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LF412-N: computing Ro from datasheeet curves

Part Number: LF412-N


Hi team,

Following Tim Green's article  in Operational Amplifier Stability, part 3 of 15,on the closed-loop output impendence VS frequency curve, FIG 20, we choose the G=10 curve and on its x-axis the point 0.8kHZ, at the intersection of 0.8KHZ and G=10 curve we see ROUT=0.5Ω.

On the open loop gain/phase vs frequency curve, Fig 15,we look at the 0.8KHZ frequency point on the x-axis and read the open-loop gain as 70dB.

According to the above formula, we calculate Ro = 158 Ω.

we choose the G=10 curve and on its x-axis the point 100kHZ, at the intersection of 100KHZ and G=10 curve we see ROUT=10Ω. On the open loop gain/phase vs frequency curve, Fig 15,we look at the 100KHZ frequency point on the x-axis and read the open-loop gain as 30dB. According to the above formula, we calculate Ro = 31.6Ω.

The calculated Ro, especially the low-frequency part, varies greatly, as shown in the figure above.

Is there any other prerequisite for the application of this calculation method? Or what causes this and which should prevail?

Best Regards,

Amy Luo

  • Amy,

    There can be a pretty significant difference just based on estimation when trying to linearize a dB value. The graphs on the datasheet appear to be scanned from physical printouts, which doesn't lend itself to the best approximations, but there exist software that you can estimate with fairly reasonable accuracy.

    Following the same procedure listed by Tim Green, and calculating the open loop output impedance at 10kHz, I got the following results based on each curve, and looked at one frequency for convenience of having the same Aol value.

    Av = 1 : 0.3*(1+398/1) = Ro = 119.4 Ohms
    Av = 10 : 1.2*(1+398/10) = Ro = 48.96 Ohms
    Av = 100 : Ro = 10.2*(1+398/100) 50.7 Ohms

    Since these are all the data points provided, you can see that the Av = 1 curve doesn't align with the other two, and most likely the Ro is around 50 Ohms.

    You can see as your closed loop gain scales down my approximation of the open loop gain value, the error decreases, so it's likely my guess for the open loop gain was not very accurate.

    Best,
    Jerry