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TLV761: 33DCYR

sandeep kothapally
Prodigy 100 points
Part Number: TLV761

Tool/software:

Below is efficiency calculation formula

I would like to measure efficiency. As per datasheet max out current is 1A. My load is 3.3V/400mA.. My adapter current rating is 1A.

1) For Pin calculation What current need to be considered(Adapter current rating or Inrush current of LDO or output current of LDO)?i.e 2A/1A/400mA as per mentioned block diagram

2)For power out calculation, have I to consider Load current or LDO maximum current?

  • 0
    TI__Genius 12460 points

    Hi Sandeep,

    For both power and efficiency calculations, the output current is the actual current consumed by your load - in this case, 400mA.

    Input current into the LDO will be the sum of the output current and the LDO's quiescent or ground pin current. For the TLV76133, ground pin current is typically 60µA. So, for your design as shown, efficiency will be (3.3V * 400mA) / (5V * (400mA + 60uA)) or 66%. 

    Your output power in this case will be the power delivered to the load - 3.3V * 0.4A, or 1.32W. 

  • 0 in reply to Alex Davis
    Prodigy 100 points

    Thanks for your replay. Power dissipation will be (5*400mA-3.3*400mA)=2W-1.32W= 0.68W wastage. Am I right?

    Actually I would like to use same series regulator to convert 12V to 3.3V.My load requirement is 600mA on 3v3.As per calculation(3.3V*600mA)/(12V*600mA)==1.98W/7.2W==>0.275W efficiency. Actual Maximum LDO efficiency is 3.3*1A=3.3W. Will I get 0.275 percentage of 3.3W from LDO? my load power requirement is 3.3*600mA=1.98W. DO I get 1.98W from LDO? I am thinking output from LDO is total efficiency(3.3W)-calculated efficiency as per load(0.275w)). Is it?

  • 0 in reply to sandeep kothapally
    TI__Genius 12460 points

    Hi Sandeep,

    It won't be possible to use the TLV76133 to provide 3.3V at 600mA from a 12V supply due to the power dissipation in the LDO.

    Regulating down to 3.3V from 12V at 600mA would mean the LDO is dissipating (12V - 3.3V) * 600mA or 5.22W. Given the TLV76133DCY's RθJA of 95.4°C/W, attempting to dissipate that amount of power will result in the LDO entering thermal shutdown (95.4°C/W * 5.22W = 500°C, well above the 150°C max junction temperature and the 180°C thermal shutdown temperature). 

    I'd recommend using a buck converter. If you need the lower noise of an LDO, you could use the buck converter to regulate down to 5V, then use an LDO to produce your final 3.3V supply. Alternatively, if your application isn't particularly sensitive to power supply noise, you could simply use a buck converter to regulate directly to 3.3V. Both of these options will operate at much higher efficiency than an LDO-only solution.

    Our Webench Power Designer tool can help you select an appropriate regulator should you choose to take this approach.

    To answer your power/efficiency question:

    1. Efficiency is the ratio of power delivered to the load compared to power consumed at the circuit input. 

    2. LDO efficiency regulating from 12V to 3.3V would be 27.5% ((3.3V * 0.6A) / (12V * 0.6A))

    3. The maximum power output (from a hypothetical LDO able to tolerate the required power dissipation, not the TLV761) would be 3.3V * 1A or 3.3W

    4. Output power into a load consuming 600mA would be 600mA * 3.3V or 1.98W. This is not 27.5% of the maximum LDO power, because efficiency is not related to the maximum output power from the LDO. 

    5. The efficiency does relate to the overall power consumed in the circuit - the 1.98W consumed by your load is only 27.5% of the overall power consumed. 

    Best Regards,

    Alex Davis

  • +1 in reply to Alex Davis
    Prodigy 100 points

    Thanks for your explanation.