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Slew rate Vs Gain

Kavindu Ratnayake
Intellectual 310 points
Other Parts Discussed in Thread: OPA378

Hi,

I'm doing some dynamic response calculations on a circuit and got the following doubt,

OPA378 data sheet says that it's slew rate is 0.4 V/us for G=1. My problem is what is the slew rate figure if the gain is (G) 20?

Thanks in advance and Regards,

Kavindu

  • 0
    Mastermind 7925 points

    You can either measure or calculate slew rate for given circuit http://www.experimentalistsanonymous.com/ve3wwg/doku.php?id=slew_rate

    Regards,
    Maciej 

  • 0
    TI__Guru 60610 points

    Kavindu,

    The calculation provided in Maciej's link above refer to a full-power bandwidth (also known as Maximum Output Voltage vs frequency-see graph below), which is the maximum frequency for given output signal amplitude without slew-induced distortion; these equations cannot be used, however, to calculate the op amp slew rate since SR is only a function of the internal bias of the op amp input stage, Itail, and Miller compensation capacitor (Cc): SR=Itail/Cc.

    All in all, a slew rate is NOT a function of a gain, and thus is the same for all gains (0.4V/us typical), but to assure that the op amp is actually slewing, a large signal (Vin>1V) must be applied to the op amp input that results in momentary large differential voltage between the input terminals before feedback drives both inputs to a virtual a short.  

    The only factor that may influence the slew rate is a change in the quiescent current, IQ (since Itail~IQ) caused by either supply voltage or temperature variation BUT judging by the graphs below in case of OPA378 these changes are no more than 10% - see graphs below.

  • 0 in reply to Marek Lis
    Intellectual 320 points
    hello sir

    dont u think slew rate indirectly related to the gain.
    i agree to the point that slew rate depends on the Iq and Cc,but in the case of op-amp if we increase the Iq then there will be decrease in the gain.
    if u decrease the Iq then gain will increase.(rds=1/lambda*Id) gain of the op-amp=gain1st stage * gain2ndstage
    hope you got my point.

    regards
    vishu