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.
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After, selecting lab2, sim_ldeal_ADC_DAC, and Schematic. the following schematic will apear.
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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.
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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.
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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.
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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.
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next, we must save check, may have error fix it, does not have error make a symbol, the symbol will apear as follow.
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next, make sure save all documents, then copy replace all the 1bit DAC just created.
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next, we check save, and then replace the ideal DAC with the new one. the new one looks like as follows
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next, save and check then running with the transient simulation for 1us, the result will show below.
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next, to get output resistacnce, using parametric sweep, by setting resistor as RL. the sircuit schematic showing below.
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next, the ADL window choosing Analyses, and the resistance assigned tor RL showing below.
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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.
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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.
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next, check and save, then run simulation, the transient response showing below
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next, by attaching RL the output magnitude should be decreased by half; that showing schemtic ans simulations showing below.
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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.
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Indeed,
combining capacitor and load resitance will effects decrease the
amplitude of the output signal and the phawe delay; showing below
simulation and schematic.
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to back up the file dragging lab2 from mobxterm to the desktop and e-mail to myself.
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