DC Filaments, any cons?

[olddawg]

Right, I've got an old Magnatone amp that runs a 12ax7(4>5), 35Z5 and a 50L6 in series with a 10w 150 ohm resistor and 6V Pilot lamp across the AC line. The thing cranks when it's turned all the way up, with surprising little hum.

Hopefully it has an isolation transformer and not a hot chassis. Could be a death trap.

[loudthud]

Looking at the schematic, the heaters look like they are all in parallel running on 6.3VAC from the green wires on the transformer. One side is grounded. You might try running a twisted pair of wires to each tube with a virtual center tap made with a pair of 100 ohm resistors.

[black_labb]

if someone wanted dc heaters on the pre getting a switch mode power supply that supplies 9v at the rated current and then use a voltage regulator, or just a 6v supply which is within 10% of 6.3v

you could just take apart one of the supplies that plug into the wall and connect that to the mains within the amp.

[olddawg]

if someone wanted dc heaters on the pre getting a switch mode power supply that supplies 9v at the rated current and then use a voltage regulator, or just a 6v supply which is within 10% of 6.3v

Or just get a small 9v transformer, a bridge rectifier, a 4700uf cap and a 6v voltage regulator. If you put a diode on the ground connection of the voltage regulator it would raise the output to 6.7vdc.

[Diablo]

Don't kill me but I want to stray this topic slightly. Rather than use DC heaters, is it a trivial idea to use diodes to drop the heater voltage on part of the amp in order to use 5V tubes in some sockets and 6V tubes in others? Could four diodes (1N5408) - two opposite polarity pairs on each side of the filament wires accomplish this? Any problems that I'm not seeing?

[loudthud]

Using diodes is not a bad idea but using a resistor will work even better because you can fine tune the resistor value. It's best to measure the resulting voltage at the tubes no matter how you drop the voltage. When diodes are used, you will need a True RMS meter because the resulting wave is not a pure sine wave.

Using a resistor you should start with a slightly higher than required resistance, then add resistors in parallel to bring the voltage up to the required voltage. Note that some "5 volt" tubes actually are rated anywhere from 4.2V to 5.4V so check a tube manual for actual voltage and current requirements. Example: 5EU8 4.7V 0.6A.

[R.G.]

Do you mean the spiral filament? And do you mean that they cancel from one half of the tube to the other?

That was the design idea, yes. The heaters in the original versions of the duotriodes and perhaps others were spiral wound and/or connected in such a way that there was some cancellation of heater induced hum from one heater spiral to the other. I've never seen any data on how effective this is versus non-centertapped and/or non-humbucking heaters. And I don't know if modern tubes have humbucking spiral wound heaters or not. It may or may not still be being done. It's very much a physical implementation thing, not an inherent part of the design.

I believe the centertapped idea (Hey! We can use this in series for 12V or parallel for 6V!) may have come first, then the cancellation idea (Yeah! And we can wind them to cancel hum!) second. This last is just supposition on my part based on reading very old technical ephemera about tube design. Kind of reading between the paragraphs as well as between the lines.

On diode drops:
Semiconductor diodes have a characteristic voltage drop that depends on the chemical/physics characteristics of the material. Silicon of any kind runs about 0.5 to 0.7V. Germanium runs about 0.2-0.3. Silicon carbide runs a few volts, as do LEDs, which are made from various mixtures of III-IV semiconductor materials.

In single-element systems (silicon and germanium) one can use a metal conductor for one side of the junction and get about half the otherwise-inevitable diode drop; these are the Schottky versions of the diodes. I don't know if this works with the Gallium Arsenide junctions and other stuff in LEDs or not. Silicon Schottky diodes are about 0.3-.5V forward drop when conducting.

