Although the output is taken from the source of the upper FET, it is operating in common source mode because the winding driving it gets it's reference (AC wise) from the output (source). Since the windings float AC wise, both top and bottom FETs operate in the same mode. Any distortion caused by non-linearity of the top or bottom FET circuits will affect the output symmetrically causing odd order harmonics.
Thank you Loudthud for your interesting analysis. It, and that of flg earlier now suggest that the upper FET is simply [and only] working as a modulated current source load to the bottom FET which is the [only] voltage gain device. Thus the math expression ZM used earlier relating voltage gain to the [bottom] FET's transconductance and load fits. This working scenario is palatable and digestible. I am happy.Although the output is taken from the source of the upper FET, it is operating in common source mode because the winding driving it gets it's reference (AC wise) from the output (source). Since the windings float AC wise, both top and bottom FETs operate in the same mode. Any distortion caused by non-linearity of the top or bottom FET circuits will affect the output symmetrically causing odd order harmonics.
yup
output Jfet is either plain stupid or just careless - where load is connected ; all he cares/reflects is own internal impedance , as function of G vs. S modulation ;
considering this - both Jfets (lower and upper one ) are behaving in identical way ......... correctly speaking - exactly opposite way - meaning - while one is decreasing internal impedance , other one is increasing internal impedance in same amount ; just look at them as on two variable resistors , working unison in opposite directions .......... and you'll forget on all common source/drain/blahblah hogwash (I'm anyway unable to remember meaning of that "common S/D/G" nomenclature .... it just don't blend with my way of thinking )
as remainder - quasi complementary is , by function , same as complementary
I'm just (or "just" ) lacking in electronic vocabulary to express that
output Jfet is either plain stupid or just careless - where load is connected ; all he cares/reflects is own internal impedance , as function of G vs. S modulation ;
considering this - both Jfets (lower and upper one ) are behaving in identical way ......... correctly speaking - exactly opposite way - meaning - while one is decreasing internal impedance , other one is increasing internal impedance in same amount ; just look at them as on two variable resistors , working unison in opposite directions .......... and you'll forget on all common source/drain/blahblah hogwash (I'm anyway unable to remember meaning of that "common S/D/G" nomenclature .... it just don't blend with my way of thinking )
as remainder - quasi complementary is , by function , same as complementary
I'm just (or "just" ) lacking in electronic vocabulary to express that
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I thought ZM's math was regarding the reflected source impeadance seen by the output FETs looking back into the transformer?
Which I thought to be about 35 ohms or so from the buffer x (the turns ratio) ^2 (1) = 35 ohms not 75? The xfrmr and the FETs probably like that?
If you are going after some gain in the xfrmr, say 1:5.6, that would be 30 X (5.6^2) = 168 ohms. Now you are starting to see some source impeadance issues but, not really enough to worry about.
This was all predicated on the fact that I was thinking there was something like a B1 arrangement on the input and I thought that would be about 35 ohms. I't not. It's a complementary buffer. So whats the output impeadance of that? 17 ohms?
Which I thought to be about 35 ohms or so from the buffer x (the turns ratio) ^2 (1) = 35 ohms not 75? The xfrmr and the FETs probably like that?
If you are going after some gain in the xfrmr, say 1:5.6, that would be 30 X (5.6^2) = 168 ohms. Now you are starting to see some source impeadance issues but, not really enough to worry about.
This was all predicated on the fact that I was thinking there was something like a B1 arrangement on the input and I thought that would be about 35 ohms. I't not. It's a complementary buffer. So whats the output impeadance of that? 17 ohms?
Sorry for injecting this confusion. ZM showed a thread even earlier than the one pertaining to the transformer [above] showing the relationship between the gain of a FET and its parameters plus load resistance etc. [e.g for for F5].I thought ZM's math was regarding the reflected source impeadance seen by the output FETs looking back into the transformer?
take it easy ;
what's output impedance of sole (say that load up is plain CCS ) buffer 2SJ74 ?
it's 1/S+Rs
what's load impedance of sole (say that load down is plain CCS) buffer 2SK170?
it's 1/S+Rs
so - when you have those two together , as complementary buffer - where each one is CCS to other one , you have output impedance as (1/S+Rs // 1/S+Rs) = (1/S+Rs)/2 , if we assume that S of both are ~ same
so - if you have , say , 50 Ohms of output impedance of buffer (lazy and don't remember S of sissy Toshiba's) , when output current capability is divided to 2 secondaries , it's logical that each secondary is seeing double that - 100 Ohms
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ZM: One liner replies may not be enough to teach DIYers and/or drive your point. Sometimes, the saying that Brevity is the soul of wit does not work well in/for technical matters. A clear math equation [like you showed earlier] may not cut it either. You pack a lot of technical/technology know-how. So please, be generous with your explanations; write, and write some more until there is no more for you to teach. You and us will feel enriched and satisfied in the end.@ Antoinel
........ blahblah ......... serve and protect
edited few lines above
ZM: That is a damn great [and clear] explanation regarding the working of the output stage of F6. ***** out of 5 for you. You are lacking nothing. Please keep up the teaching.
I would think it is logical to look at 1 secondary and say 50 ohms (1:1+1) Or the other. But a 1:2 connection (both is series) is 200 ohms???
I guess now and then, now being now, other's are dumber than ZM
OK - if S of both 2SJ74BL and 2SK170BL is declared to 22mS , meaning that 1/S=45Ohms , with (say) 10 Ohms for Rs , that means that each of them is represented with 45+10 Ohms resistor = 55 Ohms resistor , and they're working together (in parallel) to move that Archimedes Lever for Earth move...... so these two are source with 55 // 55 = 27,5R output impedance .
so - if you follow my drift - when you have water pipe giving you 100 L/min capacity , and you divide it to two equal hoses , each hose will take/will be given with 50L/min ........ it's the same - each secondary will be given with 50% of source capacity - meaning reflected/seen capacity will be 50% of
overall one ; in this case - each secondary will be fed with reflected source of 27,5R/2=55R internal impedance
why we are talking about that ...... ?
I can't remember
Well, if your not a Bottle Head, some of this is difficult to get strait. Other than my Akido(transformerless) I haven't built anything Tube in 35 years!why we are talking about that ...... ?
I can't remember
