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TPS715: TPS715, Fixed version, Internal divided resister

Tamio Yokokura
Expert 3560 points
Part Number: TPS715

Our customer ask me about the value of the internal divided resister(R1 and R2) in TPS71525.

Please let me know about two resister value.

Regards,

Tamio

  • 0
    TI__Guru 55940 points
    Hi Tamio,

    Please allow a day or two for us to lookup this information. Please keep in mind that the ratio is trimmed and ensured at final test rather than the exact resistance value.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    TI__Guru 55940 points
    Hi Tamio,

    The internal reference is 1.205 V and the internal R2 is nominally 843 kΩ. As a result R1 is approximately 906 kΩ for the 2.5 V version of TPS715. Please keep in mind that the ratio is trimmed and ensured at final test rather than the exact resistance value.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    TI__Expert 3560 points
    Hi, Ryan

    Than you for sharing your information.
    It is helpful for our customer to design his circuit.

    If you have the evaluation board which TPS71530 was mounted, please confirm the discharging time until Vo=2.0V from Vo=3.0V.
    If possible, the output capacitor is total 100uF. His target time is 10sec.

    Thanks
    Tamio
  • 0 in reply to Tamio Yokokura
    TI__Guru 55940 points
    Hi Tamio,

    The discharge of the capacitor can be calculated but it is important to keep in mind that the internal resistor divider is only one discharge path. If the load is still actively drawing current, the discharge time will change.

    Vc = V0 x e^(-t / (R x C))

    Vc is the voltage across the capacitor
    V0 is the initial capacitor voltage
    t is time
    R is the internal resistor divider
    C is the capacitance

    Assuming that the only discharge path is the internal resistor divider:

    2 V = 3 V x e^(-t / ((843 k + 906 k) x 100 uF))
    2 V / 3 V = e^(-t / (1.749 M x 100 uF))
    2 / 3 = e^(-t / 174.9)
    ln(2/3) = -t / 174.9
    t = 70.92 s

    Without another discharge path, the capacitor will not be discharged in 10 s.

    Very Respectfully,
    Ryan