150W Is it good?

[sreten]

Originally posted by Tube_Dude

What you mean by input stage balance???

That in the two colectors of the LTP pass the some current?

Yes, the two sides of each LTP have equal
current for minimum open loop distortion.

No, this only ocurrs when both LTPs are balanced since upper and lower bases are tied and current sources are of the same value

I'm sorry but as far as I can see both LTP's can be unbalanced
in the same direction, so Iq's are equal, I still cannot see how
each LTP is forced to be balanced.

As I said above slightly rephrased:

Its quite easy to build an amplifier with a single ended unbalanced LTP input, and it will quite happily still work, so why is this not possible with a double ended LTP input ?

sreten.

[Eva]

Originally posted by sreten


I'm sorry but as far as I can see both LTP's can be unbalanced
in the same direction, so Iq's are equal, I still cannot see how
each LTP is forced to be balanced.

sreten.

How both LTP could be unbalanced in the same direction?

Remember that upper lower bases are tied toghether, current sources are of the same value and transistors/resistors are matched by pairs like in a single ended LTP

To get noticeable unbalance in the same direction you need severely unmatched Vbe or resistors

[In a single ended LTP with a current mirror you would get severe offset in the same unmatched conditions]

[Tube_Dude]

Overall feedback ensures current balance...if not ,a large DC ofset will be displayed at the point of connection of the two colectors of the VAS.

[sreten]

Originally posted by Eva



How both LTP could be unbalanced in the same direction?

Remember that upper lower bases are tied toghether, current sources are of the same value and transistors/resistors are matched by pairs like in a single ended LTP

To get noticeable unbalance in the same direction you need severely unmatched Vbe or resistors

[In a single ended LTP with a current mirror you would get severe offset in the same unmatched conditions]

In real life very easily. Consider a single ended LTP driving a
VAS base-emitter to ground. Lets say Vbe of VAS = 0.7 volts.

Lets assume a LTP current of 2mA so Rc = 700R, 1mA per side.
(collector resistor only used on one side of the LTP)

A Rc of 1.4K will force a unbalance of 1.5mA/0.5mA and
this will force the Vbe's of the LTP pair to be different.

Nevertheless the amplifier will still function quite happily.
(With assymetric slewing and more DC offset and distortion).

sreten.

[millwood]

Originally posted by Eva
A balanced contribution to a little unbalanced world ...

slightly different numbers (multisim):

voltage drop upper collector resistor, voltage drop off lower collector resistor, and current through the VAS.

with no VAS emitter resistors: 728mv, 781mv, 30.8ma;
with 1ohm VAS emitter resistors: 752mv, 806mv, 27.8ma;
with 10ohm VAS emitter resistors: 853mv, 907mv, 14.7ma;
with 100ohm VAS emitter resistors: 938mv, 994mv, 2.8ma;

do you guys spot a problem?

I have no idea how the simulator calculated those numbers,

I will try two other simulators later.

[sreten]

Originally posted by millwood



slightly different numbers (multisim):

voltage drop upper collector resistor, voltage drop off lower collector resistor, and current through the VAS.

with no VAS emitter resistors: 728mv, 781mv, 30.8ma;
with 1ohm VAS emitter resistors: 752mv, 806mv, 27.8ma;
with 10ohm VAS emitter resistors: 853mv, 907mv, 14.7ma;
with 100ohm VAS emitter resistors: 938mv, 994mv, 2.8ma;

do you guys spot a problem?

I have no idea how the simulator calculated those numbers,

I will try two other simulators later.

The simulation is flawed as there are no base resistances which
allow Vbe differences to develop. With the circuit shown the the
left and right Vbe's must be equal, not the case in real life.
The circuit has no voltage input.
If Vbe's are equal then Ic must be equal as shown,
but this is putting the cart before the horse.

sreten.

[millwood]

protel dxp didn't do much better: it wouldn't simulate when the VAS emitter resistor is 0ohm, complaining about singular matrix.

again, the same formate: upper collector resistor voltage drop, lower collector resistor voltage drop, and VAS collector current.

with 0.0001ohm VAS emitter resistor: 634mv, 1089mv (Wow!), and 12.55ma.
with 1ohm VAS emitter resistor: 643mv, 1098mv, 11.82ma.
with 10ohm VAS emitter resistor: 694mv, 1142mv, 7.73ma.
with 100ohm VAS emitter resistor: 760mv, 1195mv, 1.85ma.

BTW, all of them started to oscillate after about 15us, about 1 cycle per 0.1us (10mhz?).

[millwood]

Originally posted by sreten
The simulation is flawed as there are no base resistances which
allow Vbe differences to develop. With the circuit shown the the
left and right Vbe's must be equal, not the case in real life.
The circuit has no voltage input.
If Vbe's are equal then Ic must be equal as shown,
but this is putting the cart before the horse.

sreten.

not sure which base resistance you were talking about and which "left" and "right" you were talking about. It actually had a 10k resistor from the non-inverting end to the ground. and it does have a signal source (1vp, 1000hz).

the Vbe difference you saw (on the two resistors) are due to difference between npn and pnp devices.

[sreten]

I've realised one thing :

each LTP pair in the double ended circuit does not need
to be balanced, the double ended nature of the circuit will
cancel distortion the same way a single ended LTP pair
that is balanced will.

So my original comment about "input balance" is wrong,
the input is balanced, whether each LTP pair is irrelevant.

But the balance of each pair needs to be known to
know the VAS current, this balance and the VAS current
is still indeterminate, in that its not easily calculated.

I'm sure it is deterministic, but my maths aren't up to the job.
Enter a simulator, with all relevant circuit values.

What I am sure about is a Rc value gives a VAS current
and the current balance in each LTP pair, and that one
cannot easily calculate either without assumming one.

Unless I've missed something..........

sreten.

[sreten]

Originally posted by millwood



not sure which base resistance you were talking about and which "left" and "right" you were talking about. It actually had a 10k resistor from the non-inverting end to the ground. and it does have a signal source (1vp, 1000hz).

the Vbe difference you saw (on the two resistors) are due to difference between npn and pnp devices.

In EVA's original circuit assuming output = ground (which must be
the theorectical case), each LTP pair bases summing horizontally
must have the same Vbe, (which is why Ic's are equal), and this
circuit has no voltage input, as it has no base resistances.

The value of Rc only determines the output current, theorectically.

sreten.