Understanding IGBT Switching Waveforms and how to trigger them...

[n3rdx]

Hello with reference to the attached graphic (snapshot of a section of Fairchild/On semiconductor HGTG27N120BN IGBT datasheet), I noticed this test circuit has one input (Vge) and two traces output (Vce and Ice). The graphic was black and white, and I am trying to understand which trace is doing what ...


Is Vce waveform rectangular in shape?

Is Ice different over the two cycles? Why?

In the accompanying test circuit, if the IGBT was left in an "active" state with current flowing between C and E, what are the risks in doing so?

By the way, what do the turns "Switch On of a IGBT mean: "closed" (C not connected in pathway to E) or "open" (C connected in pathway to E).

What is a typical value for Vge -- would that be 15V or can it be lower ?

Lastly, what is the recommended initial configuration of Vge signal to that IGBT in that test circuit - for safety? High, or Low - and does anyone have tips for designing  to  ensure that we don't startup in a non-safety state?



Thanks.

N3RDX

[T3sl4co1l]

I made this,



https://www.seventransistorlabs.com/Images/Diode_Recovery_Tester.pdf

which generates the double-pulse waveform used in testing transistors and diodes.  The aim was to test diodes specifically, but the transistor is being tested all the same, or if you like, the pair together is being tested (especially relevant when the diode is the co-pack or body diode of the same transistor type, as in a half bridge).

Yes, Vce is supposed to be rectangular (with sloped sides and some peaking); it's not well shown with the two traces jammed together there.

Turns?  Also, you dropped a quotation mark there, not sure what you're meaning.

Vge is default low.

Tim

[n3rdx]

Oh ... s/turns/terms .. so restating my question ...

 what does the term "Switch On of a IGBT mean?  "closed" (C not connected in pathway to E) or "open" (C connected in pathway to E).

[T3sl4co1l]

Switching on is an event:
Initially off condition: Vge < Vge(th), Ic ~= 0, Vce = (nominal supply)
Then Vge is pushed up to Vge(on).  Ic is rising and Vce is falling.

Tim

[n3rdx]

Switching on is an event:
Initially off condition: Vge < Vge(th), Ic ~= 0, Vce = (nominal supply)
Then Vge is pushed up to Vge(on).  Ic is rising and Vce is falling.

With reference to Fig. 18, if Vge (Fig 19) is not applied to Rg, is there any current flowing through the inductor ? That graphic is really confusing .. which one is the Vce line?

[T3sl4co1l]

The one in here may be clearer, https://www.infineon.com/dgdl/irg4bc30wpbf.pdf?fileId=5546d462533600a40153564014f2228d

[n3rdx]

The one in here may be clearer, https://www.infineon.com/dgdl/irg4bc30wpbf.pdf?fileId=5546d462533600a40153564014f2228d

The document you quoted shows traces that were grabbed from a scope and printed in black and white - same problem as the original document, so I looked at fig 17 of this app note.https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=19&ved=2ahUKEwiBp4bitfzcAhXRTd8KHeHAANM4ChAWMAh6BAgAEAI&url=https%3A%2F%2Fwww.renesas.com%2Fen-us%2Fdoc%2Fproducts%2Figbt%2Fapn%2Fr07an0001ej0410_igbt.pdf&usg=AOvVaw00Ks81ycGFFi0nx2E54kSG

So, Ic is flowing only when Vge is above Vge(th) ?