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The current design of the MOSFET coil drive looks like this:
The IRF740 has an input capacitance of 1400pf. The discharge time through R9 (229 ohm) is about 500ns.
The design is charging the MOSFET's capacitance through a low impedance (emitter of Q2) and discharging it through a higher impedance (R9). So we have a fast turn-on and a slower turn off of the MOSFET. But the opposite should actually happen.
The turn-on time is not critical because the current in the coil increases slowly anyway (at a rate V/L), so a fast turn-on has no advantages. The lower impedance should instead be used to drive the gate of the MOSFET low for a fast turn-off and a better target signal. This can be accomplished with the simpler circuit below:
The on-resistance of the BS170 is 5 ohm maximum, so this allows for a much faster discharge of the IRF740 gate capacitance that the original 220 ohm resistor.
P.S. I overlooked the fact that the 555 and the IRF740 have different ground references, so some level-shifting must precede the BS170 stage. But you get the idea anyway.
The design is charging the MOSFET's capacitance through a low impedance (emitter of Q2) and discharging it through a higher impedance (R9). So we have a fast turn-on and a slower turn off of the MOSFET. But the opposite should actually happen.
The turn-on time is not critical because the current in the coil increases slowly anyway (at a rate V/L), so a fast turn-on has no advantages. The lower impedance should instead be used to drive the gate of the MOSFET low for a fast turn-off and a better target signal. This can be accomplished with the simpler circuit below:
P.S. I overlooked the fact that the 555 and the IRF740 have different ground references, so some level-shifting must precede the BS170 stage. But you get the idea anyway.
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