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In Reply to: RE: sorry- posted by Ralph on August 02, 2007 at 10:28:04
"When the signal arrives from the single-ended gain stage, the cathode coupling of the differential gain stage ('LTP' using the rather quaint vernacular) causes the other half of the differential amplifier to have an inverted waveform."
Right. That's why we use it as a phase 'splitter.'
"There is an equal number of gain stages thus seen by both phases of the signal, inversion occurs in real time rather than have a propagation delay."
Sorry, I don't see the equal number of stages. Could be that you're trying to tell me something and I just don't get it, but at the moment I just don't see it. I certainly don't see how it must follow from your first statement.
"You forgot the essential capacitor that goes to ground from the grid of V2."
I didn't forget any capacitors. I designed the circuit so that the grids of the input stage are at ground potential; no DC blocking caps are required.
-- Dave
Follow Ups:
If you have a single-ended input to the differential driver, the second grid has to be a virtual ground. Even if the voltage is at or near ground, the *AC* potential cannot exist- that's what the cap is for.
That way that half of the diff amp is trying to be constant current, insuring the proper cathode coupling. A differential amplifier by the way is always considered a single gain stage- its not two.
Look carefully at the drawing. The second grid is connected directly to ground.
I think I see where the confusion is coming from. Yes, the signal at the grid of V4 comes from both V1 and V2, but the signal at V3 also comes from both V1 and V2, not just V1. This is because the signal on the grid following the 'LTP' is derived from the difference between the two plates. Hence, differential amplifier. So, the signals at both grids go through one gain stage. Because each grid signal is taken from the different plates, they have to be 180 degrees out of phase (barring distortion of course).
Chris
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