150W Is it good?

[sreten]

I'm sorry TD but you got this one wrong.

You cannot assume the LTP's are balanced.

Your basic electronics in this case are too basic.

My original point is the LTP's are almost certainly not perfectly
balanced and the VAS current is not exactly known, because
circuit values cannot be easily defined.

Assume a VAS current say 30mA with 47R.

Work backwards and this will give you an unbalanced
LTP with too much current in the collector resistor side.

You can now adjust the collector resistor to give balance again.

But note you started by assuming a current, the only way you
can set the current is by assuming balance, and you cannot
do this as the above example shows.

Your calculations assume the collector resistor is the right value
to give balance, and from this you have derived the operational
points of the circuit.

But you can derive another set of operational points by assuming
a percentage + or - imbalance.


sreten.

[Tube_Dude]

Originally posted by millwood


In a typical ltp (with resistors), the current in each arm is determined by the VAS stage. So you would know the voltage drop off R15, plus the Vbe of T7. then you can calculate the current through the 680ohm resistor for T2.

In this case, there is no way to determine the voltage drop off the 680ohm resistor, nor the idle current for the VAS.

Sorry to say...but what you are saying is not correct...do some readings about the subject.

I don't want to repeat my self...i have done the calculation you have asked...and you can be sure that they are correct.
Have you any other values?...Of your calculation!!

[Tube_Dude]

Originally posted by ChocoHolic
Hi folks!
I think Dude is right.
Depending on 47 Ohms or 100 Ohms I also end
up in about 10mA (100 Ohms) or 20mA (47 Ohms).

Thanks Markus...after all is a very easy calculation!!!

[ChocoHolic]

...well they will be more or less balanced, let's says less than
15% mismatch.
But I have to correct my values.
Red LEDs typically show a voltage drop around 1.8V (if I remember right), not 2.8V.... Coming from this we will have about 1. 2 mA in each tail. Multiplied with 680 Ohms ==> 816mV minus Ube about 200mV across the resitor.... 47 Ohms should be OK without frying the transistors....

[millwood]

Originally posted by Tube_Dude
Have you any other values?...Of your calculation!!

I cannot calculate it, because it is not calculatable as sreten has pointed out, unless you wanted me to calculate it incorrectly,

[Tube_Dude]

Originally posted by millwood


edit: tube-dude, you may wan tto look into this thread to better understand the issue here.

Unstable VAS current in amp from Slone book

In the first schematic of the link you gave the VAS current is not perfectely defined because the VAS transistor doesn't have emmiter degeneration (resistor)..

[Tube_Dude]

Originally posted by ChocoHolic
...well they will be more or less balanced, let's says less than
15% mismatch.
But I have to correct my values.
Red LEDs typically show a voltage drop around 1.8V (if I remember right), not 2.8V.... Coming from this we will have about 1. 2 mA in each tail. Multiplied with 680 Ohms ==> 816mV minus Ube about 200mV across the resitor.... 47 Ohms should be OK without frying the transistors....

Hi Markus...

2,8 volts was a arbitrary value that millwood gave us...only for calculation.

[millwood]

Originally posted by Tube_Dude


In the first schematic of the link you gave the VAS current is not perfectely defined because the VAS transistor doesn't have emmiter degeneration (resistor)..

if you are talking about Q18 on the upper side: if it had a emitter resistor, you would not have been able to calculate the current on R17.

the problem with that circuit is that while the current and voltage on the current mirror are determined, the current going through Q18 isn't determinate.

much like the 2nd half of our problem here.

[sreten]

Originally posted by Tube_Dude


In the first schematic of the link you gave the VAS current is not perfectely defined because the VAS transistor doesn't have emmiter degeneration (resistor)..

In a non balanced circuit, one LTP and one VAS transistor the
VAS current is set by a current source. The non degenerated
base emitter is treated as a Vbe volt drop virtual earth.

This allows you to set the collector resistor/ current by assuming
a Vbe drop across the resistor, the resistor is set to give current
balance in the LTP.

As the VAS current is known, degenerating the VAS transistor
doesn't cause any problems in balancing the LTP, the extra
drop across the emitter resistor just needs accounting for.

You cannot do this in the balanced 2 LTP's and 2 VAS Tr's case.

sreten.

[Tube_Dude]

Originally posted by millwood



if you are talking about Q18 on the upper side: if it had a emitter resistor, you would not have been able to calculate the current on R17.

the problem with that circuit is that while the current and voltage on the current mirror are determined, the current going through Q18 isn't determinate.

much like the 2nd half of our problem here.

In the problem of the link the VAS is drived by a high impedance point (connection of colector of Q1 and Q6)...only in this case the voltage at this point is defined by the current in the VAS.
In our case the resistor has 680 Ohms and the input of the VAS with a current gain of say 100 will be 10KOhms with 100 Ohms emiter resistor
So the resistor voltage define the operation point of the VAS and consequently the current in the VAS.