Removing the input of boost converter

[kanmaedexandzelbladex]

Hi,

I would like to ask about shutting down boost converters. Let's say I have a boost converter such as in this figure **broken link removed** . Let's say that my inductor current is at 10A and my input voltage is let's say around 12V and the converter is at steady-state. If I suddenly remove the input, then the inductor current cannot get the current anywhere and since di/dt = instantaneous (current is being forced to 0A immediately), then voltage across it is forced to be really high. This would be problematic? What could happen if I try to do this in real life?

Next question is let's say I want to remove this. I plan on putting a capacitor in the input (parallel to the input voltage source)which most DC-DC converter designs I have seen has this and they label it sometimes as Bulk capacitor. First I thought it acts as a filter only but then I thought that if the input is suddenly removed, the inductor current could get current from that capacitor and slowly drain its voltage. Is this correct? Is this a good solution?

Now the next question is how should I size this input capacitor? From theory, if inductor current is 10A, then at the removal of the input, 10A = C dV/dt. So should the capacitor be sized only with the wanted dV/dt (turn-off time)? Are there other considerations for this capacitor? I'm not too sure if a capacitor can handle 10A discharge current will it explode?

Thanks

 

[mtwieg]

A clamping diode across Vin is probably a simpler solution than a capacitor. If the source is disconnected then it will become forward biased and conduct the inductor current until it falls to zero.

 

[ch wazir]

diode must be connected across an inductor to avoid the damage since the back emf forward bias the diode when it's switched off.