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TLV755P: Efficiency of TLV75525PDBVR

Jason Ciou
Expert 3600 points
Part Number: TLV755P

Hi team,

my customer want to know the efficiency of TLV75525PDBVR,

Here is the specification of my customer's application

Vin = 3.3V

Vout=2.5V

Io=0.017A

but I never see the efficiency in the datasheet,

could you please give some information of TLV75525PDBVR.

  • 0
    TI__Mastermind 37915 points

    Hi Jason,

    With the given specification I calculate the worst-case (i.e. max ground current) efficiency as follows:

    efficiency = (powerDelivered - nonIdealPowerDissipation)/powerDelivered = (Vout*Io - (Vin - Vout)*Io - Vin*Ignd) / (Vout*Io) 

                    = (2.5V*.017A - (3.3V - 2.5V)*.017A - 3.3V*250uA) / (2.5V*.017A)

                    = 66%

    For the ground current I used Figures 17 and 18 to make an approximation. The data in Figure 17 at zero load starts at about 150uA which is close to what the data shows in Figure 18 when Vin is above dropout, and I think it is reasonable to assume that the case where Vout = 2.5V is pretty close to the case where Vout = 3.3V. 

    Does that make sense?

    Regards,

    Nick

  • 0 in reply to Nick Butts
    TI__Expert 3600 points

    Hi Nick,

    Thanks for your kindly answer, 

    I have some questions about your calculation. 

    1.)According to figure 17, when output current is 0.17A, the GND pin current is around 520uA. I do not understand why you choose the 250uA to calculate the nonIdealPowerDissipation.

    2.) How do you use Figure 18 to know the GND current in my case, this figure is based on the Vo=3.3V, in my case the output voltage is 2.5V. could you please tell me in detail.


    Jason

  • +1 in reply to Jason Ciou
    TI__Mastermind 37915 points

    Hi Jason,

    I used 250uA because in your original post you specified the Io = 0.017A. This must have been a typo. In that case I agree that the IGND should be about 500uA. 

    We do not have data for the case Vout = 2.5V. In my last comment I explained that I think it is reasonable to assume that the ground current is similar between the cases Vout = 2.5V and Vout = 3.3V. Power dissipated through the internal biasing is much less than the power dissipated through the pass FET, so approximating IGND(@Vout = 2.5V) ≈ IGND(@Vout = 3.3V) does not change the calculation much. 

    With 0.17A instead of 0.017A here is my updated calculation:

    efficiency = (powerDelivered - nonIdealPowerDissipation)/powerDelivered = (Vout*Io - (Vin - Vout)*Io - Vin*Ignd) / (Vout*Io) 

                    = (2.5V*.17A - (3.3V - 2.5V)*.17A - 3.3V*500uA) / (2.5V*.17A)

                    = 67.6%

    Just as a proof of concept, here is a calculation showing the result if the approximation for IGND was off by a factor of 2 (i.e. IGND = 1mA instead of IGND = 500uA):

    efficiency = (powerDelivered - nonIdealPowerDissipation)/powerDelivered = (Vout*Io - (Vin - Vout)*Io - Vin*Ignd) / (Vout*Io) 

                    = (2.5V*.17A - (3.3V - 2.5V)*.17A - 3.3V*1mA) / (2.5V*.17A)

                    = 67.2%

    Regards,

    Nick

  • 0 in reply to Nick Butts
    TI__Expert 3600 points

    Hi Nick, 

    Thanks, I understand.

    Jason