fishski13
Member of the Trade: SolderWorksAudio
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- Nov 2, 2004
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Quote:
thanks!
Originally Posted by runeight /img/forum/go_quote.gif Ah. I did this calculation quite a while ago and I don't remember how to do it again, but the output impedance is given by: Zo = rp / (2 * (1 + mu)) The ECC99 has an rp of 2300R and a mu of 22 so ... Zo = 2300 / (2 * 23) = 50R The 6N6P has an rp of 1800R and also a mu of 22 so ... Zo = 1800 / (2 * 23) = 39R In this configuration, with the crossed outputs from the phase splitter, both O/P triodes operate as cathode followers. So the Zo is essentially the paralleled Zo of two cathode followers. |
thanks!