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TPS7A49: Vout change rate

Gil Hershman71
Intellectual 580 points
Part Number: TPS7A49

Tool/software:

hello,

We are using the TPS7A49 in the below circuit.

When OUT=8V, we short R48, connected to R49 and the FB pin, so TPS7A49 OUT changes from 8V to 15V (R48 shorted via transistor for OUT=15V and connected for OUT=8V).

We perform this when the output is stable at 8V.

The change to 15V takes few msec.

Is a delay of few msec in OUT change normal?

 

I tried removing the NR/SS capacitor (thought the SS cap will affect the change), LDO output cap and the load’s power supply cap, but the effect was minimal.

It still takes around 1msec.

I calculated the average ΔI for this change to be 1mA (ΔI=C*ΔV/ΔT).

The LDO can easily driver higher currents than that.

 

Why is the loop so slow?

How can we make it faster?

thanks

Gil

  • 0
    TI__Mastermind 37065 points

    Hi Gil,

    I think the main issue slowing down the voltage is the feedforward capacitor.  That 10nF capacitor forms a time constant with the top setpoint resistor so if you want this to be as fast as possible then you'll want to remove the 10nF Cff.

    It is possible that the internal bandwidth of the LDO is limiting the speed.  If you don't see the improvement you are looking for by removing the 10nF capacitor, make sure you have 1mA of load current on the output and retest.

    Thanks,

    Stephen

  • 0
    Intellectual 580 points

    Thank you Stephen.

    Which resistor is the top setpoint one?

    Can you explain how the constant 1mA affects the speed?

    Can reducing the 200K and 17.4K affect the speed?

    Is there a parameter in this LDO that describes the speed of the LDO? (how fast it regulates the output).

    Thanks,

    Gil

  • +1 in reply to Gil Hershman71
    TI__Mastermind 37065 points

    Hi Gil,

    R50 is the top setpoint resistor.

    LDO's sometimes have higher bandwidth when they have some load current.  With a 1mA load the internal LDO circuitry is energized to supply this load and increasing the load usually is not as difficult for the LDO to react to, than when it has 0A load to start with.

    Yes, reducing the 200k resistor to a smaller value will reduce the time constants and increase the speed, but not as much as you would see if you eliminated the 10nF capacitor altogether.

    There are graphs in the datasheet that describe the turn on time and the transient response time, which help to understand the speed of the LDO, depending on which parameter is dominating the LDO output response.

    Thanks,

    Stephen