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TPS74401: What is the maximum efficiency which I can achieve?

Part Number: TPS74401

Hi Power Management,

I would like to use the TPS74401 in a applications, however I can only provide it with 2A of of supply current.   I need to generate a 1.2V rail with 1.8A of output current.  My input voltage is somewhat flexible.   Can you recommend an input voltage range and overall approach which will achieve the efficiency required to meet my output needs with the constraint of the 2A input limit?  If this is not practical, then I will need to look at adjusting my approach on the supply current.  

Thank you,

HSG

  • Hi HSG,

    Efficiency for linear regulators is equal to (Vout x Iout) / (Vin x Iin) and often simplified to Vout / Vin when Iout is much greater than Ignd.  As such for the maximum efficiency, you will want to set your input voltage to the minimum value at which regulation can be achieved (Vin = Vout + Vdo).

    Lets assume that you are using the VQFN package.  As specified in the Electrical Characteristics table, the maximum dropout at 3 A is 195 mV.

    As Figure 4 shows, dropout is linear with respect to Iout.  As such we can setup a simple ratio to calculate our maximum dropout at 2 A.

    Vdo(2A) / 2 A = Vdo(3A) / 3 A

    Vdo(2A) / 2 A = 195 mV / 3 A

    Vdo(2A) = 130 mV

    In order to ensure the output regulation, our calculated maximum shows that you should set Vin 130 mV above Vout or 1.33 V for your application.  You will still need to supply Vbias at a higher voltage; however, very little current flows into the BIAS pin (4 mA max) so the impact on efficiency is negligible.

    It should be noted that some performance such as PSRR will be better with additional headroom (Vin - Vout); however, if efficiency is your key concern, 1.33 V will be adequate for a regulated output.

    Very Respectfully,

    Ryan

  • Hi Ryan,


    Thank you for the additional information which answers my questions.


    Best

    HSG