Snubber Cap Selection - Page 1
Forum section: electronics / projects

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 06:55:50 am
content: Could someone please walk me through what a snubber cap is?How would I select a snubber cap for the following circuit? The IGBT is not switching.Datasheet to IGBT i am using:https://www.infineon.com/dgdl/auirgps4067d1.pdf?fileId=5546d462533600a4015355ba8107152d

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 07:05:28 am
content: What do you mean it's "not switching"?Show the rest of your circuit.Tim

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 07:09:50 am
content: I mean that the IGBT is not rapidly turning on and off. It turns on once and then gets reset several minutes later. The rest of the circuit is just a mosfet and a PSU to turn the IGBT on or off.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 07:15:04 am
content: That circuit is always on.Tim

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 07:22:21 am
content: Whoops...sorry that was dumb.

author: ElektronikLabor
author_rank: Regular Contributor
timestamp: July 19, 2017, 07:25:52 am
content: Quote from: eerocketman on July 19, 2017, 07:09:50 amI think you swapped the drain and source of the mosfet.Do you intend to switch an inductive Load (e.g. Res3)?

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 07:31:44 am
content: Yes, however I wont have specific values for the load inductance until later. I just want to be prepared for when I do.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 07:59:28 am
content: Turn-on snubbing is used to reduce dV/dt and peak current into a capacitive load, while turn-off snubbing is used to reduce dI/dt and peak voltage across an inductive load.With turn-off of a whopping ~millisecond, I would be shocked if you can ever find a load inductive enough to be a problem.I would be much more worried about the IGBT catching fire during turn-off.Tim

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 08:02:42 am
content: Is that because of a voltage spike after shut off? Could you recommend a method to prevent this?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 08:07:08 am
content: What is the load?Get as specific as you can, given that you've said you don't have exact knowledge of it right now.Tim

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 08:12:21 am
content: The load is a PTC array that will be used to bring down the 400V-600V bus of a very large battery charger.

author: Zero999
author_rank: Super Contributor
timestamp: July 19, 2017, 08:41:55 am
content: What's the load current? Switching frequency?The IGBT will dissipate a lot of power, as it switches off. The higher the current and the longer the turn-off time, the more energy dissipated at each turn off.

author: eerocketman
author_rank: Contributor
timestamp: July 19, 2017, 08:45:23 am
content: Load current is about 90 A initially, It is not switching. The IGBT turns on, the resistors dissipate power and after they cool down it turns back off.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 10:14:30 am
content: So load current is guaranteed* ~zero when it turns off? That's good.*Good idea to have an interlock, so the discharge isn't accidentally turned on while the cap bank is still powered (will dump too much continuous power into the load), or turned off under the same condition (which would occur if the discharge signal is set by a timer, rather than verifying that the capacitors have actually finished discharging).You may still want an active (or at least much lower resistance) pull-down, because the IGBT will turn on under modest dV/dt (inrush) conditions.Tim

author: jbb
author_rank: Super Contributor
timestamp: July 19, 2017, 08:17:02 pm
content: I would second T3sl4co1l's suggestion of a faster turn off to avoid dV/dt induced turn-on. I also suggest a clamp diode up to the 400V (ish) rail next to the IGBT (just like you see on relay drivers), so that the load inductance doesn't blow the IGBT up. I guess a 900V 10A fast diode would probably do the trick...A second option for this system is a thyristor a.k.a. Silicon Controlled Rectifier (SCR). These have the interesting property that you can turn them on with a gate pulse, and they latch on -even if your 16V rail is turned off. They finally turn off when the current through them falls to (near) zero, which could be just the ticket for completely discharging a DC bus.Cons: there would be some design calculations to make sure you got acceptable results, and the SCR would turn off a little before the DC bus was completely empty.

author: floobydust
author_rank: Super Contributor
timestamp: July 20, 2017, 06:07:53 am
content: If the mystery load is a big wirewound resistor, they have inductance and you will need a catch-diode or snubber. A diode is cheaper and smaller.I know the DC bus is theoretically discharged before the IGBT gets shut off, but in reality you don't want any weakness causing the IGBT to short and then resistors start on fire, if the control circuit ever glitches.The IGBT is rated VCES 600V, kinda low for a 400-600V bus?EN-DSCHG signal needs to swing up to 16V (not MCU level) to shut off the driver mosfet.25k ohms on the gate is too high. There is a lot of capacitance from collector-gate for dV/dt turn on. I wonder if push-pull drive is better.

author: bson
author_rank: Unknown
timestamp: July 22, 2017, 07:30:41 pm
content: When you turn off the P ch MOSFET (Q?) Q4's gate gets drained through R9 @ 25.5k. If its threshold is low enough this could take a while.The gate capacitance is discharged according to Voff = Von*e^(-t/RC) where R is 25.5k (+51), C is the gate input capacitance, Von is the gate voltage at the "on" state, and Voff is your off voltage. Hence, t = -ln(Voff/Von)*RC = -ln(6.5/16)*25.5k*7.8n = 0.18ms. If your switching frequency exceeds approx 5.6kHz it won't have time to ever fully discharge and hence will never turn off.

author: eerocketman
author_rank: Contributor
timestamp: August 04, 2017, 06:07:53 am
content: Hi guys so I followed the advice of reducing the pull down resistor on my initial circuit to about 500 ohms. I also added a 10A 800V diode parallel to the IGBT.I tested the circuit with a 600V 8A capable power supply. As the voltage was increasing too about 400 V the IGBT blew up along with the 500 ohm and 51 ohm resistor. When I tested the IGBT with a meter all the pins were shorted. Can anyone provide advice as to what may have cause this problem?!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 04, 2017, 06:50:42 am
content: Blew up, as in explosively?Tim

author: eerocketman
author_rank: Contributor
timestamp: August 04, 2017, 06:59:21 am
content: Well the resistors blew up. The only way I could tell the IGBT shorted was that the power supply was hitting its current limit and the gate was low

author: floobydust
author_rank: Super Contributor
timestamp: August 04, 2017, 07:26:52 am
content: The diode was to go across the resistor-bank load, not the IGBT.The IGBT already has a C-E diode.This is to clamp any +ve spikes over Vcc due to inductance in the load and wiring.You have a heatsink, right?Only other reason it shorted is excessive gate voltage or wimpy gate voltage.