Converting Filaments AC/DC

[lowell]

So my, fairly small, knowledge of filaments and wattage are telling me I can convert a typical 6.3vac heater supply to 6vdc just fine WITHOUT changing the PT. I'm toying with this for a tube mic preamp design to reap the benefits of reduced hum injection. Gonna use a full wave bridge. Only real issue I forsee is the need for either a series resistor, series diodes, or high power zener. I'm thinking resistor or diodes to avoid extra loading on the PT from a zener. I dont believe this changes the wattage required of the PT filament winding....as it's supplying the same wattage DC or AC. I also am doing this so I can use a PT I already have. (no 5v rectifier winding) I know a 5v winding would actually work perfectly without series diodes as in the following math. I will use a full wave bridge with a 4700uf capacitive input. (more than 2,000uf per amp)

2 12ax7 = 600ma
1 6v6 =900ma
Total = 1.5A

my math
Secondary voltage 6.3vac
Target DC Voltage 6.3vdc
6.3rms=8.9vPEAK=rectified dc voltage
Accounting for diode drops (.7v) 8.2vdc

If i add 3 series silicone diodes I should be at 8.2-2.1 = ~6.1vdc.

I'm probably missing something....what is it?

[MarkusBass]

You are missing that with full bridge rectifier you have two diodes in series and not one as you think. Also the assumption that after rectifying 6.3VAC you get 8.9VDC is optimistic. The bridge is loaded with 1.5A current and the DC voltage will be lower. The rest seems to be OK. Just try it and tell us what you get. Several companies do it (e.g. Mesa) but the results are not always perfect.

Mark

[Dave H]

If you bridge rectify and smooth the 6.3V rms winding it will probably be close enough to 6.3V DC (It is on my Valve Junior) but the rms current in the 6.3V winding will be greater. When I tried it in psud2 for 1A @ 6.3V DC output the rms transformer curent was nearly 2A.

[nickb]

Also, to get the ripple down to useful values you'll need a BIG capacitor e.g. 20,000uf will get you to about 0.5Vpp ripple and 2.6Arms in the transformer. The bridge rectifier will be dissipating around 3 watts and that may need a heatsink. If you were to use DC on just the input stage the hum reduction should be still be substantial but the transformer and diodes will be much less stressed. The capacitor can be smaller since you only need 300mA instead of 1.5A. The ripple is roughly proportional to current / capacitance.

[Leo_Gnardo]

A 2 stage filter would be a treat and allow you to select the DC filament voltage to taste with the R between C's.

If using DC filament supply on the same transformer that's supplying filament current to other circuits, sometimes you'll find the current spikes from that supply show up in your audio. Most of the time, no problem. I find it's a big problem in the Ampeg SVP bass preamp, making it useless as built due to buzz noise induced by the filament DC supply, very frustrating. Best to breadboard your circuit & make sure it's not a problem for you before committing to drilling up metal & fabricating a circuit board.

[loudthud]

Your transformer heats up according to the RMS current flowing in it's windings, not the average current. To illustrate how RMS and average are not the same, consider the following. If 1 Amp is flowing all the time in a wire, the average is 1 Amp and the RMS is 1 Amp. If 2 Amps is flowing half the time, the Average is still 1 Amp, but the RMS is 1.414 Amp. If 3 Amps is flowing one third of the time, the average is 1 Amp but the RMS is 1.732 Amp. See the spreadsheet posted below. In the typical full wave bridge rectifier circuit, the transformer peak current is 3 to 10 times the average load current depending on how low you try to get the ripple by jacking up the filter capacitance. You won't get an accurate picture of what the RMS current will be without using a modeling program that takes into account the source impedance to the transformer, the rectifier type and the ESR of the filter cap.

Note: Load power in the spreadsheet is calculated based on a 1 Ohm load.

[lowell]

Markus I dont see 2 diodes in series. To me it looks like each end of the secondary only pass thru one diode. Same goes for the negative swing. I'm sure you're right...but I'm missing it, please clarify.
https://upload.wikimedia.org/wikiped..._alt_1.svg.png

[nickb]

Markus I dont see 2 diodes in series. To me it looks like each end of the secondary only pass thru one diode. Same goes for the negative swing. I'm sure you're right...but I'm missing it, please clarify.
https://upload.wikimedia.org/wikiped..._alt_1.svg.png

Does this help you?

[J M Fahey]

You are seeing it:
start with the red left AC terminal, and voltage at the positive peak, current goes left to right through one diode , loses 0.7V (1 diode) , passes through the filament, comes back (blue) through a second diode, loses another 0.7V , returns to other end (bottom) of filament winding.

