Buck Converter for battery charging

[dhodge]

I am charging a bank of batteries with a buck converter cct, L=700uH, c = 3600uF.
My question is probably very simple if the input to the converter is 350VDC and the output is 58VDC at 80A, what is the current going through the IGBT.
I probably stupidly assumed that it would be similar to a transformer as in Power in = Power Out + Losses, so I In=(VOUT * IOUT)/VIN

But strangely enough I am blowing 75A IGBT, I was running at 20KHZ and I am going to slow this down a bit maybe to 10KHZ but I cant seem to get my head around the switch current. If the current is the same as the output then that explains it, but if not then why am I blowing them up.

Thanks in advance

 

[sreein]

your circuit?

 

[dhodge]

This is basically it, 350VDC Input, 58VDC Output, depending on current limit of 80A.

 

[sumanthpala]

sir i am doing the same project ,can u tell me how to proceed with closed loop control circuit of buck converter.

 

[dhodge]

I basically read in input voltage, output voltage and output current.
My input source comes from a wind turbine so I have to limit the charging current according to how much power is available from the turbine.
I have a software PID loop which adjusts the modulation of the igbt to limit the current into the batteries and or the voltage, the set points vary according to which state of charging we are in, I.e bulk, is high current followed by lower current higher voltage followed by float charge.
The available power from the turbine is boosted to give a dc bus voltage of 350v which is then bucked by the charger.
In the event of no wind the batteries supply the dc bus for an inverter stage which feeds the load.

 

[uoficowboy]

I am charging a bank of batteries with a buck converter cct, L=700uH, c = 3600uF.
My question is probably very simple if the input to the converter is 350VDC and the output is 58VDC at 80A, what is the current going through the IGBT.
I probably stupidly assumed that it would be similar to a transformer as in Power in = Power Out + Losses, so I In=(VOUT * IOUT)/VIN

But strangely enough I am blowing 75A IGBT, I was running at 20KHZ and I am going to slow this down a bit maybe to 10KHZ but I cant seem to get my head around the switch current. If the current is the same as the output then that explains it, but if not then why am I blowing them up.

Thanks in advance

Hi - the problem is that your switch will see a peak current equal to your output current + half your ripple. By my calculations your ripple is about 3A. So your switch will see a peak current of about 81.5A. Considering that it's a 75A part, you are probably melting the poor thing. I would recommend putting in an IGBT rated for maybe 160A. Also, how are you driving the gate of the IGBT?

Please note that the average current through your switch is equal to approximately (Vin/Vout)*Iout (about 13A), but your peak currents are probably destroying the part. Either that or you aren't switching it on/off fast enough. Either way, a larger part and/or more heat sinking should improve things a lot!

Hope this helps!