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TPS7A90: I want to give 5V as input and get 1V as output and wants to draw a current of maximum 400mA

Part Number: TPS7A90

I want to give 5V as input  and get 1V  as output and wants to draw a current of maximum 400mA.So is it practically advisable to drop 4V and draw 400mA.Will there be heating or other issue if done So.

  • Hi Ram,

    Linear regulators are power dissipative products.  The power that is dissipated will become heat as you mentioned; therefore, we do need to consider the thermal implications.

    Layout and ambient temperature will impact the thermal performance.  The primary heatsink for TPS7A90 is the GND copper in the PCB local to the LDO.  As such, in order to achieve the best thermal performance, you should maximize the GND copper local to the LDO.

    Although layout will impact the thermal performance, we can do an early calculation based off of the Thermal Information table of the datasheet to understand where the application stands thermally.  The Thermal Information table metrics come from modeling the LDO on the JEDEC Hi-k board in order to give a standardized layout for comparing two LDOs.  We have seen that board layouts that maximize the GND copper can improve the thermal metrics by 20 to 50%.

    Tj = Pd x Rtheta_ja + Ta

    Your application calls for 1.6 W to be dissipated in the LDO (Pd = (Vin - Vout) x Iout).  TPS7A90 is designed, tested, and characterized for a maximum operating junction temperature of 125 C.

    125 C = 1.4 W x 56.9 C/W + Ta

    125 C = 91.04 C + Ta

    Ta = 33.96 C

    As you can see, in order to remain within the operating temperature range of TPS7A90, you should limit your ambient temperature range to 33.96 C for the JEDEC Hi-K board layout.

    Very Respectfully,


  • Hi Ryan,

    Thanks for the reply.Is there any tutorial for limiting the ambient temperature for this device.