KISS 115 by Andre Jute
This text is copyright Andre Jute 1998, 2004 and may not be reproduced
except in the thread KISS xxx on rec.audio.tubes
THE VOLTAGES IN THIS AMP WILL KILL YOU. GET EXPERIENCED SUPERVISION IF
IT IS YOUR FIRST TUBE AMP.

Allowing for rectification effects and calculating distortion more
closely
An advanced look at operating points and distortion
by Andre Jute

Distortion is the ratio of harmonic to fundamental amplitude.

For triodes, it is usually adequate, unless an amplifier is
incompetently designed (like the Bubbaland 300B where the transformer
is fundamentally mis-specified for the tube) to calculate 2nd
harmonic and forget about 3rd harmonic distortion because it will be
too small to bother about.

To do the calculation, we require to know the plate current of the
tube at quiescence, that so-called design centre which we call Iq, and
the current drawn at the extremities of signal swing, Imax and Imin.
See the article on selecting an operating point about marking these
off on the tube's Ia-Eb-Eg curves. We use D for distortion and add a
number for the harmonic whose distortion we are measuring.

D2 = (C2/C1)

where

C1 = (Imax - Imin)/2

C2 = (Imax + Imin - 2Iq)/4

or combined

D2 = (Imax + Imin - 2Iq)/(2(Imax -Imin))

or, perhaps easier to visualize

D2 = ((((Imax + Imin)/2)-Io))/ (Imax-Imin)

Multiply by 100 to find percentage. C1 or C2 may be negative but this
has no bearing on D2.

Generally speaking, it isn't worth bothering about rectification
effects for triodes either, as measuring on the graphs is never that
accurate, and anyway a tiny fluctuation in the mains voltage will
make such niceties irrelevant. If you're keen, Cdc for triodes
amounts to C2, and is merely added to Iq to establish a new current
centrepoint Id. A line projected from Id on the current axis to the
original bias line establishes a new quiescent point, Q1 (in the
literature usually rendered as a Q with a tick of a single quote mark,
which the net does not reproduce consistently), and a new loadline can
be drawn through it parallel and to the right of the original
loadline.

For pentodes and transistors the position is different. A very
substantial third harmonic element must be expected and a five- point
schedule is used to calculate it. This is used for triodes only if, as
in the Bubbaland 300B already referred to, they are suspected of bad
design.

It is first of all necessary to determine two further current draws,
I1 and I2. They are respectively the projections of the electrical
(rather than geographic) midpoints of the signal swing between Iaq and
Imax and Iq and Imin. Thus if the signal swing above Iq is 40V, I1
will be the current drawn at a signal swing of 20V above Iq and I2 the
current draw at a signal swing of 20V below Iq.

First calculate the direct current distortion

Cdc = ((Imax + 2I1 + 2I2 + Imin)/6) - Iq

If Cdc is an appreciable amount, an operating point shift may occur
and must be drawn to discover the new operating point Q1 and its
associated current draw Id. The procedure is the same as for triodes,
described above.

Now calculate the components

C1 = (Imax + I1 - I2 - Imin)/3

C2 = (Imax + Imin - 2Iq)/4

C3 = (Imax - 2I1 + 2I2 - Imin)/6

and combine them to get 2nd, 3rd and total harmonic distortion:

D2 = C2/C1

D3 = C3/C1

Dt = SQRT((D2*D2) + (D3*D3))

Good luck!

© 1998 and 2004 Andre Jute

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