Determining total resistor error - Page 1
Forum section: electronics / projects

author: thexeno
author_rank: Regular Contributor
timestamp: October 11, 2017, 03:44:45 pm
content: Hi all,I was reading an application note, for which I was trying to understand the various steps, where this involves also the mismatch error calculation. I know this might be a naive question, sorry about that.In this picture is the input/output voltage relation:and this is their relation for the output error due to the components:(for a better picture attachment let me know)How they have found the relation of R1, R2 and R3?

author: TimNJ
author_rank: Super Contributor
timestamp: October 11, 2017, 05:52:34 pm
content: Do you have a link to the application note or the circuit they are referencing?

author: thexeno
author_rank: Regular Contributor
timestamp: October 11, 2017, 10:37:33 pm
content: Sure, it is from FTM magazine, page 10http://www.my-ftm.com/2017/06/ftm-june-2017-emea-edition/

author: StillTrying
author_rank: Super Contributor
timestamp: October 12, 2017, 12:54:45 am
content: web versionhttp://www.my-ftm.com/2017/06/stmicroelectronics-how-to-make-a-high-voltage-high-side-current-sensor-with-a-low-voltage-precision-op-amp/

author: orolo
author_rank: Frequent Contributor
timestamp: October 12, 2017, 09:45:04 am
content: It's a bit of calculus mixed in with a bit of malevolence.The calculus part:The output voltage is a function of the five resistances (R1-R4 and Rshunt). Let us forget from now on of the proportional term I_sense that will play no part from here on. \$V_o = \frac{R_sR_4}{R_1R_3}\left(R_1 + R_2 + R_3\right)\$.To know what happens when one resistance varies a little, you must compute the partial derivative with respect to that variable. Let us just do that, and rearrange things afterwards:\$\displaystyle \frac{\partial V_o}{\partial R_1} \quad = \quad \frac{R_sR_4}{R_1R_3}\cdot\frac{-R_2-R_3}{R_1}\$\$\displaystyle \frac{\partial V_o}{\partial R_2} \quad = \quad \frac{R_sR_4}{R_1R_3} \$\$\displaystyle \frac{\partial V_o}{\partial R_3} \quad = \quad \frac{R_sR_4}{R_1R_3}\cdot\frac{-R_1-R_2}{R_3}\$\$\displaystyle \frac{\partial V_o}{\partial R_4} \quad = \quad \frac{R_sR_4}{R_1R_3}\cdot\frac{R_1+R_2+R_3}{R_4}\$and (this we will treat apart):\$\displaystyle \frac{\partial V_o}{\partial R_s} \quad = \quad \frac{R_sR_4}{R_1R_3}\cdot\frac{R_1+R_2+R_3}{R_s}\$Now, note that the resistor error is relative to the value of the resistor. That is, if resistor precision is \$\epsilon\$, then the worst error of say R1 will be \$\epsilon R_1\$. Meaning that the error contributed by R1 will be:\$\displaystyle e_1 = \epsilon R_1 \cdot \frac{\partial V_o}{\partial R_1} \$and so on.So the errors contributed by each resistor are:\$\displaystyle e_1 \quad = \quad \epsilon \cdot \frac{R_sR_4}{R_1R_3}\left(-R_2 - R_3\right) \$\$\displaystyle e_2 \quad = \quad \epsilon \cdot \frac{R_sR_4}{R_1R_3}R_2 \$\$\displaystyle e_3 \quad = \quad \epsilon \cdot \frac{R_sR_4}{R_1R_3}\left(-R_1 - R_2\right) \$\$\displaystyle e_4 \quad = \quad \epsilon \cdot \frac{R_sR_4}{R_1R_3}\left(R_1 + R_2 + R_3\right)\$and:\$\displaystyle e_s \quad = \quad \epsilon \cdot \frac{R_sR_4}{R_1R_3}\left(R_1 + R_2 + R_3\right)\$The malevolent part:If we are to assume the worst, we should not add these errors,but their absolute values. So forget about the signs in e1, e2, e3 and e4, and add them up. Notin that they have all the fractional term RsR4/R1R3 in common, they add up to the expected:\$\displaystyle e \quad = \quad \epsilon\cdot \frac{R_sR_4}{R_1R_3}\left(2R_1 + 4R_2 + 2R_3\right)\$And now it's very easy to get an expression for error. Write:\$\displaystyle V_{altered} \quad = \quad V_o + e \quad = \quad \frac{R_sR_4}{R_1R_3}\left(R_1 + R_2 + R_3\right) + \epsilon\cdot \frac{R_sR_4}{R_1R_3}\left(2R_1 + 4R_2 + 2R_3\right) \quad = \$\$\frac{R_sR_4}{R_1R_3}\left(R_1 + R_2 + R_3\right) \cdot\left[1 + \frac{2R_1 + 4R_2 + 2R_3}{R_1 + R_2 + R_3}\cdot\epsilon\right]\quad = \quad V_o \cdot\left[1 + \frac{2R_1 + 4R_2 + 2R_3}{R_1 + R_2 + R_3}\cdot\epsilon\right]\$Doing just the same for the shunt resistance gives the shunt term.

author: thexeno
author_rank: Regular Contributor
timestamp: October 13, 2017, 08:57:38 am
content: Quote from: orolo on October 12, 2017, 09:45:04 amYes, and \$ R_{shunt} \$ error contribution has the same coefficient of the nominal voltage, so that in the end you just simply obtain\$\displaystyle V_{altered} \quad = \quad V_o \cdot\left[1 + \frac{2R_1 + 4R_2 + 2R_3}{R_1 + R_2 + R_3}\cdot\epsilon + \epsilon_{s}\right]\$Thank you, this is a basic but fundamental piece of information. I save it somewhere in my notes for future reference. Cheers!