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IR2110 at 500Khz

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rdpdo2002

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Hello all,

I would like to use a IR2110 at 500Khz, and I found some informations about how to use it here : https://tahmidmc.blogspot.com/2013/...howComment=1603911775348#c2042051763996131553

However, the values of the capacitors used for bootstrap are done for few 100th of Hz to 50kHz. For 30-50Khz, values must be between 4.7uF to 22uF.

Does anyone has a idea of the values I must use for 500 Khz ?

Thanks in advance.

Regards.
 

Hi,

There are application notes for IR2110 and it's bootstrap circuit.

Never heard that bootstrap capacitor needs to be bigger with higher frequency. I'd rather expect it the other way round.

Klaus
 

500kHz at more than 50V high side bus is asking to cook the device - it was really only designed for 100kHz max and then with buffers on the outputs ....
 

Ok thanks you for answer.

I know that capacitor must be smaller when frequency is bigger, but I do not found any example in the app note at 500KHz... I have the formula but it seems a litle complex because many values are unknow (page 6 of the app note AN978 RevD).

In the app note AN-978 RevD, page 25, they say that at 500Khz with IRF830 there is no problem of heating for the IR2110... But the capacitors they use (0.47uF) on figure 28A page 25 is at 100KHz). Voltage is 160V.

Did you know a device like the IR2110 which can handle 500Khz at 50-100V ? Thanks !
 

Hi,

The AN also says:
For more detailed information on bootstrap component selection see DT98-
2a “Bootstrap Component Selection for Control IC’s.”

Did you follow this?

"...Many values are unknown.."
Means exactly which one? We don´t know them. Some can be found in the according datasheets, some in your circuit/requirements.

The formula says ">=" this means you may use bigger ones as you like.
So if 0.47uF is used for 100kHZ you should be on the safe side with 0.47uF for 500kHz.

Klaus
 

Thanks you very much for the link I will follow this :) thanks a lot. Regards
 

Bootstrap capacitance value necessary is a function of external MOSFET gate capacitance drive charge requirement. Since higher freq operation generally requires faster slew rate, therefore higher drive current, the bootstrap capacitor may need to be larger. The driver must be capable of the higher drive current requirement which might be an issue depending on output device capacitance.

The high side driver floats above chip Vss with highside driver input drive signal from push-pull current sources. The internal capacitance at higher frequency will limit speed with the given amount of current source input stage drive.

For high freq operation the slew rate to output MOSFET has to be faster requiring higher driver current capability to push the gate charge. If external device is small (low current, low voltage) there might be enough drive current.

You will have to verify the turn on and turn off delay for high side verses low side drivers.

 

470nF is more than enough for HF operation - it is a question of the losses in the 2110 @ 500kHz - which will be large for fets with a Ciss of > 1nF and for HVDC bus > 50V @ 500kHz ... ...
 

Thanks you very much, i have use the formula to design my boostrap circuitry and I am trying now to simulate it on pspice. For 24V DC BUS, with 15V VCC, 500 kHz, 10A, using a MOSFET IRLU3410 and a diode MURS120T3G I found 2uF - with a x15 factor included)

I have another question : At 500Kz, with a 50% duty cycle, do I need to add some dead time on the HIN/LIN command (full bridge - 2x2110 and 4 MOSFET ILRU3410) ? If yes, how much do you think I need to use ?

Thanks again !
 

50nS clear dead time should be OK .... where is the diode used ?
 

Hi thanks you for answer :) The diode is with the bootstrap capacitor.

I am currently trying to simulate my project with pspice but I have some difficulties... I will create a new post for this problem.

Thanks again.
 

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OK - the diode needs to be extra-ultra fast for 500kHz, Trr < 30nS
 

Ok, trr for MURS120T3G is 25ns :) Thanks
 

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