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Better Math

Posted by Lynn Olson (M) on December 5, 2003 at 21:01:31

In Reply to: Assorted Thoughts posted by Lynn Olson on December 5, 2003 at 14:21:20:

Got a little sloppy about the math, not calculating the series and parallel output Z correctly. Let's start with the quiescent output Z, assuming Rp of 750 ohms, a primary of 6000 ohms, and a secondary of 8 ohms. That's a turns ratio of 27.3861, or for a half-primary, 13.6931.

Series connection:
750 + 750 = 1500 ohms.
1500 * (8/6000) = 2 ohms source, or in terms of damping factor, (8/2) = 4.

Parallel connection:
Impedance of half-primary = 1500 ohms
750 * (8/1500) = 4 ohms, but that is parallel with the other half-primary, so (4 || 4 ) = 2 ohms, and the damping factor is (8/2) = 4, same as before.

But things change when the two tubes become unequal in instantaneous output impedance.

Series connection:
200 + 3000 = 3200 ohms.
3200 * (8/6000) = 4.667 ohms source

Parallel connection:
Impedance of half-primary = 1500 ohms
200 * (8/1500) = 1.667 ohms
3000 * (8/1500) = 16 ohms

(1.667 || 16 ) = 1.509 ohms

This is rough back-of-the-envelope stuff, but it gives a flavor of what happens with series and parallel connections.

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