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TIPD135: Vout(min)

Gabor Kurucz
Prodigy 100 points
Part Number: TIPD135
Other Parts Discussed in Thread: PGA281

Hi,

I'd like to ask about Vout(min) in tidu033.pdf on page 4 .

Why is it 10mV ? I think I understand that Vout(max) is 4.9V  because the PGA821 output can swing to VSOP (5v in this case) - 0.1V . But I just can't find how come this Vout(min)

Regards,

Gabor

  • 0
    TI__Guru 50846 points

    HI Gabor,

    On the PGA281, shunt measurement,  high side current sense design, the calculations are used to set the maximum gain and ensure full utilization of the linear operating range of the device, where the gain(s) required for this design depend on the maximum output swing of the amplifier, shunt resistor, and the load current range.

    As you have mentioned, to calculate the maximum gain for the maximum load current, the VOUT(max) is chosen to be 4.9 V, which is close the maximum output voltage swing of the amplifier when using VSOP=5V, VSON=GND.

    To calculate the maximum gain for the minimum load current, the Vout(min) voltage is chosen to be a small voltage of only a few millivolts; and in this particular case the author chose ~10mV.  Essentially, the minimum output voltage needs to be small in the range of 10s of millivolts to ensure the full utilization of the linear operating range of the device.  There is not an exact or strict minimum differential voltage value.  However, the minimum voltage can not be chosen to be much smaller than a 10 millivolts, or the measurement will start to be heavily dominated by other errors such as offset and noise of the PGA281 making the measurement unreliable.    

    Thank you and Regards,

    Luis