• State
  • Locked Locked
  • Replies 3 replies
  • Subscribers 63 subscribers
  • Views 594 views
  • Users 0 members are here
Support feedback
Options
Options
Related

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.

TPS7A16: TPS7A1601 with Vout lower than feedback voltage

Gabriel Barbat
Prodigy 130 points
Part Number: TPS7A16

Hi,

Is it possible to obtain an output voltage lower than Vfb (feedback voltage) by connecting R2 to a voltage Vdac higher than Vfb instead of GND?

If so, this should be true even for Vdac > Vfb:

Vout = Vfb * (1+R1/R2) - Vdac * R1/R2

Thanks,

Gabriel

  • 0
    TI__Guru 55940 points
    Hi Gabriel,

    I believe that there is an error in your equation. Perhaps I am missing an assumption you made.

    Vfb = I * R2
    I = (Vout - Vdac) / (R1 + R2)

    Vfb = [R2 * (Vout - Vdac)] / (R1 + R2)

    Vfb * (R1 + R2) = R2 * (Vout - Vdac)

    Vfb * [1 + (R1/R2)] = Vout - Vdac

    Vout = Vdac + Vfb * [1 + (R1/R2)]

    One way to accomplish regulation at a lower voltage than Vfb is to create an artificial ground. Since all internal voltages are referenced to the ground pin, if you were apply a negative rail at the ground pin, you could reduce the output voltage relative to the actual ground on your system. If you choose this route, you will want to be sure to reference your input capacitors, output capacitors, delay capacitor and resistor divider to the artificial ground as well. It is also important to keep in mind that accuracy, all abs max and operating voltages are relative to the artificial ground.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Prodigy 130 points

    Thanks for your reply, Ryan.

    The topology I am questioning is different.

    What I mean is to connect the ground pin of the LDO to 0V as usual, not to Vdac. Only R2 is connected to Vdac, then:

    I = (Vout - Vfb)/R1 = (Vfb - Vdac)/R2

    That results in the equation I sent before:

    Vout = Vfb * (1+R1/R2) - Vdac * R1/R2

    Moreover, I'm assuming Vdac being possitive.

     How low can Vout be in this case?

     

  • 0 in reply to Gabriel Barbat
    TI__Guru 55940 points
    Hi Gabriel,

    Okay, I see what you are doing now. Your theory does make sense and a quick bench test I was able to reduce Vout to 13mV.

    We have not extensively tested or characterized using TPS7A16 in this way; therefore, some specifications in the Electrical Characteristics table may not be met. In particular an additional source of inaccuracy is introduced with the DAC voltage. Additionally the output rail will be subjected to any fluctuation on Vdac. Some internal features looking at the output voltage such as PG may not function properly. Finally, even when Vin is 0V and the LDO is disabled, there could be a voltage on Vout through the resistor divider.

    Very Respectfully,
    Ryan