I have a buck converter that is delivering 6A to a load of 3ohm load.
I'm measuring the output current and stop increasing the dutty cicle when the current reach 6A, at this moment the output voltage will be 18V. A few second after reaching this point the resistance will break and I will have and open circuit at the ouput.
My concern is the following, the inductor's current can't change inmediatly, I guess I'll go to the capacitor increasing the output voltage, It is a risk to damage the power supply or the driver circuit because of a backward current?
I'm using a IR2110 to drive the transistor.
R1 is a heating wire, I'll use it to control the ignition of a small rocket, so at the moment of the ignition the wire will break and the inductor will be is series with the capacitor.
Now, I just made this calculations:
$$ capacitor \ energy = 0.5*C*V^2 $$
$$ inductor \ energy = 0.5*L*I^2 $$
$$ L = 43uH \ , \ I = 6A $$ $$ \ C = 800uF $$
$$ inductor \ energy = 0.001548 $$
for 0.001548 joules I get:
$$ \Delta_V = 2V $$
Is It that an accurate estimation?
Is this scenario of removing the load different than operating in discontinuous conduction mode? where the current ripple is high and adsorbed for the capacitor.