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Original question:

LM5145: When TI FET CSD19534Q5A is used with LM5145, is there any shoot through problem?

LM5145: Do you have any SOA curve in case of 5.5us single pulse?

Kazuya Nakai59
Guru 21505 points
Part Number: LM5145
Other Parts Discussed in Thread: CSD19534Q5A,

 Hello guys,

 One of my customers is evaluating CSD19534Q5A for a buck switching regulator.

 They have 2 questions about SOA as the follows.

1. Do you have any SOA curve in case of 5.5us single pulse? If answer is no, could you tell me any equation for calculation?

2. In case of this FET using for low side FET of buck switching regulator, the inductor current of the regulator flows through

    parasitic diode of the FET during dead time period (low side FET off period).

    Do you have any SOA data for the parasitic diode. Or do you have any equation for the SOA calculation?   

 Your reply would be much appreciated.

 Best regards,

 Kazuya Nakai.

 

  • 0
    TI__Mastermind 23845 points
    Hi,

    This seems like like a FET question, i will forward the information with the FET group.

    Thanks
    -Arief
  • 0 in reply to Arief Hernadi
    TI__Mastermind 37950 points
    Nakai san,
    Thanks for the questions. Please see this blog on SOA: e2e.ti.com/.../understanding-mosfet-data-sheets-part-2-safe-operating-area-soa-graph. The curves in the datasheet are generated from actual test data - not calculated. We test to failure and derate from there. For (1), the SOA curve in the CSD18543Q5A datasheet only goes down to a 10us pulse width. The customer should be safe using 10us pulse width. As I pointed out above, there is no calculation. For (2) we do not test SOA of the body diode and do not have a calculation for this. The peak forward current rating of the body diode is less than the peak drain current specified in the datasheet for the FET because the forward voltage drop, VF, across the body diode is greater than ID x rds(on). Given the very short dead time of the LM5145, SOA is not a concern for the body diode of the FET.
  • 0 in reply to John Wallace1
    Guru 21505 points
    Hello John,

    Thank you very much for the prompt reply.
    I could understand well because of your reply.

    Could I ask you a few additional questons?
    Q3. How much current can be flowed through the body diode of CSD19534Q5A the customer uses
    during the device dead time (14ns typ)?
    Is it 10A because I can see 10A as a condition of "DIODE CHARACTERISICS" in Electrical Characteristics table?

    Q4. Can the consumed power at the body diode be calculated by the following equation?
    Body diode power (W) = VSD (V) * ISD (A) * total dead time at turning on/off (ns) * switching frequency (Hz).

    Thank you and best regards,
    Kazuya Nakai.
  • 0 in reply to Kazuya Nakai59
    TI__Mastermind 37950 points
    Nakai san,
    The body diode continuous and peak current ratings are different than the FET ratings specified in the datasheet because the diode VF is typically greater than ID x rds(on). It all comes down to power dissipation (IDS^2 x rds(on) vs. ISD x VF) and thermals. On page 1 of the datasheet, the continuous current rating for the FET is specified at 10A on a 1 sq. in. , 2oz. copper pad with typical Rtheta(j-a) = 40degC/W @ 25degC. This equates to 3.2W power dissipation: PD = (Tj(max) - 25degC)/Rtheta(j-a) = (150 - 25)/40 = 3.125W which is rounded up to 3.2W. For the same dissipation thru the body diode, the current is ISD = PD/VF = 3.2W/1V = 3.2A. That's the continuous current rating of the body diode at 25degC. Similarly, the peak current rating can be calculated from the peak power dissipation: ISDpeak = 63W/1V = 63A.

    In a buck converter, during the dead time periods the diode will carry the same peak and valley current as the FET. When the high side turns off, the inductor current is at its peak and when the low side FET turns off, the inductor current is at its valley. The 10A you see in the datasheet is a test condition for measuring forward voltage and reverse recovery and is not the current rating of the diode.

    Please see this app note: www.ti.com/.../slyt664.pdf. It has a good explanation of the body losses on page 4. The body diode conducts the inductor current during the dead time after the high side FET turns off and before the low side FET turns on and during the dead time after the low side FET turns off and before the high side FET turns on.