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TPS2121: How to achive fast switchover in manual swichover configiration

Hidekazu Someno
Expert 5130 points
Part Number: TPS2121

Hi Application tean,

We would like to know how to achieve fast switchover time(5us_typ) in Figure 23 application(manual switchover) on datasheet(SLVSEA3B).

In Figure.23 configuration, selecting VIN1 or VIN2 as input sourve voltage should be done by driving(toggling) CP2 by MCU. 

But,  there is the description that  the CP2 voltage needs to be higher than VREF(1.06 TYP)  to enable fast switchover time.

How must we drive CP2 pin from MCU to have fast switchover time?

(Does it mean that  low input level of CP2 driven from MCU be higher than VREF?)

Best regards,

H.Someno

  • 0
    TI__Genius 10855 points
    Hi Someno-san,

    Thanks for reaching out on E2E!

    Your assessment is correct. For manual switchover on the TPS2121, you can toggle the CP2 pin by the MCU. In order for fast switchover to be enabled, the voltage on the CP2 pin must be higher than VREF.

    Therefore, you need to toggle CP2 above and below the PR1 voltage to switch between IN1 and IN2, while keeping the voltage above VREF.

    Let me know if you have any additional questions! For reference, can you comment on the end application?

    Thanks,
    Arthur Huang
  • 0 in reply to Alpha H
    TI__Expert 5130 points

    Thanks for your support.

    I have an another question about Figure.27(TPS2121 Switchover from IN2 to IN1) on datasheet, page-24.

    I understand that this is the operation shifting VOUT from IN2(3.3V) to IN1(5V) with normal switchover time(100ns_typ) by toggling CP2 from 3.3v  to Low(0v).

    But there is no voltage dip on VOUT during the switchover time.   

    Why is that?     Is it the waveform in no-load condition?

    Best regards,

    H. Someno

  • 0 in reply to Hidekazu Someno
    TI__Genius 10855 points
    Hi Someno-san,

    Yes, those waveforms were taken without the load on the output. With a load connected, the voltage will dip depending on the following equation:

    VDIP = tsw * (Iout / Cout)

    Where:
    VDIP = Amount of voltage dip before switchover occurs
    tsw = switchover time (100us for slow switchover), (5us for fast)
    Iout = Load Current
    Cout = Load Capacitance

    Thanks,
    Arthur Huang