to see what the voltage and hum level will be with VAC in parallel. If, too much hum then rectify, smooth, and drop it if need be to wire in series.

On AX84 a guy name Doug H built a Magnatone Melodier 109 (self-split) clone several years ago using either a 1:1 or an old transformer he had on hand. (IIRC) He used 25w variable resistor to control heater voltage and he said hum was noticeable but not excessive.

[DWhiteDenim]

Large watt low ohm resistor is one option.

Simple is better. A 2.3Ω resistor will drop 4V at 1.75A.

FWIW, this would be my first choice, too. I'd get the biggest one that will fit, at least 10W. It will get hot and the higher the wattage rating the better. I'd make sure it is reasonably isolated from other parts or wires due to the heat.

I am ordering parts through Mouser and the nearest is a 2.2 Ohm/25 watt in either a chassis mount wire-wound ($5.25) or thick-film type ($3.63) that can be mounted on a small terminal strip. Both are 5% tolerance. The thick film type max heat is 155 C and the chassis mount is 200 C. I going with the chassis mount for the cool (no pun) factor in the amp guts.

While on this build subject anyone opposed to using a through-hole 1000v/4A bridge rectifier in place of four separate 1N4007 or UF4007 diodes in a bridge configuration for the secondary one HV rectification?

[Phil_S]

I'd get the 200C resistor. The price difference is insignificant. There is nothing wrong with 4A bridge rectifier if you have a way to mount it appropriately. It will be a package with 4 leads.

[TUBEDUDE]

An SiC bridge is mighty efficient and generates lower noise, a short Trr etc.

[DWhiteDenim]

An SiC bridge is mighty efficient and generates lower noise, a short Trr etc.

Thank you. I will purchase a couple.

[DWhiteDenim]

I'd get the 200C resistor. The price difference is insignificant. There is nothing wrong with 4A bridge rectifier if you have a way to mount it appropriately. It will be a package with 4 leads.

Buying the 200C chassis mount wire-wound. I'm going to mount the bridge rectifier on a terminal strip. I bought an amp from a guy several years ago that he built and it uses a through-hole bridge rectifier. Noise is very minimal with that amp.

[R.G.]

If you have 4V to drop, there is a trick that may be simpler than some of these.
If you use a full wave bridge rectifier as a series AC element (that in at one AC terminal, out at the other), then whatever you put between the + and - terminals acts like it is an AC element outside the bridge rectifier. So you can put a zener between the + and - terminals of a bridge, and at the output, you get the AC voltage minus two diode drops plus the zener voltage. Example: 120Vac has a peak value of 169.7V from line to neutral, and in fact has peaks of +/- 169.7V on each side of neutral. Putting the AC terminals of a FWB in series with the line and just shorting the + and - terminals of the bridge subtracts two silicon diode drops of about 0.6V from this, so the voltage from the other AC terminal of the bridge has 1.2V subtracted from it and therefore has peaks of +/- 168.5V and is still roughly sinusoidal.
That's not a big issue on the AC line, but you can do things like inserting a zener, or an amplified zener between the + and - terminals of the bridge too. This subtracts the two diode drops plus the zener (or whatever) from the AC line voltage too. So you can literally dial down the AC mains voltage with a supposedly DC-only element like a MOSFET amplified zener.

For a heater winding at 10.5V, the peak is 10.5 * 1414 = 14.8V peak. A 6.3V AC winding has a peak of about 8.91V. Knocking about 6V peak off the 10.5Vac's 14.8 makes it look a whole lot more like 6.3Vac. Two diodes are already gone with the bridge, so it's 1.2V down already. Inserting a 5V zener gets the peak down by 6.2V, very close.
It's not quite the same as a real sinusoidal 6.3Vac, as it introduces "crossover distortion" with a dead spot around 0V, and the heating value of a cored-out sine is different from a real sine. But for heaters - it would probably heat just fine. And it's adjustable with a pot if you use an amplified zener approach.

[DWhiteDenim]

If you have 4V to drop, there is a trick that may be simpler than some of these.
If you use a full wave bridge rectifier as a series AC element (that in at one AC terminal, out at the other), then whatever you put between the + and - terminals acts like it is an AC element outside the bridge rectifier. So you can put a zener between the + and - terminals of a bridge, and at the output, you get the AC voltage minus two diode drops plus the zener voltage. Example: 120Vac has a peak value of 169.7V from line to neutral, and in fact has peaks of +/- 169.7V on each side of neutral. Putting the AC terminals of a FWB in series with the line and just shorting the + and - terminals of the bridge subtracts two silicon diode drops of about 0.6V from this, so the voltage from the other AC terminal of the bridge has 1.2V subtracted from it and therefore has peaks of +/- 168.5V and is still roughly sinusoidal.
That's not a big issue on the AC line, but you can do things like inserting a zener, or an amplified zener between the + and - terminals of the bridge too. This subtracts the two diode drops plus the zener (or whatever) from the AC line voltage too. So you can literally dial down the AC mains voltage with a supposedly DC-only element like a MOSFET amplified zener.

For a heater winding at 10.5V, the peak is 10.5 * 1414 = 14.8V peak. A 6.3V AC winding has a peak of about 8.91V. Knocking about 6V peak off the 10.5Vac's 14.8 makes it look a whole lot more like 6.3Vac. Two diodes are already gone with the bridge, so it's 1.2V down already. Inserting a 5V zener gets the peak down by 6.2V, very close.
It's not quite the same as a real sinusoidal 6.3Vac, as it introduces "crossover distortion" with a dead spot around 0V, and the heating value of a cored-out sine is different from a real sine. But for heaters - it would probably heat just fine. And it's adjustable with a pot if you use an amplified zener approach.

A 5V zener is all I need? Watts?