Resistors matching calculation

resistor matching layout

The sheet resistance is 1000 (Ω/Square), R1/R2=10kΩ/5kΩ,
The matching relationship is σ(ΔR/R)=a/sqrt(W*L), a=0.5%*um

If the matching requirement is 1%, Could someone please tell me how to design the W, L and the modules of R1 and R2.

resistor matching

use common centroyed technique

Added after 1 minutes:

In ur case it is pretty easy;
keep ur 5k as oefingerin the middle and 10k on bothsides.
donot forget to put dummies

individual resistor calculations

lqy said:

The sheet resistance is 1000 (Ω/Square), R1/R2=10kΩ/5kΩ,
The matching relationship is σ(ΔR/R)=a/sqrt(W*L), a=0.5%*um

If the matching requirement is 1%, Could someone please tell me how to design the W, L and the modules of R1 and R2.

Click to expand...

Pretty nice homework ;-)

sqrt(W*L) = a/σ(ΔR/R) = 0.5%*um/1% = 0.5µm
W*L = 0.25(µm)^2 . . . . . . . . (1) valid for both R1, R2
W1/L1 = 0.1 . . . . . . . . . . . . . (2) for R1=10kΩ
W2/L2 = 0.2 . . . . . . . . . . . . . (3) for R2= 5kΩ

multiply equs. (1) * (2) :
(W1*L1) * (W1/L1) = (W1)^2 = 0.025(µm)^2 --> W1 = 0.158µm ; L1 = 1.58µm

multiply equs. (1) * (3) :
(W2*L2) * (W2/L2) = (W2)^2 = 0.050(µm)^2 --> W2 = 0.224µm ; L2 = 1.12µm

For good matching use the larger of {W1, W2}, for practical purpose round up to:
W1 = W2 = 0.23µm
R1=10kΩ : W1=0.23µm ; L1 = 2.3 µm
R2= 5 kΩ : W2=0.23µm ; L2 = 1.15µm

Split R1 into 2*R2 and create layout as sudheerkm told above (and don't forget the side dummies) ;-)

matching resistors 10k

lqy said:

The sheet resistance is 1000 (Ω/Square), R1/R2=10kΩ/5kΩ,
The matching relationship is σ(ΔR/R)=a/sqrt(W*L), a=0.5%*um

If the matching requirement is 1%, Could someone please tell me how to design the W, L and the modules of R1 and R2.

Click to expand...

What do you exactly mean under "matching requirement"?

If this is a requirement (1% for one standard deviation, σ) for each individual resistor (R1 and R2), then Erik's calculations give the answer.

If, however, you require that distribution of the ratio, R1/R2, has a sigma of 1%, then each resistor should have sigma of 1%/sqrt(2) or so, i.e. ~0.71%. The reason is that if resistors R1 and R2 are not correlated (and I believe this is the case, for that model of σ), and have variations (σ) of σ1 and σ2, then their ratio R1/R2 has a variation (standard deviation) of \sqrt{\sigma_1^2+\sigma_2^2}.

calculations of resistors

erikl said:

lqy said:

The sheet resistance is 1000 (Ω/Square), R1/R2=10kΩ/5kΩ,
The matching relationship is σ(ΔR/R)=a/sqrt(W*L), a=0.5%*um

If the matching requirement is 1%, Could someone please tell me how to design the W, L and the modules of R1 and R2.

Click to expand...

