Slow turn off / 2 level turn off for isolated igbt drivers - Page 1
Forum section: electronics / projects

author: MarkoAnte
author_rank: Regular Contributor
timestamp: November 03, 2015, 11:10:30 am
content: Hi,I have noticed that that most manufacturers of isolated igbt drivers with integrated protection offer 2 variants of how the driver reacts when the overcurretn, desaturation fault is triggerd. (Eg. TI5851vsTI5852S1ED020l12-F2 vs 1ED020l12-FT)Taken form the TI datasheet:"During IGBT overcurrent condition, a Mute logic initiates a soft-turn-off procedure which disables, OUTH, andpulls OUTL to low over a time span of 2 ?s." (OUTH is for +15V positive drive and OUTL is for -8V negative drive).All additional info I fould was: " In addition, soft turn-off, which slowly cuts off the short circuitcurrent, is necessary as abrupt turn-off leads to a large dv/dt, which increases the possibilityfor a latch-up." inhttps://www.fairchildsemi.com/application-notes/AN/AN-9020.pdfSo the slow turn off in fault is good or bad? Do you guys(that have experience with this) use it or not? I was always told to hard turn it off.Best regards,Marko

author: retrolefty
author_rank: Super Contributor
timestamp: November 03, 2015, 12:10:24 pm
content: QuoteI would suspect that depends on the load type / application. These are often used to drive inductive motors where a slow turn-off can aid in dealing with sudden collapsing magnetic fields and their higher voltages created?

author: MarkoAnte
author_rank: Regular Contributor
timestamp: November 03, 2015, 12:22:33 pm
content: The driver would be for a PFC --> DC/DC --> DC/AC (HF) feed into a resonant LC circuti. (For induction heating). The DC voltage would be 1KV. The most common for transistors is to have 1.2 KV rating so I'm close. I'm considering the soft turn off because of that. But I don't know about the potential down falls because of slow turn off.Regards,Marko

author: rs20
author_rank: Super Contributor
timestamp: November 03, 2015, 12:26:20 pm
content: Turn off too quickly --> inductive effect (voltage spikes) might cause damage.Turn off too slowly --> Fault is cleared more slowly, fault has more "chance" to cause permanent damage.

author: nuno
author_rank: Frequent Contributor
timestamp: November 03, 2015, 12:30:46 pm
content: I have no hands-on experience with this, but when the desaturaton protection kicks-in, you may have a very big current flowing through the IGBT, maybe significantly higher than the peak current your circuit it was designed to handle. If you turn it off too quickly, it will suffer from the inductive kick-back of the "load circuit"'s inductance, maybe even if it's just parasitic inductance - depends on the combination of loop inductance, turn-off time and current at turn-off.p.s. According to an expert in motor drivers with IGBTs, you should turn it off in under 10us. I think he mentioned also that you can get that info from the datasheet, but I've never looked.

author: MarkoAnte
author_rank: Regular Contributor
timestamp: November 03, 2015, 12:38:35 pm
content: So basicly its:Do I over build it for current (slow shut down) or over build in for voltage (quick shut down)?Maybe its better to build it for over current because I'm thinking of using a boost based PFC and buck based dc/dc converter so I will have a lot of inductance that can somewhat limit the current (or kick up voltage).

author: MarkoAnte
author_rank: Regular Contributor
timestamp: November 03, 2015, 12:41:34 pm
content: Then again, if I build my pcb smartly I could get away with making it universal for the 2 versions of the driver.

author: nuno
author_rank: Frequent Contributor
timestamp: November 03, 2015, 06:00:30 pm
content: v = L di/dt is your friend (although not enough for your analysis).