My discrete opamp

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Discrete Op-Amp.

O.k. I give up - can someone tell me how the two diodes, at the current through them, can lift the bias voltage sufficiently to achieve 17mA quiescent current. There are two diodes “feeding” two transistor junctions; so isn’t the net voltage across the two Re’s nominally zero? Where does the extra voltage come from to provide the 340mV required? A 33R resistor in series with the diode string would do it, or an extra diode, as has been suggested, but then Re would have to go down to about 4R7 for the required 17 mA.

Or am I totally wrong?

Brian.
 
Re: bump

AndrewT said:
Udip's schematic shows output Re=10 Ohm.
How do you calculate the output impedance?
For both the open loop and closed loop (gain =A) would be useful.
AndrewT said:
Hi,
Can anyone help with my post41 question

Hello my mate AndrewT

Somebody can inform you! I am sure!

Personally I never bother at all to find output impedance,
not within my voltage+power amplifier, inside the feedback loop.

I know, if I did design my amplifier properly according to my knowledge,
it will have a figure that is good enough for the job.
And output Z wil be fairly constant,
when used with intended loads (loudspeaker Z variations in audiable band)
and within its restricted power bandwidth (=not too high).

This is maybe why I never bother to calculate, estimate
'damping factor' of my amplifiers, either.
If an amplifier is good and properly design for its task
then wont matter much with trying to catch such numbers.


I know Nelson Pass has expressed similiar ideas.
In this forum and at his websites.
This makes me feel totally relaxed, about this issue.
:cool:
 
Hi,
I don't think so, maybe high impedance speakers over 300ohms but even then I think current ability will be struggling.
This circuit can be scaled to drive 600r with a tiny bit of parasitic capacitance, but asking it to drive a reactive load is a different ball game.

Low impedance! how low? 8ohm, 16ohm? 32ohm?
 
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