## 5.2 Process Integration

### 5.2.1 Chips on Wafers

| We now have a crude idea of what we want to make. The question now is how we are going to do it.
|  | We start with a suitable piece of a Si crystal, a Si wafer. A wafer is a thin (about 650 µm) round piece of rather perfect Si single crystal with a typical diameter (in the year 2000) of 200 mm. Nowadays (2007) you would build you factory for 300 mm.
|  | On this wafer we place our chips, square or rectangular areas that contain the complete integrated circuit with dimensions of roughly 1 cm2.
|  | The picture below shows a 150 mm wafer with (rather large 1st generation) 16 Mbit DRAM chips and gives an idea about the whole structure.

| The chips will be cut with a diamond saw from the wafer and mounted in their casings.
|  | Between the chips - in the area that will be destroyed by cutting - are test structures that allow to measure certain technology parameters.
|  | 'The (major) flat" of the wafer is aligned along a <110> direction and allows to produce the structures on the wafer in perfect alignment with crystallographic directions. It also served to indicate the crystallography and doping type of the wafer; consult the link for details
|  | Don't forget: Si is brittle like glass. Handling a wafer is like handling a thin glass plate - if you are not careful, it breaks.

| How to get the chips on the wafer? In order to produce a CMOS structure as shown before, we essentially have to go back and forth between two two basic process modules:
| Material module
|  | Deposit some material on the surface of the wafer (e.g. SiO2), or
|  | Modify material already there (e.g. by introducing the desired doping), or
|  | Clean the material present, or
|  | Measure something relative to the material (e.g. its thickness), or
|  | - well, there are a few more points like this but which are not important at this stage.
| Structuring module
|  | Transfer the desired structure for the relevant material into some light sensitive layer called a photo-resist or simply resist (which is a very special electronic material!) by lithography, i.e. by projecting a slide (called a mask or more generally reticle ) of the structure onto the light sensitive layer, followed by developing the resist akin to a conventional photographic process, and then:
|  | Transfer the structure from the resist to the material by structure etching or other techniques.
| Repeat the cycle more than 20 times - and you have a wafer with fully processed chips.
|  | This is shown schematically in the drawing:
|  | For the most primitive transistor imaginable, a minimum of 5 lithographic steps are required. Each process module consists of many individual process steps and it is the art of process integration to find the optimal combination and sequence of process steps to achieve the desired result in the most economic way.
| It needs a lot of process steps - most of them difficult and complex - to make a chip.
|  | Even the most simple 5 mask process requires about 100 process steps.
|  | A 16 Mbit DRAM needs about 19 masks and 400 process steps.
| To give an idea what this contains, here is a list of the ingredients for a 16 Mbit DRAM at the time of its introduction to the market (with time it tends to become somewhat simpler):
|  | 57 layers are deposited (such as SiO2 (14 times), Si3N4, Al, ...).
|  | 73 etching steps are necessary (54 with "plasma etching", 19 with wet chemistry).
|  | 19 lithography steps are required (including deposition of the resist, exposure, and development).
|  | 12 high temperature processes (including several oxidations) are needed.
|  | 37 dedicated cleaning steps are built in; wet chemistry occurs 150 times altogether.
|  | 158 measurements take place to assure that everything happened as designed.
| A more detailed rendering can be found in the link.
| Two questions come to mind:
|  | How long does it take to do all this? The answer is: weeks if everything always works and you never have to wait, and months considering that there is no such thing as an uninterrupted process flow all the time.
|  | How large is the success rate? Well, lets do a back-of-the-envelope calculation and assume that each process has a success rate of x %. The overall yield Y of working devices is then Y = (x/100)N % with N = number of process steps. With N = 450 or 200 we have
x | Y for N = 450 | Y for N = 200
--- | --- | ---
95% | 9,45 · 10–9 % | 3,51 · 10–3 %
99% | 1,09 % | 13,4 %
99,9% | 63,7 % | 81,9 %
|  | N = 200 might be more realistic, because many steps (especially controls) do not influence the yield very much.
| But whichever way we look at these numbers, there is an unavoidable conclusion : Total perfection at each process step is absolutely necessary!
Questionaire
Multiple Choice Questions to 5.2.1

© H. Föll (Electronic Materials - Script)