[SOLVED] Maximum number of mosfet driven by IR2110

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hello

i am using a mosfet with gate charge of 550nC , switching frequency is 16khz to 20khz , gate resistor about 10ohm

the iR2110 specifies 2A current capability , the bootstrap capacitor is 1uF (non polarizied) ,

1-how many mosfets in parallel this chip can drive ?? ( is 4 a resonable number ??)

2-do i have to increase the bootstrap cap (duty is from 0-to 95%)?? do i have to added 10uF polarizied as i have seen in some posts online ??

3-and if it works in halfbridge configuration so: 8 mosfets and 1 driver , how much maximum dc current should the 12v regulator supply ??

thanks and regards , any info, answer or guidence to an app note is appreciated
 

thanks for your helpful replies !!!

i have some answers for what i was looking at : AN799 microchip :Matching MOSFET Drivers to MOSFETs

and irf appnote : HV Floating MOS-Gate Driver ICs

i will mark as solved , although i still have no answer for question 2- and 3
 
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I believe that the design equations that answer your questions are derived in many application notes, e.g. from IRF.

Gate driver supply current (ignoring static quiescent current) is e.g. simply Qg*Vdd*fsw.

Minimum boostrap capacitance can be calculated from Qg and acceptable Vdd voltage drop. Some circuits have a larger static current consumption that must be considered additionally.
 

thanks , but just a small question about the equation : Qg*Vdd*Fsw ,
Qg is for 1 transistor ?? , example : H bridge 4 mosfets each have 550nC , so total current =(4x550n)*Vdd*Fsw ???
I think this might be wrong because not all transistors are switched i the same time , assume sign magnitude drive or locked antiphase
please clarify
 

I think this might be wrong because not all transistors are switched i the same time , assume sign magnitude drive or locked antiphase
Click to expand...
The calculation applies to all transistors driven by the respective high side driver and it's bootstrap capacitor.
 

are you sure about this equation Qg*Vdd*fsw? because it has not current dimension.
 

Qg*Vdd*fsw (ampere*t)*vdd*1/t= i*vdd=p
still beat me!
could you plz introduce a reference for this equation?
 

I have looked to IXYS and fairchild ANs, as I thought this is gate power not gate current. as it evidence it has power dimension not current (ampere).

FvM said:
I believe that the design equations that answer your questions are derived in many application notes, e.g. from IRF.

Gate driver supply current (ignoring static quiescent current) is e.g. simply Qg*Vdd*fsw.
Click to expand...
 

FvM said:
You are right. Qg*Vdd*fsw is power, not current.
Click to expand...

I rechecked : page 2 in AN799 microchip . power due to switching is : P=C*V*V*F , can be reduced to P=Q*V*F

then I is independent on Vdd , and is : Q*F only right ?????

- - - Updated - - -
 

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