This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TPS7A24: Why LDO's input current is equal output current?

Part Number: TPS7A24

Hi,

I am using LDO regulator in my design. I measured input voltage 5V and input current 20mA also measured output voltage 3.3V and output current 19mA. I think, It shouldn't be same current input and output. 

I looked quiescent current LDO's gnd pin, it is 1mA. I think in ideal form, total power will be 100mW. Output power should be same input power. So, i need to measure output current about 30mA.

Why LDO's input current is equal output current?

Best Regards.

  • Hi Ugur,

    Are you using the adjustable version or fixed version?

    In general, the output current is not equal to the input current, but with many of our low-quiescent current devices they are very close. The total current is:

    I_IN = IOUT + IGND

    The total power dissipated in the LDO is calculated as follows:

    PD = (VIN - VOUT) * IOUT + (VIN * IGND)

    Normally the IGND term is small compared to the IOUT term and we can neglect it. From Figures 11 and 12, the IGND should be on the order of 10's of uA with your operating conditions, so 1mA IGND that you measured seems very high. 

    So, the power dissipated in your application is (to close approximation):

    PD = (5V - 3.3V) * 20mA = 34mW

    Regards,

    Nick

  • Hi Nick,

    I am using fixed version.

    My desing is using 5V*20mA in input. It is totally 100mW. In output, my design should use about 30mA because total power is 100mW.

    Lossing power on LDO is PD = (5V - 3.3V) * 20mA = 34mW. Lossing power will flow through gnd. I am not measuring lossing power's current.

     Thanks for answer.

  • Hi Ugur,

    I think you have a fundamental misunderstanding of how LDOs operate. Power lost in an LDO (most of it anyways) is due to dropping the voltage from VIN to VOUT, but the current is not lost to GND (the exception is the GND or quiescent current that biases the internal circuitry, but this current is typically much smaller than the load current). Any current that goes to the output also goes through the LDO, so if you want 30mA at the output there will also be 30mA through the LDO. Power lost in the LDO if you are driving 30mA is (5V - 3.3V) * 30mA = 51mW, and the power delivered to the load by the LDO is 3.3V * 30mA = 99mW. Please check out this app note that discusses the basic operation of an LDO:

    https://www.ti.com/lit/eb/slyy151a/slyy151a.pdf

    Regards,

    Nick

  • Thanks for answer. I am going to look into documentation.