u can refer to design rule wihch may include some of the key parameters related to mismatching.
Depending on these, the size of MIM cap can be determined.
Hope you have used unit caps of same size with common centroid arrangement ?
the ratio match between cap can be calculated by checking the extracted cap values between them.
The mismatch ratio can be estimated with the Pelgrom equation and a mismatch coefficient to be provided by the foundry. For the calculation, see the following (resistor) mismatch examples.... the ratio match between cap how to calculate. What tool can verify it? or it can be calculated
The mismatch ratio can be estimated with the Pelgrom equation and a mismatch coefficient to be provided by the foundry. For the calculation, see the following (resistor) mismatch examples.
Extraction tools (Diva, Assura, Calibre) can only find additional associated (parasitic) values due to routing and possible environment differences.
The mismatch ratio can be estimated with the Pelgrom equation and a mismatch coefficient to be provided by the foundry. For the calculation, see the following (resistor) mismatch examples.
Extraction tools (Diva, Assura, Calibre) can only find additional associated (parasitic) values due to routing and possible environment differences.
Hi Amy,How to make A cap and B cap to get good matching? Besides common centroid arrangement and add dummy cell.
Unit caps usually are provided with the foundry's PDK - often several units depending on the cap type: MOSCAP, PP, MOM, MIM, fringe, ... - always square.Actually I want to ask you how to choose unit cap value? Please give some advices.
Hi Amy,
seems you already know all the necessary tricks ;-) . Try to get a square block, see the PDF below, p. 19 : View attachment 49938
Unit caps usually are provided with the foundry's PDK - often several units depending on the cap type: MOSCAP, PP, MOM, MIM, fringe, ... - always square.
Mine were in the order of 10×10(µm²), around 100..350 fF each (0.18µm process).
Best regards, erikl.
Sure it can. See this PDF: View attachment MIM-cap_mismatch.pdf... unit cap can not be designed for square.
Sorry:no. The unit cap layout - with which you build the caps you need - has nothing to do with your available real estate on the chip.Because my room is rectangle (width=40um length=500um) not square. so unit cap is not fit for square. Do you think so?
Unfortunately not: how can you get picoFarads from a multiplied length? On the other hand: 12 MIM caps with an edge length of 0.18µm impossibly can create a capacitance of ≥ 2pF !I have selected unit cap equal 0.18um each. A and B MIMCAPs value is separate (0.18umx12)=2.16Pf. (0.18 UMC process). Do you think it is a good choice?, By the way ,hope you understand my expression.
Thanks for your professional suggestions.Sure it can. See this PDF: View attachment 49956
.
Unfortunately not: how can you get picoFarads from a multiplied length? On the other hand: 12 MIM caps with an edge length of 0.18µm impossibly can create a capacitance of ≥ 2pF !
Better square than rectangular! Max. edge length depends on the process.... for the MIM-CAP, the shape of the unit cap should be either square or rectangular. But the size should be kept less than 30um per side. Yes?
Right!For the width of metal lines which are used to connect top-plates and bottom-plates, I think minimum width(such as 0.3um) is enough, because wider width will cause more parasitic capacitor. Yes?
Unfortunately this isn't possible: The above PDF page has been extracted from our foundry's process analog characterization docu, which of course is confidential info (that's why I've made it irrecognizable by the yellow top & foot notes ;-) ). Hope for your comprehension!I found the file you provided to me is very useful. Could you do me a favor sent the whole pdf file to me for my reference.
By the way: For me, (9um x 20 um )x0.98fF/(um)*2 x 12= 2 * 2.12pFOK 0.18 process :MIM capacitance =0.98fF/(um)*2. Length was selected randomly Or by my feeling to choose. It is not any basis. By the way, total MIMCAP value= (width x length )x0.98fF/(um)*2 x 12=(9um x 20 um )x0.98fF/(um)*2 x 12= 2.16Pf.
Right so, thank you, same to you! - eriklHope you happy every day.
Amy.
Unfortunately this isn't possible: The above PDF page has been extracted from our foundry's process analog characterization docu, which of course is confidential info (that's why I've made it irrecognizable by the yellow top & foot notes ;-) ). Hope for your comprehension!
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