Re: EL84 SE TRANSFORMER



flipper wrote:

On Sun, 7 Dec 2008 14:13:17 +0200, "Bar Nash" <therightinfo@xxxxxxxxx>
wrote:

Hi all

I need to order a SE EL84 transformer .
Does any one knows the ratio of turns primary/seccondary for 8ohm load ?

Thanks
EC


Depends on how you want to operate it.

For typical values and curves get the tube datasheet from

http://www.mif.pg.gda.pl/homepages/frank/vs.html

If you want typical pentode mode then about 5k PP with 250V B+ and
screen for 5-6 Watts out, less OPT losses.

If you want triode mode then 3.5k PP with 250V B+ for about 2 Watt
out, minus OPT losses.

Turns ratio is the square root of the impedance ratio. I.E. 5K/8 is
25/1.


Correction to my replies to the OP.

I mis-read the original post and gave load values for PP applications.

For SE applications The load value for a power pentode = 0.9 x (Ea / Ia)

The tube will always operate in class A and lets assume the anode idle
disipation = 12 watts.
So Ea x Ia = 12 watts.

If you have Ea = 300V, then Ia = 12 / 300 = 40mA, so RL for maximum
pentode power = 0.9 x ( 300 / 0.04 ) = 6,750 ohms.

The transformer has an impedance ratio of 6,750 : 8, ie, 843:1, so the
turn ratio = 29:1.

But you may find that such a transfromer is unavailable and you can only
find one rated for 5,000 : 8 ohms.

Working backwards, 5,000 = 0.9 x ( Ea / Ia ).

Pda = 12 watts = Ea x Ia, then Ea = 12 / Ia.

Then 5,000 = 0.9 x [ (12/Ia)/Ia ] = 0.9 x ( 12 / Ia squared ).

Ia = sq root of ( 0.9 x 12 / 5,000 ) = 46mA.

if Pda = 12 watts, then Ea must be 12 / 0.046 = +258Vdc.

For triode use of most negatively biased pentodes, the load for maximum
useful power and symetrical clipping = ( Ea / Ia ) - ( 2 x Ra ). The
same reasoning for pentodes can be used for triodes, but with EL84, you
don't want to use an Ea which is +/- 33% from the usual 250vdc voltage
mentioned in the data pages for this tube.

Triode connection works best with ea about 33% higher than the centre
value of 250Vdc, and with Ia at 36mA. From this you can work out
RL.
For SE pentode use, Ea from 220 to about 300V is the range.

Maximum power output with symetrical clipping = ( 0.707 x Ia )squared x
RL.

So for the 6,750 ohms expect 5.4 watts of audio power at the anode.
At the secondary load you might only get 5 watts because of power lost
in the transformer winding resistances.

Patrick Turner
.

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