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SwitcherPro Transistor Power & Temp Calculations Seems Wrong

Wing Poon
Prodigy 20 points
Other Parts Discussed in Thread: CSD16410Q5A, TPS40192, SWITCHERPRO

I just created a buck-regulator design using the TPS40192, using one each of CSD16410Q5A, which has Ron=10mOhm@4.5V, for high and low-sides.

I_rms(Q1) was calculated to be 4.05A. The Q1 (high-side) Power Dissipation is 637mW according to SwitcherPro. I don't see how this can be the case because for a MOSFET in saturation, P=I^2.R = 4.05^2*0.010 = 164mW ?? Qg=3.9nC, so that's not contributing much to power dissipation either.

The calculated junction temperature seems to be based upon the 637mW number, so that too seems inflated.

-Wing

  • 0
    TI__Expert 3875 points

    Hi Wing,

    For each FET in the database, SwitcherPro stores 3 R_ds_on values:  R_ds_on at 2.5V,  R_ds_on at 4.5V, and R_ds_on at 10V. SwitcherPro uses a part specific equation to calculate the value of V_gs and, then, uses linear interpolation to calculate R_ds_on. Unfortunately, as we all know, the relationship between V_gs and R_ds_on is not linear (it is more like exponential decay). This can result in mis-calculation of power dissipation in the FETs if the V_gs voltage falls between 2.5 and 4.5.  In your case, you were looking at the CSD16410Q5A, which has the following values in SwitcherPro:

    R_ds_on at 2.5 = 10 Ohm
    R_ds_on at 4.5 = 10 mOhm
    R_ds_on at 10V = 6 mOhm

    Since there are three orders of magnitude between R_ds_on at 2.5 and R_ds_on at 4.5, the linear fit will do poorly in predicting the correct R_ds_on.

    Hope this explanation helps.

    Regards,
    Michael

  • 0 in reply to Michael Krasnicki
    Prodigy 20 points

    1). Looking now at TP40192 datasheet and looks like there's an error there. On Pg#17 it says "The LDRV driver  switches between VDD and GND". Since VDD can go up to 18V for this part, this gave me a scare at first, because most FETs have a Vgs(max) < 10-15V! But looking closely at the block-diagram on Pg#10, I think the above statement is wrong -- you can see that the LDRV driver is powered off of the internal 5V rail (and not "VDD"), so LDRV in fact switches between GND and 5V. Pls pass this on to the person responsible for this datasheet.

    2) From the above, therefore Vgs of the low-side FET (when ON) should be 5V (I told the tool that my Vin= 11.5-12.5V > 5V). Assuming the switching time is negligible (it should be, otherwise it shouldn't be called a switching regulator!), then Vgs does NOT "fall between 2.5 and 2.4V", certainly should not for any appreciable amount of time.

    3) Since HDRV is driven by Vboot, it stands to reason than Vgs of the high-side FET (when ON) is ~4.3V. So once again, Rds of 10mOhm should have been used. In fact, if you look at Fig#7 of the CSD16410Q5A datasheet, at T=25C, Rds_on(Vgs=4.3V) is between 9-10mOhm.

    So, it sounds to me like this is a serious bug in SwitcherPro that should be fixed. Otherwise, you'll be doing your customers a disservice by making them spec a much bigger (and costlier) FET(s) than they would actually need to.

     

    -Wing

  • 0 in reply to Wing Poon
    TI__Expert 3875 points

    Hi Wing,

    Could you please post your design, including the schematic and BOM.

    Thanks,
    Michael