Lab 2 - ECE 421L 

Authored by Silvestre Solano,

Email: solanos3@unlv.nevada.edu

9-22-2014

  

Lab description

 

Prelab

After login in and unziping the lab2.zip file, the schematic for the ADC/DAC will be selected as shown below.

 

http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/1.JPG

 

After clicking on "schematic", the schematic shown below should appear.

 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/2.JPG
 

After launching the schematic, the ADE L will be loaded with the save state in cellview as shown below.

 

http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/3.JPG

Then the simulation parameters will appear and the green button will be pressed as shown below. The simulation results are shown following the picture below.

 

http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/4.JPG

 

 

http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/5.JPG

 

The LSB is found  by the equation 1LSB=VDD/(2^n) were n is the number of bits of the DAC.

 

 

 

 

Postlab

The design of a 10-bit DAC using an n-well R of 10k

In this lab, we had to implement our own design for a 10-bit digital to analog converter (DAC). This design is shown in the picture below.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/31.JPG

The first step in acomplishing this task is to first continue were the prelab left off and copy "sim_ideal_ADC_DAC" into a new file called "1_sim_ideal_ADC_DAC." After the copy is complete, the schematic shown below will be used.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/30.JPG

 Using the "Descend Edit" option on the DAC part of the above schematic reveals the inside structure (shown below) that will be edited in order to create the resistor based DAC.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/8.JPG

 
Everything shown above will be completely erased with the exception of the B0-B9 input pins in order to make room for the resistor based DAC. Unfortunately, this is the part were I screwed up the first time I did the lab and I had to start all over with a new file called "2_sim_Ideal_ADC_DAC."
 
In a seperate file called "R_R2", I created the simple resistor network as shown below. For some strange reason, the 2R resistor had to implemented as two seperate resistors in series. Pins are added in order to be used with the symbol view.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/9.JPG

After creating the resistor network, a new symbol view will be created for it. The new symbol view that results from this is shown below.

http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/10.JPG
 
In order to implement the 10-bit DAC, 10 of these resistor cells will have to be strung up together in series within the original 10-bit DAC. The completed interconnected resistor cell network is shown below.
This is how you create a symbol view for the design that has the exact same footprint as the Ideal_10-bit_DAC symbol view.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/16.JPG
 
Simulations to verify your design functions correctly.
 
In order to illustrate that the resistor based DAC works correctly, I will run a simulation of its output and display it below this sentance.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/13.JPG
 
Originally, I could not get my simulaion to work. I spent the first week of this lab stuck in the same spot because my simulation would not run. On the second week, I was still working on this problem until I decided to ask professor Baker for help. He spent a good 5 minutes trying to understand what was wrong with my schematic. In the end, he somehow fixed it, but I doubt he even knew what exactly was wrong with my schematic.
 
When the circuit drives a 10pF load (shown below), the following output voltage is shown on the graph that follows the schematic.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/23.JPG
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/22.JPG
 
When the circuit drives a 10pF load  and a 10K resistor (shown below), the following output voltage is shown on the graph that follows the schematic.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/24.JPG
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/25.JPG
 

When the circuit drives a  10K resistor (shown below), the following output voltage is shown on the graph that follows the schematic.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/20.JPG
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/21.JPG
 
When the DAC drives a 10k load, the output voltage is cut in half. This is because the output resistance of 10K works with the load to form a voltage divider that essentially cuts the outgoing voltage in half. In the above grapy, the input voltage is 5 while Vout is 2.49 volts..

Delay, driving a load.
 
Ib order to simulate the delay of a 10pF capacitor, the B0-B8 inputs of the DAC will be grounded and a 1V pulse source will be added to the B9 input, as shown below.

http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/27.JPG
 
The simulation results are shown below for the above circuit.
 
http://cmosedu.com/jbaker/courses/ee421L/f14/students/solanos3/lab2/29.JPG

Because the output voltage follwows the equation 1LSB=(VDD/(2^n)), which in this case n=1, the output voltage will only be half of VDD. The capacitor will only have a maximum charge o 0.5V, and the above simulation result shows that the capacitor will reach half of its maximum voltage (250mv) in about 75.194ns. Using the equation td = 0.7RC, where R is the total output resistance of 10k, the calculated delay is 70ns. The simulated time delay and the calculated time delay are very, very close.
 
How to determine the output resistance of the DAC (answer: R) by combining resistors in parallel and series
 
The output resistance is determined by simply starting at the bottom of the resistor network and combining the parallel resistors of 2R. This lead to a simplified result of a resistance R, which adds with the other R in series adding up to another 2R. This process continues until the B9 resistors. Eventually, it all simplifies to an output resistance of R.
 
 

Discuss what happens if the resistance of the switches isn't small compared to R
 
If the resistances of the switches isn't low, then the resistance of the switches may or may not cause an undesired voltage drop across the switches.
 
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