# Tag Info

16

Holy Carp! You're trying to do 10's of nsec switching on a solderless breadboard? And you don't have a flyback diode on your transformer? If you're going to do this stuff, you've got to learn to respect fast switching and inductive parasitics. Go to a ground plane and make all your switching paths as short as possible. Also, put a 100 uF cap (tantalum for ...

9

A combination of the IRF840's reverse transfer capacitance (120pF), the dv/dt of the drain voltage and the rather weak driver (MCP1402) is my best guess. Starters, read the data sheet on the driver - it says on page 3 that the "latch-up protection withstand reverse current" is typically greater than 0.5 amps - that is a clue to why that device might be ...

9

Multiple problems. It sounds like you don't really understand HV electronics, and I suggest you stop before you get injured or worse. Your circuit is dangerous (to you, your Arduino and your computer). Your 1N4001's are only rated to 50 V -- you should use 1N4007's if you need diodes. You have nothing to limit inrush current when plugged in to the mains -- ...

9

Sometimes... Assuming the point of interest is power MOSFETS and not small signal MOSFETS and silicon (as oppose to SiC, GaN) The first characteristic to check is the output voltage. For power devices they should be 0V to 12-15V (acpl-312T) to cater for gate thresholds around 4V (as well as being able to drive to -15V if miller turn-on is a concern). As ...

8

Provided you respect its safe operating area and thermal limitations, the IGBT itself will have no trouble with 100% duty cycle. The gate driver will need some attention. Some types of gate drive circuits or ICs aren't going to be compatible with 100% duty cycle, e.g. those using gate drive transformers or certain types of "bootstrap" power supply. But ...

7

I think there may be two reasons. First, here's a transistor that is specified for use in ignition systems and note that it has a protection circuit built in that will turn the transistor back on (thus protecting itself) if the voltage at the collector exceeds 350V. Normally, car ignitions won't generate much more than a 300V spike and to demonstrate this ...

6

Yes with ... considerations. Can you stack them in series? Yes especially in theory. In theory the devices in a series stack would have the same leakage (in fact no leakage) and thus share the voltage perfectly. In theory they would all switch at the same rate (in fact instantaneous switching ) and in theory the devices in the stack would receive the ...

6

Most likely, the IGBT was killed by inductive kickback from the coil. Most of the energy from the primary should have been transferred to the secondary, but there is always some leakage inductance. This leakage inductance is the inductance of the primary that is not coupled to the secondary, so looks like a plain inductor in series with the part of the ...

6

Andy's on to something, I believe, with the drain-gate capacitance. But also: Measure what this is doing to the 12V supply. That would be an alternative path for spikes through the gate driver. Currently you're showing a single 0.1uF capacitor as decoupling, and I suspect that isn't enough. You may need a broad spectrum of decoupling from 10nF up to 100 uF ...

6

As you may know, semiconductor devices are fabricated doping a very pure silicon (or other, less common, semiconductor materials) substrate using various kinds of ions. Doping different zones of the semiconductor with different types and concentrations of dopants produces the different kinds of semiconductor devices you are accustomed to (diodes, BJTs, FETs) ...

6

suitable IGBT gate driver And the key to your question is "suitable". The short answer is yes you can. The IGBT combines an isolated-gate FET for the control input and a bipolar power transistor as a switch in a single device (wikipedia). Your question already contains the appropriate considerations, "threshold, plateau, and turn on voltage ratings,...

6

Current draw is determined by the heater power you need. If you can reduce the heater watts, do that. I'll assume that you have already sized the heater. Both triacs and IGBTs have relatively high losses, and you can expect roughly 1W/A of losses for a triac. The IGBT may have a bit more, and the required bridge will have still more. Your best bet is to ...

6

The gate-emitter voltage is just that, the voltage that happens to be across the gate to emitter terminals. The threshold voltage is that gate-emitter voltage where the device starts to conduct, where "starts" is usually defined as some minimum collector current at some applied voltage. The datasheet should define what criteria exactly are used to define "...

5

Yes it can be used to switch 220V. Viable? Well, cost and complexity could be issues for static switching applications. It's rarely done, which may be an indication. Yes, you can detect and prevent destruction with a short (even if the short occurs at peak mains voltage) however you need a short-circuit-rated IGBT and you need to detect and turn off the ...