Pretty nice homework ;-)

sqrt(W*L) = a/σ(ΔR/R) = 0.5%*um/1% = 0.5µm
W*L = 0.25(µm)^2 . . . . . . . . (1) valid for both R1, R2
W1/L1 = 0.1 . . . . . . . . . . . . . (2) for R1=10kΩ
W2/L2 = 0.2 . . . . . . . . . . . . . (3) for R2= 5kΩ

multiply equs. (1) * (2) :
(W1*L1) * (W1/L1) = (W1)^2 = 0.025(µm)^2 --> W1 = 0.158µm ; L1 = 1.58µm

multiply equs. (1) * (3) :
(W2*L2) * (W2/L2) = (W2)^2 = 0.050(µm)^2 --> W2 = 0.224µm ; L2 = 1.12µm

For good matching use the larger of {W1, W2}, for practical purpose round up to:
W1 = W2 = 0.23µm
R1=10kΩ : W1=0.23µm ; L1 = 2.3 µm
R2= 5 kΩ : W2=0.23µm ; L2 = 1.15µm

Split R1 into 2*R2 and create layout as sudheerkm told above (and don't forget the side dummies) ;-)

Click to expand...

Thx for a nice explanation... can u help me wiht some more information??
1) what is this "a" ?? is it defined or we have to calculate it?
2)how we calculated that W1/L1 = 0.1 for R1.....

any document giving details of matching calculation weill be highly appreciated...

thx..

matching resistors in layout

deepak242003 said:

1) what is this "a" ?? is it defined or we have to calculate it?

Click to expand...

"a" is a process & technology dependent parameter, i.e. it depends on the structure size as well as on the type of processing technology used for the production, e.g. photolithography (direction & angle dependency), implant (angle), ion etch (angle), temperature gradients during high temperature processes (direction) and more.

"a" is to be evaluated by the foundry and (should) be published with the PDK.

deepak242003 said:

2)how we calculated that W1/L1 = 0.1 for R1.....

Click to expand...

This of course is the aspect ratio of the resistors, calculated from the given resistor values (R1=10kΩ;R2=5kΩ) and the given sheet resistance of the resistor layer used (1000 Ω/Square).

deepak242003 said:

any document giving details of matching calculation will be highly appreciated...thx..

Click to expand...

Pls. find here Pelgrom's basic paper about mismatch calculation, and another one on layout strategies for better matching.
Cheers, erikl

layout matching resistor

thx a lot erik.

resistor calculations

timof said:

lqy said:

The sheet resistance is 1000 (Ω/Square), R1/R2=10kΩ/5kΩ,
The matching relationship is σ(ΔR/R)=a/sqrt(W*L), a=0.5%*um

If the matching requirement is 1%, Could someone please tell me how to design the W, L and the modules of R1 and R2.

Click to expand...

What do you exactly mean under "matching requirement"?

If this is a requirement (1% for one standard deviation, σ) for each individual resistor (R1 and R2), then Erik's calculations give the answer.

If, however, you require that distribution of the ratio, R1/R2, has a sigma of 1%, then each resistor should have sigma of 1%/sqrt(2) or so, i.e. ~0.71%. The reason is that if resistors R1 and R2 are not correlated (and I believe this is the case, for that model of σ), and have variations (σ) of σ1 and σ2, then their ratio R1/R2 has a variation (standard deviation) of [tex:5025d88d2c]\sqrt{\sigma_1^2+\sigma_2^2}[/tex:5025d88d2c].

Click to expand...

Thank you for all your help, My requirement is for the distribution of the ratio, R1/R2, Not for each individual resistor.
and for the yield consideration, the 1% should be 3σ. Could you please give the detail caculation as Erik's answer. Thank in advanced.

resistor match 0.1%

lqy said:

The sheet resistance is 1000 (Ω/Square), R1/R2=10kΩ/5kΩ,
The matching relationship is σ(ΔR/R)=a/sqrt(W*L), a=0.5%*um

My requirement is for the distribution of the ratio, R1/R2, not for each individual resistor.
and for the yield consideration, the 1% should be 3σ.

Click to expand...

