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UCC28730-Q1: HV Supply above 700Vdc as in UCC28740

Part Number: UCC28730-Q1
Other Parts Discussed in Thread: UCC28740, UCC28730, UCC28740-Q1

Hi,

the data sheet of the UCC28740 suggests to use two capacitors to allow start up from high voltage for voltages above 700Vdc. Why is this not included in the data sheet of the UCC28730? Is it possible to use the same setup with the UCC28730?
How would you calculate the size of these capacitors?

Kind regards

  • Thanks for asking.

    Basically, the HV design of UCC28740 and UCC28730 are the same. 

    Would you please let me know which figure or descriptions that you mentioned in UCC28740's datasheet?  It would be helpful to clarify your questions. 

    Regards, 

    Wesley

  • Hi, 

     I am sorry I just look up the UCC28740's datasheet. That's the reason I did not find this before. I found this figure in UCC28740-Q1's datasheet. When the Vbulk is over 700VDC, which is over HV pin's spec, CIN2 and CIN1 could be a voltage divider to provide a voltage which is below 700V. It could be calculated by a simple equation by KVL. The equation is as below. 

    VHVPIN = VBULK * (CIN1/(CIN1+CIN2)) < 700VDC.

    The HV start circuit design of UCC28730 and UCC28740 are the same. Thus, this circuit is also able to applied to UCC28730.

    Please let me know if you need any assistance.

    Regards,

    Wesley

  • Hi,
    thank you.
    Concerning the values of the capacitors I was looking for a guideline to find the absolute values of the capacitors, so that they are able to supply the neceassary start up current. Do you have something like that? As for the relative values I would look at two equal capacitors, since this would also result in the lowest necessary voltage rating of the components.

    Regards
    Ludwig

  • Hi, Ludwig:

    I have no this guideline but I thought it could get the capacitance value by equation Q=CV. This approach is as below. 

    The 1st step is to design the CVDD. You may get this value based on the calculation tool of UCC28740 or UCC28730. 

    The VDD(on) is noted on datasheet. It's around 23V(max)

    Thus, the total Q to charge up CVDD is Qtotal = CVDD * VDD(on). which is the min value that CIN2 has to provide to charge VDD.

    Assume CIN1=CIN2, VHVPIN=VBulk/2. 

    then we could get  Cequal = Q/VHPIN

    For example, 

    If Vbulk is 700V. CVDD is 10uF.  The necessary Q would be Qtotal=10uF*23V= 230uC. 

    Since CIN1=CIN2, VHVPIN would be VBulk/2=350V. 

    C2 = 230uC/350V=657nF (min.)  . Consider the 20% tolerance of this capacitor, CIN2=CIN1=1.2*657nF=788nF ==> 820nF should be okay to use. 

    This approach may be not so accurate but I think it is still helpful to choose the capacitor value.

    Please let me know if you have any thoughts about this. We could discuss about it. 

    Regards, 

    Wesley 

  • Hi, 

    I actually thought of a similar approach, but was using an energy approach instead of charge. From that you could calculate the voltage drop in the low side capacitor and adapt the capacity so that the top side voltage would still be within range. I also used a conservative efficiency coefficient.

    From your knowledge of the insides of the IC is the charge approach more accurate? This would lead to much higher capacitance values than the energy approach.

    Regards

    Ludwig 

  • Hi,

     I consults my member who have ever used this circuit. It is still needed to add balance resistors in parallel with CIN1 and CIN2. 

    The current would flow through CIN1 to charge CIN2 and CVDD.  Thus, the capacitance only impact charging time of CVDD. 

    So if you choose higher value, the charging time would be faster.  Please note the IHV is limited by IC internally. The  IHV is 250uA(typ.) and 500uA(,max.).

    So I would suggest you to choose the CIN1/CIN2 which can provide >500uA current to avoid longer CVDD charging time.