# Remove the load of buck converter

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.

simulate this circuit – Schematic created using CircuitLab

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.

Edit:

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?

Edit 2:

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.

• What exactly is the problem? What part is failing? What is backward current? Do you mean back-EMF? Your synopsis is a bit confusing. Please clarify your question, or if more than one question please list them. – Sparky256 May 18 '16 at 1:28
• Do you realize that R1 is dissipating 108 watts. You may need to use a 300 watt wire-wound resistor for R1. Was that the only issue you had? – Sparky256 May 18 '16 at 1:33
• @Sparky256 I just edit the question to clarify that. I'm using kanthal wire, it puts red hot. – Luis Ramon Ramirez Rodriguez May 18 '16 at 1:54
• What do other rocket people do? I think I would prefer to use a linear current regulator (with a few transistors), or even a big power resistor. You could still control the timing with a FET. Even though the linear circuit will dissipate a lot of power, it doesn't have to do it for very long, so big heavy parts will probably last through many cycles. Also, if you lowered the supply to 18V, you wouldn't need any current limiting circuitry... – mkeith May 18 '16 at 5:25
• IRF2110 should be OK, based on quick look at the datasheet. Now that I think about it blowing the wire quickly is really not the goal. You want to heat up the grain until it ignites. So limiting the current could be a very important requirement because it will allow the hot wire to transfer heat for a longer duration. I still think PWMing the output without the LC could work, but if you want DC current, then your original approach is probably still the best, considering everything. Good luck! – mkeith May 20 '16 at 3:27

In general, $\Delta V = -V +\sqrt{V^2 + I^2(\frac{L}{C})}$