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TIDA-00666: LM5165

jay chen
Prodigy 150 points
Part Number: TIDA-00666
Other Parts Discussed in Thread: LM5165

Hi buddy,

In TIDA-00666 designed for BLE transmitter with looped power, could you please help me understand how to provision the power to BLE with less than 3.6mA loop current? in the design guide, it is simple to highlight this point but I have no idea how to understand it, could you please help on it? thank you.

best regards,

Jay

  • 0
    TI__Intellectual 1320 points

    Hello Jay, thanks for your interest in TIDA-00666. A colleague will get back to you shortly. Regards, Alex

  • 0
    TI__Prodigy 610 points

    Hello Jay,
    the statements in the mentioned TIDA-00666 design guide are related to the fact that input current I_IN of a low quiescent current LDO equals roughly its output current I_OUT.  In reality it will be slightly higher by the amount of quiescent current. For an exemplary 1mA of I_OUT the respective I_IN would be slightly more than 1mA and is almost independent from the applied V_IN of the LDO.


    Switching DC/DC converters (like the used LM5165 synchronous buck converter) have ideally  a 100% efficiency (neglecting all their losses) which means that their input power P_IN equals their output power P_OUT. In reality their PIN will be higher and can be described by P_IN=P_OUT/η. Their real efficiency η can best taken from datasheet curves or from specific ones as shown in "Figure 12. Typical Efficiency of LM5165 Buck Converter (Output Rated for VIN = 8 V)" of the TIDA-00666 design guide. The latter one shows for example roughly 87% for 1mA of I_OUT.
    This would result in
    P_IN = (3.3V x 1mA) / 0.87 = 3.8 mW and an I_IN of the of the buck converter of 3.8mW / 8V = 0.47mA. 

    The I_IN of the buck converter depends on the voltage conversion ratio and is at the assumed step-down from 8V to 3.3V with 0,47mA significant lower than the roughly 1mA  I_IN of a LDO.

    Best regards

    Juergen

     

  • 0 in reply to BertW
    Prodigy 150 points

    Thank Juergen.

    For our 4-20mA loop application, we mostly care for the total current through the loop+/loop- as you know this current should be less than 3.6mA (3.2mA is preferred).So if we select DCDC buck to replace LDO, it sounds the loop current still is 1mA based on about case. it looks this DCDC solution can't help us reduce the loop current, is it correct? 

    Thanks. Jay 

  • 0 in reply to jay chen
    TI__Prodigy 610 points

    Hi Jay,
    let us rearrange the equation given in my former message from 

    P_IN=P_OUT/η

    to:                     I_OUT = η x V_IN x I_IN / V_OUT = 0.89 x 8V x 3.2mA / 3.3V = 6.9mA    (this is for the preferred 3.2mA loop current)

    or for the case of 3.6mA loop current:       I_OUT = 0.89 x 8V x 3.6mA / 3.3V = 7.8mA

    For both cases I have increased the efficiency form 87% to 89% due to the increased I_OUT in the 7mA range compared to the exemplarily used 1 mA I_OUT in my former message.

    I_OUT stands in the equations above for the output current available from the DCDC buck. The calculated values are now close to the roughly given value of 7.5 mA  (TIDA-00666 design guide: Table 2. Beacon Event, State Analysis). It need furthermore assumed, that part of those short high current pulses are supported by the COUT of the DC/DC buck and by the bypass cap directly at the respective pins of the devices.

    Best regards
    Juergen

  • 0 in reply to BertW
    Prodigy 150 points

    Thank Juergen.

    So the I_OUT current does NOT flow into the loop path, saying the LOOP- pin？I'm confused why the final current through LOOP- pin is NOT 7.8mA but 3.6mA. based on the loop circuit design, could you please help me understand the 7.8mA current path and the 3.6mA loop current path? thanks.

