IGBT circuit design

Thread Starter

hensem

Joined Sep 19, 2006
7
hello everyone,may i ask about IGBT.

i just do a simple testing,i expect the output is like the graph i drawn,the load i put is a 800W resistive heater.but i burn that IGBT in 1 second after switch on.may i know what is the mistake? i had turn off the IGBT before switching it into 240Vac,so i expect the initial state is no output at all.before i turn on the 240Vac switch,the gate terminal of IGBT i leave it float.

the IGBT is SGP20N60,it claim that it could carry 20A.here is the datasheet:
http://www.farnell.com/datasheets/42.pdf

may i do switching with IGBT without snubber circuit?

now after it burn,the IGBT become short circuit in every terminal,regardless any direction.

may i know what is my biggest mistake please?i had try to switch 20Vdc device,it work ok when 20Vdc,but when i try to switch 240Vac it spoil in one second.infact i instantly smell very little burn smell,then i discover its already spoiled,spoiled in the sense that every leg of the IGBT is shorted.

can anyone kindly share your experience in IGBT please? i couldn't afford to try and error anymore because IGBT is quite expensive.

i thank everyone in advance.
 

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beenthere

Joined Apr 20, 2004
15,819
Hi,

Your IGBT is rated at 178 watts. You have it pulling 20 amps at 240 volts. That works out to 4800 watts. You have to keep the duty cycle down to where the device dissipation is under 178 watts (around 3%).

Minimising turn-on time and turn-off time is also necessary. That way the device is in the linear region where heating is worst for the least amount of time.

It is possible to run several IGBT's in parallel to spread the heat load, but you always have to keep individual device dissipation under the spec sheet figure. Using heat sinks and fans are sometimes necessary.
 

Thread Starter

hensem

Joined Sep 19, 2006
7
thank you [beenthere],

but i dont understand,when the max rating of the device is 20A,i mean idealy it should able to let DC 18A flow thru it right?

the load i put in my case is a 800W resistive heater,if assume power factor is 1,then current drawn should 3.3A ,this is far below 20A limit right?

is that mean i cannot use IGBT to switch on my heater like that?i cannot do anything to duty cycle because i must fully on the heater to boil water,a 3% duty cycle wont produce heat at all.

or may i ask a direct question:
the IGBT i use (SGP20N60) to switch 10A DC current?

actually i didnt short circuit the IGBT to 240Vac,there is a resistive heater at collector.

i dont understand the calculation of power dissipation.because according to datasheet,the maximum Vce is 2.9V at 150 degree celcius,at 24 degree celcius the Vce = 2.4V ,why not calculate the power dissipation of switch by 2.4 X current?
or my understanding is wrong?

thanks
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

We might need to see your circuit to give better answers. You will need to heat sink the IGBT, though. 3.3 amps times 240 volts is 792 watts. Some heating will still occur in the IGBT. The fact that it shorted suggests that it got too hot.
 

n9352527

Joined Oct 14, 2005
1,198
I don't think that IGBT is able to withstand peak reverse CE voltage of 340V exposed by the mains. Check the maximum reverse breakdown voltage of CE and if it has internal diode or not.

Any particular reason why you don't want to rectify the line voltage first?
 

Thread Starter

hensem

Joined Sep 19, 2006
7
I don't think that IGBT is able to withstand peak reverse CE voltage of 340V exposed by the mains. Check the maximum reverse breakdown voltage of CE and if it has internal diode or not.

Any particular reason why you don't want to rectify the line voltage first?
heat dissipation is not calculated by Vce X current? why you all use 240Vac X current?

i already put a big heatsink(as long as second finger)
the breakdown is 600V,actually i connect it back to back,it actually is a AC converter,but i just experiment with single IGBT first,i didnt expect it to burn on such small load.i prodive the datasheet in the first post,it doesnt have internal reversed diode.

the picture i attach is my circuit,the square block above collector is a 800W resistive heater.
 

n9352527

Joined Oct 14, 2005
1,198
Are you sure the reverse breakdown is equal to the forward breakdown of 600V? They might differ depending on the junctions. Have you also consider the reverse emitter-gate/gate-collector breakdown and all other things related to the operating the device under reverse condition?

I've checked the datasheet and it doesn't say the reverse CE breakdown voltage, unless I missed it. If it doesn't and nowhere else it says that the reverse CE equal to forward CE or specifies that the device is capable of operating in reverse condition, then I'd assume the worse...
 

Thread Starter

hensem

Joined Sep 19, 2006
7
Are you sure the reverse breakdown is equal to the forward breakdown of 600V? They might differ depending on the junctions. Have you also consider the reverse emitter-gate/gate-collector breakdown and all other things related to the operating the device under reverse condition?

I've checked the datasheet and it doesn't say the reverse CE breakdown voltage, unless I missed it. If it doesn't and nowhere else it says that the reverse CE equal to forward CE or specifies that the device is capable of operating in reverse condition, then I'd assume the worse...
mean probably it spoil because the reverse voltage?i didnt consider that point before i test it.
i was assuming forward and reverse breakdown is same,then adding a diode should solve this problem is it?
 

akhil k k

Joined Jan 23, 2012
5
the part number SGP20N60 is not an igbt,it is a mosfet with free wheeling diode.then how can you use this device in ac.?because it is not possible to control a mosfet with free wheeling diode in ac.
 
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