• State TI Thinks Resolved
  • Locked Locked
  • Replies 3 replies
  • Answers 2 answers
  • Subscribers 63 subscribers
  • Views 1371 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

UCC21750: How to decide the capacitor (CBLK) and resistor (R) for the DESAT function?

UCC21750: CBLK value for DESAT

Masayuki Uchihara
Intellectual 1686 points
Part Number: UCC21750

Hi team,

I have question about the DESAT solution. According to the EVM of UCC2732, the CBLK is 27pF. 

- how do you calculate this ?

- I feel very small capacitor, is it no problem about surge even such a small capacitor? 

Thanks,

Uchihara

  • 0
    TI__Expert 4875 points

    Hello Uchihara,

    The small CBLK used on the EVM is geared towards a Silicon Carbide (SiC) based application. SiC MOSFETs switch at a very fast speed, and have a smaller short circuit withstand time to survive a short circuit event. Thus, this requires very fast DESAT detection and reaction. If an IGBT is being used, then the blanking time can be longer due to slower switching speed and longer short circuit withstand time.

    Please find more information on DESAT and how to calculate the blanking time in this e-book.

    If the value of CBLK is too small to suppress the noise in a given system, then CBLK can be increased while adjusting the blanking time via another component. For example, by increasing RDESAT, adding a Zener diode in series with the HV diodes, or by adding more HV diodes in series.

    Regards,

    Audrey

  • 0 in reply to Audrey Dearien7
    TI__Intellectual 1686 points

    Hi Audrey

    Can you check my calculation? If I set the Tblk=500ns, Cblk should be 10 pF order. Is it correct? The pF order capacitor is not too small right?

    Thanks,

    Uchihara

  • 0 in reply to Masayuki Uchihara
    TI__Expert 4875 points

    Hi Uchihara,

    Your calculations are correct. It may be more susceptible to noise because it is a small CBLANK. One thing you could do is to increase CBLANK to 100pF, then add a resistor between OUTH and the DESAT pin to provide more current to charge during turn-on. You could use a 30-kOhm resistor between OUTH and DESAT with a diode in series to block current flow back to OUTH from the DESAT pin. Using 100pF from DESAT to COM and 30kOhm + diode between OUTH and DESAT you will get around 600ns blanking time.

    Regards,

    Audrey