Jan,
In the case of class A, the quiescent DC current is the same as maximum load current.
Same amount of fixed quiescent DC current (determined by the biasing arrangement, say 2A) flows through the FET ceaselessly. The FET converts a portion of the quiescent DC current to AC signal current for the load. The rest of the quiescent DC current flowing through the FET (or any DC current flowing anywhere in the circuit, for that matter) will be dissipated as heat, so the dissipation is highest under no signal conditions.
In the case of class A, the quiescent DC current is the same as maximum load current.
There is no 'fixed' DC current through the FET, except when there's no signal.
There's only a fixed 2A as the sum of the FET and load current. That's what Norton tells you.
You know that, come on! Stop playing games, please. I'm sure you have better things to do with your time.
jan didden
That's because you're wrong. Sorry, don't take it personal, happens to the best of us.
jan didden
Its a funny beast this one...if you look at it from the outside. The amplifier is a voltage source..and would have low output impedance (apart from the output cap), but at the same it works by diverting currents around the load..
Still think that a two string arrangement driven by a center tap biased interstage transformer and with the load placed between the two strings (takes the coupling cap out) would be a fantastic performer...
Still think that a two string arrangement driven by a center tap biased interstage transformer and with the load placed between the two strings (takes the coupling cap out) would be a fantastic performer...
Yes, you have not understood the operation and thus cannot express your viewpoint in words.
It works.
The problem is:
Is there any change in performance?
Is it better or worse?
Does the better or worse result depend on a number of operational conditions/restrictions?
Are we wasting our time?
Andrew,
the common drain / common collector amplifier is basically an impedance converter. Driving full range speakers necessitates low output impedance. Yet, sometimes we might consider a true transconductance output stage.
the common drain / common collector amplifier is basically an impedance converter. Driving full range speakers necessitates low output impedance. Yet, sometimes we might consider a true transconductance output stage.
My aim is to learn all the time. Maybe you could kindly explain the operation...
That's one of the weak points of getting your knowledge from the 'net. If you are faced with conflicting opinions, how do you decide who's right and who isn't? You really have to have *some* knowledge about the subject yourself - a sort of chicken and egg situation.
Of course, if you get one or two books and do some studying it's much easier, but that takes an effort.
jan didden
Andrew.. for sure we're not wasting our time...
If this amplifier works the way the simulation suggest it does, then it's the only concept I know of that isolates current drive from the power supply caps...as the signal does not modulate them in any way... the current loop is tight and purely local...Seems like a very pure approach. Inefficient yes, but also pure...
If this amplifier works the way the simulation suggest it does, then it's the only concept I know of that isolates current drive from the power supply caps...as the signal does not modulate them in any way... the current loop is tight and purely local...Seems like a very pure approach. Inefficient yes, but also pure...
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You get the same result using a shunt regulator. The sum of the current through the load and the shunt is constant so also in this case the power supply sees a constant DC load current.
jan didden
I have been toying a bit with my dual shunt amplifier....looks rather promising...
at small signal levels where the fets are most linear they simply cancel all distortion...
at levels close to clipping at -+3.8 amps the distortion is mainly 3. order (-70dB) and some much lower higher order..
at small signal levels where the fets are most linear they simply cancel all distortion...
at levels close to clipping at -+3.8 amps the distortion is mainly 3. order (-70dB) and some much lower higher order..
OK, so the supply cap has ripple. Now if we go around in the fig B circuit, where is the ripple seen? You basically have a series circuit of the supply cap, the CCS, and the FET//load, right.
The FET will not react to the ripple, agreed? Since the CCS has much higher resistance than the load, most of the ripple will be across the CCS (just a simple attenuator).
BUT. Look at the ground of the signal source and knowing that the ripple is across the CCS. You will see that the ripple comes into the signal anyway.
No such thing as a free lunch I'm aftraid.
jan didden
Yes, I see that. I should say, "Who would like their speakers to sit at supply rail potential?"
That "plus" rail potential will be connected to device's case, and "0 volt" will be floating. For removing all the confusions, somebody should redraw the schematics with minus rail at the top, and apply the cap-multiplier based on p-channel MOSFET (in this case it will be sitting on the negative rail). In this case everything will be understandable forever.
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