From mxfle–(at)–bm.net Mon Jan 30 10:30:12 CST 1995
Article: 33688 of alt.guitar
Path: geraldo.cc.utexas.edu!cs.utexas.edu!howland.reston.ans.net!swiss.ans.net!newsgate.advantis.net!news-m01.ny.us.ibm.net!news
From: mxfle–(at)–bm.net
Newsgroups: alt.guitar
Subject: Re: Phase Inverter Question
Date: 28 Jan 1995 18:16:21 GMT
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In <3gbfo7$k--(at)--ewsbf02.news.aol.com>, omarnep–(at)–ol.com (OmarNepo) writes:
>In a 59 Tweed Bassman, why are the resistors feeding the plates of the
>phase inverter not equal (ie both 100k)? One is 82k and the other is 100k.
>Is this due to the difference in the transformer windings of side of the
>push-pull. Gerald Weber mentioned something about the difference in his
>article on x-formers.
>
>Later
>Omar
Transformer are different for Class A, but that is not the reason for the
plate resistor difference. For a phase inverter of this type,
the half driven from the tone stack sees an input signal equivalent to the
tone stack output minus the cathode resistor feedback. The other half sees
just the signal from the cathode feedback, which is a slightly smaller
signal, about 20% less (its inverted as well, since the signal is on the
cathode rather than the grid). This makes the current swing in the driven
half 20% larger. Since the current through the tube sections is relatively
a function of the input and independent of the plate resistors, the plate
resistor values are adjusted so the the voltage swing in each section is
the same.
The unbalanced output effect of a phase inverter is greater with a small
cathode feedback resistor, which the Bassman’s 10k ohm is. Some Vox amps
have a large 47k ohm cathode feedback resistor. The cathode feedback is
high enough to neglect the unbalance, so the plate resistors are
equal (100k). In fact one input drives one side of the inverter, the other
input drives the other. The high amount of cathode feedback does limit the
voltage gain available in the inverter.
Regards,
Max
