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TPS730: How to calculate the power consumed for a regulator for its working

Harikrishnan k
Intellectual 455 points
Part Number: TPS730
Other Parts Discussed in Thread: MSP430F2416, , ADS1292

Hi all,

How can I calculate the power consumed by a low dropout regulator for its own working? In my circuit I have two TPS73033DBVT regulators for powering MSP430 and a class2 bluetooth input to the regulators are from a  3.7V 400mAh LI-Ion battery,help me to calculate the power consumption of these regulator

  • 0
    TI__Guru 55940 points

    Hi Hari,

    The power available at the input to the LDO will be equal to the power delivered to the load plus the power consumed by the LDO under your application's conditions.  In other words, we can simply calculate your application's efficiency (n):

    n = (Pout / Pin) x 100

    n = ([Vout x Iout] / [Vin x Iin]) x 100

    n = ([3.3 V x Iout] / [3.7 V x (Iout + Ignd)]) x 100

    At this point we will assume that your Iout is 100 mA for this example.  You can adjust based on your actual application.  Ignd comes from the datasheet:

    n = ([3.3 V x 100 mA] / [3.7 V x (100 mA + 170 uA)]) x 100

    n = ([3.3 V x 100 mA] / [3.7 V x 100.17 mA]) x 100

    n = (330 mW / 370.629 mW) x 100

    n = 89.04%

    Commonly you will see this calculation simplified for LDOs when Iout >> Ignd to

    n = (Vout / Vin) x 100

    n = (3.3 V / 3.7 V) x 100

    n = 89.19%

    These calculations show that 89% of the power from battery is being delivered to the load.  This means that the LDO is consuming 11%.

    Using the simplified efficiency equation, you can see that the smaller the headroom (Vin - Vout) is in the application, the more efficient the application will be.  The minimum headroom (Vin - Vout) for an LDO to be able to regulate the output voltage is equal to the LDO's dropout (Vdo).  As such, the LDO will be most efficient (consume the least amount of power) when Vin is equal to Vout + Vdo.

    Very Respectfully,

    Ryan

  • 0 in reply to Ryan Eslinger
    Intellectual 455 points
    Hi Ryan,

    Thank you very much for the detailed reply as per the above method I have tried to calculate the power consumption of LDO ,

    Battery power :1.48W
    Vi: 3.7V
    Vout: 3.3V

    efficiency = Vout/Vin *100 (Iout >> Ignd,Iout=30mA)
    = 3.7/3.3 *100
    = 89.19%
    so the power consumed by LDO is , 1.48*10.81/100 = 0.1599W

    Is this interpretation is true?
    and also in case of MSP430f2416 the data sheet says the Active mode current is 365uA @1MHz ,2.2V how can I calculate the active mode current at 16MHZ @3.3V VCC?
  • 0 in reply to Harikrishnan k
    TI__Guru 55940 points
    Hi Hari,

    You correctly corrected the efficiency in your application as ~89%. The LDO is consuming the additional 11%; however, I neglected to point out that the majority of this is simply heat dissipation which occurs in the LDO due to your application conditions.

    The LDO's consumption in order to function is simply Vin x Ignd. In your application this would be 3.7 V x 170 uA = 629 uW is being consumed by the LDO in order for the LDO to function.

    The LDO is dissipating (Vin - Vout) x Iout as heat. In your application this is (3.7 V - 3.3 V) x 30 mA = 12 mW.

    When you combine these two together you see that 12 mW + 629 uW = 12.629 mW is being consumed within the LDO.

    As you can see this is different from your calculation. The reason is that just because the battery can supply 1.48 W does not mean that it is in your application. Your load is only 30 mA; therefore, your input power estimation should be 3.7 V x 30 mA = 111 mW.

    11% of 111 mW is approximately equal to 12.6 mW.

    Unfortunately I am not familiar enough with the MSP430 products to help you with your second question. I would recommend posting the MSP430 question as a separate thread so that the team more familiar with the product can help you.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Intellectual 455 points
    Hi Ryan,
    Thank you for the quick reply, I am currently working on the development of wearable device and I want to optimize the circuit for longer working time without increasing the battery capacity,what are the steps that I should take to increase the battery life of my proto type?
    I am using an MSP430 microcontroller,ADS1292,TPS730 series of regulators and a LI-Ion battery charger,bluetooth(classic)?
  • 0 in reply to Harikrishnan k
    TI__Guru 55940 points

    Hi Hari,

    In order to increase the battery life, you will need to increase the efficiency of your application.  For a linear regulator the output voltage and output current are fixed requirements driven by the application.  This only leaves two variables to increase efficiency: Vin and Ignd.  You have already selected TPS730 as your LDO and as we discussed previously Ignd is relatively negligible.  If you are interested in reducing Ignd please consider our low Iq LDOs.

    The biggest impact for increasing efficiency will be to reduce your input voltage to the LDO.  Often this is done by using a switching regulator before the LDO.

    We also have reference designs for wearable applications that you may be interested in which may give you additional ideas:

     

    Very Respectfully,

    Ryan