2
\$\begingroup\$

I'm studying the implementation of resistors in integrated technology. In particular the books deals with the accuracy of the obtained resistors. It says:

Integrated resistor formula

Then it says:

"We observe that the value of the resistor depends on four parameters that are achieved with independent technological steps. Therefore, to estimate their accuracy we assume them to be statistically independent." and it comes out with the following equation:

Integrated resistor total error

The question is: what is the meaning of this equation? How is it mathematically derived?

\$\endgroup\$
  • \$\begingroup\$ simple application of the chain rule dy/dx = dy/du x du/dx \$\endgroup\$ – Scott Seidman Aug 1 '19 at 13:56
4
\$\begingroup\$

\$R\$ depends on \$L\$, \$W\$, \$\overline{\rho}\$ and \$x_j\$.

If the errors of those 4 parameters are statistically independent, then the variance of \$R\$ will be the sum of the variances of those 4 parameters, hence the summation.

Another way to look at it is by taking partial derivatives of \$R\$ with respect to each parameter, then dividing by \$R\$ and squaring. Then you can add it up all together because all variations are statistically independent. Otherwise you couldn't just add them up, you should take into account any cross-correlation between parameters.

And just in case you're wondering how it's possible to add errors that are in the denominator of the equation (parameters \$W\$ and \$x_j\$): we can do it because there is a first order approximation involved.

$$ \frac{1}{W+\Delta W} = \frac{\frac{1}{W}}{1+\frac{\Delta W}{W}} \approx \frac{1}{W} \left( {1-\frac{\Delta W}{W}} \right) \text{ if } \vert{\frac{\Delta W}{W}}\rvert \ll 1 \text{ thus}\\ R+\Delta R \big\vert_{\text{due to }\Delta W} = \frac{L \overline{\rho}}{(W+\Delta W)x_j}= \frac{L \overline{\rho}}{Wx_j}\left( {1-\frac{\Delta W}{W}} \right) = R - R\frac{\Delta W}{W} \\ \Delta R \big\vert_{\text{due to }\Delta W} = - R\frac{\Delta W}{W} \\ {\frac{\Delta R}{R}} \Bigg\vert_{\text{due to }\Delta W} = - \frac{\Delta W}{W} $$

After squaring, the minus sign disappears. The same applies to the error contributed by \$x_j\$.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you for your answer. I've never had a statistic course at University, then I have some doubts. I looked at the definition of variance, and the concept is clear: it means in average how far we are from the average. Then is (deltaR/R)^2 just a notation to express the variance (the variance is typically expressed with σ)? \$\endgroup\$ – Stefanino Aug 1 '19 at 11:10
  • \$\begingroup\$ Yes, it is a way to express the variance. \$\endgroup\$ – Enric Blanco Aug 1 '19 at 11:13
  • \$\begingroup\$ Your equation is not making sense to me. How can you equate a number and its negative? You should go for the binomial expansion but that is \$\frac{1}{1+x} = 1-x\$, if \$|x|<<1\$. \$\endgroup\$ – sarthak Aug 1 '19 at 11:13
  • \$\begingroup\$ You're right @sarthak , it's W+deltaW, sorry. I'll correct that. \$\endgroup\$ – Enric Blanco Aug 1 '19 at 11:16
  • 1
    \$\begingroup\$ Now I've corrected it, @sarthak. Thank you for pointing that out, thus contributing to enhance my answer! \$\endgroup\$ – Enric Blanco Aug 1 '19 at 16:22
2
\$\begingroup\$

To follow up on the excellent answer of Enric Blanco, I provide the adage of my semiconductor layout mentors: "The layout is the real schematic."

To improve the matching of Rs and Cs and transistors, do the following:

1) have identical layouts, including what is surrounding the component; in other words, do not expect matching between two resistors embedded in a random bucket of other implants; the lateral diffusion of those other implants will affect the carrier density of the dopants in the two (we need these to match) resistor volumes.

2) do not expect matching across the wafer or across the die or even from adjacent doped or etched regions; if you expect matching, then interdigitate the pieces of the resistors (or capacitors, or transistors).

3) for successful matching, you need to tolerate the edge-etching and edge-doping variations; this requires significant widths in resistors and in capacitors and in transistor active regions; do not expect minimum-width structures to match; thus in a 0.1 micron process, do not use that minimum width (or length) for precision components

4) I've seen people's circuits provide excellent matching of very small components in over-sampling ADCs, because during the over-sampling behavior the capacitors were pseudo-randomly switched in and out

5) in capacitors, the fringing fields must also be matched; that requires "dummy" capacitors be used to surround your capacitor-array

6a) you must match the temperature environment; learn to create thermal shorts on silicon (heavy metal) to minimize the temperature gradients

6b) learn to map the heat flows; be responsible; put the high-heat components ON ANOTHER CHIP; don't punt and hope --- have test chips to learn of what is possible

6c) learn to create thermal opens on chip, at least at the surface

6d) expect to have to match the METAL environment; this requires AREA to play with, around the must-match region; as you might expect, precision will cost you

6e) if you do not know the lateral thermal-resistance of your various metal layers (standard 35 micron PCB copper is 70 degree Centigrade per watt per square, any size square; 1 micron copper will be 35*70 = 2,400 degrees per watt per square), then you are not a serious matching-is-my-goal designer

7) circuits can make decisions very quickly; do you know how the thermal transients affect the matching? 1 micron cubes of silicon have 11.4 nanosecond thermal time constants.

"The layout is the real schematic" because the physical-layout lets the system designer and the circuit designer and the layout designer manage the variances.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.