v4 pin 6 is this cathodyne phase inverter supposed to chop off the top of the wave form on the positive side? it doesnt do it to the negative side. seems odd to me and I dont see a lot of these. It also chops off the neg bottom of the waveform on pin 8
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This guy has a this exact circuit described. Everything seems normal except my grid measures 21V not 55V (meter error??) https://www.tdpri.com/threads/fender...erters.755580/
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I guess this IS normal for this design, so never mind. https://www.youtube.com/watch?v=IjAVqBRT92g
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The Wiz covers the Cathodyne Phase Splitter pretty well.
http://www.valvewizard.co.uk/cathodyne.html
It is Not a good choice for amps where the power amp is going to be overdriven although there are things which can be done to improve it.
21V on the grid is NOT good - is the correct tube fitted? Has the circuit been modified?.
The cathode should sit at 1/4 to 1/5 of B+ and the grid will then sit at a few volts (the bias volts level) below that.
Cheers,
Ian
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Yes, the resistance of the volt meter will form a potential divider with the 1M grid leak resistor and draw current through it, pulling the grid voltage down whilst it’s being measure thus. Evidence of that will be a thump from the speaker, and a second volt meter will reveal that the cathode and anode voltages change.
To correctly measure the grid voltage of such an arrangement, measure the voltage to 0V at the other end of the grid leak, then measure the voltage across the grid leak, and then add those together.
Generally there’s only a few mV across the grid leak, so only the 1st measurement (0V to the ‘non grid’ end of the grid leak) is necessary.My band:- http://www.youtube.com/user/RedwingBand
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We have discussed this elsewhere. Your meter loads teh grid voltage so it appears low. The thing to do is measure the cathode voltage, then measure voltage from cathode to grid.Education is what you're left with after you have forgotten what you have learned.
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