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OPA810: amplifer error calculation

liping pan
Intellectual 460 points
Part Number: OPA810
Other Parts Discussed in Thread: THS4551, TPS7A20

Hello,

Based on below table from datasheet.

Let's take typcial value.

- Voffset = 50uV

- Ioffset = 10nA

So, total error = Voffset + Ioffset * Rf = 50uV + 10nA * 1k ohm = 60uV.

Pls check if this calculaiton is correct.

  • 0
    TI__Expert 6050 points

    Hi,

    The error of a fully differential amplifier is contributed to by six factors:

    1. Input offset voltage

    2. Input bias currents

    3. Mismatch between input and output common-mode voltages

    4. Common-mode rejection ratio

    5. Power supply rejection ratio

    6. Resistor mismatch

    I believe you will find this article helpful in calculating the total error of the amplifier.

    Please let me know if you have any other questions.

    Thanks,

    Nick

  • 0 in reply to Nick Nauman
    Intellectual 460 points

    Hi Nick,

    Thanks! Let me take below ths4551, power supply tps7a20 as example for this calculation.

    1. Input offset voltage,  v1 = 375uV * (1k+499+499)/(499+499) = 750uV

    2. Input bias currents,  v2 = 2uA * 1kohm / 93db CMRR = 22uV

    3. Mismatch between input and output common-mode voltages,   v3 = Vocm offset / CMRR = 6mV/90dB = 67uV (no sure if CMRR should use db or log data)

    4. Common-mode rejection ratio, ??

    5. Power supply rejection ratio,  take from power supply datasheet noise voltage is 10uVrms, v5 = 2 * (2.5V/93db CMRR + 10uV/93PSRR) = 53mV

    6. Resistor mismatch, not sure if use formula 27 in the above linked document, but how to find Vocm and Vicm?

    7. Input offset current, v7 = 78nA * 1k = 78uV

    power supply tps7a20

  • 0 in reply to liping pan
    TI__Expert 6050 points

    Hi Liping,

    To answer your questions:

    3. CMRR and PSRR should be log data (so instead of dividing by 93dB, you would divide by 211.35 V/V). This also affects your calculation for 2 and 5)

    4. CMRR is a factor that has an affect on the error from input bias currents, mismatch between the input and output common-mode voltages, and the incomplete rejection of the input common-mode voltages. 

    6. For your application, there is no consideration that needs to be taken for resistor mismatch because your circuit is kept symmetric

    The rest of your calculations look correct.

    I hope this clarifies how to calculate the total error. Please let me know if there are any other questions.

    Thanks,

    Nick

  • 0 in reply to Nick Nauman
    Intellectual 460 points

    3) where does  211.35 V/V comes from? for example 93dB CMRR, I dont' know how to get your numbers with log?

    6) Even my circuit kept symetric, I still think there should be some error due to resistor tolerance. It can not perfect match. Can you explain how to calculate it?

    one more , in the document, it mentioned Vocm and Vicm, in datasheet, only 1 pin called it, how to match both in and out?

  • +1 in reply to liping pan
    TI__Expert 6050 points

    Hi Liping,

    3) The relationship between dB gain and V/V gain is represented by the equation: G_dB = 20*log(G_v/v) or G_v/v = 10^(G_dB/20)

    Your confusion is probably a result of an incorrect calculation that I made, for which I apologize. The gain should actually be 44668.36 V/V for a 93dB gain

    6) You are correct! In the article, section 4.2 gives an example where there is a circuit designed symmetrically and then tolerances of the gain resistors are accounted for. I would recommend taking a look at this section again for more clarity. Essentially, the formulas you used above made the assumption that the resistors matched. In order to account for differing values of resistors, you must calculate beta_1, beta_2, R_eq1, and R_eq2. These values should then be used in the general expression for each error calculation rather than the asymptotic form. Section 4.3 provides a nice summary of all these formulas

    In the configuration that you have, since Vocm is open (for DC signals) the output common-mode voltage will be equal to midsupply, or 2.5 V in this example. That means Vicm would also have to be 2.5V in order to match Vocm to Vicm. With that said, the error due to the mismatch between the input and output common-mode voltages is small and the total error will be dominated by the input offset voltage.

    Thanks,

    Nick