Tube heaters combination?

borislavgajic · Post by **borislavgajic** » 23 Sep 2009, 07:38

is it possible to connect two different tubes....ecc83 and ef86....on 12,6V heater voltage?

and how...if that can be done.....?

thanks
boris

briggs · Post by **briggs** » 23 Sep 2009, 09:47

You would just place them in parallel across the 12.6V supply

himister · Post by **himister** » 23 Sep 2009, 09:54

borislavgajic wrote:is it possible to connect two different tubes....ecc83 and ef86....on 12,6V heater voltage?

and how...if that can be done.....?

thanks
boris

Hi. Boris.
Try serial connection with this two. Also connect 2x100R resistors between filaments and ground to even the current flow on both sides and prevent hum. You can use a pot instead of resistors.
"It's a fairly common noise-reducing trick to elevate the heater winding's CT a bit."

Cheers bro.

ECC83-EF86-HEAT.GIF

borislavgajic · Post by **borislavgajic** » 23 Sep 2009, 10:19

thank you very much !!!!!!!

it was quick answering....

boris

marshmellow · Post by **marshmellow** » 25 Sep 2009, 15:26

Hey Boris,

don't do it like that!! The ECC83 will be extremely underheated,
while the filament of the EF86 will glow like a christmas tree.
Just calculate for yourself...

And you can't heat the EF86 with 12,6V. If you really have to
use that voltage: ECC83 in series and in parallel a voltage regulator
(only with DC source of course) or a resistor with the EF86.

Volker

himister · Post by **himister** » 25 Sep 2009, 15:53

Well I'm also interested in that..
Could you be more specific or draw some schematic...

Thanks

Cheers

frequencycentral · Post by **frequencycentral** » 25 Sep 2009, 16:33

ECC83: 6.3 volts @ 300ma
EF86: 6.3 volts @ 200ma

Therefore run the two in series from 12.6v, but with a 63 ohm 1 watt resistor in parallel with the EF86 heater. That should give each heater what it needs. You can measure where they meet to confirm that it's 6.3v right in the middle.

So:

- ECC83 pins 4 and 5 to ground,

- ECC83 pin 9 to EF86 pin 4,

- EF86 pin 5 to 12.6v,

- 63 ohm / 1 watt resistor from EF86 pin 4 to EF86 pin 5.

You should be able to measure 6.3v at ECC83 pin 9.

I think I got the math right! http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

lolbou · Post by **lolbou** » 25 Sep 2009, 17:15

frequencycentral wrote:I think I got the math right!

+1 Rick!

Calculations checked OK. Please accept this A+ grade from an humble physics teacher!

May I explain to all:

R(ECC83) = 6.3/.3 = 21 ohms

R(EF86) = 6.3/.2 = 31.5 ohms

To get a proper 1/2 voltage divider (giving 6.3V to both heaters), you need to get 21 ohms out of a parallel combination of the EF86 heater and a resistor (to compute), which I call R.

You get:

1/R +1/31.5 = 1/21

Maths gives:

R= (31.5 x 21) / (31.5-21)
R= 63 ohms

That's the hard trick (or shall I say tRICK)...

....

A simpliest way is to think "well, I need a resistor paralled with R(EF86) that would suck the extra 100mA that the heater don't need"...

R = U x I = 6.3 x 0.100 = 63 ohms...

.....

Choose your prefered method!

marshmellow · Post by **marshmellow** » 25 Sep 2009, 17:32

EF86 needs 200mA heater current and you want to drop half of 12,6V:

R = 6,3V / 0,2A = 31,5 Ω

It needs to dissipate 1,26W so make it at least 2W.

Connect it like that.

/edit: of course it would work with the parallel one as well. Didn't even think about that.

lolbou · Post by **lolbou** » 25 Sep 2009, 17:39

Well, not really...

frequencycentral wrote:Therefore run the two in series from 12.6v

The heaters are in series in my mind as Rick mentioned, so this way the resistors gets 0.1 mA, and only dissipates 0.63W (P=RxI²)...

Paralleled resistors is a trick to dissipate more power with lower rated components.

edit: just seen your edit, and do agree about your set up too

, but you need a bigger resistor...

frequencycentral · Post by **frequencycentral** » 25 Sep 2009, 17:47

lolbou wrote:
frequencycentral wrote:I think I got the math right!
+1 Rick!

Calculations checked OK. Please accept this A+ grade from an humble physics teacher!

Hahaha! I got some maths correct! Maths is not my strong point. Thanks for the A+.....!!

briggs · Post by **briggs** » 25 Sep 2009, 18:16

Ahhh, very nice solutions! I apologize for my earlier BULLSHITE answer when I hadn't bothered to check datasheets. I hope it caused no distress!

