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Boot-strap components

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mido46

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bootstrap minimum duty

Hi,
I have to design a H-bridge inverter and I need to understand the meaning of using a boot-strap diode and capacitor with the driver of the switch?
Thanks in advance,
Mido
 

ir2184 bootstrap

In full (H) and half bridge converters you need to drive gate(s) of upper transistor(s) with voltage exceeding the DC-link voltage to get them saturated. Getting such rised voltage for the driver circuit is the role of bootstrap components. Another way for properly driving the transistors is using a trigger transformer.
 

boost strap capacitor application

the app note of the chip should tell you.

theyhave limitations dont they....what happens when you go on no load with those bootstrapped jobbies?
 

yes so anyway, those bootstrap high and low side drivers- they are bad news at times............because when you are on no load....you are not switching......and the problem is the bootstrap capacitor must remain charged up and it cannot charge up when there is no switching going on............................so maybe you are talking about getting your pulse transformer hi side driver out of the cupboard and using that now?
 

You can use also a charge pump to keep the bootstrap cap always charged even when no load is present
 

paurelian...but the charge pump action is gotten by the switching of the fets....and if you are having a green" psu then on no load you are not switching the fets....and the blasted charge pump cap is not charged up, it discharges ....so this is your problem.

because it will at times possibly trying to switch on the upper fet when it doesnt have the voltage to properly turn on the upper fet.........

and BANG!!!!!!!!!!!!!

seriously, i was at an electic drive company and the IR2110 (?) or similar was used.

the fets kept on blowing.

The inverter needs not only to be on load, but there needs to be sufficient duty cycle to keep that boost cap charged up..which aint easy.

if your on full load all the time then no worries.

For low freq inverter u can get away with opto driver.......few KHz.

Otheriwse Laszlo Blogh has the document for you.....on mosfet driving....search www

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Gate drive transformers are no joke though and need lots of components to make them efficient...like turn off BJT's etc
 

I don't doubt, that these guys have managed to blow their FETs by inappropriately using bootstrapped drivers.

But it's not generally true, that you can't use a IR2110 driven H-bridge with no load. Yes, you have to keep a minimum duty cycle for the low-side switches to keep the bootstrap capacitors charged. You even can shut down the inverter completely, and you can start at full current, if you follow a specific turn on-procedure, to charge the bootstrap capacitors first.

As an additional remark, the said drivers also have a low voltage lock, so they don't cause MOSFET desaturation even with discharged bootstrap capacitors. But I can imagine, that you get problems in some cases, e.g. with a transformer load and no effective over-current protection.
 

OK thanks FvM,

sorry it was "Laszlo Balogh" who does a very good article on high and low side gate drivers.

if you google it you find it.
 

I made some trials with ir2184 and ir2104 gate drivers and these drivers are working fine also with no load. regarding the boot cap, depending on your total gate input cap of your mosfet you will choose the value for the boot cap. 100nF will be a prefered value. with such boot cap you will keep the high side on for ~8msec using a mosfet with ~1400pF input cap. to charge such cap you will need max 10usec. having in mind the above results, you can make some calculation to decide what will be the max switching frequency and duty cycle. also you can increase this boot cap to allow a longer time for the high side to be on. i tried also wth 2.2uF and I got ~170msec. l6384 datasheet have a good description about how to calculate the boot cap according with mosfet specifications.
I hope that you undesrtood a little bit the phenomen.
 

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