🚀 go-pugleaf

roy@alanine.phri.nyu.edu (Roy Smith) writes:
>
>	I'm stumped solving the following DC circuit to find the bias points
>of all the transistors.  It's the final audio amp for one channel of a
>Pioneer SX-770 reciever, after a few unimportant simplifications.
>
>...
>	I don't know how to attack this.  I could do a straight forward mesh
>analysis, but I'd end up writing KVL around 12 loops and I don't have any
>plans on solving 12x12 matricies manually, at least not in this life.  Are
>there some tricks I can use to break this into several simplier circuits?  My
>first thought was to slice it into 3 vertical sections and assume that the
>base drive for section N+1 is negligable when solving section N, but I can't
>get the voltage at point V1 without knowing the collector currents for the
>later sections.  In practice, I can cheat because the schematic lists the
>nominal V1 as 20.2V, but if I didn't know that, how would I solve for it,
>other than building the circuit and measuring it with a voltmeter?
>
First, I suspect you haven't drawn the circuit quite right:  I'm surprised
to find the bottom of the 1.3v thing connected directly to the
(apparent) output.  As drawn, the upper "half" of the circuit is
independent of the lower half, except in that it "sees" voltage change
at v1.  A first order approximation is that the base-emitter voltage of
each upper NPN is 0.6 volts, leaving 0.1 volts across the 0.5 ohm
emitter resistor ... normally, there would be a way to adjust this (I
suspect the 110k has something to do with this if properly drawn...but
this is supposition only at this point).

But answering your question (which was "how" not "do it"):  I would
never think about setting up a set of node or loop equations for such a
circuit, because there are too many variables: not too many loops, but
too many things that are heavily temperature dependent, dependent on
exactly what parts were put in in the first place, etc.  Instead, I use
some simple rules of thumb to look at it: things like b-e junctions of
silicon transistors drop about 0.6 volts when forward biased, and they
should be temperature compensated somehow to stabilize bias currents
(possibly through cancellation, possibly through feedback...).  Things
like emitter resistors reflect back to the base terminal as about beta
times the resistance, but beware of beta changes because of saturation
(corner cases).  I'd suspect the two rheostats are used to set the
quiescent currents in the output stages; the bias point is probably set
by yet another pot in the feedback circuit (which presumably isn't
shown), or just by design of the feedback ckt, no pots needed.

An example:  the collector current of the lower-left NPN is set by the
voltage at the top of the "110k":  the emitter resistor reflects as
_about_ 15k ohms to the base, in parallel with the 6.8k.  That parallel
combination forms a voltage divider with the 110k.  That sets a base
voltage for the transistor.  That base voltage minus 0.6 volts appears
across the 150 ohm emitter resistor, establishing its current.  So long
as the 0.6 volts is small wrt the base voltage, the 15k assumption was
OK (it's not in this case, so adjust the effective resistance up
accordingly, or alternatively, figure the base current in perhaps by
writing a little more complicated voltage divider equation for this
little subloop).  The collector current is about the same as the emitter
current:  if the beta is truly 100, then collector current is about 99%
of emitter current.  Repeat this procedure to find the collector current
of the PNP.  Note that by the time you get to the output NPN, the
current is highly dependent on the values you actually put in for the
110k and the 25!  It's also pretty dependent on the value of b-e voltage
you assumed; that's a big part of the reason for the pots.

Even more fun than analyzing such things is designing them so that they
really work right.  The ckt shown looks like a minimum cost to get the
job done (I'm quite surprised at the lack of a bootstrap to the base of
the upper-middle NPN's base to maintain current when the output swings
high, and also at the apparently large required drop across the bottom
section when the output swings low.  This thing must clip well inside
the power rails...).