So it passed first through one diode, then through another (I don't care what else is in the path) you lost two diode drops ... diodes are in series.

Same on the negative swing.

EDIT: simulposting

[lowell]

Ok. Seems that the second diode happens AFTER the filament though. It's diode - filament - diode....so isnt the second one post heater therefore not affecting the heater? This is where electronics get hairy sometimes!! I think your answers will be that it doesnt matter the order since it's a circuit (loop).?

[J M Fahey]

Suppose it's DC just to go a little easier on the Math.
The transformer winding gives you 6.3V "DC".
If you connect it to the filament, it gets full 6.3V "DC".
Do we agree so far?

* Now you add one diode to the upper (red) wire and nothing else.
How much does the filament receive?
6.3V "DC" minus 0.7V "DC" or: 6.3-0.7=5.5V "DC".

* repeat but now the diode is on the lower leg, the blue line.
How much does the filament receive?
Again 6.3V "DC" minus 0.7V "DC" or: 6.3-0.7=5.5V "DC".
So now we know that it does not matter whether the diode is, as long as current has to go through it.

* repeat but now we use two diodes in series on the upper line.
How much does the filament receive?
Now 6.3V "DC" minus (2 x 0.7V) "DC" or: 6.3-1.4=4.9V "DC".

* and if we put one diode above, one below?
How much does the filament receive?
Again 6.3V "DC" minus (2 x 0.7V) "DC" or: 6.3-1.4=4.9V "DC" ..... we already know that each diode will drop 0.7V ,no matter where it is, as long as it is in the path.

Which is what we are trying to explain.

Note: I made a very simplified assumption because at this very moment we are not trying to actually design a DC filament circuit but to make a concept clear, that's the main point.

[lowell]

Thanks JM. I still dont get why this is... But clearly I'm lacking in understanding of a fundamental rule in electronics.

Here's an example. Say we have a voltage regulator...as in your typical PSU. Then we put a diode, cathode to ground between the regulator center pin and ground. Does this raise the voltage of the regulator .7v? Or reduce the voltage by.7v? Im pretty sure it raises it. Maybe I have it backwards.

[MarkusBass]

Your example with voltage regulator is not good because voltage regulator has three pins, is a highly non-linear component and you cannot easily apply basic formulae in this case (like Ohm's law). In fact the output voltage will be increased because you change the reference voltage for the regulator. But this has nothing to to with your initial question. If you have such a problem, you should always bring the problem down to something simpler, and not to something that is more complex (and is much more difficult to undersand).
I suggest that you consider two circiuts:
- the most simple one: a diode in series with 1k resistor, connected to 10V DC power supply. The voltage on the resistor is 10V - 0.7V = 9.3V. You measure the voltage between ground and the top pin of the resistor. But the ground is the same as the lower pin of the resistor (this is important). To be correct you should measure the voltage directly on the resistor.
- slightly more complex circuit: a diode in series with 1k resistor and again in series with another diode , connected to 10V power supply. On the resistor you have 10 - 0.7 - 0.7 = 8.6V. And this is exactly the case you have (only you have AC instead of DC and filaments instead of resistor). Which part of the explanation you don't understand?
Your intuition that the voltage will be increased is misleading you. It will be exactly the oposite.
You are making two mistakes:
- you assume that additional component (like the lower diode) behaves linear as resistor. But this is not the case with diode. With diode, not matter what voltage you apply, you get voltage between 0.6 and 0.7 Volt. This is also not the case with any other component like Zener diode, capacitor, inductor.
- you consider the voltage on the resistor is being measured between the upper pin of the resistor and the ground. But when you add a diode, the voltage should be measured between the upper and the lower pin of the resistor (and not to ground).

Mark

[lowell]

Perfect! Thanks Markus. Heres how I see it now

The voltage on the seconday, end to end, is whatever it is, MINUS 2 diode drops. So measuring across the secondary only, "inside the diodes" its the secondary voltage minus 2 diodes.

Using the resistor between 2 diodes gives my visual brain a perfect explanation. Thanks!

[Enzo]

An analogy: Imagine you are 6 feet tall, standing in a room with 8 foot ceiling. You could stand on a 1 foot box, and hold another 1 foot box on your head, and fill the 8 feet exactly. Or you could stand on two boxes with your head on the ceiling, or hold two boxes on your head and stand on the floor. In any case, you and the two boxes add up to 8 feet, no matter how you stack them.

A transformer or a battery, whatever the power source, and you stack a load (resistor, tube heater) and two diodes across it, each will have its own voltage drop, and you can order them any way you like, they still add up the same. Remember, this is a circuit, so there is no pre or post heater.