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TPS7B87-Q1: TPS7B8750QDDARQ1

Mohamed Suhail
Prodigy 40 points
Part Number: TPS7B87-Q1

Tool/software:

Hi, we have procured this above device part numbers and implemented into our PCB design, during power on this chip is heating very too much. We have provided the input voltage of 12volts nominal and taken 5 volts as the output consuming less than 200mA, now my entire design and 100 Nos' of PCB got wasted , we got loss of around 1680$, Due to this rubbish design.

So, we want compensation for this, TI have wasted our full time.

we want the full solution on this.

Regards,

Suhail.

R&D design.

  • 0
    TI__Expert 5555 points

    Hello,

    Can you provide some details of what is not working? Is the load current typically 200mA or is it more? What is the expected ambient temperature?

    ~ Aaron

  • 0
    Prodigy 10 points

    we had purchased TPS7B87-Q1 for converting 24v to 5v (as mentioned in datasheet TPS7B87-Q1 is rated for 36V input voltage) hence we used this part number to step down from 24V to 5V. As given its giving 5V output on 24V input and our load is consuming 100ma only the IC is heating and going beyond 50C (50 degrees). Our ambient temperature is 34 degrees. Does using LDO (linear regulator for such step down is it bad...? if so then why such LDO has high input voltage option)

  • 0
    Prodigy 40 points

    Hi,

    My load current is less than 200mA only , expected ambient temperature is 45 degrees C.

  • 0 in reply to bobby bibby
    TI__Genius 14245 points

    Hi Bobby and Mohamed, 

    Converting 24V to 5V at 100mA will dissipate 1.9w of power which is a lot for any LDO and will require the PCB to be designed and laid out very thoughtfully to maximize heat dissipation. The PCB is the primary heatsink for any IC or LDO which does not have a separate dedicated heatsink attached to it. 

    You can calculate the junction temperature by using the ambient temp, Power Dissipation, and Thermal Resistance using Tj=Ta+Pd*RθJA. However RθJA is VERY dependent on the PCB layout and the values given in the PDS assume a JEDEC High-K layout. The JEDEC High-K layout is a reasonably good layout but your actual layout could be substantially better or worse depending on the amount of metal you have connected to the IC and around the IC.

    For this package and your stated power dissipation that would indicate that your junction temperature on a JEDEC layout is likely reaching Tj=34+1.9*41.8=113.42C. I'm unsure what your requirements are but the LDO should be operational at this point since the thermal shutdown doesn't kick in until approximately 175C. If the LDO is going into thermal shutdown that would indicate that either the PCB isn't able to dissipate the heat similarly to the JEDEC High-K board or you're dissipating more power than expected. 

    Below is a quick powerpoint with some more info on the JEDEC standard layout and how to maximize the thermal dissipation of a PCB and we have a an FAQ on this topic as well. 

    As for your other question about why the input voltage is rated for 40V, this is primarily to handle transient voltage spikes on the input rail. Note, that this device is used in a large number of automotive designs by many different customers to convert from the 12V car battery down to 5V so we are confident in it's ability to do so however thermal management is something the user must always design properly (selecting the right package, laying out the PCB, etc...). If you have challenging heat dissipation or efficiency requirements, you may want to consider a the TO-252 package that this device is also offered in (RθJA=29.7C/W) or a switching regulator for your application. 

    5466.LDO Thermal Performance.pdf