ADC
Module-4
INTRODUCTION
ANALOG TO DIGITAL CONVERSION
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1V
1000 Discrete Levels
1 mV
A number is assigned to 1V------- 1000
Dividing an analog voltage into discrete levels is termed a QUANTIZATION and the error involved is known as QUANTIZATION ERROR.
When 1V is divided into 1000 discrete levels, it is said to have a RESOLUTION of 1 in 1000 or 1mV
ANALOG TO DIGITAL CONVERSION
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1V
100 Discrete Levels
10 mV
A number is assigned to 1V------- 100
Dividing an analog voltage into discrete levels is termed a QUANTIZATION and the error involved is known as QUANTIZATION ERROR.
When 1V is divided into 100 discrete levels, it is said to have a RESOLUTION of 1 in 100 or 10mV
An ADC quantizes and converts it into binary code (consisting of zeroes and ones)
ADC
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Input Voltage
Digital Code | Analog Voltage | ||
| | | |
0 | 0 | 0 | 0V |
0 | 0 | 1 | 1V |
0 | 1 | 0 | 2V |
0 | 1 | 1 | 3V |
1 | 0 | 0 | 4V |
1 | 0 | 1 | 5V |
1 | 1 | 0 | 6V |
1 | 1 | 1 | 7V |
There is no binary code for 8V level
Next higher code after 111 is 000 or 1000
A 5- bit ADC has a higher precision than a 3-bit ADC
A 10-bit ADC has a higher precision than a 5-bit ADC
LSB and MSB
Digital Code | Analog Voltage | ||||
| | | | | |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 31.25mV |
0 | 0 | 0 | 1 | 0 | 62.5mV |
0 | 0 | 0 | 1 | 1 | 93.75mV |
0 | 0 | 1 | 0 | 0 | 125mV |
0 | 0 | 1 | 0 | 1 | 156.25mV |
0 | 0 | 1 | 1 | 0 | 187.5mV |
0 | 0 | 1 | 1 | 1 | 218.75mV |
0 | 1 | 0 | 0 | 0 | 250mV |
0 | 1 | 0 | 0 | 1 | 281.25mV |
0 | 1 | 0 | 1 | 0 | 312.5mV |
0 | 1 | 0 | 1 | 1 | 343.75mV |
A 500mV level is to be converted into a 7-bit digital code. Determine the resolution of the conversion, the analog levels represented by the LSB and the MSB, and calculate the analog level represented by 1111111
DIGITAL TO ANALOG CONVERSION
DAC
Analog Output
SETTLING TIME
MONITONICITY
ACCURACY
WEIGHTED RESISTOR DAC
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| | | | Analog Voltage |
0 | 0 | 1 | 0.25mA | 0.25mA*4kΩ= 1V |
0 | 1 | 0 | 0.5mA | 2V |
0 | 1 | 1 | 0.25mA+0.5mA=0.75mA | 3V |
1 | 0 | 0 | 1mA | 4V |
0
0
5V
5kΩ
10kΩ
20kΩ
0.25mA
4kΩ
0.25mA
WEIGHTED RESISTOR DAC
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| | | | Analog Voltage |
0 | 0 | 1 | 0.25mA | 0.25mA*4kΩ= 1V |
0 | 1 | 0 | 0.5mA | 2V |
0 | 1 | 1 | 0.25mA+0.5mA=0.75mA | 3V |
1 | 0 | 0 | 1mA | 4V |
0
5V
0
5kΩ
10kΩ
20kΩ
0.5mA
4kΩ
0.5mA
WEIGHTED RESISTOR DAC
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| | | | Analog Voltage |
0 | 0 | 1 | 0.25mA | 0.25mA*4kΩ= 1V |
0 | 1 | 0 | 0.5mA | 2V |
0 | 1 | 1 | 0.25mA+0.5mA=0.75mA | 3V |
1 | 0 | 0 | 1mA | 4V |
0
5V
5V
5kΩ
10kΩ
20kΩ
0.25mA
0.5mA
4kΩ
0.75mA
WEIGHTED RESISTOR DAC
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| | | | Analog Voltage |
1 | 0 | 0 | 1mA | 5V |
1 | 1 | 0 | 1.5mA | 6V |
1 | 1 | 1 | 1.75mA | 7V |
5V
0
5kΩ
10kΩ
20kΩ
1mA
0
1mA
WEIGHTED RESISTOR DAC
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| | | | Analog Voltage |
1 | 0 | 0 | 1mA | 5V |
1 | 1 | 0 | 1.5mA | 6V |
1 | 1 | 1 | 1.75mA | 7V |
5V
5V
5kΩ
10kΩ
20kΩ
0.25mA
0.5mA
1mA
5V
1.75mA
Advantages of Weighted Resistor DAC
Disadvantages of DAC
R-2R DAC
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R-2R DAC
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R-2R DAC
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R-2R DAC
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R-2R DAC
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10k Ω
R-2R DAC
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5 kΩ
R-2R DAC
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10 kΩ
R-2R DAC
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R-2R DAC
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R-2R DAC
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001----🡪 0.25mA*5kΩ =1.25V
=0.25mA
0.25mA
R-2R DAC
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011----🡪 0.75mA*5kΩ =3.75V
=0.75mA
0.5mA
0.25mA
R-2R DAC
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111----🡪 1.75mA*5kΩ =8.75V
=1.75mA
1mA
0.5mA
0.25mA
If the R-2R DAC circuit is expanded into a 5-bit DAC, calculate the analog output voltage when the inputs are (a) 00001�(b) 10000 (c) 11111
Integrated Circuit 8-bit D/A
Linear Ramp ADC
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Ramp Generator
Clock Generator
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Register
H
0
0
1
1
1
Reset
Linear Ramp ADC
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Ramp Generator
Clock Generator
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Register
L
0
0
1
1
1
No Clock Pulses
0
0
0
Reset=0
Comparator
output
Clock pulses
A linear ramp ADC uses a 10-bit counter and a 15kHz clock frequency. The counter output is to be 1111111111 when the input voltage is 100mV. Calculate the required ramp rate of change and the ADC conversion
SUCCESSIVE APPROXIMATION ADC
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DAC
Clock Generator
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Register
H
0
0
H
1
0
0
0V
Reset
Control
H
0
0
1
8V
SUCCESSIVE APPROXIMATION ADC
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DAC
Clock Generator
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Register
L
0
H
0
0
10V
Reset
Control
H
0
1
8V
0
SUCCESSIVE APPROXIMATION ADC
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DAC
Clock Generator
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Register
H
0
H
0
0
0V
Reset
Control
H
0
1
8V
0
1
SUCCESSIVE APPROXIMATION ADC
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DAC
Clock Generator
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Register
L
0
H
0
0
8V
Reset
Control
H
0
1
8V
0
1