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.

enter image description here


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.

  • \$\begingroup\$ 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. \$\endgroup\$ – Sparky256 May 18 '16 at 1:28
  • \$\begingroup\$ 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? \$\endgroup\$ – Sparky256 May 18 '16 at 1:33
  • \$\begingroup\$ @Sparky256 I just edit the question to clarify that. I'm using kanthal wire, it puts red hot. \$\endgroup\$ – Luis Ramon Ramirez Rodriguez May 18 '16 at 1:54
  • \$\begingroup\$ 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... \$\endgroup\$ – mkeith May 18 '16 at 5:25
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    \$\begingroup\$ 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! \$\endgroup\$ – mkeith May 20 '16 at 3:27

Your formulas are correct, but I'm not sure where you are getting your numbers from.

Based on the 18V output voltage, the capacitor energy at the time of the break is 129.6mJ and you are adding 0.77mJ to it, so the capacitor voltage will change from 18V to 18.054V if all the energy of a pulse finds its way into the capacitor. Of course if more pulses come along that will add up quickly at 100kHz.

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

  • \$\begingroup\$ The OP's stated output voltage is 18 volts @ 6 amps, until R1 intentionally blows. \$\endgroup\$ – Sparky256 May 18 '16 at 2:08
  • \$\begingroup\$ @Sparky256 Okay, will be even less. I'll go get my specs and recalculate. \$\endgroup\$ – Spehro Pefhany May 18 '16 at 2:13
  • \$\begingroup\$ @SpehroPefhany The numbers are from the circuit diagram, I don't have any issue if vo reach 30V when there is no load, I can turn off the converter manually, my concern is regarding the transient state just before R1 blows up, there is any change of Vo been bigger than Vi? or any other side effect that could damage the circuit? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez May 18 '16 at 2:20
  • \$\begingroup\$ There should be no bad effect provided the converter shuts down immediately, which it should, assuming voltage feedback from the output capacitor C1. C1 absorbs the inductor energy from one cycle easily. It may spark due to the energy in the stray inductance on the output- it would be good to minimize that loop area. \$\endgroup\$ – Spehro Pefhany May 18 '16 at 2:30
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    \$\begingroup\$ @LuisRamonRamirezRodriguez, OK, but put a resistor in parallel with cap to make sure it discharges over time. Calculate the dissipation to make sure it is OK (don't exceed the resistor limit). \$\endgroup\$ – mkeith May 18 '16 at 14:40

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