I wont make a bridge with IR2101 to drive a BLDC motor, but the IR2101 doesn't polarise the up-side mosfet (Vgs only 3.5V). I use a capacitor between Vb and Vs of 3.3uF (I had also used 200uF but the voltage of polarization is grown but of little). I use as Vcc=15V end the scheme of data sheet. I'm in despair!!! I don't know to do!!!
I excuse for my bad English.

 

Welcome to the forums.

I put your post in a new thread, because the original was about a year old - http://forum.allaboutcircuits.com/showthread.php?t=23392

 

Between Vs and Vb, try using a 0.1uF (100nF) metallized poly film or ceramic AND a 10uF aluminum electrolytic capacitor. Use the same (0.1uF/100nF and 10uF caps) between Vcc and COM (ground).

The diode between Vcc and Vb must be a fast-recovery type, and be rated for at least Vcc + your MOSFET drain supply voltage.

You will not be able to keep the high-side MOSFET turned ON for very long (maybe 50mS-100mS) at a time, because the capacitors will need to be re-charged. The only way to re-charge the capacitors across Vb/Vs is to turn the high-side MOSFET off for a moment.

 

You will not be able to keep the high-side MOSFET turned ON for very long (maybe 50mS-100mS) at a time, because the capacitors will need to be re-charged. The only way to re-charge the capacitors across Vb/Vs is to turn the high-side MOSFET off for a moment.

Thank you for the answer!!! I believed that the driver IR201 could polarise the mosfet for as long as I wont because I don't understand how work the CI. In the data sheet there isn't so mach. Thus this driver can't be used for to pilotage a CC motor for long time. I made the tests with a cc motor and after a determined time his speed fell (due the charge or discharge of cap). Then I must provide input to Hin or Lin a impulsive signal, right?
Thanks again for the answers!!

 

You can keep the low-side MOSFET turned on for as long as you would like to without problems.

But as you discovered and as I tried to explain in my first reply, the capacitor(s) between Vcc and Vb begin to discharge as soon as you turn the high-side MOSFET on. After a period of time, the charge on the capacitors is too low to keep the high-side MOSFET turned fully ON. This will make the high-side MOSFET begin to dissipate a lot of power as heat.

If your high-side MOSFET will have a high duty cycle (>94% or so) then you should consider using a much larger or more capacitors between Vcc and COM. That/those capacitor/capacitors is where the caps between Vs and Vb get their charge from.

Have a look at the attached simulations.
The white trace represents the charge on C1, the capacitor between Vcc and COM.
The red trace represents Vb.
The green trace represents the charge on the cap between Vcc and Vb.

R1 is just an example of the load that the IR2101 represents. The exact value is not important here.

Notice that the first time C2 gets charged, it only gets up to roughly 6v. This is because C1 and C2 are about the same value, and it takes time for the current to start flowing from the power supply V1 through the inductance of the wiring and it's own internal resistance.

If two capacitors of equal value are connected in parallel, the resulting voltage will be the average of the initial voltage of both capacitors.

The difference between the two schematics is C1. Note that in the 1st simulation, C1 is 10uF; in the 2nd simulation it is 50uF. Notice that C2 gets charged more completely because there is a larger store of charge in C1 in the 2nd simulation.

 

Thanks again for the answers and the simulation!!! Now I understand better how it works. Tomorrow I'm trying to pilot the tree complete bridge with a PIC (PWM output)end I will tell you if the BLDC begin to turn!!! I had calculated that when the motor turn at 500rpm (min speed) the hall sensor change 6*3=18 times per rev so the pwm output change 60/(500*18)=6,6ms end the capacitor doesn't have to discharge so fast. Thank again...

 

One item that I omitted from my prior post is that not only does the high-side MOSFET have to be turned off (Vgs=0), but the lower side MOSFET has to be turned ON in order to charge the boost capacitor.

This is not really obvious, but once you think about it a bit, you'll realize that without the low side MOSFET being turned on, there is no path to ground to charge the boost cap.

 

Now the bridge works as I wont but I have a problem with the motor but maybe I create a new discussion.
Thank again for the help...