On Jun 7, 2:11*pm, flipper wrote:
On Wed, 6 Jun 2012 15:58:03 +0000, TerryG
wrote:
snip, for brevity,
My only other question at this point is does anyone know a way of
calculating the primary impedance of the output transformer other than
multiplying the plate voltage by to itself and then dividing by the
plate dissipation wattage?
Steve sure has a lotta useful insight.
I can say, to begin with, Pda = 24 watts for 1 x EL34, and if you try
Ea at +400Vdc, then Ia = 24 / 400 = 60mAdc.
This is for pentode, triode, UL or Acoustical. But its not 100%
correct because really, Ig2 might be 5mA, so to keep
the tube cool, we might use Ia = 55mA and Ig2 = 5mA, where Eg2 is
close to Ea.
For calculating loads for maximum PO for class A1 SE pentodes, RLa =
0.9 x Ea / Ia. For approximate load calculations, neglect Ig2 in power
tubes.
For pentode operation where the knee of the Ra curve for Eg = 0V is
well above 2 x Ia at idle, for max PO for class A1, SE, RLa = 0.9 x
Ea / Ia = 0.9 x 400 / 0.06 = 6,000 ohms.
If you chose Ea t 300V, then Ia = 24 / 300 = 80mAdc, so RLa = 0.9 x
300 / 0.08 = 3,375 ohms.
You can develop an an equation where you have Pda and Ia or Pda and Ea
only as the variables and I leave you to think about that.
Pentodes have dreadful spectra when in SE mode, and piles of THD which
can only be cured with NFB applied externally.
In PP mode, the RLa experienced by one tube becomes a curved line
because the current change and voltage change in one tube affects the
same in the other, so I find the composite PP load line to be
confusing, and a trifle simplistic, so I never bother with it, and
just concentrate on what is happening in ONE tube of the pair in PP,
and then settle for approximate estimations. The real outcome is
always what I measure. Most PP amps are class AB so when one tube is
cut off, the other sees 1/4 of nominal RLa-a, and most PP power is
effectively produced by SE action of each tube. Because any pair of
average PP tubes are not perfectly matched, there will aways be 2H and
4H in the PP THD. Mostly you get 3H and 5H.
Triodes are different, and the PO is limited in AB1 by the Ra line for
Eg1 = 0V having no knee, and having the similar shape to all other Ra
lines for negative values of Eg1.
Ra at a given Ia varies slightly for most triode conditions, and for
EL34 can vary between say 1k5 for Ia = 50mA down to 1k2 for Ua =
100mA.
RLa load is calculated for maximum PO, triode class A1 = ( Ea / Ia )
- ( 2 x Ra ), so if you have Ea = 400V and Ia = 60mA, and Ra = 1,350
hms, then RLa = ( 400 / 0.06 ) - ( 2 x 1,350 ) = 6,666 - 2,700 = 3,966
ohms, ie, about 4k0, and if the winding losses were 10%, load is 4k0
-10% = 3,600 ohms. Of course speakers present a wide range of loads
above and below the calculated RLa value. Tubes and any other devices
are expected to perform with a dynamically changing load for every
microsecond of their existance when being used.
Devices with adequate open loop gain and adequate capability to
produce a useful amount of power at less than 10% THD routinely used
in amplifiers with adequate NFB used to straighten out what is a usual
mess to begin with, especially when penodes are used. While a trioded
SE EL34 can produce 7.2 watts at about 5% THD, RLa = 4k0, at all other
RLa values away from 4k0 the max PO will be less because either grid
current or cut off cause assymetical clipping before 7.2Watts is
produced. If we insist on having an ability for say 3 Watts at less
than 5%, then load values can range between say 2k0 and 8k0, but you
should be able to work all this out yourself with load line analysis.
SE Pentode operation with RLa above or below RLa for max PO also shows
the limited range of RLa for a useful PO.
See my pages,
http://www.turneraudio.com.au/loadma...amtetrodes.htm
Follow your nose to many other pages, and most useful is to scurry
away to your workshop and get busy with measuring gear and
oscilloscope.
For working out the class A PO for any RLa value which is less than
that for max PO, PO = 0.5 x Ia dc squared x RLa.
Eg, for 2k0, Ia = 60mAdc, PO max at clip = 3.6Watts.
For working out the class A PO for any tube value which is higher than
that for max PO, PO = ( peak negative Ea swing x 0.707 ) squared /
RLa.
Eg, for 8k0, triode, Ea peak swing might be 300V, so PO max = ( 300x
0.0707 ) squared / 8k0 = 5.6Watts. There are formulas to find
theoretical peak Ea swing using Ea, Ra, Ia and RLa etc, but just
drawing a line across the Ra curves is less confusing.
Patrick Turner.