• State Resolved
  • Locked Locked
  • Replies 7 replies
  • Answers 2 answers
  • Subscribers 63 subscribers
  • Views 181 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

LMR38020-Q1: Iq current and input current

monica bagade
Prodigy 110 points
Part Number: LMR38020-Q1
Other Parts Discussed in Thread: LM5012, LMR38020

Tool/software:

IC shows Iq current of 40uA at No load. At low load current (<1mA) as well as all at full load operating current will it maintain same? Or it's Iq active current will change? By How much?  Not able to find any graph on that. Can you please confirm?
Is my understanding correct as per below?
Input current =  Iq+ (Vout* Iout/n* Vin)
For 1 ma at 12VOutput and 48V input?
Input= 40uA + 450uA = 490 uA?  (efficiency considered 60% for PFM variant?)
Also, Datasheet mentions 40uA Iq only in PFM variant. As per my understanding, both these variants support PFM,and If I want automatic switching of PWM and PFM, I should use highlighted part?
Please confirm
Thanks,
Monica
  • 0
    TI__Expert 5680 points

    Hi Monica,

    We do not have data about Iq with load since it is difficult to differentiate what is strictly Iq and what is switching current. Also, it will be much smaller than the switching current taken at any significant load.

    FPWM variant works in PWM throughout the load range. So even at no load it will still be switching, making it it difficult to measure non-switching Iq. This is why only the PFM variant Iq is mentioned. The current taken from input even at no load will be in few mA range for the FPWM device.

    Only the non-FPWM variant supports PFM. So, if you need automatic switching between PWM and PFM you should go with the non-highlighted part (the one without F in its name).

    Regards,

    Niranjan

  • 0 in reply to Niranjan Nair
    Prodigy 110 points

    Dear Niranjan,

    Thanks a lot for Reply

    I am currently using LM5012 which consumes 400uA at light load even it has Iq=10uA at no load. I want to evaluate LMR38020, but not sure about calculating current at 2mA light load condition assuming 40uA at 0A load current only?

    Doe it mean,

    For 1 ma at 12VOutput and 48V input?
    Input= 40uA + 450uA = 490 uA?  (efficiency considered 60% for PFM variant?)

    Then it's performance will be same as LM5012?

    Please clarify.

  • 0 in reply to monica bagade
    TI__Expert 5680 points

    Hi Monica,

    You don't have to consider Iq and efficiency separately in the input current equation. The Iq will already be factored in the efficiency measurement. So (Vout* Iout/n*Vin) should give a decent estimate of the input current taken.

    Looking at LM5012 and LMR38020 datasheet, LM5012 seems to have better light load efficiency. So you will probably get a higher input current with LMR38020. 

    From the LM5012 datasheet, 10uA is not the no load Iq but the Iq in sleep mode. In actual no/light load operation, the LM5012 will turn on for a short time (where it takes much more than 10 uA current) then remain off for a long time where the device goes into sleep mode (where it takes 10 uA typ. current). This is why you are seeing 400 uA average current at light load.

    The best way to calculate input current is using just the efficiency : Iin = (Vout* Iout/n*Vin). What is your spec for the input current?

    Regards,

    Niranjan

  • 0 in reply to Niranjan Nair
    Prodigy 110 points

    Okay, thanks. 

    I am designing it for my BMS application and for sleep mode implementation, I need current<1 mA at 48V.

    I am using Below circuit for My BMS application.
    MCU TMS320F28P659 100 PIN MCU.
    LM5012 for converting 48V to 12 V
    TPS62901QRYTRQ1 for converting 12V to 3.3V
    TPS78501BQWDRBRQ1 for 3.3v to 1.2V
    I want to calculate current drawn by MCU.
    Currently MCU is in Unprogrammed mode(New board with controller not programmed  - idle mode)
    When I am not turning on 1.2V (core supply of MCU, current consumption at 48V input is 0.15mA)
    LM5012 must be in sleep mode current
    As soon as I turned on 1.2V, current consumption at 48V input is 2.33mA. In this, 7.12mA is flowing thorough 12V output path which is contributing 1.78mA at 48V (7.12*12/48). Remaining 0.55mA should be drawn by LM5012.
    As I can't reduce LM5012 current and it consuming half of current, I am exploring If I can change IC  with input voltage range of 80V and high efficiency. Max load current 1.5 to 2A
  • 0 in reply to monica bagade
    TI__Expert 5680 points

    Hi Monica,

    Among the devices in this voltage and current range, LM5012 and LMR38020 seems to have the best efficiency.

    Out of these two, LM5012 seems to be better and has more factors you can tune to get higher efficiency.

    The way LM5012 works at light load is, it turns the switch on once to charge the output capacitors then goes into sleep mode (where it takes very less current) until the capacitors are discharged to a certain value. To improve the efficiency, the device should operate in the sleep mode for longer. Few things you can try are :

    1) Increase Ron to increase the switch on time (increasing the voltage the capacitors are charged up to)

    2) Reduce the inductor (again, increasing the voltage the capacitors are charged up to)

    3) Increase feedback resistor values to reduce the current consumed by feedback network

    However, 1 and 2 will increase the output voltage ripple and 3 will increase noise susceptibility (better to keep the series resistance below 1 MOhm).

    Regards,

    Niranjan

  • 0 in reply to Niranjan Nair
    Prodigy 110 points

    Thanks a lot for the detailed answer

  • +1 in reply to monica bagade
    TI__Expert 5680 points

    Let me know if you have any more questions. If not, you can close the thread by clicking the resolved button.

    Regards,

    Niranjan