    Best regards,

    Jay

  • 0 in reply to jay chen
    TI__Prodigy 610 points

    Hi Jay, 
    the LM5165 buck converter is configured to operate in PFM Mode, see Fig 7-3 "PFM Mode SW Node Voltage, Feedback Voltage, and Inductor Current Waveforms" as shown in the datasheet of LM5165 https://www.ti.com/lit/ds/symlink/lm5165.pdf .
    The inductor waveform ramps up when the device's internal HighSide FET is ON - current flows from the device VIN pin over the High Side FET , the SW-pin, the inductor , the output capacitor and output load to the circuit GROUND - and via the device's input capacitor back to the  VIN-pin. Please see the Functional Block Diagram of the device in the datasheet section 7.2 to identify the described path.
    When the Inductor current reaches the selected peak current limit value (typ 60mA), the High Side FET is switched OFF and the Low Side FET is switched ON - causing a falling slope of the inductor current. The current will now be driven solely from the the inductor through the output capacitor and output load , the circuit GROUND and back to the inductor via the Low Side FET and SW-pin. The device's input capacitor is not involved in that current flow.
    Important to note: the average inductor current equals the output current when the device switches continuously. With the 60mA peak current limit the average current would be 30mA. If you need just 7.8mA, then the device need to stop operation in between - going into SLEEP (fig 7-3) - after a certain number of switching pulses (ACTIVE) . 

    The same happens at the input of the DC/DC converter where the input capacitor is doing the averaging. The input capacitor does see (will be discharged) only (by) the rising slopes of the inductor current. There will be no current demanded from the input capacitor during the falling slope of the inductor current nor during its sleep state. This will reduce the average current the device's input capacitor draws from the Loop+  and is feeding into Loop-  to the 3.6mA.

    Best regards
    Juergen

  • +1 in reply to BertW
    Prodigy 150 points

    Thank Juergen very much for your kind explanation, I get to know some here. 

    In order to make sure the current through LOOP- less than 3.2mA, the buck DC-DC can help provide more current for the loop's devices.

    best regards,

    Jay

  • +1 in reply to jay chen
    TI__Prodigy 610 points

    Hi Jay,
    I think, the current consumption of all the components powered by the 3.3V VOUT_DCDC needs to be corrected by the factor

    VOUT_DCDC / (VIN_DCDC x Efficiency_DCDC).

    This is because the output current of the DC/DC is during the OFF-time of its internal High-Side FET driven by the inductor of the buck converter, but not provided by the buck converters VIN.

    Imagine, that the current "starts" during the OFF-time of the High-Side FET at the output capacitor connected side of the inductor , flows through COUT and all the 3.3V load circuits  and returns back to the SW-pin connected side of the inductor via the Low Side FET. The buck converter's input  is not involved in that current flow. Consequently there is no direct load caused current flow into the VIN of the buck nor a return current to the input ground during that time. "No direct current flow" is related to the fact that there is still a neglectable quiescent current, but also a recharging current of the Buck's CIN which had been slightly discharged during the High-Side FET's ON-time.

    Best regards

    Juergen

  • +1 in reply to BertW
    Prodigy 150 points

    Thank Juergen for your kind explanation. Now the challenge is how to get VIN_DCDC? it sounds this VIN_DCDC is varied if the ILOOP current is varied from 4mA to 20mA.

    Assuming this VIN_DCDC is varied (referred to GND), the VOUT_DCDC is stable and fixed at 3.3V (referred to GND), the LM5165's VOUT_DCDC is stable always regardless of the variable input votlage? thanks.

    best regards,

    Jay

  • +1 in reply to jay chen
    TI__Prodigy 610 points

    Hi Jay,
    you are correct, the VOUT_DCDC is a stable 3.3V even when VIN_DCDC is changing. The larger the VIN_DCDC, the larger the available IOUT_DCDC at the "magic" roughly 3 to 3.6mA input current . Some customer using this and increase their load current at such conditions, for example to transmit messages with a higher repetition rate.
    Some increase the load current also when the loop current increases within the 4 to 20mA loop current range.

    Best regards

    Juergen

  • 0 in reply to BertW
    Prodigy 150 points

    Thank Juergen, it is clear now.

    best regards,

    Jay