HUUUUUMMY new build

[bnwitt]

Loudthud is saying the resistor will do both.

[loudthud]

The purpose of the resistor is to drop the voltage. If you had 10VDC and the line voltage dropped 10% you would have 9VDC. With 10VDC you need to drop 3.7V at 900mA that's 4.111 ohms. (note 6.3V/.9A = 7ohms) At 9VDC you would have 9V*(7/(7+4.111))=5.67V (90%). If you had diodes dropping the 3.7V, 9V-3.7V=5.3V (84%) It's true that diodes don't drop a constant voltage but a resistor provides a little more cushion.

Plan on this though. I just did a quick experiment with an EH 12AX7. The cold resistance was about 10 ohms but at 6.3V it is closer to 20 ohms. Any circuit needs to be designed to withstand a 2X current surge at startup.

If you have a full wave bridge rectifier with caps on plus and minus to ground you might want to have half the resistance on each side. There will be some ripple that may cause a buzz type hum. A C-R-C filter would reduce the ripple.

[dehughes]

Wow. Okay. I'm going to have to re-read that one...thanks man. I wish I had a greater understanding of all this...

[dehughes]

So, I think I'm getting it:

If I have 9.56v as my supply DC voltage right now (without tubes installed), a 3.6k 5w resistor would, with a .9A draw (one 12AX7 and one 6SN7), drop about 3.24v, right? This, combined with the drop caused by installing the tubes, should put me under the 6.3v mark...probably just under 6v once the tubes have warmed up, right?

If I ordered some 3 ohm and 3.6 ohm 5w resistors, I'd probably be set, yes?

[loudthud]

As I suggested in post #44, you'll need to trim the resistor to get the voltage up to where is needs to be. Once you see what the heater voltage is with 3 ohms, solder some 10 or 22 ohm resistors in parallel with the 3 ohm (reducing the effective resistance) to bring the voltage up. If you get to three 10 ohm or six 22 ohm, you might as well solder another 3 ohm in parallel after removing the others. Keep trying until you get to atleast 6.3V.

[bnwitt]

Keep trying until you get to atleast 6.3V.

I would think that anything between 4.5 and 6 volts DC would be good enough considering that it's the AC peak to peak being measured at 6.3VAC. With the equivalent rms/DC being 4.5 VDC for the same power, anything in that range should be ok, right? I believe the Marshall DSL401 I just had in here measured 4.5 on the DC heater supply.

[loudthud]

The equivalent power of 6.3VDC is 6.3VAC RMS by definition! If you want I can explain the calculus. You can run the heaters on less but your gain will go down (not always a bad thing). RMS is used because it make ohms law and power equasions work. 1 VDC across 1 ohm causes 1 amp to flow and 1 Watt to be dissipated. 1 VAC RMS (2.8V pk-pk)causes the same 1 Watt to be dissipated in a 1 ohm resistor.

[bnwitt]

The equivalent power of 6.3VDC is 6.3VAC RMS by definition! If you want I can explain the calculus.

Look, there is no need to shout or be condescending. That must be where "Loudthud" comes from. I'm just kidding, as I hope you were as well.

On the technical side, it is my understanding that the 6.3 vac design voltage listed for tube amp filaments is a peak to peak value. Not RMS. Maybe I'm wrong there( and if so please correct me as I certainly don't know everything and always want to learn), but most AC meters being used today and certainly VTVM's from the era of tube amps do/did not read accurate RMS voltage. That being said (and if it's correct cause I'm not really sure what the original designers were referencing), the actual RMS value or DC voltage equivalent of that common peak to peak value is 4.5 volts which generates the same resistant heater power in the tubes. There is certainly no need to explain the cosine function or root mean square to me either as I understand that just fine as well as the other trigonometric functions. It's not calculus by the way. If it was I'd have no clue.

[erndawg]

6.3 VAC filament voltage is an RMS voltage. In fact, unless explicitly stated otherwise, AC voltages are assumed to be RMS. If it is a peak or peak-to-peak voltage, it is referred to as such.

If you place your Digital Multi-Meter leads across the filament pins, you are reading VAC RMS. If you place an o'scope across those same pins, you will read 17.82 VAC pk-pk.

