I notice this statement in
the manual for the CAD200:
Short circuit protection activates if load impedance is about 1.6 ohms or less.
I'm not sure how the protection circuit would sense what the load impedance is, but presumably it would do it by some sort of measurement of the relation between output voltage and output current. If the load is highly inductive, the current drawn by the speaker will lag the voltage by a substantial fraction of 90 degrees (exactly 90 degrees for a pure inductor, that has no resistance).
Therefore at the instant that current reaches its peak, the voltage will be at a considerably smaller value than the instantaneous voltage which caused that peak, which occurred somewhat earlier in the sine wave cycle. The protection circuit may interpret, however, that the peak current was caused by the instantaneous voltage occurring at the exact time of the current peak, in which case it would "think" that a lower impedance is present than is really the case (recall Ohm's Law, resistance = voltage/current; therefore a lower voltage producing the same current implies lower resistance).
When you run the speakers full range, as I said earlier, the capacitance which is likely to be a significant part of the impedance of the mid/hi drivers is placed in parallel with the woofer and its crossover elements, which would partially negate the inductive component of the loading presented by the woofer (the effects of inductance and capacitance on the phase relationships between voltage and current work in opposite directions).
So I think the statement in the manual about the protection mechanism reinforces the theory I presented earlier. It could be that the design of the amp's protection mechanisms is incompatible with your woofers, when they are driven alone. While you are waiting for the measurements you may want to ask Cary if a heavily inductive load, such as the woofer section of a speaker connected by itself, can trigger protective shutdown of the amp.
Regards,
-- Al