How does the optimal load resistance change when you strap a pentode into a
triode in an SE design, and is there a rule of thumb for calculating the
load resistance of the resulting triode?

I've got some small 5K/8 ohm SE transformers for 6T9 Compactrons as well as
some for 7591s and my goal is to build an all-triode headphone amp.

TIA,

Jon

Jon Yaeger wrote:

How does the optimal load resistance change when you strap a pentode into a
triode in an SE design, and is there a rule of thumb for calculating the
load resistance of the resulting triode?

There is a lot about the ideal load for a pentode in DH4, and suppose
the idle current is 70mA, and Ea = 350v, then Ea / Ia = 5,000 ohms,
and RL should be 0.9 x ( Ea/Ia ) = 4,500 in this case,
which is about right for an EL34.

Now when the EL34 is trioded, the Ra reduces from 12k to about 1.3k,
and as a rule of thumb the load would be about 3 to 4 x Ra, so
that's about 3.9k, or similar to the pentode case.

Triode loading depends a lot on the Ea and Ia of the tube,
and you could have 420v x 52 ma and RL would be high, say
between 5k and 7k, and you get the highest efficiency,
but the load will be greater number of Ra than with
say Ea = 300, and Ia = 73 mA.
The efficiency is down a bit and RL will be
a smaller number of Ra.

If the EL34 was set up a s a pentode with 300v x 73 mA, then
its RL = 0.9 x ( 300/0.073 ) = 3.68 k,
and perhaps this would also be fine as a triode load.

But all this needs confirmation by load line analysis, very simple,
and able to be done in 5 minutes on the back of an envelope
when you have learned how to do it, and you should, because then you
are not just taking my word for all this, and you will see on the graph
you draw exactly what the max/min voltage swing limits will be.

You do at least have to know what the Ra is for the triode concerned,
and the line of Ra where Eg1 = 0 volts is the important one,
because that is a boundary for grid current for class A,
and what you end up with for voltage swings is deternimable between
the boundaries of the horizontal axis and the Eg1 = 0v line.

You don't need all the curves to work all this out, just Ra.
that can be drawn as a line from the 0.0 vertical / horizontal axis points,
and it should be stepped over about 10% to make it more realistic.
Take a look at all the triode Ra curve graphs.
Look at their Ra lines for where Eg1 = 0V, and you'll see what I mean.
If you choose a point on the Ra line where Ia = say 50mA,
then the slope of the Ra line at that point is the Ra for that condition of Ea
and Ia.

The small voltage change divided by the small plate current change
will give you what the Ra is for that line position.



I've got some small 5K/8 ohm SE transformers for 6T9 Compactrons as well as
some for 7591s and my goal is to build an all-triode headphone amp.

Try setting up the tube in a socket with a proposed transformer.
Have a 1 kHz signal into the grid of the triode under test.
Make sure the output level is much lower than what is expected,
say 10vrms will do.

Keep the input grid voltage constant.

Try connecting 5k across the primary of the transformer, and measure the
output voltage; it will be less than with no load, or with just the primary
only. The primary inductance at 1 kHz should be a much higher impedance than
the load value at 1 kHz, and it may able to be neglected in your
Ra estimations.

Then record the voltage change with 5 k RL.
( in fact a 5 watt x 4.7k wire wound R will do fine for a test RL )

The Ra is the dynamic resistance of the triode.

Say you had 10v without an RL.

Say you had 8vrms with 5k RL.
the load current is thus 8 / 5k = 1.6 mA.

The the change of load current is 1.6 mA, because there
was 0.0 mA of load current with no load.

The change of voltage between no load and load = 10v - 8v,
so you have 2v load voltage change.

Now the Ra = load voltage change / load current change,
and its valid for all tubes, and in this case = 2 / 0.0016 = 1,250 ohms,
and you should about observe these results with a test with EL34.

if you find Ra = 1,250 ohms, and you say you want a load
for fidelity, rather than lots of power,
then RL should be over 4Ra so
between 5k and 7.5k would be about right,
which will allow for large swings in the load impedance.

Once that decision is made, the triode is set up with Ea x Ia =
about 2/3 the maximum allowable dissipation,
so Ea and Ia have to be chosen to suit the load intended,
and to get the idle DC condition such that the AC operation
is centred for the most linear part of the curves for the triode.

Graphical loadline anaysis is the simplest way to see in the dark
with choosing a load, and hence the turn ratio of the transformer,
or to see if the turn ratio of a tranny you have will
give you an anode load which is what is best for the triode.

Patrick Turner.