Transofrmer secondaries

[jeffdavison]

I have a transformer with dual 6.3vac secondaries, each is rated at 5 amps.

If I were to wire them in series I'd have 12.6vac.. Would the amperage available be 10 amps?

What I'd like to do is get 6.3 vdc. after rectification and regulation of the 12.6v. I'd like to use this to put DC on my amps tube filaments instead of AC to help rid all the vestiges of possible hum.

Current requirements for the filament total 4.7 amps (2 x 6550, 1 x 6SN7, and 1 x 6H30).

I plan on using Schottkies to keep dreaded hash out of the voltage, perhaps even shunting them for good measure. then regulate down with an LT1083 7.5a adjustable voltage regulator.

Just need to be sure that the amperage from the series secondaries adds rather than divides.

JD

[Geek]

Hi,

Secondaries in series, doubles the voltage but current remains the same. You'll have 12.6V @ 5A

Paralleling the secondaries keeps the voltage the same, but doubles the current. You'll have 6.3V @ 10A.

There's no free lunch I'm afraid

Cheers!

[Yvesm]

Hi Jeff,

Why the hell do you want to vaste some 30 Watts in a pass regulator while well balanced wiring technique will do the job ?

Not to tell about the huge current peaks needed to relaod huge smoothing caps 120 times per second.
This is a serious noise generator

Yves.

[duderduderini]

Hi
If you rectified the 6.3V you would get around 1.4 times that value.. why dont you try say low dropout regulators to achieve your target end voltage? A lot of heat will be produced coming down from 12.6 to 6.3 thats all.
There are some good low dropout regs available and you could parallel the supply giving you some headroom instead of sailing close to the limit.
Cheers
Nick

[leadbelly]

I would NOT do it like that, and I speak from experience as I built one like that for my PP amp and ripped it out right away because of the considerable heat generated.

Rectify the 12.6 and use that for the 6550 filaments in series, either with a regulator or CRC filter. Then use a separate regulator to supply the small signal tubes.

[jeffdavison]

What I'm 90% sure that I'm goig to do is leave both 6550's have their 6.3v on one of the secondaries.

The 6SN7 and 6H30 (.6A and .9A) 1.5 amps in total, will be fed from a voltage doubler which will give 1/2 the amperage or 2.5a at the double voltage.

The doubler will be Schottky and BlackGate filtered into an adjustabe LDO LT1084 voltage regulator to get to the 6.3 VDC that both the input / driver will draw from.

I don't really need DC on the power tube filaments as it's very quiet as it is (with the output of the input / driver muted to ground), just a bit of just audible tube rush.

Hopefully the voltage drop down from the rectified voltage to the regulated voltage won't bring a heat issue. Anyone know where the calculations can be found for voltage drop / amperage used for power dissapation? Can't seem to recal that particular Ohms Law to help me calc the size of the heat sink I'll need.

Jeff

[hirafu_boarder]

I think you should review the latest TubeCad on the Janus regulator.
http://www.tubecad.com/2008/09/blog0149.htm

With a doubler and regulator, I estimate your 5A AC will result in about 1.4A DC (just a bit less than your desired 1.5A for your small tubes).

Cheers,
Jeff

[jeffdavison]

Can you show me your math?


I really dont have room for the Janus under my amps chassis and will have to configure for my space available.

JD

Originally posted by hirafu_boarder
I think you should review the latest TubeCad on the Janus regulator.
http://www.tubecad.com/2008/09/blog0149.htm

With a doubler and regulator, I estimate your 5A AC will result in about 1.4A DC (just a bit less than your desired 1.5A for your small tubes).

Cheers,
Jeff

[hirafu_boarder]

I still very new to all this, but I'll take a stab at it.

In the linked article, Mr Broskie wrote:

So how much current can the voltage doubler circuit wring from a transformer secondary? Well, we can start by halving its nominal AC rating and then dividing by 1.8. thus, for example a 1A current rating will yield 0.28A into a resistive load; 3A, 0.83A. In other words, the 6.3Vac @ 3A winding, once run through the voltage doubler, will not power four 6SN7 or four 12SN7. It will, however, power four 12AU7s or 6DJ8s quite nicely—powering four 6SN7s will require a 6.3Vac @ 4.5A winding.


So given your 5A AC, from the above you will have 5 /2 /1.8 = 1.4A DC after doubler and regulator.

His statement that he could power 4 6SN7 from a 4.5A winding is a bit odd. 4 6SN7 in parallel would require 2.4A , which is not possible if he doubled/reg (to 6.3) the original secondary. So I think he doubled/reg (to 12.6) then ran 2 6SN7 in series and paralleled that with 2 6SN7 in series. This arrangement uses 12.6VDC at 1.2A which is just right.

However in your situation, you cannot run your 6SN7 in series with 6H30 (dissimilar current needs) so parallel it must be at 6.3V.
If you forgo the doubler, then with Schottky and LDO regs, you might be able to get 6.3DC from 6.3AC.

Cheers
Jeff

[jeffdavison]

yep.. just went to Broksies page and did the math....

My current needs total 1.45 amps and the math says from a 5 amp 6.3vac secondary, I can expect the circuit to allow 1.3889 amps as the actual current deliverable.

However, with an LDO LT 1084, I can get the voltage of 14.3164 dc (peak = (rms x 1.414 x 2))- diode voltage drop ( 2v) - Regulator voltage drop (1.5v) down to the 6.3vdc needed with that regulator, tho I can go as low as 6v to give a bit more life to the filaments.

the amperage deficit is only .06 amps and should be "just" ok for both tubes. There will be ample capacitor uf's for filtering ripple.

Another alternative scheme I've been thinking about is to put DC only on the 6SN7... after taking .845a away from the ac traffo, this leaves 4.55a left and doing the math should net out at 1.1542 amps for the 6SN7. This gives me more than enough and since this is where I believe the most sensitve to hum tube is (input stage) this is where the DC filament would have the greatest effect. I have a mute switch that grounds the input signal between the 6SN7 and 6H30, and when muted there is no discernable hum, only tube rush noise - the 6H30 and 6550 are basically WOT. Also this give me more than enough headroom to use an ECC34 which has a .95a filament requirement.


JD

Originally posted by hirafu_boarder
I still very new to all this, but I'll take a stab at it.

In the linked article, Mr Broskie wrote:

So how much current can the voltage doubler circuit wring from a transformer secondary? Well, we can start by halving its nominal AC rating and then dividing by 1.8. thus, for example a 1A current rating will yield 0.28A into a resistive load; 3A, 0.83A. In other words, the 6.3Vac @ 3A winding, once run through the voltage doubler, will not power four 6SN7 or four 12SN7. It will, however, power four 12AU7s or 6DJ8s quite nicely—powering four 6SN7s will require a 6.3Vac @ 4.5A winding.


So given your 5A AC, from the above you will have 5 /2 /1.8 = 1.4A DC after doubler and regulator.

His statement that he could power 4 6SN7 from a 4.5A winding is a bit odd. 4 6SN7 in parallel would require 2.4A , which is not possible if he doubled/reg (to 6.3) the original secondary. So I think he doubled/reg (to 12.6) then ran 2 6SN7 in series and paralleled that with 2 6SN7 in series. This arrangement uses 12.6VDC at 1.2A which is just right.

However in your situation, you cannot run your 6SN7 in series with 6H30 (dissimilar current needs) so parallel it must be at 6.3V.
If you forgo the doubler, then with Schottky and LDO regs, you might be able to get 6.3DC from 6.3AC.

Cheers
Jeff