Authored by Jonathan Young on September 21, 2015
Email: youngj1ATunlv.nevada.edu
This lab, in conjunction with Lab 2, went through the creation of a 10-bit Digital to Analog converter (DAC). While Lab 2 focused on the schematic layout using the R-2R topology; Lab 3 focused on taking that schematic layout and turning it into a layout using the 10k n-well resistors.
1. Back up all of your previous work from the lab and the course.
2. Finish Tutorial 1.
Continuing where we left off in Lab 1, after the simulation of the schematic and creation of the symbol for a voltage
divider, a layout using the n-well will be created. This process will be explained below, though not in minute detail as
described in Tutorial 1, which can be found by clicking here.
To create an n-well
resistor, such as the 10k n-well resistors needed for our 10-bit DAC, a new cell view is created that is titled
"R_n_well_10k" to denote that it is resistor based on the n-well and that it is of 10k value. This layout is what is
used to physically build the 10k resistor using the CMOS process, since we are using a C5 process, the correct size
dimensions must be realized. To determine the value, read length and width of our n-well, a calculation must be
performed as well as knowing which design rules are being used.
In this course and lab, the MOSIS submicron
(SCUBM) design rules are used as this is how integrated circuit chips will be produced at the end of the semester. To
produce such chips, design rules from the manufacture must be followed. These rules specify the minimum well width,
minimum spacing between wells at different potential, minimum spacing between wells at the same potential, and minimum
spacing between wells of different types (p-well for example). These rules and sizes are displayed out in the table
below:
Note: CMOSedu is also present as the course textbook uses this scale as well, which is half that of the SUBM that MOSIS
uses. The lambda next to each number indicates a scaling factor.
Since the C5 (300nm=0.30um) process is being
used, the minimum width of an n-well using this process is calculated by the following: 12lamda*0.3um=3.6um. Using this
width, the length must now be determined. An n-well resistor is created with the following formula: r=length/width*sheet
resistance (ohm/square). Sheet resistance for this process is roughly 800 ohm/sheet. Thus, using the above formula,
including the width calculated above and the known resistor value needed (10k), the following will determine the length
of the n-well needed: Length=Resistor/Sheet Resistance * width=10k/800 * 3.6um=46um.
Thus, to create a 10k
resistor using an n-well with the C5 process, the length must be 46um and the width must be 3.6um. However, Tutorial 1
is using a 4.5um width (56.25um length - this is valid, since this is greater than the 3.6 minimum this process allows)
which I will be using to layout my n-well as described below. The reason for this variance is to give a little extra
error as the MOSIS process is not 100% perfect, thus if the minimum value was used the chip may not fabricate correctly
as it could end up with a 4.0 minimum length. Thus it is best to design a little over the exact minimum values.
This image also shows how a circuit can be measured to verify that the length and width of the n-well resistor was done correctly. Using the Ruler Tool in Cadence, allows portions of the nwell to be measured thus aiding in the verification of the length and width of the n-well resistor to match the hand-calculations.
1. Use the n-well to layout a 10k resistor as discussed in Tutorial 1.*
2. Discuss how to select the width and
length of the resistor by referencing the process information from MOSIS.*
3. Use this n-well resistor in the layout
of the 10-bit DAC.
4. Discuss how the width and length of the resistor are measured.*
5. Ensure that each
resistor in the 10-bit DAC is laid out in parallel having the same xposition but varying y-positions (the resistors are
stacked). Note: All input and output pins should be on metal 1.
6. DRC and LVS, with the extracted layout, the
10-bit DAC design. Include the final design directory as a compressed file and provide a download link.
Note: *
indicates this task was performed as part of the pre-lab above.
This layout was done in the pre-lab, as discussed above as well as how to select the width and length of the resistor using the process information from MOSIS. Since this information has already been discussed previously in the report, the use of the 10k n-well will be used for the 10-bit DAC.
To create the 10-bit DAC, the use of three 10k n-well resistors will be used. This creates the 2R to R resistor topology
seen in lab 2, figure 10.
The Top and Bottom pins are bi-directional pins, to allow for easier chaining of this circuit for the remaining 9-bits
of the DAC to create the final design of a 10-bit DAC. This is just a small subset of that circuit, a single bit. The
distance shown above, 5.4um, was calculated using the MOSIS table from the pre-lab. Since the potential (voltage)
changes between the n-wells because of the voltage divider, the calculation formula is: 0.3um*18=5.4um. This is the
minimum distance that must be between the three n-wells, as displayed above.
To create the 10-bit DAC, the above 1-bit DAC will be duplicated ten times and inter-connected. Lastly, at the
end due to the total resistance of the 2R to R topology having the resistance of R. To realize this circuit, we must add
another 10k n-well to the ten bit DAC, as explained in the previous lab (lab 2).
The following is the final 10-bit DAC design.
In all, the simulations above show very minor changes from the schematic simulated layout versus the extracted layout
simulations. The changes that were encountered were do to the small change in the actual resistance when no load was
attached.
The final design compressed file can be downloaded by clicking here. This file
contains the 10- bit DAC layout, simulations, as well as schematic and symbols. This file is provided for grading and
verification purposes only.