sizing transformer for heater current

[skal1]

ok  just bought a couple of these for my DC  filament current  for project ,http://www.ebay.co.uk/itm/LT1084-5A-Voltage-Regulator-Power-Supply-Module-Kits-for-Tube-Amp-Filament-CDROM/131692953391?_trksid=p2047675.c100005.m1851&_trkparms=aid%3D222007%26algo%3DSIC.MBE%26ao%3D2%26asc%3D41433%26meid%3D50fe4ebfd2ca443eb5e51a095ec28c4a%26pid%3D100005%26rk%3D2%26rkt%3D6%26sd%3D131692953739 i was wondering how size the transformer to deliver 6.3v @4.6AMPs to the filament's of the tubes , i had a couple of days  of looking on net , but haven't seen anything mentioning how to size the transformer for this application.

any help on this would be nice

cheers

skal

 

[johnheath]

How many and what tubes are you going to use for each heater supply?

Best regards

/John

 

[skal1]

HI johnheath  ,

13 TUBES  [email protected] EF93  1)@0.300Ma 12AX , 1 )@.600Ma12BH7  2)@0.450Ma  6973.


regards

skal1

 

[moamps]

Do you really need DC stabilized power supply for heaters?
If yes, IMO you should use a 9Vac/8A transformer with secondary distributed to two or (better) three boards you linked. The reg. chips should be mounted separately on bigger heatsinks.

 

[ruffrecords]

The LT1084 is a very good low dropout regulator. I used it in my lunch box power supply to generate 12VDC for heaters. With  analogue power supplies you need to carefully about the dissipation in the regulator. to work this out you need to know the likely mains voltage variation, the transformer efficiency, the size of the smoothing cap, the drop out voltage of the regulator and its thermal properties, the thermal properties of the heat sink and lastly the load current. this is a non-trivial exercise, especially with large load currents.

Therefore, as others have advised, with the large heater current you have you need to think very carefully about whether it all needs to be dc.

Cheers

Ian

 

[skal1]

Hi Ian,

is their some kind of step by step guide on net show me how to workout these variables.

regards

skal

 

[mjrippe]

13 TUBES  [email protected] EF93  1)@0.300Ma 12AX , 1 )@.600Ma12BH7  2)@0.450Ma  6973.

I do not understand your math.  You say 4.6A and 13 tubes, then you list 12 tubes with a total draw of 3A (the 2x6973 should be .9A).

 

[ruffrecords]

I do not understand your math.  You say 4.6A and 13 tubes, then you list 12 tubes with a total draw of 3A (the 2x6973 should be .9A).

[email protected] is 9.6A by my reckoning, never mind the rest.

Cheers

Ian

 

[skal1]

your right lads 12 tubes 8 6ba6 @ 1.2AMP

Skal

 

[skal1]

i was thinking about another tube  total current  is 4.2amp , sorry for confusion..

regards

Denis

 

[skal1]

just on the RS web now going to order a big 150VA  2x6v  that should be big enough to run those heater, what do think,

 

[mjrippe]

[email protected] is 9.6A by my reckoning, never mind the rest.

Cheers

Ian

EF93/6BA6 is .15A *each*, so 1.2A total

However, I still get 3A as the total for those tubes combined, not 4.2A

 

[PRR]

Gawd, remind me not to split the check with you guys.

Even the tube count is wrong.

EF93 is 0.3A each
12AX? is probably 0.3A (unless this is Rectifier 12AX4)
12BH7 is 0.6A
6973 is 0.45A

""13 TUBES""
8 @ 0.300 AMP EF93  ===============  2.4A 
1 @ 0.300 AMP 12AX ===============  0.3A 
1 @ 0.600 AMP 12BH7 ============== 0.6A 
2 @ 0.450 AMP  6973 =============== 0.9A     
12 tubes ============================== 4.2 Amps

 

[skal1]

Ok  sorted out that mistake, cheers we can move on .

as i was saying in the beginning  i don't know how to  pick the right transformer for the job, so i have ended up getting a 150va@2x6 tx, by my reckoning thats over 12 AMP per 6V winding  yes. This should be ample enough current for the regulator to do its thing, i am i wrong?

regards

Skal

 

[PRR]

If you are going for DC (wise in light of what you are doing), aim for *12VDC*. This tube complement can easily be wired 12V. Losses at 6VDC are huge, 12V less.

_I_ would get a 12V 5A switcher DC supply and be done with it. (Yes, the running current will only be 2.1A but cold heaters suck hard.)

 

[sahib]

The transformer secondary current should normally be  about1.6 times the required DC current. For 4.2ADC you will need 7.2A.  So, as Moamp's suggestion of 8A is good.

But you have other worries with that regulator board.

1). The rectifier diodes on that board are rated at 3A. Even at 3A they will heat a lot.

2). With 6VAC after rectification you will have about 7.8V before the regulator. You'll drop 1.6V over the regulator and that's almost 7W of heat. That heatsink will never handle that.

Also, the transformer is rated at 6VAC but that is under load. No load voltage will be much higher than that (ignore the transformer regulation figure). What that means is that if the transformer has unnecessarily higher current output then it will hold the no load voltage, which is high. This in return will give you higher rectified voltage and increase the voltage drop over the regulator, hence increased heat dissipation.


Edit: I have just noticed that you might be using two boards. In which case you might be ok but the heatsink size will still be an issue.

 

[skal1]

ok  sahib,  yes i think the heat sink will need changing i will look at changing that and the diodes on the board maybe Fairchild SB560 Schottky Diode, 60V 5A, 2-Pin DO-201AD or  do i need a diode which can switch more current.


regards


Skal

 

[sahib]

You don't need Shottky.  I'd say use a bridge rectifier (though you'll twist the leads a bit to fit) which allows you to fit a heatsink. Alternatively you can mount it onto the chassis and hard wire to the board.

If you are going to order the heatsink for the regulator IC, don't. I have gazillions and would happily send you one. I'll stick in a bridge rectifier too. Just pay the postage.