how slow can i switch this IGBT ? - Page 1
Forum section: electronics / projects

author: CopperCone
author_rank: Super Contributor
timestamp: April 07, 2018, 01:52:53 am
content: I have these IGBThttp://www.mitsubishielectric.com/semiconductors/content/product/powermod/powmod/igbtmod/tgatef/cm200du-12f_e.pdfI have 2.5 amp gate driversI calculated my charge time for the gate with my drivers will be approximately 500ns. I estimate the switching frequency will be between 20-60KHz.Is there a safe collector current I can switch the IGBT with a 'puny' 2.5a gate drive signal?Does efficiency just go down?Is there a hard limit for switching time that causes them to blow up by phenomena not detectable by measuring external temperature, some kinda abrupt failure? (i.e. some kinda spot heating etc).I want to use my iso5500 drivers.

author: jbb
author_rank: Super Contributor
timestamp: April 07, 2018, 03:24:09 am
content: I'm not a semiconductor expert, so I could be wrong, but I think you'll be OK. There will belotsof switching losses, so calculate your losses and cooling needs. Also make sure your turn-off resistor is small enough to prevent dV/dt induced turn on of the 'other' IGBT in the half bridge.I think I heard a semiconductor scientist say it's important to make sure the on (and off) pulses are long enough to fully switch the device (including tail current, if any) and flush out the relevant charges, so check your minimum pulse widths.Have you considered adding a current buffer stage to the output of the ISO5500? It's not hard to do (1x NPN, 1x PNP + some decoupling caps) and very standard in higher current drivers.

author: jmelson
author_rank: Super Contributor
timestamp: April 07, 2018, 04:28:26 am
content: Quote from: CopperCone on April 07, 2018, 01:52:53 amThe problem is IGBTs must never be operated in the linear region, because current sharing is only assured when fully saturated. In the linear region, the center of the die (usually) pulls all the current, and fails shorted, destroying the whole device.Note the datasheet spec for Cres. This is the "Miller" capacitance, or capacitance from the collector to gate. You need to calculate this additional charge into your charging time. As the gate voltage rises, the collector voltage falls through a much larger swing, coupling to the gate. The gate driver needs to supply charge during this time until the collector stops dropping. This Miller charging can be several times the plain gate charge when the collector is not powered. Note that total gate charge is 1240 nC. So, maybe your 500 ns calculation is right -- if you believe their spec. If that is the real turn-on/turn-off time, then that probably is OK for this device.The reverse scenario occurs on turn-off.Jon

author: CopperCone
author_rank: Super Contributor
timestamp: April 07, 2018, 01:48:56 pm
content: Well the ISO5500 driver does +15/-15V.I wanted to put something together for an induction heater. I could leave blank footprints for a standard current buffer, but I just wanted it to work. I don't really wanna design custom hardware or pay money for this right now. I am setup to do 2.5A drive with ISO5500.But what I am trying to get at, are switching loss behavior:1) is it purely a thermal limit?2) where do the thermal start to behave weird? Like, when is running enough current in the linear region, maybe the thermal transfer is not good enough at some particular 'regime' , so even with heavy cooling its gonna cook itself due to thermal time constants on start up? I want to work in a region where I can put a thermocouple on the bottom and decide to turn the system off if it gets too hot. Or even a PTC feedback switch.you wroteQuoteHow fast is this failure mode? Can a thermal over ride shut it down or is it going to run away inside the case before there is any indication outside?\I am sure with the right signals its possible to do this, but roughly where can I expect this to happen?And yes, I am also interested in measuring the EMI of something like this with slow gate times, with my new LISN project. More to get a feel of things.This is more of a test bed then any kind of real project.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 07, 2018, 03:26:37 pm
content: Just do what the datasheet says. 3.1 ohms, so you need 15V/3.1ohm = 5A or better drive current. Use bipolar (+/-15V) drive.Tim

author: CopperCone
author_rank: Super Contributor
timestamp: April 07, 2018, 03:28:27 pm
content: wow I missed that in the datasheet. I guess that answers my questions.31 ohms must be the drive current limit where going any slower causes it to enter some kind of other failure mode.and 3.1 ohm is the minimum. Why? Gate drivers can go alot faster. Ringing? Bond wires going bad?

author: mk_
author_rank: Regular Contributor
timestamp: April 08, 2018, 02:35:47 pm
content: Quote from: CopperCone on April 07, 2018, 01:52:53 amWe used the ISO5500 with a IGBT of much lower Cies (20nF) at lower frequencies and finaly ended with boosters as the 5500 runs really hot (up to selfdestruction) when running at 20kHz.Read the datasheet of the ISO5500 up to every detail... for 2,5A Gatecurrent you end up with 12Ohm Gateresistor... which is way to much (The Iso must lift the Gate from -15 up to +15V, so to keep the current within specs 12Ohm are ok... minimum will be 10Ohm)So - boost the ISO as shown in fig. 72 "Current Buffer for Increased Drive Current" and the ISO will be happy... and stay cool.You can also change symetry of the ISO-Powersupply as those boosters reduce U_gate. So.. setting V+ to 17V and V- to maybe -12V is within the specs of the ISO and allows enough Gatevoltage....