5

I'd suggest your problem is nothing to do with inrush current. You have no protection for the G20N50C, and are driving a highly inductive load. Put commutation protection across the motor: What is the driving PWM schematic? I hope the R4_1G is not in any way defining the turn on charge timing for the G20N50C gate. For applications such as you have you ...

5

Yes, the same amplifying classes applies to FETs or other amplification devices. What defines the amplification class is the conduction angle and the presence (or absence) of switching during operation (like PWM modulation, etc), not the amplification device (transistor). However, voltage-controlled devices like FETs are generally better suited for ...

5

I have designed a few boards like this in the past, and the first thing that I would do, if possible, was to stagger the middle pin inwards into the PCB, in order to increase the clearance to the outer pins. This is not always possible, so the next step is to cut slots between the pads. This article explains quite well the issue of creepage One thing that ...

5

It's a 600V 40A IGBT from Fairchild. FGH40N60UFD.

5

I want to estimate the power dissipation of this IGBT SGL160N60UFD but I can't find any Rdson Well, it's an IGBT and it doesn't have a drain (d) or source (s) pin. They are actually called collector and emitter like an ordinary BJT. It does have a gate pin (like a MOSFET) of course. can this TO-264 package handle currents of 100A? From the mouser data ...

5

There is no Rds(on) on an IGBT because it does not behave like a MOSFET- it has a Vce(sat). According to the SOA curve, 100A continuous is not safe even at Tc = 25°C. The limitation is thermal. Note that the Vce(sat) is not guaranteed above 80A so there is uncertainty in the maximum power dissipation at 100A. 300A is the upper limit for a single short pulse ...

4

There are a few mistakes in the question: The original reference design called for 4K7 (4.7K) ohm resistor for R1. This should fix your problem. Note that the specs for hfe (DC current gain) are typical and not guaranteed, so you need to design in some margin (as was done in the reference design). There was a mistake in your equation: vc=(12.38−25m∗47k) ...

4

Before the masses close this question, I think it deserves an answer. If you refer to the datasheet, you will see it clearly states the operating limits of the IRS2113. Second line on page 1: "Fully operational to $+600\mathrm{V}$". So basically if you try and use it in a circuit to drive an IGBT with a high side voltage of $+800\mathrm{V}$, ...

4

A IGBT for the low side switch makes no sense here. It's easy enough to drive a low side NPN directly. The FET is more in the way than doing anything useful for you. It also increases the overall resulting saturation voltage. If you're going to have a PNP doing the high side switching, it's probably easier to drive it directly too. You don't say ...

4

It's not a MOSFET, it is an IGBT, so doesn't have an Ron resistance in the way a MOSFET does. An IGBT functions like a combined device with a low power MOSFET driving a bipolar transistor. Because the operating current density for bipolar transistors is higher than FETs an IGBT can carry higher currents than the same size FET. The data sheet gives the ...

4

No worries if you put a 10 ohm resistor on each gate driver output.

4

Fundamentally yes but not quite You have captured the basic concept $P_{total} = P_{cond} + P_{sw} + P_{leak} \approx P_{cond} + P_{sw}$ I have added in leakage losses for completeness but this only becomes of significance one you get to series stacked devices. Conduction loss This is the subtle difference. Conceptually it is \\$ P_{cond} = V_{ce} \...

4

With these types of things the devil is in the detail. 1EDI60I12AF : Very nice little driver chip, I am making use of a selection of these for a SiC inverter (except the wide body variant due to a few hundred volts ) SIM-0512D : Not used this exact DC:DC but SIL DC:DC with suitable rating and rails are my goto for lab lash-ups (newport: bad experience, ...

4

These are used for measurement purpose and driver reference. Due to the high current and high transients you can have already quite noticeable voltage difference between the auxilliary emitter and the power emitter.

4

To measure the voltage at the junction without the lead voltage drop. This is used so that the gate voltage can better be controlled. Since IGB can have pretty high current, even a small lead resistance can cause significant voltage drop, you can make sure to drive the IGBT at its maximum switching capacity, without the risk of destroying it by having a ...

4

The current through the capacitor is nearly steady since it's related to the (assumed-constant) current that the gate driver can supply. Assuming that other parasitics are insignificant, the gate driver current cannot avoid the plateau, only shorten it. If the driver output current is able to source more current than the Cgd capacitor is able to ...

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