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OPA211: OPA211 Offset Voltage and Bias Current error calculations

N S
Intellectual 365 points

Part Number: OPA211

Hi Support

I am currently going through the online video training series TI Precision Labs - Opamps

In the video for offset and bias current, some calculations are done for typical & Maximum cases for OPA211 with input resistance of 5K as shown in the picture below:

I have done calculations below using the method suggested in the first video of same series as below:

For OPA211:

For OPA211

Typical:

Maximum

Vos = +/-30uV

Vos = +/-125uV

Stage-1:

Stage1:

Gain = 1+100/1 = 101

Gain = 1+100/1 = 101

Vo_os = (+/-30uV) * 101 = +/-3.03mV

Vo_os = 101*(+/-125uV) = +/-12.6mV

Ib = +/-60nA

Ib = +/-175nA

Rf = 100K

Rf = 100K

Rin = 5K

Rin = 5K

Vo_ib = [100k * (+/-60nA)] + [5k * (+/-60nA)] = +/-6mV +/-0.3mV = +/-6.3mV or +/-5.7mV (Use largest as +/-6.3mV)

Vo_ib = [100K * (+/-175nA)] + [5k *(+/-175nA)] = +/-17.5mV+/-0.875mV

= +/-18.375mV (largest)

Stage-2:

Stage2:

Vos = +/-30uV

Vos=+/-125uV

Total Vos = Vo_os(Stage1) + Vos(Stage2) = +/-3.03mV +/-30uV = +/-3mV or +/-3.06mV (Use largest as +/-3.06mV)

Total Vos = Vo_os(Stage1) + Vos(Stage2) = +/-12.6mV +/-125uV =

 +/-12.725mV  (largest)

Gain (Stage-2) = 1+20/1 = 21

Gain(Stage2) = 1+20/1 = 21

Total Vo_os = 21* (+/-3mV or +/-3.06mV)

= +/-63mV  or  +/-64.26mV (Use largest as +/-64.26mV)

Total Vo_os = 21*(+/-12.725mV OR +/-12.475mV)

 = +/-267.225mV OR +/-261.975

For Stage-2, ib = +/-60nA

For Stage2, Ib = +/-175nA

Rf = 20K

Rf = 20K

Total Vo_ib = Vo_ib_stage1 * Gain + Vo_ib_stage2

Total Vo_ib = Vo_ib_Stage1 * Gain + Vo_ib_stage2

Gain = 1+20/1 = 21

Gain = 1+20/1 = 21

Vo_ib_stage2 = 20k * +/-60nA = +/-1.2mV

Vo_ib_stage2 = 20k * +/-175nA = +/-3.5mV

Total Vo_ib = [(+/-6.3mV * 21] + [+/-1.2mV] = +/-132.3mV +/-1.2mV

= +/-133.5mV (largest)

Total Vo_ib = [+/-18.375mV*21] +[+/-3.5mV] = +/-385.875 +/-3.5mV =

+/-389.375mV (largest)

Grand Total Error

Grand Total Error

Vo_error = Total Vo_os + Total Vo_ib

Vo_error = Total Vo_os + Total Vo_ib

Vo_error =  [+/-64.26mV ] + [+/-133.5mV]

Vo_error = [+/-267.225] + [+/-389.375mV]

Vo_error = +/-197.76mV

Vo_error = +/-656.6mV

 

 

And there is a huge difference between my calculated values and the ones in the training series (as shown below).

Typical Training  Typical (My Calc) Max Traning  Max (My Calc)
(+/-827mV) (+/-)197.76mV (+/-2492mV) (+/-656.6mV)

Could someone help me in clarifying this and indicate where I am  making errors in my calculations.

ManyThanks

  • 0
    TI__Mastermind 23050 points
    Hi,

    I will try to clarify this as best I can. The first error I see is in your calculation of Vo_ib in the first stage. Multiplying the bias current of the inverting pin by the 100kOhm feedback resistance results in the output voltage as a result of this bias current. Multiplying the bias current of the noninverting pin by the 5kOhm resistance to ground is the offset produced by the bias current referred to the input, and thus you still need to multiply this result by the noninverting gain. You cannot add a voltage that is referred to the input to one that is referred to the output. It is easiest to calculate everything referred to the input first, and then just add the offsets and multiply by the gain as is done in the slides above. To do this, you actually need to multiply the bias current of the inverting pin by the parallel combination of the input resistor and the feedback resistor, which is Req in the slides.

    You are technically correct to add the offset errors from the second stage to the output of the first opamp, but it is useful to notice that because the offset and bias currents of the second opamp do not see significant gain, their impact on the overall output offset will be next to negligible (hence why they are not accounted for in the slides).

    If you are going to add the offset contribution of the second stage bias current to the output offset of the first opamp, you need to refer the offset from the bias current to the input of the second stage in the manner mentioned above, unless you are going to add it separately to the total output at the end as you have done, but again it is easiest to refer everything to the input.

    Hope this helps!

    Regards,
    -Zak Kaye
    Precision Amplifiers applications
  • 0 in reply to Zak Kaye
    Intellectual 365 points

    Thanks for the help and clarification. I am now getting reasonable results that are similar to the ones given in the slides :)