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Original question:

UCC24624: Whether the gate Vg waveform is correct?

UCC24624: Design Questions

Brian Wang0928
Genius 14080 points
Part Number: UCC24624
Other Parts Discussed in Thread: UCC24610, UCC24612

There are three tech. questions about the UCC24624 design. please help on this. thanks. 

1. How to disable the UCC24624? Can we put down the REG pin to Ground when it's disabled? Or what is the best way to do? Please offer workaround solution.

2. If the inside sourcing and sinking driving current of this device is not enough, do you suggest to add one more buffer or totem pole to enhance the driving capability? What existing part number of the totem pole you suggested?

3. In this design, there are 3pcs MOSFETs connected in parallel for higher current rating. how to optimize the layout routing for these MOSFETs in parallel? Please offer the user guide. 

Information:

Application: Isolated DC/DC module for Motor Drive

LLC topology:

Controller: MCU

SR IC: UCC24624

Input Voltage": 820Vdc, typ

Output Voltage: 27V, typ

Output Current: 120A

Regards

Brian W

  • 0
    TI__Guru* 86266 points

    Hello,

    1. Reg does have short circuit protection of 9.5 mA.  If you pull it to ground you will disable the gate drive. The UCC24624 does disable itself when the switching frequency is below 9 kHz.  Please review section 8.3.6 for details.

    2. The UCC24624 regulates the gate voltage based on the voltage across the Rdson.  Please review section 8.3.2.  If you feed the gate driver into a buffer it will not function correctly.

    >If you need to parallel FETs and use a buffer you might consider using the UCC24610.  This is VDS sensing and is not a dual driver.  However, it does not have the proportional gate drive and can work up to 600 kHz and the output can be fed into a gate driver buffer.

    >If you decide to use this device and questions for it.  Please post a new e2e thread with the UCC24610 in the title.

    3.  Both the UCC24624 and UCC24610 give guide lines for layout.  I know that you are using three in parallel which is a little more challenging.  However, these guidelines should apply for 3 FETs in parallel as well.  Please review and let me know if you any questions.

    Regards,

  • 0 in reply to Mike O'
    Prodigy 30 points

    Hello,

    1、Can the ucc246624 drive three parallel FETs IPB027N10N5? The Rds(on) of this FET is 2.7mohm.

    2、If the ucc24612 cannot drive the FETs who has small Rds(on)? The Rds(on) of three parallel FETs is 0.9mohm.

  • 0 in reply to Andyzhao
    TI__Guru* 86266 points

    Hello,

    I have not driven three FETs in parallel with a single driver.  However, I looked at the FETs you have chosen and with evaluating the drive current, and Qg, Vgs curves and power dissipation.  I believe you might be able to use the UCC24624.

    The output current of your design is 120A which means the peak LLC current going through the design is 170A.  So with a sense resistor of 0.9 ohms your peak Vds when the FET is on -153 mV.  So sensing will be O.K.  I believe the FETs turn off with a 10 mV in both ICs.

    The UCC24624 has 1.5A of gate drive current and 4A of sink current.

    The UCC24612 has 1A source and 4 A sink.

    The IPB027N10N5 has  Vgsth of 3 V typical an according the FET Vgs Qgate curve it will take 72 nC to fully engage the FET.  Using dt = Q/I and estimated that you have 1/3 the current to drive each FET.

    UCC24624

    Ton = 144 ns

    Toff = 54 ns

    UCC24612

    Turnon = 216 ns

    Turnoff 54 ns

    So you could possibly drive three of those FETs with the UCC24624. 

    Next lets estimated the power dissipation of driving 6 FETs with the UCC24624.  You will need one for each leg.   I estimate driving 6 FETs at 100 kHz.

    I don't know the frequency of your design but you can go through the estimates yourself based on your actual switching frequency (fsw).

    Pdrive = Number of ParallelFets*Qg*Vg/fsw = 6*72nc*9V/100kHz =0.389W 

    You would have to drop VDD to less than 26V.  If you used and average circuit and dropped VDD down to 15V.  The power dissipation would be

    PUCC24624 = VDD*Irun + Pdrive = 15V*1.5mA + 0.389W = 0.415W.

    Tjmax = 125C

    Next calculate the maximum ambient teampature.

    Tamb_max =  Tjmax - PUCC24624*Rja = 125C - 0.415W*108.4C/W = 80 deg C.

    Regards,