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Old 26th May 2006, 12:22 PM   #1
jarthel is offline jarthel  Australia
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Default regulated DC heater supply

thinking of something like this =>

http://www.national.com/images/pf/LM317/00906301.jpg

If I do use a similar circuit in the schematics, what goes to the heater pins? say for a 6922, pins 4 and 5 are the heater pins.
maybe the +V goes to pin4 and the ground goes to pin5?

also what happens to the regulated DC heater supply ground? would I be connecting it to the ground of the B+ supply?

Are there benefits to a regulated DC heater supply? Looking at the schematic above, the cost doesn't seem to be that much.

thanks for the help.

ps. Is the schematic I mentioned above suitable for a heater supply? if not, please point me to the right direction. Thanks again.
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Old 26th May 2006, 01:09 PM   #2
jarthel is offline jarthel  Australia
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I forgot to ask if I could use the heater secondary as my power source for the regulated DC supply?.

like heater secondary going to a diode rectifier then the schematic I mentioned above.

Thanks again.
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Old 26th May 2006, 01:52 PM   #3
Gluca is offline Gluca  Italy
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Not sure I understood perfectly your Qs. But I am using LMs quite everywhere to heat tubes.

That circuit, yes it works, but you need to adjust the resistors R1 & R2 to give 6.3V out (or whatever V you need). I am sure you looked at the NS datasheet for details.

I refer the heaters to the cathode of the valve and not to the B+ ground. It will not make a great difference, you just need to be sure not to exceed the heater-cathode max V diff (apllicable if you are using SRPP etc...). The circuit needs only one connection to ground or cathode!

You want to install the LM on an heatsink and make sure it is not dissipating too much ... I mean it should not drop much more than 3V. So if you need 6.3V at the output you will need 9V or so at the input. Can your 6.3AC be rectified to give 9V? Probably not under load. Just give it a try.
Ciao
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Old 26th May 2006, 01:57 PM   #4
jarthel is offline jarthel  Australia
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Quote:
Originally posted by Gluca
Not sure I understood perfectly your Qs. But I am using LMs quite everywhere to heat tubes.
Do you use the heater secondary as power source? or do you have a separate transformer?


Quote:
Originally posted by Gluca

I refer the heaters to the cathode of the valve and not to the B+ ground. It will not make a great difference, you just need to be sure not to exceed the heater-cathode max V diff (apllicable if you are using SRPP etc...). The circuit needs only one connection to ground or cathode!

could you explain what you meant by this? I can't understand what you mean.

Thanks for the help and patience.
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Old 26th May 2006, 02:22 PM   #5
Gluca is offline Gluca  Italy
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I use separate trannies (torroidal 12V for a number of reasons)

1) solder one leg of a 100R resistor to the first heater pin
2) solder one leg of a 100R resistor to the other heater pin
3) solder the remaining legs of the Rs togheter
4) connect that point (ie the center of the two Rs) to the tube cathode.
5) forget the ground connection you see on that LM circuit. The whole stuff is now referenced via the 2 R's to the cathode.

If you are using a double triode, just tie the heater to one cathode only and make sure the other cathode-to-heater voltage is not exceeding the max rating.

I hope it helps.

Ciao
Gianluca
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Old 26th May 2006, 02:35 PM   #6
jarthel is offline jarthel  Australia
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Default just remembered that...

I can I can combined secondary 2 and 3 to get 12.6 (xformer datasheet => http://euphoniaaudio.netfirms.com/ea...g/pdf/928P.pdf)

which should be enough to feed an LM317 (possibly lm350 for higher current rating).

My question is : If I combined sec2 and 3, what would be their current rating?

If could please have a look at the datasheet I mentioned above to see how sec2 and 3 are combined.

Thanks again for the help
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Old 26th May 2006, 02:43 PM   #7
jarthel is offline jarthel  Australia
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Quote:
Originally posted by Gluca
I use separate trannies (torroidal 12V for a number of reasons)

1) solder one leg of a 100R resistor to the first heater pin
2) solder one leg of a 100R resistor to the other heater pin
3) solder the remaining legs of the Rs togheter
4) connect that point (ie the center of the two Rs) to the tube cathode.
5) forget the ground connection you see on that LM circuit. The whole stuff is now referenced via the 2 R's to the cathode.

If you are using a double triode, just tie the heater to one cathode only and make sure the other cathode-to-heater voltage is not exceeding the max rating.

I hope it helps.

Ciao
Gianluca
steps 1 to 5 is somewhat clear. with step 4, the combined end of the 100ohm resistors is connected to the 6922 cathode (using 6922 as an example)?

Also where/how do I connect transformer output? I would assume it goes to the LM regulator but how is the LM regulator connected to the heater pins?

also, would it be same steps if I decided to use the heater supply secondary instead of a separate transformer?

what capacitor values do you use for the LM circuit?

thanks again for the help
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Old 26th May 2006, 03:10 PM   #8
Gluca is offline Gluca  Italy
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Sec2 and sec3 are rated at 1.5A. If you series connect them you can get 1.5A out of them (and 6.3+12.6V).

Sec3 is rated for 12.6V that, once rectifed, will give you 16V or so: it is too much for the LM and you need to drop the extra voltage through a resistor or use an heavy heatsink under the LM. So you can use sec3 (forget its CT).

You need to connect Vout (the upper leg of the regulator in your first jpg) to first pin of the heater and the lower leg to the other pin. Order doesnt matter. Combined end of the 100R resistor tied to the cathode.

I do use LM1084adj and 10 or 22uF electrolytic for local bypass. Diodes are followed by 10000uF, 1.5uH, 10000uF.

Ciao
Gianluca
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Old 26th May 2006, 03:18 PM   #9
jarthel is offline jarthel  Australia
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Quote:
Originally posted by Gluca
Combined end of the 100R resistor tied to the cathode.
1 last question: Is there a reason for this?

thank you very much. i really appreciate the help
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Old 26th May 2006, 03:49 PM   #10
Gluca is offline Gluca  Italy
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OH well! Doing so the heater will stay at a voltage close to the voltage of the cathode and will not float around.

You want to use 100R resitors or bigger as they are in parallel to the heater but only a little current (coming from you regulated supply) will flow through them as their resistance is much bigger than the heater resistance. Most of the current will go into the heater.

Keep in mind the whole stuff will cost you > 20 euros (not including the trafo you have already in your hands): rectifiers, large can C, inductors, PCB, heatsink, LM, resistors. And you are populating the amp with more compenents and increasing the probability to have something going wrong.

The advantages (for a indirectly heated valve) are often questionable. If you do not have any issues with hum (coming from heaters) ... well ... it's up to you ... just experiment and have fun.

Ciao
Gianluca
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