Lab
X - ECE 421L
Authored
by Worku, Yetneberk
Email: workuy@unlv.nevada.edu
NHSE ID 1007208058
09/08/2014
Pre-lab Two
first, Back-up all of your work from the lab and the course.
second, Download lab2.zip to my desktop.
third, Upload this zip file to the design directory on the server that i am running Cadence., and Unzip this directory and add, to my cds.lib in the design directory.
finally,Start Cadence from the design directory, and Open the schematic view.
After, selecting lab2, sim_ldeal_ADC_DAC, and Schematic. the following schematic will apear.
now,
it is ready to make run simulation. After simulation the following grap
should be apear: if it important we can chage the back groung color.
post lab
The
objective of this lab is to design to a 10-bit DAC using an n-well R of
10k, determine the output resistance of the DAC, and calculating the
delay. Our design is based upon the topology seen in Fig. 30.14, which
showing blelow.
next,
create a symbol view for design, which exact the same footprint as the
Ideal_10-bit_DAC symbol, so to do that need to make copy of the
schematics already provided in Lab2 before drafting design. the
following fig showing that copy.
next,
creating a voltage divider for the circuit in figure 30.14 in R2Rbit
newly file, then place the pin on branch of schematic as seen below.
next, we must save check, may have error fix it, does not have error make a symbol, the symbol will apear as follow.
next, make sure save all documents, then copy replace all the 1bit DAC just created.
next, we check save, and then replace the ideal DAC with the new one. the new one looks like as follows
next, save and check then running with the transient simulation for 1us, the result will show below.
next, to get output resistacnce, using parametric sweep, by setting resistor as RL. the sircuit schematic showing below.
next, the ADL window choosing Analyses, and the resistance assigned tor RL showing below.
next,
Vin=Vout/2, this happen when Rout is 10k. this means the output drop by
half of its original output, the DC response showing below.
next,
we can expected that the time taken to reach half the value is 0.7RC
(0.7*10k*10p=0.7us). this delay happen when the output signal by
replaceing the 10k resistor with 10pF, the fig showing below.
next, check and save, then run simulation, the transient response showing below
next, by attaching RL the output magnitude should be decreased by half; that showing schemtic ans simulations showing below.
next,
by attaching capacitor the signal will delay, and signal have phase
shift. also, the capacitor are often used to smooth out singnal, that
showing schematic and simulation below.
Indeed,
combining capacitor and load resitance will effects decrease the
amplitude of the output signal and the phawe delay; showing below
simulation and schematic.
to back up the file dragging lab2 from mobxterm to the desktop and e-mail to myself.
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