KISS 115
KISS 115
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 described in Section 114) 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 Iq or Io of the tube at quiescence, that so-called design centre which we call Q, 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!
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All text and illustration is Copyright © Andre Jute 1998 and 2004
and may not be reproduced except in the thread KISS xxx on rec.audio.tubes
