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TPS92692: Heat problem of major components

JAEWOONG CHUNG
Prodigy 140 points
Part Number: TPS92692

I made the Boost LED Drive Board using TPS92692.

The target specification is to turn on the LEDs of the 8X12 array. (max 200W, max 6.0A, input 24V)

The temperature of the major components was measured in the recent 5.2A output.

The LED array was turned on for about one minute, and the temperature is as follows.

1) LED Drive IC: 48.6°C
2) Input Scotty diode: 102.6°C
3) N-MOSFET: 67.1°C
4) Inductor: 45.6°C
5) P-MOSFET: 55.1°C
6) Output Scotty diode: 91.9°C
*Not saturated temperature.

Among the above components, the temperature of the Schottky diode is very worrisome.

I have attached a circuit diagram and layout, so please advise me on how to lower the heat.

  • 0
    TI__Expert 3630 points

    Hi

    Your question will be answered soon.

    Thanks

    Sumeet

  • 0 in reply to Shashank Kulkarni
    TI__Mastermind 18890 points

    Hi,

    With 12V input and at such a high output of 200W your thermal design is of the most important not only from choosing the components to handle power level but proper heat sinking has to be done to get the heat out.  For example, the input diode which has about about 9.4A thru in based on about 87%-88% efficiency and assuming it has about 0.6V drop across it...That will be about 5.6W that you have to get out.  This is a huge amount of current and you need to verify that you diode can handle this current first and then proper heat sinking either conduction or force air.  Note that this is at your stated 24V input at the 12V input that your schematic shows the input current will double.

    Just based on your question and design you will not be able to get there and you will have to educate yourself on fundamental of design.  At that power level you might have to use a FET in place of the input diode and parallel 2 switching FETs and output diodes or bigger diode first then properly design for thermal.  Below

    http://www.ti.com/lit/an/snva419c/snva419c.pdf?&ts=1589898001408

    training.ti.com/thermal-design-managing-thermal-performance

    http://www.ti.com/lit/an/snva419c/snva419c.pdf?&ts=1589898001408

  • 0 in reply to Tuan Tran74
    Prodigy 140 points

    Thank you for your reply.

    I'm going to review the change to a product with a larger capacity of Scottky diode first and try to make a parallel connection at the same time.

    But if my board has a heat problem again, I'm thinking of designing and installing additional heat sink as the next step.

    I'll look at the educational materials you attached. This is my first time to do this kind of application, so I think I will definitely need to study.

    Thanks again!

  • 0 in reply to Tuan Tran74
    Prodigy 140 points

    Oh, and two of the three links you gave me are the same. Did you intend it or did you write it wrong?

  • 0 in reply to JAEWOONG CHUNG
    TI__Mastermind 18890 points

    Hello Jaewoong,

    That was a typo there should only be 2 links.  Just a few words of recommendation since base on your design you will not get there.

    1.  At 200W you will need heat sinking and in addition maybe force air cooling.  The heat being dissipated on your board will be greater than 20W

    2.  If you do not need reverse battery protection then you do not need D1.  If you need reverse battery protection then use a PFET in place of D1

    3.  You will need to reduce your switching frequency to about 100KHz-200KHz to reduce switching loss since you will need to have a big FET for Q2 or parallel more than one FET there.  You should short out L4 on your schematic and drive the FET directly.

    4.  When changing to lower switching frequency, you will need to change the inductor to a higher value along with the appropriate saturation level

    5.  You definitely need a larger output diode for D2 or parallel some

    Thanks Tuan

  • 0 in reply to Tuan Tran74
    Prodigy 140 points

    Hello Tuan Tran74. 

    Thank you for your reply again. 

    Below is a reply to your comment.

    1.  At 200W you will need heat sinking and in addition maybe force air cooling.  The heat being dissipated on your board will be greater than 20W

    --> I will consider applying Heat sink and air cooling.

    2.  If you do not need reverse battery protection then you do not need D1.  If you need reverse battery protection then use a PFET in place of D1

    --> Input power is supplied from the 600W class SMPS. D1 will be deleted because it is not a battery.

    3.  You will need to reduce your switching frequency to about 100KHz-200KHz to reduce switching loss since you will need to have a big FET for Q2 or parallel more than one FET there.  You should short out L4 on your schematic and drive the FET directly.

    --> Please check if what I understand is correct.

    1st) The switching FET shall be connected in parallel.
    2nd) The switching frequency must be lowered as the FET increases.
    3rd) The inductor value must be increased because the frequency must be lowered.

    So all three of the above have to be carried out?

    --> I thought switching FET was a very large enough capacity, but can you explain which part of the current product specification is insufficient and needs to be changed?

    --> I will delete L4.

    4.  When changing to lower switching frequency, you will need to change the inductor to a higher value along with the appropriate saturation level

    --> Can I increase the value of input capacitance instead of increasing the inductor value? Or do I have to raise the price of the inductor only?

    5.  You definitely need a larger output diode for D2 or parallel some

    --> I will definitely choose a larger D2 and connect it in parallel.

    Thank you. 

  • 0 in reply to JAEWOONG CHUNG
    TI__Mastermind 18890 points

    Hi Jaewoong,

    There are big FET's that can handle these current level but the FET you are using cannot handle this current , power level and dissipate the heat.  You can parallel FET's or use bigger FET's and that is your choice but you need to decrease the switching frequency because of gate drive limitation for driving larger FET's.

    For power supply design the switching frequency determine the inductor and it's peak current for a given output current.  You will need to follow the data sheet to help you with the design.  You will also need to increase the output capacitance along with the inductance.

    Thanks Tuan

  • 0 in reply to Tuan Tran74
    Prodigy 140 points

    Hello Tuan

    I rewrote the circuit diagram according to your advice.

    Please check one more time to see if there is any problem.

    Thanks always

  • 0 in reply to JAEWOONG CHUNG
    TI__Mastermind 18890 points

    Hi Jaewoong,

    From a design point of view, the paralleling of FET's and diodes will help dissipate the heat but again you still need to get the heat out with heat sink/force air cooling.  I notice that you are still trying to switch at 380KHz with Rt being 20.5K.  You will need to lower the switching frequency here since you are paralleling FET's  and we mentioned this before.

    What is your input voltage?  On the original post you had 24V as your requirement and on the schematic you have 14V.  This will change your inductor value and calculations.  If you have controlled input voltage less you can use lower voltage rating input caps and with the same package have more capacitance.

    If you do not have EMI requirement then you can remove L2 and L3 also.

    Thanks Tuan

  • 0 in reply to Tuan Tran74
    Prodigy 140 points

    Hello, Tuan.

    I forgot to change the switching frequency. I will change 380kHz to 200kHz. (RT=40.2 kohm)

    And the input voltage is still 24V.

    I will leave L2 and L3 because I have to check EMI through future evaluations.

    Thank you again. 

  • 0 in reply to JAEWOONG CHUNG
    TI__Mastermind 18890 points

    Hi Jaewoong,

    Ok...I will close this post.

    Thanks Tuan