ENERGY AMPLIFICATION

[Magluvin]

I think the answer may be the statement from datasheet :   Repetitive Avalanche Allowed up to Tjmax
I had to place it on radiator and now it runs cold.
Attached actual circuit with the freewheeling diode which seems to suck current from battery (12V7Ah)

Like I said earlier, maybe not clear enough, its possible that the on/off timing is too fast for the coil used.  Like if you dont use the diode, the coils collapse happens quick and the coil is settled down and ready for the next on cycle. But, with the diode, the coil holds its field longer, and possibly long enough that the field never really dies out, considering 15us sq input. 

So without the diode, coil field at 0, then transistor goes 'on' and initially, the impedance of the inductance limits the current.  So its possible that the diode is keeping the coil freewheeling and the impedance of the coil remains low because the field is continuously there due to freewheeling. So the coil passes 4A every 30us for 15us. 

Soo, maybe the coil inductance is too high for the freq used.  15us, the coil should be pretty fast reacting. Lower inductance and retry.

mags

[MarkE]

Like I said earlier, maybe not clear enough, its possible that the on/off timing is too fast for the coil used.  Like if you dont use the diode, the coils collapse happens quick and the coil is settled down and ready for the next on cycle. But, with the diode, the coil holds its field longer, and possibly long enough that the field never really dies out, considering 15us sq input. 

So without the diode, coil field at 0, then transistor goes 'on' and initially, the impedance of the inductance limits the current.  So its possible that the diode is keeping the coil freewheeling and the impedance of the coil remains low because the field is continuously there due to freewheeling. So the coil passes 4A every 30us for 15us. 

Soo, maybe the coil inductance is too high for the freq used.  15us, the coil should be pretty fast reacting. Lower inductance and retry.

mags

If the operating frequency is high enough to sustain continuous conduction in the coil, then with the diode in place the current scales with the duty cycle.  This is easy to test:  Alter the duty cycle.  If the on time is fixed, just drop the repetition rate to one fourth and see if the current falls to 1/4th or less.

[Magluvin]

If the operating frequency is high enough to sustain continuous conduction in the coil, then with the diode in place the current scales with the duty cycle.  This is easy to test:  Alter the duty cycle.  If the on time is fixed, just drop the repetition rate to one fourth and see if the current falls to 1/4th or less.

Or, lower the freq and reduce the on time duty cycle to 15us, if 15us is what is required as a pulse.

Mags

[Magluvin]

Or, lower the freq and reduce the on time duty cycle to 15us, if 15us is what is required as a pulse.

Mags

15us is quite a short time period. Adding the diode may increase the on time beyond 15us.  There may be a resistor of a particular value that may suffice instead of the diode. The higher the resistance, the quicker the field drop, where the diode would provide a longer collapse similar to a very low ohm resistance. Could add the resistor in series with the diode to reduce the collapse time.

Mags

[MarkE]

15us is quite a short time period. Adding the diode may increase the on time beyond 15us.  There may be a resistor of a particular value that may suffice instead of the diode. The higher the resistance, the quicker the field drop, where the diode would provide a longer collapse similar to a very low ohm resistance. Could add the resistor in series with the diode to reduce the collapse time.

Mags

Whether 15us is short or long depends on the components used.

He stated 15us as his pulse width.  He did not mention his repetition rate.  If the L/R time constant of his coil is longer than his repetition interval then current will build up cycle to cycle until it equalizes the V*T product of the on pulse with the V*T product of the off time.  The former is the product of almost the supply voltage and the 15us.  The latter is the sum of the diode drop and I*R drop (which changes as the current decays ) in the coil multiplied by the off time.  So if that is what is going on, then all he has to do is slow down his repetition rate.  When a switching power supply detects over current they often "hiccup", that is drop their operating frequency to limit the current build-up.