There have been posts to this list in the past asking about measuring large
DC currents without using a shunt resistor. The latest issue of Nuts & Volts
magazine (April 2004) has an article on page 38 describing the construction
of a transformer based DC current meter with full scale up to 100 amps (and
no, this is not an April Fool's joke).
   The sensor is a toroid wound with 200 turns which acts as one half of a
transformer. The other winding is a single wire passing through the center of
the toroid, carrying the DC current to be measured. The main (first) winding
is placed inside the feedback loop of an oscillator made from a voltage
comparator, three resistors, and a power amplifier which boosts the output
of the comparator, driving the coil. An additional resistor in series with the
main coil measures the DC current in this coil. The DC voltage across this
resistor is the product of the resistance of the measuring resistor and the
DC current in the coil.
   With no "unknown" current, the circuit oscillates freely. An adjustment
allows setting the duty cycle of the oscillation to exactly 50 percent. This
makes the DC current in the coil zero, giving zero DC output voltage.
(There will be a large AC component on this signal, and this may have
to be filtered out if the measuring meter has a problem with it.) The
frequency is determined by the voltage trip points of the comparator and
the number of ampere-turns which drive the core to saturation. The
circuit requires a core with high permeability and a square loop character-
istic. Sources for these are given in the article.
   The presence of current in the single input turn will shift the B-H loop
to one side or other. This will change the trip points of the oscillator,
making one half of the cycle longer than the other. This change in duty
cycle will produce a net DC current in the main winding. The author
claims that this current, multiplied by the number of turns in the main
winding must exactly cancel the unknown current times its one turn.
If so, the output voltage will be V = R * I / N, where R is the resistance
of the sense resistor, I is the unknown current, and N is the number of
turns on the main winding. I say "if so", because I am not sure of the
argument, but I can see that there will be a DC of some kind produced.
   This is certainly one of the most original concepts for measuring
current that I have seen in a long time, and seems worth checking out
if only for the novelty of it.

John Power

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