• State Resolved
  • Locked Locked
  • Replies 4 replies
  • Subscribers 48 subscribers
  • Views 288 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

OPA4170: How to think the output voltage vs ouput current with Negative feedback loop

Kazusa Suzuki
Prodigy 580 points
Part Number: OPA4170

Hello, 

When I think output voltage vs output current for the OPA4170 with Negative feedback loop...

Should we refer the Voltage output swing from rail in Electrical Characteristics?

Or is it for open loop and should we refer Figure 9. Output Voltage Swing vs Output Current (Maximum Supply)

Best regards,

Kazusa Suzuki

  • 0
    Guru 140200 points
    Hi Kazusa,

    both specifications are assuming an OPAmp in normal operation and both are showing the maximum available output voltage.

    Kai
  • 0 in reply to kai klaas69
    TI__Prodigy 580 points
    Hi Kai,

    thank you. To confirm, I do not need to think "negative feedback loop" and "open loop" if think "voltage swing" and "output voltage swing vs output current". Is it correct?

    Kazusa
  • 0 in reply to Kazusa Suzuki
    TI__Mastermind 30600 points
    Kazusa-san,

    The output limitations of the amplifier as defined in the electrical characteristics table or in the figure you mentioned do no necessarily require that the op-amp be in negative feedback. The information in the electrical characteristics table on output swing is just one point on the "output swing vs output current" plot.

    One way to test the output swing vs output current limitations for an op-amp is to put it in open loop mode change the amount of current the op-amp is providing. However, if the op-amp is open loop, then the output will always be at either its positive or negative limit. Negative feedback allows the op-amp to output voltages and currents which are "inside" the claw curves shown in the plot.

    I hope that makes sense.

    Best,
    Paul
  • 0 in reply to Paul Goedeke
    TI__Prodigy 580 points
    Hi Paul,

    thank you, I totally understand.