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Lab 2: Design of 10-bit digital to analog converter (DAC) - EE 421L Author: Mario Verduzco Email: Verdum1@unlv.nevada.edu 09/13/17
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Pre-lab:
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Lab description:
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Experimental Results:
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· Back-up all of your work form the lab and the course. · Read through this entire lab write-up before doing the pre-lab · Download and upload the Ideal ADC and DAC from CMOSedu · Simulate and describe the operation of both the ADC and DAC |
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The objective of this lab is to implement a 10-bit DAC using n-well resistors using a design topology from the CMOS book. |
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Exercise #1 - Design the 10-bit DAC using 10k Resistors and determine the output resistance
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DAC topology from the CMOS book |
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10-bit DAC using 10k resistors |
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· In a real circuit the switches seen above (the outputs of the ADC) are implemented with transistors (MOSFETs). Discuss what happens if the resistance of the switches isn't small compared to R. |
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10-bit Ideal ADC to DAC |
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The Analog to digital converter works by taking a smooth analog signal as an input and converting it to digital steps that are each related to a bit. The Digital to analog converter works in the opposite direction by taking a digital signal as an input and outputting an analog signal. This is done by using a series of voltage dividers that are in parallel with each bit being an input. |
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10-bit Ideal ADC to DAC Simulation showing the least significant bit |
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10-bit Ideal ADC to DAC Simulation |
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10-bit Ideal DAC Least significant bit |
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Exercise #2 - Delay through the 10-but DAC
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10-bit DAC driving a 10pF load |
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10-bit DAC driving a 10pF load Simulation |
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Exercise #3 - Simulation of ideal ADC to designed DAC driving a variety of loads
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10-bit ADC to DAC driving no load |
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10-bit ADC to DAC driving no load simulation |
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10-bit ADC to DAC driving a 10pF load |
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10-bit ADC to DAC driving a 10k load simulation |
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10-bit ADC to DAC driving a 10k load |
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10-bit ADC to DAC driving a 10pF load simulation |
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10-bit ADC to DAC driving a 10pF and 10k load |
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10-bit ADC to DAC driving a 10pF amd 10k load simulation |
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Since the load on the DAC is a 10k load and that is equal to the output resistance, the Vout will be subjected to a voltage divider cutting the voltage in half. |
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In a real circuit if the resistance of the switches are not small compared to R then the parallel combination of the Voltage dividers will not be accurate therefore not displaying the right voltage respective to the bit inputs. |