LOG112: Compensating for Errors

Part Number: LOG112

Hello,

I would like to use the LOG112 in a new design.  I need best possible accuracy.  I have a few questions about how to accomplish that.

1.  I plan to measure over 5 decades, 1nA - 100uA.  I plan to use Iref = 3.16uA (center of decade range).  Is that the best value for Iref?

2.  To measure voltage offset, I plan to drive I1 and I2 with 3.16uA, measure the voltage at VLOGOUT, and use that value to compensate for offset.  Is that a good plan?  Should I use a different value of input current?

3.  To measure gain error, I plan to drive I1 with 31.6uA and I2 with 3.16uA, read the voltage at VLOGOUT, compensate it for previously measured offset, and divide it by nominal gain = 0.5V/dec.  I don't know if that will work because Log Conformity error will be present (over 1 decade).  Is there a better way?

4.  I plan to generate Iref from a 2.5V reference.  The 2.5V reference in the LOG112 is +/-0.5%.  I plan to use a 0.05% reference like the LM4132A-2.5 to generate Iref.  I will also use it to generate the test currents mentioned above.  How does reference current error affect overall accuracy?  In particular, what term should be added to Eqn. (9) on page 12 of the datasheet?

Thanks for your help!

Richard

5 Replies

  • Hi Richard,

    Regarding your questions:

    I would like to use the LOG112 in a new design.  I need best possible accuracy.  I have a few questions about how to accomplish that.

    1.  I plan to measure over 5 decades, 1nA - 100uA.  I plan to use Iref = 3.16uA (center of decade range).  Is that the best value for Iref?

    Setting Iref in the range of 1 to 10 uA produces just about the best overall dc, ac and log conformity performances. We use 1 uA as the reference current level for much of the testing. An Iref = 3.16 uA is close and you shouldn't see much difference from the 1 uA results

    2.  To measure voltage offset, I plan to drive I1 and I2 with 3.16uA, measure the voltage at VLOGOUT, and use that value to compensate for offset.  Is that a good plan?  Should I use a different value of input current?

    Setting I1 = I2 would ideally result in 0 V at VLOGOUT. Whatever current levels you set I1 and I2 to will result in the offset at that particular current level; therefore, it won't zeroed out at all current levels. Again, I would use an IREF in the 1 to 10 uA, range, either 1 uA or 3.16 uA should be fine.

    3.  To measure gain error, I plan to drive I1 with 31.6uA and I2 with 3.16uA, read the voltage at VLOGOUT, compensate it for previously measured offset, and divide it by nominal gain = 0.5V/dec.  I don't know if that will work because Log Conformity error will be present (over 1 decade).  Is there a better way?

    The gain error is the average of the difference between the actual output and the desired output at the maximum positive and negative output values at VLOGOUT. Therefore, the current ratio should be set up such that the maximum values are attained.  The output referred voltage offset (Voso) from your question 2 needs to be factored into the gain error calculation.

    4.  I plan to generate Iref from a 2.5V reference.  The 2.5V reference in the LOG112 is +/-0.5%.  I plan to use a 0.05% reference like the LM4132A-2.5 to generate Iref.  I will also use it to generate the test currents mentioned above.  How does reference current error affect overall accuracy?  In particular, what term should be added to Eqn. (9) on page 12 of the datasheet?

    Using a higher accuracy voltage reference will directly and positively affect the accuracy of the input current levels. The more accurate and precise I1 and I2 are the closer VLOGOUT should be to he ideal value (less the other errors).

    It looks like the EQ. 9 terms affected by higher accuracy reference voltage are I1 and I2.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • In reply to Thomas Kuehl:

    Hi Thomas,

    Thanks for the quick response.  I have a followup question on my Item 4 (pasted in below).

    I did some calculations of error using the Vlogout equation (eqn. (8) on datasheet p12).  I assumed a 1% error in Vref, which causes a 1% error in Iref (aka I2).  My Iref was 0.99uA instead of 1uA.  I found that the resulting error is a constant 2mV at Vlogout.  I don't see how I can measure that error directly, unless I have a very accurate 1uA current source to drive I1.  I'll plan to use a 0.05% Vref if I need to avoid Iref errors.

    4.  I plan to generate Iref from a 2.5V reference.  The 2.5V reference in the LOG112 is +/-0.5%.  I plan to use a 0.05% reference like the LM4132A-2.5 to generate Iref.  I will also use it to generate the test currents mentioned above.  How does reference current error affect overall accuracy?  In particular, what term should be added to Eqn. (9) on page 12 of the datasheet?

    Using a higher accuracy voltage reference will directly and positively affect the accuracy of the input current levels. The more accurate and precise I1 and I2 are the closer VLOGOUT should be to he ideal value (less the other errors).

     

    The other question I have is about Eqn (9) on page 12.  It seems to me that the Log Conformity error term, "Nm", should be multiplied by the Gain (0.5V/dec) in order to convert it to volts.  If you look at the LOG114 datasheet page 21, in the section "Log Conformity", you see this equation:

    VLOGOUT (NONLIN) = 0.375V/decade • 2Nm

    That equation makes sense to me.  A little further down on the same page, eqn. 20 drops the Gain as a factor for the Log Conformity error in volts.   I don't understand which is correct.  Can you explain?

    Also, sometimes there is a factor of "2" in front of "Nm", sometimes not.  What does that mean?

    Thanks,

    Richard

  • In reply to Richard Leath:

    Hello Richard,

    The LOG112 datasheet output voltage including error terms is listed in Eq. 9 as:

    While for the LOG114 it listed in Eq. 20 as:

    The LOG112 gain term is 0.375, while the LOG114 gain is 0.5 which is correct for each model. The LOG112 includes the input bias current term, while the LOG114 doesn't. The +/-5 pA could be significant

    The real issue is the absence of the "2" in "Nm" term in LOG112 Eq. 9. Its absence in some of the equations doesn't make sense.It appears, disappears and reappears in the subsequent LOG112 error equations.

    One of my colleagues found that the original log amplifier individual error components explanation goes back to the LOG101 datasheet, and I found it in an even earlier LOG100 datasheet. For both of these current ratio log amplifiers the "2Nm" appears in all of the individual error component equations. It looks like the errors first occurred in the LOG104 datasheet and were carried through to the LOG112 datasheet. So do include the "2" in all of the associated LOG112 error equations.

    Sorry for the inconvenience this caused, but I hope this correction helps.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • In reply to Thomas Kuehl:

    Thomas,

    Thanks for the reply concerning the factor of "2" in the Log Conformity error.  However, you didn't answer my first question:  It seems to me that the "2Nm" term should be multiplied by the Gain, 0.5V/dec, in order to convert it to volts instead of a % of full scale.

    What do you think?

    Thanks,

    Richard

  • In reply to Richard Leath:

    Hi Richard,

    It appears in the first and third terms of the VLOGOUT Eq. 9 they have units of V, while the second term the log nonconformity is in percent. In the example they simply convert the "n" term from a percentage to a unitless number. However, the resulting 2Nm second term has to have units of volts for the equation to be correct. I hope I am not stating the obvious.  

    Each individual term in the LOG112 VLOGOUT equation 9 is output referred, and the factors that produce their errors have already acted upon them. In the case of the 2Nm factor whatever contributed to the nonconformity error shows up in the output. The first gain term in Eq. 9 likely influences the 2Nm term and its effect is seen in the output. Therefore, I don't believe it is necessary to multiply 0.5 V/dec factor.

    Regards, Thomas

    Precision Amplifiers Applications Engineering