Hi Jeff,
Thank you for posting on the Infineon community.
Bootstrap circuit:
Initially the low side gate signal should be turned ON to charge the bootstrap capacitor through the low side switch. Please refer the application note for all the calculations.
During this condition the diode should be forward biased and placed in line with the low side and high side supply.
In IR2110, the bootstrap diode should be placed between the pins VB(high side floating supply) and VCC(low side floating supply) if it is used in half bridge configuration. (Attached for reference)
In your application I assume it is a buck converter. Hence please go through the following explanation and follow the connections (especially the diode) as shown in the below figure,
Below figure shows a typical implementation of a buck converter with the high-side drive function. The diode connected on COM prevents the negative spikes from affecting the operation of the IC and provides an extra measure of noise immunity. COM should not be connected together. At start-up the bootstrap capacitor is discharged and, in most applications would charge through the inductor and the filter capacitor. The same is true under no-load conditions, when the freewheeling diode may not conduct at all. This alternative path works, as long as the filter capacitor is at least 10 times larger than the bootstrap capacitor. The Q of this resonant circuit should be low enough to insure that the bootstrap capacitor does not get charged beyond the
limits of VSS (20 V). If this is not so, a zener in parallel with the bootstrap capacitor would take care of possible overvoltage events.
As mentioned please follow the application note for detailed calculation. Please do let me know if you need any further clarifications.
Regards,
Abhilash P
Hi Jeff,
Thank you for posting on the Infineon community.
Bootstrap circuit:
Initially the low side gate signal should be turned ON to charge the bootstrap capacitor through the low side switch. Please refer the application note for all the calculations.
During this condition the diode should be forward biased and placed in line with the low side and high side supply.
In IR2110, the bootstrap diode should be placed between the pins VB(high side floating supply) and VCC(low side floating supply) if it is used in half bridge configuration. (Attached for reference)
In your application I assume it is a buck converter. Hence please go through the following explanation and follow the connections (especially the diode) as shown in the below figure,
Below figure shows a typical implementation of a buck converter with the high-side drive function. The diode connected on COM prevents the negative spikes from affecting the operation of the IC and provides an extra measure of noise immunity. COM should not be connected together. At start-up the bootstrap capacitor is discharged and, in most applications would charge through the inductor and the filter capacitor. The same is true under no-load conditions, when the freewheeling diode may not conduct at all. This alternative path works, as long as the filter capacitor is at least 10 times larger than the bootstrap capacitor. The Q of this resonant circuit should be low enough to insure that the bootstrap capacitor does not get charged beyond the
limits of VSS (20 V). If this is not so, a zener in parallel with the bootstrap capacitor would take care of possible overvoltage events.
As mentioned please follow the application note for detailed calculation. Please do let me know if you need any further clarifications.
Regards,
Abhilash P