Ok - now here is another question for the heater calculators! I'm going to need to run a 6112 submini a 7327 submini and an ECC99 off of either 18.9V or 12.6V, how do you lot think it would be best to do this

frequencycentral · Post by **frequencycentral** » 25 Sep 2009, 18:22

briggs wrote:Ahhh, very nice solutions! I apologize for my earlier BULLSHITE answer when I hadn't bothered to check datasheets. I hope it caused no distress!

Ok - now here is another question for the heater calculators! I'm going to need to run a 6112 submini a 7327 submini and an ECC99 off of either 18.9V or 12.6V, how do you lot think it would be best to do this

Easiest way would be to run the 6112 and the 7327 in series from 12.6v. And the ECC99 from 12.6 volts, not using the centre tap. That way the 6112 and the 7327 draw 300ma combined, and the ECC99 draws 400ma.

briggs · Post by **briggs** » 25 Sep 2009, 19:02

Having thought about it again - here is the situation I find myself in (see the attached picture). Now I need to power this setup from the 15Vdc supply I have. Confusion ensues especially after a few beers on a friday eve! Legitimate values for R1 & R2 are required!

frequencycentral · Post by **frequencycentral** » 25 Sep 2009, 19:11

briggs wrote:Having thought about it again - here is the situation I find myself in (see the attached picture). Now I need to power this setup from the 15Vdc supply I have. Confusion ensues especially after a few beers on a friday eve! Legitimate values for R1 & R2 are required!

Why not just use a 7812 to regulate the 15v down to 12v and dispense with the resistors?

frequencycentral · Post by **frequencycentral** » 25 Sep 2009, 19:16

Or place a 16 ohm / 5 watt resistor between the 15v and the tube heaters. Again, dispense with the two resistors to ground.

EDIT: 16.8 ohm

briggs · Post by **briggs** » 25 Sep 2009, 19:23

frequencycentral wrote:
briggs wrote:Having thought about it again - here is the situation I find myself in (see the attached picture). Now I need to power this setup from the 15Vdc supply I have. Confusion ensues especially after a few beers on a friday eve! Legitimate values for R1 & R2 are required!
Why not just use a 7812 to regulate the 15v down to 12v and dispense with the resistors?

Indeed that is an idea I shall use. I (again!) am guilty of not checking datasheets and assumed that the regulator wouldn't be able to provide enough current. I WAS WRONG. God, today has been a poor one! Cheers again Freq. I'll buy a large batch and just keep trying them unitl I find one that puts out ~12.3-12.6V

EDIT - When I calculated it like that Rick I got 18 ohms, I must have done something wrong>! What method did you use? I could do with some practice!

frequencycentral · Post by **frequencycentral** » 25 Sep 2009, 19:31

briggs wrote:EDIT - When I calculated it like that Rick I got 18 ohms, I must have done something wrong>! What method did you use? I could do with some practice!

DC voltage = 15 volts

Tube heater's (combined) power requirement = 12.6 volts @ 700ma

Voltage drop required = 2.4v (15 - 12.6 = 2.4)

2.4 (volts) x 7 (hundred ma) = 16.8 (ohms)

2.4 (volts) / 0.7 (amps) = 3.4285 (watts)

Therefore 16.8 ohm 3.4285 watts. In reality a difficult value to find, so go for the nearest 'easy' value, 18 ohm 5 watt. Always better to have the wattage over-spec'd.

If Loubou agrees, I might get another A+......!!

briggs · Post by **briggs** » 25 Sep 2009, 19:42

frequencycentral wrote:
briggs wrote:EDIT - When I calculated it like that Rick I got 18 ohms, I must have done something wrong>! What method did you use? I could do with some practice!
DC voltage = 15 volts

Tube heater's (combined) power requirement = 12.6 volts @ 700ma

Voltage drop required = 2.4v (15 - 12.6 = 2.4)

2.4 (volts) x 7 (hundred ma) = 16.8 (ohms)

2.4 (volts) / 0.7 (amps) = 3.4285 (watts)

Therefore 16.8 ohm 3.4285 watts. In reality a difficult value to find, so go for the nearest 'easy' value, 18 ohm 5 watt. Always better to have the wattage over-spec'd.

If Loubou agrees, I might get another A+......!!

When I go back over my working I've done the same thing. I must have pushed the wrong buttons somewhere - it's like being back at school - "check your workings!", I never did it then and I guess I've not learnt

EDIT-
Oh ho! Hang on - I've found it, the error that is, I had each of the submini tubes requiring 300mA while you have them as only needing 150mA each.

lolbou · Post by **lolbou** » 25 Sep 2009, 20:00

Well, it turns to be A+ for everyone here

... Try and learn, again and again! I sometimes wish my pupils were as good willing in trying as you are...