Getting down to 0.1 requires a germanium schottky, I think. I've never seen one of these commercially produced, so it may not be practical. But you *can* get lower than that with a synchronous rectifier. This uses the drop across an active device, bipolar or MOSFET, to get down to very low voltages. There is no particular limit to the saturation voltages on these things, so you can have forward drops of under 100mV if you want, but it requires being able to drive the base/gate of the synchronous rectifier devices with enough drive to saturate them and at the right time. This technique is used in high current switching power supplies, but it's otherwise quite expensive and carries a lot of overhead.

As a practical matter, a 6.3Vac filament supply produces the actual AC voltage times 1.414 (conversion to peak voltage) minus twice the drop of the rectifiers used (for full wave bridge connections) minus any ESR losses in the filter caps, minus any ripple voltage.

6.3Vac comes with a tolerance. It's generally +/-10% from the wall voltage tolerances. But depending on the age of the amplifier, the "6.3Vac" MAY have been designed to produce 7Vac under light load, and sag to 6.3V nominal under maximum filament loading. And the transformer MAY have been designed for as little as 110Vac in and be running on 125Vac in, producing a "6.3Vac" output on the heater winding of as much as 7.875Vac with little or no load.

That makes the peak DC voltage out be (7.875*1.414)-1.4= 9.73Vdc from which the ripple voltage comes. This can also be as little as (6.3*0.9*1.414)-1.4 = 6.61Vdc minus ripple voltage for a heater winding that's really putting out 6.3V with a 10% low AC line and very little transformer losses. A standard 7800 series regulator needs 2V to regulate from, so the high voltage version of this can, possibly, work a regulator, the low voltage version cannot.

The ripple voltage is also an issue. You can reduce ripple voltage to any amount you like as long as you're willing to keep buying capacitors. The ripple voltage on the filter caps is Vripple = I*dt/C where I is the DC load voltage on the filter cap, dt is half the power line period and C is the filter cap. To keep ripple down to 5% or so, for each 300ma duotriode, at 60Hz, you need C = I*dt/Vripple = 0.3*0.0086/(0.05*8V) = 0.00645F or 6450uF per 300ma heater load.

As you can see, keeping ripple down can require massive filter caps.

Low dropout regulators (LDOs) offer some help. These things generally only have a 0.2V voltage overhead requirement, so you can regulate with only 0.2V higher than the regulated output. That converts the voltage budget to needing 6.3V(dc output)+0.2V(dropout overhead) +Vripple +Vrectifiers all divided by 1.414 to change peak DC back to ACrms. We can get 10% ripple easily enough, and 0.4V schottky rectifiers, so we could get down to (6.3+0.2+0.63+0.8)/1.414 = 5.6Vac, and that is then practical to run on a nominally 6.3Vac heater winding, even sagging a little and running on a lowish AC power line.

This analysis is why there is controversy on making DC filament supplies - the devil is in the details. It may or may not work, depending on how you set it up and what you power it with.

[Diablo]

Using diodes is not a bad idea but using a resistor will work even better because you can fine tune the resistor value. It's best to measure the resulting voltage at the tubes no matter how you drop the voltage. When diodes are used, you will need a True RMS meter because the resulting wave is not a pure sine wave.

Using a resistor you should start with a slightly higher than required resistance, then add resistors in parallel to bring the voltage up to the required voltage. Note that some "5 volt" tubes actually are rated anywhere from 4.2V to 5.4V so check a tube manual for actual voltage and current requirements. Example: 5EU8 4.7V 0.6A.

Thanks Loudthud, you da man again.
My brainstorm is to build a double deluxe using 5V6GT tubes.

[guitician]

The spirals in the heater filaments I'm looking at (TeleFunken 12ax7)are going in the same direction. I always thought that the coil just reduced the stray capacitance from heater to cathode. I've seen two types of heater spirals. One has the coating completely covering the coil, the other, larger coils, are open with just the wire being coated with insulating material. Running then at 6.3VAC has them 180 degrees out-of-phase inside the cathode, but they aren't wrapped together, like wires from the transformer to the sockets, usually are.