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best way to elevate heater?

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greetings

i have two B+ lines, 630V & 320V and one 6V heater. i want to lift the 6V 100V above ground. does it matter if i use the 630V or the 320V B+? they share a common ground. can't see that it should matter and in the past i've never given it much thought, but what do you think? thanks in advance

berry
 
I can tell you what I do -
You have two resistors in series from HT to earth, and the junction is where you take the DC reference. If the HT is 300v I use approx 300K in total for the two resistors, i.e. 1K of resistance for every volt of HT. For 300k that's 0.3 watts so use minimum 1 watt resistors.
To calculate voltage out, start with the ratio of bottom resistor to top. If the top resistor is 250k and the bottom is 50k, then ratio = Rbottom/Rbottom+Rtop. In this case .17. Voltage out is HTxratio, so 300x.17 = 50v. You can get other voltages out by changing the ratio. I do this in Excel.
Then from the junction I take a 330 ohm resistor to each leg of the AC heaters, and from the same junction a cap rated higher than the HT (300+ volts) to earth. Cap size is as big as I can squeeze in - probably 22uF to 47uF or something like that

Hope this helps - seems to work.
 
Here are another 1000 words !

Check this drawing. Hope the connections are clear.
 

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Hello,

I drew this before I saw that others have already posted a schematic, but here it is anyway:

An externally hosted image should be here but it was not working when we last tested it.


This circuit elevates the heater to 0.13x the B+ voltage. So if you have 360V B+ the heater will be at 47V. The cap is sometimes required to prevent hum, you have to try this in your individual circuit.

Best regards
Michael
 
Hello,

There is a simple formula for that:

Looking at Tubemaster's drawing:
- assume that the 470K resistor is R1, the 68K resistor is R2 and the voltage (i.e. B+) is U:
The voltage Ux at the junction of the two resistors is calculated as followed:

Ux= U(R2/R1+R2)

If U= 400V then Ux would be: 400(68/68+470)= 50.55V


Hope this helps!🙂
 
After everything has been explained quite succinctly, just a cent's more:

If used to eliminate diode emission effects from heater to cathode e.g. in low signal stages, no more than 15 - 20V positive on heaters are needed. On the other hand, if there is say an ECC81 Schmitt (ltp) phase inverter with some 130V on the common cathodes preceded by an EF86, some calculation for a proper tap voltage will be necessary. My tube manuals give a max. value of 60V (heater positive) for the EF86 and 90V (heater negative) for the ECC81. Thus very close to 50V only will have to be provided by the "tap" if voltage variations are also to be allowed for, as well as manufacturing tolerances. This just as an example to what degree one needs to keep tube ratings in mind as suggested earlier.

Then I always use a bypass capacitor (tap-common) - some junk from the mains can find its way into signal circuits via heater-cathode capacitance, e.g. in the above example. It is cheap enough.
 
And to complicate matters even more, instead of resistors to each 'leg' of the filaments, connect your reference voltage to the center tap of the heater winding. That is if you have a CT! And if you do and you use the other schemes make sure it's not connected to ground i.e. your filaments are "floating", not grounded that is. 🙁
Except AC/DC ground referenced by the reference voltage point bypass cap/resistor of course!

Wayne 😉
 
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