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OPA2320: Output Voltage swing versus Output Current test conditions

Sal LaRussa
Genius 9730 points
Part Number: OPA2320

Hi,

the open loop output impedance of the OPA2320 is 90 Ohms at 1 MHz.  In the data sheet, there is also a curve that shows Output Voltage Swing versus Output current.  Any give point on the curve indicates a much lower Zo than the 90 Ohms specified.   For example, at 40 mA output current, the output voltage is 2V while the rail voltage is 2.75V.    The 0.75V drop at 40 mA indicates an output impedance of 18.75 Ohms.   

Why is there such a discrepancy from the 90 Ohm measurement?    At what gain, frequency, etc. was the voltage vs.current plot taken?  

Thanks for any insight you can offer.

Best,

Sal

  • 0
    Guru 140200 points

    Hi Sal,

    you must distinguish between "open-loop output impedance" and "closed-loop output impedance" (or simply "output impedance") :-)

    The closed-loop output impedance is usually much smaller than the open-loop output impedance. See this nice training video:

    Kai

  • 0 in reply to kai klaas69
    TI__Genius 9730 points

    Hi Kai,

    Rout = Ro/loop gain.  In my application, we're using the amp as a unity gain buffer.  I'm not sure if I can assume I can get 40 mA out with only a 0.75V drop across Rout.     I was curious about the conditions that were used to obtain that "Output Voltage Swing vs Output Current" curve. I'm not sure if this curve would apply to my application.

    What I see is -> At TA = 25°C, VCM = VOUT = mid-supply, and RL = 10 kΩ (unless otherwise noted).  No mention of gain, etc.

    Isn't the test condition important to understand Rout given that Ro (open loop) is 90 Ohms?

    Thanks,

    Sal

  • 0 in reply to Sal LaRussa
    TI__Genius 9730 points

    Hi again Kai,

    I think the answer may be I can't get the current I need give I'm in a unity gain configuration.

    I suppose given that I know Ro is ~ 90 Ohms and I'm seeing 0.75V drop at 40mA current (which looks like an Rout of 18.75 Ohms) , can I assume that the curve was taken with a loop gain of ~ 5 V/V?

    If you can confirm that, I think I'll be all set.

    Thanks again,

    Sal

  • 0 in reply to Sal LaRussa
    TI__Genius 9730 points

    Hi Kai,

    I realized that it's the "loop gain" that I need to factor in so I was off by a factor of ~100 dB (i.e. Aol).   So I think I'm all set.

    Sorry for my misunderstanding - the presentation you sent was very good BTW!

    Thanks,

    Sal

  • 0 in reply to Sal LaRussa
    TI__Guru 64680 points

    Keep in mind that the graph of the output voltage swing vs output current is given only under DC condition, and not any other frequency, and thus it is mainly a function of the output transistor Ron resistance as the output transistor gets pushed into triode region (non-linear operation of the output stage).  Thus, Vout vs Iout graph has nothing do to with the open-loop output impdance, Zo vs frequency, which is taken for linear operation of the output stage.