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Halfbridge driver IC working help please

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themaccabee

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Hi,

I'd like to learn about half bridge driver ICs that is being offered from International rectifier or similar ones IRF211XX series..
I ve got two N channel MOSFETs , high side MOSFET Drain connected to Vdd=28V ,its Source connected to Low side MOSFETs drain & Lowside MOSFET source is grounded.
The load is driven from the point at source of high side FET( where drain of lowside is connected).

I d really like to know how the IRF Ics drive the high side FET or how the VGs for high side is being supplied..?I think the High side FET gate should be supplied at least by a voltage =28V+Vgs threshold Isnt that right??Will these ICs generate that by themselves or shall i do something to set the Gate voltage?(wondering since the Vgs for different mosfets changes right?)
Can somebody explain me or help me to find a tutorial about the same..
Thanks & Regards
 

leo_o2

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Yes. It needs 28V+vgs.
In most of case, it can be generated by itself in the principle of boot-strap.
 

themaccabee

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In principle of bootstrap.
.Can u please explain a little bit..
This is a circuit i obtained from IR2110 datasheet..
there is a bootstrap capacitor as u ve mentioned but i dont understand its working..
Also the load is taken from lowside FETs drain.. I d like to short the Lowside drain with high side source and control it with a TTL signal..I only want to switch the device ..ie either ON or OFF not like from a fast switching waveform..

Thanks For any help..

Regards
 

leo_o2

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The capacitor between VB and Vs is boot-strap capacitor. When Vs is pulled low, the capacitor will be charged thru the diode between Vcc and VB. Then when Vs is pulled high, VB will be coupled to Vs+Vc. Vc is the cpacitor voltage that is charged when Vs is low.
 

themaccabee

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I just added the functional block diagram with that too..
Sorry i coulnt still get the complete picture...
As Q2 is ON the cap charges to Vcc through the diode.At the same time i believe that HO & Vs will be connected together and hence Vgs for the Q1 will be zero & Q1 will be OFF.But how to explain the next high side FEt ON procedure..?Im confused ..
can someone help
thanks
 

themaccabee

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Okie ... So while the low side Q2 is ON Vs in pulled to ground...& hence Bootstrap cap charges and that cap voltage appears across the VB. Now Q2 is turned OFF & HO is connected to VB ie the bootstrap CAP.
Now Vs is Floating right??
Then VB is applied to the HO & a potential drop arises between the floating source & Q1 gate.Will this turn ON Q1??
if it turns ON Q1 the source of Q1 (Vs) will slowly acquire the drain voltage i.e here 28V..this will pull up Vb which was sitting on the Vs, to 28V+Vb, Enough to keep the Q1 ON.
Is this the correct procedure..?
Please correct me if im wrong.
Thanks
 

leo_o2

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Okie ... So while the low side Q2 is ON Vs in pulled to ground...& hence Bootstrap cap charges and that cap voltage appears across the VB. Now Q2 is turned OFF & HO is connected to VB ie the bootstrap CAP.
Now Vs is Floating right??

Leo: When HO is connected to VB, Q1 will be turned on. So Vs will be pulled high. It is NOT floating. As Vs is pulled up, VB will be shifted up by the cap between VB and Vs.
 
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