> I have an application where I need to monitor and
> digitize the current waveform output of an audio power
> amplifier into a subwoofer (speaker).
> Previously, I used a hall-effect based sensor, but for
> cost purposes (consumer product), it is impractical.
> So instead, I have decided to use a 1 milli-ohm
> current sense resistor. The problem now is creating an
> electrical interface to convert the instantaneous
> current (0 - 50 amps) into a proportional
> instantaneous voltage (0-2.5v). Since the audio output
> is not referrenced to ground, some type of
> differential circuit must be used. But although the
> voltage across the resistor will always be small, the
> voltage at each of the 2 resistor taps, with respect
> to ground can swing as high as much as +/-100 volts.

Forget the ground referenced diff amp.  You need too much common mode
rejection for that.  You are trying to measure 50mV on a 100V signal.  Thats
20 * Log10(100V/50mv) = 66dB common mode rejection just to get the noise
signal the same size as the signal of interest.  Let's say you want at least
1 part in 100 accuracy (a little less than 7 bits), then you need another
40dB of common mode rejection for a total of 106dB.  Clearly you need to
attack this in a different way, and this doesn't even get into the very high
common mode range required of the diff amp.  I can think of two approaches
off the top of my head:

One way is to float a regular amp with one end of the current sense
resistor.  This would require a separate small power transformer to make the
floating supply voltages.  This floating section would digitize the signal
and pass back the result to the rest of the world via opto isolators.

Another way which I like a little better for this problem is to use a small
transformer which you might construct yourself.  The primary would only be a
turn or two of thick wire, and wouldn't add significant inductance in series
with the speakers.  The secondary can be more turns of a fine guage wire
because there will never be current thru it.  Your ground referenced circuit
measures the open circuit voltage produced by the secondary.  You will want
an electrical ground shield between the primary and everything else to
eliminate capacitive pickup problems.  The output will actually be
proportional to the current and its frequency, but a single pole filter
should be able to compensate for this over the frequency range of interest.
Most of the tricky part will be making it work at the low end of the
frequency range.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olin@embedinc.com, http://www.embedinc.com

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