How does it work? IGBT Gate Driver - Desaturation Fault Detection - Page 1
Forum section: electronics / projects

author: applicanon
author_rank: Contributor
timestamp: September 10, 2020, 09:13:04 pm
content: I'm planning on implementing gate drive for large IGBT's in a project and found many similar projects using gate driver IC's such as the HCPL-316J.It boasts about its ability to provide desaturation fault detection, which basically should protect from the dreaded shoot through of an H-Bridge configuration..When one IGBT stays on a little longer than intended and another turns on a little sooner than intended, an unintended short to ground through the two IGBT'sbegins to blast massive amounts of current through them, causing them to quickly self destruct. That is, unless some sort of fault protection is in place to sense sucha thing and put a stop to it before things get nasty.The HCPL-316J claims to do this by detecting runaway current at pin 14 (DESAT). When pin 14 detects 7 volts or greater, it triggers a fault and the IGBT gates are closed.I've looked over the datasheet for this gate drive IC as well as some similar ones, but I can't seem to understand how the thing ends up with 7 volts or greater at pin 14 whenthe current through the IGBT begins to surge as it would during a short to ground. There is a resistor and a diode connected between pin 14 and the IGBT collector, and a capacitor connecting pin 14 to pin 16 (Ve), which is also connected to ground.I'll just show the schematic and maybe someone can give me the run down as to how we end up with the trigger voltage of >7V at the DESAT pin when current begins to go nuts through the IGBT.I'll also provide a link to the datasheet so if anyone wishes to do so they can give it a read and see if they can figure it out any better than I failed to.Hope someone can fill me in on this! I know I can just simply wire it up as the datasheet recommends, but really I'd like to understand how the circuit functionsif I'm going to commit the time and dollars to such an endevour.

author: capt bullshot
author_rank: Super Contributor
timestamp: September 10, 2020, 09:26:25 pm
content: IGBT desaturation means the C-E voltage of the IGBT rises (usually by too high a collector current).When the IGBT is turned on, its collector voltage is in the 2..3V ballpark. D_Desat pulls Pin 14 of the driver down to that voltage plus one diode drop. Inside the IC there's a current source that supplies current from V_cc2 When the IGBT desaturates, its collector voltage rises quickly way above 7V. D_Desat blocks that voltage, it is reverse biased now. The current source pulls pin 14 to a voltage near V_cc2, which is higher than 7V.Bridge shoot through "never" happens in a correctly designed IGBT bridge circuit, desat detection is rather used for output short circuit detection.

author: langwadt
author_rank: Super Contributor
timestamp: September 11, 2020, 12:10:58 am
content: maybe easier to get the idea of how it works the data sheet for IR2127

author: applicanon
author_rank: Contributor
timestamp: September 11, 2020, 04:26:20 am
content: is there a voltage divider being implemented somewhere that I'm not seeing which is sensing the current through the IGBT as a voltage on pin 14?I took a look at the datasheet for IR2127 and have some confusion right off the bat.. It says that pin Vb is "High Side Floating Supply Voltage" with a max value of 625v. If it's the high side floating supply voltage, I would expect it to be connected to the collector of the IGBT? Instead it's connected to the emitter side of the IGBT through a capacitor.K so maybe I am clueing into something here.. So when the IGBT is fully saturated (fully gate turned on) and there is no resistance to current flowing through the IGBT straight to ground, this causes the voltage at the collector to be lower than the high side output voltage, making the diode forward biased... but how does this cause a current sense to occure at Vcs?

author: applicanon
author_rank: Contributor
timestamp: September 11, 2020, 04:34:28 am
content: wait wait wait... where is the load located in IR2127? I assumed the load was after the IGBT.. so the high voltage supply (600v) would be connected at the collector, then the load connected at the emitter, then the other side of the load to ground. Is this incorrect? is the IGBT and the load swapped places? is this why im confused

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 11, 2020, 05:16:35 am
content: They're rather terse about it, but it looks like they're doing it simply by sourcing a current out of the DESAT pin (see the I_DESAT spec and figure) and monitoring the voltage there. When the voltage charges up, collector voltage must be high; when it's pulled low, collector is saturated.It's not clear to me if I_DESAT is gated by V_GE, or the monitoring is. Probably, it's gated and has an internal switch, pulling DESAT low (to V_E) when V_GE is low (off), so that when the switch releases, it takes some time for the capacitor to charge up, thus implementing the blanking period (letting you set the period with an external cap).Tim

author: Berni
author_rank: Super Contributor
timestamp: September 11, 2020, 05:24:48 am
content: This is a low side IGBT driver. One leg of the IGBT touches ground while the top (Where the ... go) is the load you are switching to ground.The Ddesat diode is there to prevent the potentially high voltage (for example 600V) that is present there when the IGBT is in the off state from getting to the driver chip and likely blowing it up. But when the IGBT turns on the voltage across it will quickly drop down to a low voltage (0.2 to 3V). The more current the IGBT is conducting trough itself the more voltage drop it will have. If this current becomes too large the IGBT will get out of its saturation region and this causes the voltage to start rising quickly, so suddenly there might be 50V across the IGBT. If there is say 30A flowing trough it this would make the IGBT disipate 50V*30A=1500W and for a small TO-220-ish package IGBT means that it will spectacularly self destruct in the next few milliseconds with a loud bang, sparks flying, fire, and IGBT pieces raining down all over the room. Since this is generally considered a "very bad situation" the driver chip tries to detect it and quickly turn the transistor off before it happens.So if the chip sources a bit of current (few mA) into the forward direction of the diode the voltage you see on the other side of the diode becomes the IGBT voltage drop + 0.6V of the diode. This lets the chip sense the voltage drop on the IGBT and if it becomes too high it knows the IGBT is about to explode.

author: applicanon
author_rank: Contributor
timestamp: September 11, 2020, 08:11:35 am
content: The IGBT modules I've got for this project are CM600HA-24H.. Yup, 1200V, 600A. All 4 IGBT's create an H bridge mounted on a big thick slab of aluminum which has passages drilled throughout which water is pumped through to keep things nice and chillified. There's also already a really nicely designed snubber network which bolts on top of this h-bridge panel which will no doubt deal with many of the dreaded dv/dt spikes caused by sudden fault triggered gate cutoffs.

author: KT88
author_rank: Frequent Contributor
timestamp: September 11, 2020, 11:15:43 am
content: Besides the DESAT question you have another issue: you need more gate drive current than the 316j can deliver - without having had a closer look you would need at least 5..8Amps.With only 2.5A of the 316j, your switching times will be longer than nescesary, causing more looses and longer dead time requirements.Although DESAT protects the IGBT from blowing up immediately you have a limited amount of DESAT events the part can survive.With a too weak gate driver the uncertainty in timeing gets larger thus increasing the risc of shoot-through events causing the part to fail early even with a proper implementation of the DESAT function...