Tube heaters combination?

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briggs
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Post by briggs »

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Post by himister »

Thanks guys. :wink:
That's why I like so much this place.
Someone would think these are trivial things, but as we all see they're not.
Thanks for calculus and physics hour and sorry for the wrong approach in drawing. :oops:

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Post by JiM »

frequencycentral wrote:Why not just use a 7812 to regulate the 15v down to 12v and dispense with the resistors?
I agree with the regulator idea.
I've used a 7806 (up to 1A in TO-220 package) to power the heaters of a 6N3P out of a 12V supply : it needs 300mA @ 6.3V, there's no series heaters option with this tube. It works fine, but you may want some kind of heatsink for the regulator. Screwing it on the enclosure should be enough.
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Post by briggs »

I've got some conductive grease and these little clip-on TO220 heatsinks. Hopefully that'll be enough, even though it is going to have to provide 700mA eek!
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Post by flood »

ah! i had a problem with this with a 6U8/6C4 amp-as-stompbox build i attempted. i'm going to power that from a 7806.

it took my some time to remember out that unequal loads would have unequal voltage drops :oops: and i studied engineering!

one filament was underheated, the other well above...

BUT: two EF86/EL84/any identical 6V tubes in series could be heated from a 12V supply, right? i'm thinking of building eric barbour's tube VCFs/VCAs and am planning to use a 12V supply back to back.
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Post by frequencycentral »

flood wrote:BUT: two EF86/EL84/any identical 6V tubes in series could be heated from a 12V supply, right? i'm thinking of building eric barbour's tube VCFs/VCAs and am planning to use a 12V supply back to back.
That's correct, I've used that method many times.

Caveat: From a 12 volt supply (as opposed to 12.6v), and with tubes with 6.3v heaters, there will be a slight 'offset' as there's 0.6v astray - but it's close enough to be within the tolerance of the heaters.

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Post by marshmellow »

flood wrote:ah! i had a problem with this with a 6U8/6C4 amp-as-stompbox build i attempted. i'm going to power that from a 7806.

it took my some time to remember out that unequal loads would have unequal voltage drops :oops: and i studied engineering!

one filament was underheated, the other well above...

BUT: two EF86/EL84/any identical 6V tubes in series could be heated from a 12V supply, right? i'm thinking of building eric barbour's tube VCFs/VCAs and am planning to use a 12V supply back to back.

When DC powering with 6,3V, use a 7805 and two diodes in series between the ref pin and ground. For 12,6V it's one diode to ground. That lifts up the reference of the voltage regulator and you have the right output.

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Post by briggs »

marshmellow wrote: When DC powering with 6,3V, use a 7805 and two diodes in series between the ref pin and ground. For 12,6V it's one diode to ground. That lifts up the reference of the voltage regulator and you have the right output.
Genius! So I'm presuming with a 7812 that a 1N4001 diode in series with the 7812 ground pin (with the 1N4001's cathode orientated towards ground) will suffice in this solution?
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Post by marshmellow »

Yes. Watch out for the power you burn up in the regulator. There are different rated versions you can choose from, 78S12/78T12/78H12 which can stand different maximum currents. By any means use an appropriate cooler. For more current you can use an additional power transistor, look in the datasheet for the circuit.

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Post by borislavgajic »

thanks very much to you all !!! :thumbsup

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Post by frequencycentral »

Can someone just check my maths (Loubou?)?

I want to run a 6112 heater in series with a 5840 heater from a 12 volt supply.

6112 heater 6.3v 300ma

5840 heater 6.3v 150ma

I've worked out I need a 42 ohm 1 watt resistor in parallel with the 5840 heater. Is this correct?

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Post by lolbou »

Well, with 12V you won't get the full current... A half voltage divider would give 2x6.0V.

Then, the current will be smaller, and so will be the heat, and the heater resistance will be lower too...

Can't you measure the heater current (and compute the heater resistance) at 6.0V for each?
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Post by frequencycentral »

Thanks, Well a quick bit of breadboarding shows that 50 ohms is about spot on for a half voltage divider at 12 volts. So is my wattage calculation still ok at 1 watt?

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Post by lolbou »

Yes you're correct: P=U²/R, which equals 0.72W...

Any info on the current in the 6112 heater (to compute the two heaters resistances at lower voltage)? Just curious...
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Post by frequencycentral »

I didn't calculate! I just played around with resistor values until I got a perfect 6 volts in the middle - which turns out to be 50 ohms, ie two 100 ohm resistors in parallel.

50 ohm 1 watt resistors seem unavailable, only 47 ohm or 56 ohm - so can I use two 100 ohm 0.6 watt in parallel? Would the wattage rating then be 1.2 watts? Or should I use two 100 ohm 1 watt instead?

Thanks for your help!

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Post by lolbou »

Two 100 ohms/0.6W is ok (power is divided by two in this case as you mentioned)... :D
frequencycentral wrote:I didn't calculate! I just played around with resistor values until I got a perfect 6 volts in the middle
That's nice, each tube is treated equally, but you can't know if the flowing current is within specs (except by looking at the dim glow)... Knowing the current generated by your power supply in this circuit would let us know what the current in each tube and the resistance of the heaters in these low-condition... You can't rely on ohm's law when you're out of nominal values (the heater is not linear at all, as any light bulb)...
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