I have not been following this thread closely, but here are a few comments
on the latest schematic ( https://www.circuitlab.com/editor/#4275j4 ):

1. Because the load is "ground" referenced, you're going to get quite a
bit of current through the voltage divider formed by R17, R18, C7, and C8.
Since you have a DC blocking capacitor in the output (C5), there's nothing
preventing you from running the op amp single supply. To do this, I'd
ground the negative side of the battery, then connect the junction of R17
and R18 to the non-inverting input of the op amp. C7 and C8 can then be
substantially reduced in size. In fact, I'd probably just do a 100nF from
the non-inverting to ground and 100nF from the positive supply to ground.
The inverting input will be biased to 1/2 supply by the feedback resistor.
A single capacitor from the non-inverting input to the input resistors
(R24 and R27) can provide DC blocking for the two ground referenced
inputs.

2. The net named combined will ideally have no AC voltage. It will be the
output voltage of the circuit divided by the open loop gain of the op amp,
which is normally quite high.

3. Is there a reason to use two resistors (R12 and R29) instead of just one=
?

4. The load resistance here is quite low (less than 10 ohms). The
frequency response on the low end will be limited to 1/(2*pi*R*C) where R
is the load and C is C5.

5. The low load resistance will probably cause current limiting of the op
amp, clipping the output voltage. The datasheet shows a current limit in
the 20 to 40mA area, so the peak output voltage would be in the 200 to
400mV area before the amplifier current limits. There are amplifiers with
higher current limits. There are also amplifiers that are designed to
drive speakers that have higher output limits. It's also possible to add a
buffer at the output. This could be a complementary emitter follower with
the feedback coming from the two emitters. Another clever current boost
circuit I saw years ago used a PNP transistor on "top", and an NPN
transistor on the "bottom". The top transistor had its collector going to
the output, the emitter to positive supply, and the base the the op amp
positive power pin. There was also a resistor between emitter and base. As
the op amp drew current through its positive power pin, the current
through the emitter-base resistor of the top transistor would eventually
reach Vbe of the top transistor, causing current to go through the top
transistor, pulling up the load. A similar circuit handles pulling down
the load.

6. One way to get rid of C5, it cost, board space, and frequency response
limits, is to use a "bridge tied load." A pair of amplifiers are built,
one outputting the inverse of the other. With no AC input, the DC outputs
are the same. The load is tied ("bridged") between the two outputs. As AC
is applied, one output goes up, the other goes down, putting double the
voltage across the load.

7. 9V batteries have very low capacity. It won't drive this for very long.

Good luck!

Harold



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