ELI_HAIT Guest

Mar 27, 2011 10:34 am 


Hello.
OK, so if the current which charged to the inductor, multiplied by discharging resistor value results with high voltage peak which is larger than the MOSFET breakdown voltage  the MOSFET will introduce low DS impedance, what will discharge the coil very quickly quickly until the voltage will drop below the breakdown value and the MOSFET will recover... right?
I wander why the charge current/resistor multiply should be so high in the first place?
Would the performance of this PI metal detector be less if the charge current value will be calculated to reach the MOSFET breakdown voltage when it switches from ON to OFF, but not to exceed it? 