If 3σ|ΔR/R| = 1% then σ|ΔR/R| = 1/3 %

As your requirement is for the distribution of the ratio (R1/R2), you must use timof's calculation, i.e.

sqrt(W*L) = a/σ(ΔR/R) = 0.5%*µm / ( 1/3 % / sqrt(2) ) = 2.12 µm

or ;-)

\[\sqrt{W*L}=\frac{a}{\sigma(\Delta {R/R})}=\frac{0.5% * micron}{\frac {1 %}{3*\sqrt{2}}}=2.12micron\]


W*L = 4.5(µm)² . . . . . . . . . (1) valid for both R1, R2
L1/W1 =10 . . . . . . . . . . . . . (2) for R1=10kΩ
L2/W2 = 5 . . . . . . . . . . . . .. (3) for R2= 5kΩ

divide equs. (1) / (2) :
(W1*L1) / (L1/W1) = (W1)² = 0.45(µm)² --> W1 = 0.67µm ; L1 = 6.70µm

divide equs. (1) / (3) :
(W2*L2) / (L2/W2) = (W2)² = 0.90(µm)² --> W2 = 0.95µm ; L2 = 4.74µm

For good matching use the larger of {W1, W2}, for practical purpose round up to:
W1 = W2 = 1µm
R1=10kΩ : W1=1µm ; L1 =10 µm
R2= 5kΩ : W2=1µm ; L2 = 5 µm

As the numbers presented look a bit like homework, I doubt I would give a detailed response if the posting were new.

It appears that the calculations given previously result in resistors that are substantially larger (a factor of 2.667 before oversizing) than needed for the problem (as posed). On the other hand (given that the resistors are very small) the 3-sigma proposal will probably cost more in additional test requirements than it saves in wafer costs. The eventual 5-sq-um area for the 5k resistor corrsponds to a 5.2-sigma margin, so in practice this would probably be more appropriate than the 3-sigma of the problem.

That said, it is probably better to start with an appropriate tolerance and calculate it accurately. So what are the issues?
First, we note that the ‘A’ coefficient is drawn from Pelgrom (the second of erikl's references) - and Pelgrom specifies the mismatch value when each device has unit area (see for example his equation 5). This is a factor of sqrt(2) larger than the assumed two-sample deviationfigure, which means that this factor has effectively been used twice in previous calculations.
We can also take into account the tighter tolerance that would be available from the two-area to shrink the smaller device.

Now to my attempt at a correct calculation:
(I choose to do this in the smallest chunks I can.

First, I convert from Pelgroms ‘A’ coefficient to the two-sample deviation, confusingly usually called Adev.
A=0.5 => Adev(resistor)=0.5/sqrt(2)/sqrt(resistor-area).
We can now use Adev to calculate the contribution of each resistor.
For my convenience, I call the 5k resistor R5k and its area A5, and I assume this will be a single resistor. In order to match this the 10-k resistor will need to be built from 2 series 5k resistors so its area is 2*A5 (areas in sq. um. – not always stated)
Sigma_mismatch = sqrt(Adev(R5k)^2 + Adev(R10k)^2)
= 0.5/sqrt(2)*sqrt(1/A5+1/(2.A5))
= 0.5/sqrt(2)*sqrt(1.5)/sqrt(A5)
So, for 1% tolerance at 3-sigma, we have:
1/9 = sigma_mismatch^2 = 0.25/2*(1.5/A5) = 3/A5/16
i.e. A5 = 27/16 = 1.69 sq-um

We can now calculate some initial nominal dimensions. Rsheet = 1k, so we need L=5*W, or
W(R5k) = sqrt(A5/5) = 0.59-um (already rounded up)
L(R5k) = 5*W(R5k) = 2.45-um
This is as far as we can go based on the data provided.
For a real case we might estimate the resistance variation due to contact mismatch and adjust the nominal tolerance and area accordingly. We would oversize to achieve the required area at minimum process dimensions – and then shape the resistor for the typical deltaL and deltaW, oversizing to match the grid.
We will of course need to place the 5k resistor centrally between the two components of the 10k resistor - and place the resistors as close together ar possible and include surrounding dummies - the bathtub effect means that four dummies (two each side) are advisable.