150W Is it good?

[millwood]

once you take that an LPT will not split the current evenly between its two legs, you will not be able to calculate the current for the input stage (the legs) thus the current for the VAS.

That is like if I don't tell you what I2 is in my simulation. what do you do then?

[Eva]

But I know that Iq of the right leg of the upper LTP has to be EQUAL to Iq of the right leg of the lower LTP

This is the trick

What implies the fact of having equal 'Iq' in upper and lower right legs?

Oh yeah, the LTPs have to be balanced to get that

[ChocoHolic]

..if you try you simulation complentary doubled then
you should not only find symmetric current sharing in the input stage.
You can also skip the current source in you output stage...
the output stage may run with 3..4mA...


Or first try Evas proposal with R6 175 Ohms...
should also end up in good balanced currents.


To All:
Have fun!
*leaving_soon_for_real_live_evening_activities*

[Tube_Dude]

Originally posted by millwood



Dude, you are going in circles, .

That's the single biggest flaw in this discussion, and the simulation was to show you that an ltp will not by itself automatcially balance leg current.

The LTP can not balance 100 % perfectelly...but with emiter resistors will be close enough...
Also the DC off set of the amp...will never also be 0 ...but close enough !

What is 100% correct???

[millwood]

Originally posted by Tube_Dude
The LTP can not balance 100 % perfectelly...

there is no mechanism for the LTP to balance. as you can see in the simulation, the leg current is determined by factors outside of the LTP.

Otherwise, people wouldn't be inventing current mirrors to balance leg current in LTPs, .

[Eva]

Why are you avoiding the complimentary LTP and VAS simulation ?

Maybe because this topology balances itself due to interaction between the upper and lower sections?

Don't fear, try it...

[Tube_Dude]

Originally posted by millwood


Otherwise, people wouldn't be inventing current mirrors to balance leg current in LTPs, .

Current mirrors are used for increased current drive for the VAS...because with CM there are no current lost in the colector load resistor of the LTP and all current from it, can be used for driving the VAS...

[ChocoHolic]

*lastkissestoEva*

CU all (in cyberheaven)

[pjacobi]

I don't want to disturb the discussion about the input and VAS stage (only noting that the analysis in Unstable VAS current in amp from Slone book seems be extensive and accurate), but add a point about the EF output stage.

In several places it is recommened not to tie the driver (and pre driver) transistors' emitters to the output node, but connect them with a single resistor additionally paralled by a 1uF capacitor.

Interestingly even the authors suggesting this in their "theory" chapters, don't (always) follow their own advice in the circuits presented in the "practical" chapters.

So can anybody shed light on this theory/practice split? Would the performance of the original poster's circuit be enhanced by observing the theoretical advice?

Regards,
Peter Jacobi

[Eva]

A balanced contribution to a little unbalanced world ...

PD: Ideal devices with Vbe=750mV and beta=100
PD2: Well, not so ideal... VAS actually show something like 844mV VBE