Regarding modern and older vintage meters for measuring AC, they do in fact measure RMS. And regarding accuracy, I guess it depends on what you consider accurate enough. Even the 1950s Heathkit VTVMs had a 5% tolerance. Inexpensive meters today are typically less than 1% tolerance.

As far as I know, there is no calculus involved in deriving equivalence in thermal heating (how the NBS measures power) between RMS and DC. I'm an EE with two engineering degrees and took 6 calculus courses. But maybe I slept through that class; I certainly slept through many. ; )

[bnwitt]

Hmmm, interesting, the math I am familiar would take 6.3 VAC RMS times 1.414 and come up with 8.9 volts peak voltage. What am I missing? Are you simply saying peak to peak instead of peak?

[loudthud]

Most tube data sheets specify heater voltage "AC or DC" and give a voltage. To get the same heating value, RMS voltage should be used.

Without getting into a really long winded explanation of how we get the numbers, I'm just going to say that using the RMS value of a sine wave is handy because those numbers make Watts law work. For DC, 1 volt across 1 ohm causes 1 amp to flow and 1 Watt to be dissipated. For AC, 1V RMS across the same 1 ohm resistor causes 1 amp RMS to flow and 1 Watt to be dissipated. In the AC case, the instantanous power is 2 Watts at the peak of the wave (1.414V*1.414A) but zero at the zero crossing. So using the peak value of an AC voltage doesn't give a true picture of how hot things are going to get.

Most VTVMs are "peak responding RMS calibrated". That means they rectify the incomming waveform, and store it in a capacitor exactly the way a power supply does but the reading is calibrated as if the input was an ideal sine wave and the load is resistive. Many inexpensive DVMs do the same thing. Higher priced DVMs have circuitry that actually takes the waveform into account and gives a True RMS indication. It tells us the DC voltage that would produce the same amount of heat in a resistive load as the incomming waveform.

Here's a little more in depth, but prehaps confusing, answer:

http://music-electronics-forum.com/s...ead.php?t=4590

[bnwitt]

Loudthud,
thanks for that meter explanation. I wasn't aware of that info.

[dehughes]

Sweet. Great info. Thanks guys...much appreciated.

So, I'm going to string a few 1N4007 diodes together and get a voltage reading (hopefully about 1.5v less...), then wire the supply up to the sockets and see how much the tubes drop. Then, I'll order resistors to replace the diodes accordingly...

[erndawg]

Hmmm, interesting, the math I am familiar would take 6.3 VAC RMS times 1.414 and come up with 8.9 volts peak voltage. What am I missing? Are you simply saying peak to peak instead of peak?

Yes, I am saying peak to peak (pk-pk) just as you'll note in my quote below.

If you place your Digital Multi-Meter leads across the filament pins, you are reading VAC RMS. If you place an o'scope across those same pins, you will read 17.82 VAC pk-pk.

An o'scope shows the true waveform; it does not do any mathematical manipulations such as convert to RMS. It is the absolute waveform. If you hook up that same scope to a wall outlet, you will see 339.36 VAC pk-pk. A Digital Multi-Meter (DMM )will read 120 VAC, which is RMS.

Most DMMs just rectify AC, filter it with a cap, (just like any AC to DC power supply does) and calibrate it with some resistors.

[dehughes]

So, here's what I've found...

I strung three 1N4007 diodes together in series after the DC cap(s), and with the tubes in and warmed up, I'm getting about 8.5vDC before the diodes (this being down from 9.56vDC...thus indicating the load of the tubes), and about 6.07vDC on the tube filament pins. SO, this is great! That's just where I want to be.

Regarding hum, it was a bit hummy still, so I put in ANOTHER 10000uf cap, and boom...all I hear is the sound of the PT humming along, just like when the 6v battery was installed. Perfect! All it took was 30000uf and DC rectification....

So, I assume then I'll need to find a resistor that would drop the same amount of voltage as the diodes. If memory serves, the diodes drop .7v each, so I'd need a resistor in the range of 2.1 ohms, yes? Probably about 5w?

Lemme know if I'm doing the math correctly...I'll place the order from Mouser tomorrow. I'm thinking nice silicone coated wirewound power resistors....

Also, any recommendations on a regulator (just in case)? I know people have recommended some previously, but figured others might have opinions. What about the LD1085 type? Or is there a better model?