Le Monstre MOSFET

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Just toying with different pieces of circuits and here’s an utter heresy schematic I came up with - Hiraga’s Le Monstre front end with lateral MOSFET output. Sims OK but I am have no clue what the output impedance of this configuration could be?
Can someone please advice how to calculate for Z Out or guess the proximate range of it with this output configuration?



Regards

Argo
 

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MikeB said:
Hi !

As this circuit is that "simpel", why not just trying it ?
Numbers can't tell you really how it will sound...
Have you made AC-sweep ? The jfet-input might need
a small cap (~100pf) to ground. (Just connected to gates)

Mike

Well, I have tried to many bad sounding MOSFET circuits lately. But seriously, even if the numbers can’t please ears they can calm down a restless spirit, as I am to not reach for soldering iron too hastily.
I mean if that circuit would have output impedance of 20 Ohms? Not good for driving even 16 Ohm speakers.


Argo
 
Argo: The output impedance is secured through the feedback loop.
After all the output impedance of the original bipolar collector's is also quite high, was it not for the feedback loop.

I think this design will work just fine, and probably you will get very good results from it. You can even use IRF540 / 9540 (much cheaper) with i would say - same result.

You don't need the 330 mOhms 'emitter' resistors when using a MOSFET. Just connect the sources directly to rails.

Le Monstre is one of the few truely innovative amplifier designs seen the last 20 years. Good idea to upgrade it with MOSFET's Argo! :)

Best regards

Lars Clausen
 
MikeB: Before i started experimenting with MOSFET amps (in the real world) i would have been inclined to agree with you on this point.
However if you look on the datasheet (Vgs / Id curves) you can see the source already has an internal resistance of about 1 Ohm @ 0.5 A (which as i remember is the idle current of the Monstre)

So 1 Ohm or 1.33 Ohm ... das ist vohl egal?

If you need to parallel several devices, then of course it's another ballgame.

Best regards

Lars Clausen
 
Hallo Argo,

following your posted schematic it is doubtful if this amplifier will perform well. It is simple. But probably too simple. The main drawback is the high impedance of the driver stage (Q3, Q4). That feeds the large input (Cgs) capacitor of the MOSFETs. Even worse these capacitors vary with the signal amplitude. So you will have a significant pole with varying corner frequency. That is bad for stability. And it could cause ringing if the amplifier is stimulated with pulses. And music consist of pulses... So I would recommend looking for a more sophisticated schematic. Or you might enjoy the special sonic colour of instabel running power MOSFETs...

Arcolette
 
Lars Clausen,

The source resistors in this particular circuit seem to help maintain the nice dominating even harmonic distortion pattern, produced by input cascades, through the output stage. Without the source resistors, the odd order harmonics tend to dominate as in usual push pull circuit. But this again is only in my simulations- in the real world this may be different though.

Arcolette,

I am aware of the shortcomings you mentioned, the thing what worried me more was an output impedance.

Mike,

Exactly, those resistors also became handy for bias current measurement ;)


Argo
 
Arcolette: You are right in theory. However you are not fighting against the input capacitances of the MOSFET's rather you are only fighting the reverse (Miller) capacitances. And for an IRF540N they are only like 50 pF at 25V and 750 pF at 1 V, so 1.5 kOhm gives you a corner frequency of the MOSFET related pole of something like 140 kHz. You should be able to get a stable amp out of that.

Even at these small powers, you might as well use an even smaller MOSFET like the IRF520 / 9520, they have about half to 1/3 of the Crss, and so the pole is even higher up. I think it's quite nice still.
And if not, you can always reduce the 1k's in the top of the cascodes.

Argo: If so it only shows just how well thought through this design is. Jean Hiraga did his homework!

Maybe here is a better alternative to GC and ZEN amplifiers?

All the best from Lars Clausen
 
Can someone please advice how to calculate for Z Out or guess the proximate range of it with this output configuration?

Maybe I can..

I never did it, but I thought about it while driving home... The amp is actually a current output amp (open loop) .. voltage in is converted to ampere out.. transductance.. you can calculate the total transductance of the amp.. first of the JFETs (24millimhos?) which due to the resistors (1k) makes a voltage out. gain=25x ... and then that voltage to the MOSFETs transductance ..(4mhos?) which give a ampere out .. 100Ampere / volt ...

The gain of the amplifier is set to 11x .. so for every volt you get 11 volt out .. in 8 ohm you need 1.375A.. but have have 100 ampere availeble for 1 volt in.. so you have a feedback factor of 72x...

Is your output impedance now 1/72 Ohm?

Well ..somebody can correct this , I'm sure.. and I am interested in the correct calculation...

cheers,
Thijs
 
tschrama said:
first of the JFETs (24millimhos?) which due to the resistors (1k) makes a voltage out. gain=25x ...

Hi tschrama

You forget that the input Fet have half of the 500 Ohms pot in this source then the voltage gain of one half of the input stage must be near 1k/250 = 4

If the output mosfet have a transconductance of 4 S then for a 1volt input we have 4 amps output...with a load of 8 Ohms this make 4X8 = 32

Then we have a input stage with a voltage gain of ~4 and a output stage of gain~32 this make a total voltage gain of 4X32=128.

As the voltage gain defined by the feedback resistors is 11.

The feedback factor is ~11,6

The output impedance with the 8 Ohms load will be 8 / 11,6 = 0,69 Ohms

Regards
 
Thanks for the corrections! Mind you, it was my first ever attempt to calculate such a thing ...

I think we both forgot that the circuit is a symtric circuit.. with twice the transductance (if in class A) ... open loop gain is doubled.. output impedance is halved/


regards,
Thijs
 
tschrama said:


I think we both forgot that the circuit is a symtric circuit.. with twice the transductance (if in class A) ... open loop gain is doubled.. output impedance is halved/

Yes you are right...i remenber at first.... see:



the input Fet have half of the 500 Ohms pot in this source then the voltage gain of one half of the input stage must be

But after...the other half is...."gone with the wind"...:D

Cheers
 
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