Transcript A/D and D/A - University of Detroit Mercy

A/D and D/A Conversion
Shuvra Das
University of Detroit Mercy
Flowchart of Mechatronic
Systems
Objective
• The objective of this section is to describe
how the D/A or the A/D converter blocks
function.
• And highlight important issues an user
should be aware of.
Analog to Digital Conversion
A/D conversion
• Sensors usually receive analog data, e.g.
sensor signals from thermocouples or strain
gages are voltages.
• These signals are converted to digital form
by A/D converters.
• Modern A/D converters are available as
single IC chips (integrated circuit).
Types of A/D converters
• The tracking ADCs (Analog to Digital
converters)
• The integrating ADCs
• The flash ADCs
• successive approximation ADCs
Quantization
• The process of A/D converting an analog voltage (or
current) to digital form requires that the analog signal be
quantized and encoded into binary form.
• Quantization consists of subdividing the range of the signal
into a finite number of intervals.
• Each quantized level is then assigned a binary word.
• Usually if “n” bits are available in the binary word one
employs (2n -1) intervals. (e.g. if n = 4, no. of intervals =
15, if n = 8, no. of intervals = 255, etc).
Quantization:Example
vd
0
1
2
…..
14
15
B3
0
0
0
B2
0
0
0
B1
0
0
1
B0
0
1
0
1
1
1
1
1
1
0
1
• Let va represent analog
voltage and vd its
quantized counterpart
• Let the analog voltage
range be 0-16 V
• Then for all va lying
between 0 and 1 V, vd = 0.
• For all va between 0and 1
vd=0, 1 and 2, vd = 1, etc,
for 15-16 vd= 15.
Quantization:Example
vd
0
1
2
…..
14
15
B3
0
0
0
B2
0
0
0
B1
0
0
1
B0
0
1
0
1
1
1
1
1
1
0
1
vd
5
4
3
2
1
0
1 2 3 4 5 6 7
va
Quantization
• As the table shows each quantized voltage
level may be represented by a unique binary
counterpart.
• There is always some error associated with
quantization. As the number of bits
increases the error decreases.
Quantization Error
• The difference between actual analog
voltage and the level to which the
quantization assigns the value.
• Example: If 5 bits are used to represent a 50
V range, what is the quantization error for
representing 10 V?
Quantization Error
• Example: If 5 bits are
used to represent a 50
V range, what is the
quantization error for
representing 10 V?
• Repeat for 12 V , 31 V,
etc for practice.
• Q = 50V/(25-1)= 1.6129V
per level.
• kQ <= 10 V
• k needs to be an integer
number of levels
• k <= 10/1.6129=6.2
• k=6
• quantized value for 10 =
6*1.6129 = 9.6774
• error = 10-9.6774 =
0.3226V
Quantization Error
• Example: If 8 bits are
used to represent a 50
V range, what is the
quantization error for
representing 10 V?
• Repeat for 12 V , 31 V,
etc for practice.
• Q = 50V/(28-1)= .19607 V
per level.
• kQ <= 10 V
• k needs to be an integer
number of levels
• k <= 10/0.19607= 51.002
• k = 51
• quantized value for 10 =
51*0.19607 = 9.99957
• error = 10-9.99957 =
0.00043V
Quantization Error
• Quantization error may be reduced by using
higher number of bits in the A/D converter
• Typical A/D converters have 8, 12 or 16
bits.
• In the scheme described earlier Maximum
quantization error thus is the length of the
quantized level. (Max error: voltage
range/(2n-1))
Saturation Error
• A/D converters have specific upper and
lower voltage ranges. Typical values used
are 0-10V or -10 to 10 V.
• If the input signal is higher than the upper
limit or lower than the lower limit the
converter saturates. This can be prevented
by appropriate signal conditioning.
Conversion Time
• The time required for A/D converter to provide the
digital equivalent of the analog output.
• What happens if analog value changes (time
varying signal) before conversion is complete?
• Sample and hold-maintains constant input to A/D
while conversion is taking place.
Conversion Time: sampling rate
Conversion time: Sampling Rate
• Sampling rate must be at least at twice the
frequency of the maximum frequency of interest
in the analog signal. This critical sampling rate is
called the Nyquist frequency (2fmax). If sampling
is not done properly aliasing will occur.
• Aliasing: Form of signal distortion as a result of
improper sampling.
• In practice about 5-10 times frequency of the
signal is used.
Sampling Rate: Example
• Datasheet for ADC
574 (a particular ADC
IC) says that the
maximum conversion
time is 35micro-sec.
What is the highest
signal frequency that
can be converted with
this ADC?
• Highest conversion
frequency for this ADC :
fmax = 1/(35 * 10-6) =
28.57kHz
• Therefore the maximum
sampling frequency can be
28.57khz.
• Therefore the maximum
frequency of the signal
should not be more than 1/2
of this, i.e. 14 kHz.
• In reality it should be even
less.
Digital to Analog Conversion
D/A conversion
• Takes a binary word and converts it into an
analog signal.
• Binary word is represented by 1s and 0s
where typically 0 corresponds to 0 V and 1
corresponds to 5 V.
Example
• Consider 4-bit binary
word representing a
positive number
• Binary number =
(b3b2b1b0)2 =
(b3*23+b2*22+b1*21
+b0*20)10
• The analog voltage
corresponding to the
binary word B is
• va = (b3*8+b2*4+b1*2
+b0*1)dv
• where dv is the
smallest step size by
which va can
increment.
D/A conversion
• The step size is determined by the number
of bits in the digital word to be converted.
• Extending the previous example:
• vamax = (2n-1) dv ==> vamax/(2n-1) = dv
• This is the smallest change in voltage when
the least significant bit changes from 0 to 1.
D/A conversion process
• A summing amplifier is
used. Each bit is
represented by a 5 V
source and a switch.
• If bit is 0 switch is off and
if bit is 1 switch is on.
• The output va is
proportional to the binary
word.
• Va = -(S(RF/Ri)biVin)
• Ri=R0/2i
D/A conversion process
• Va = -(S(RF/Ri)biVin)
• Ri=R0/2i
• Va = -RF/R0(2 n-1 bn-1
+2n- 2 bn-2 +..+ 20b0)Vin
Example
• Consider a 4-bit DAC with 015V range (I.e. va max = 15)
• R’s chosen are 10kohm, 5, 2.5 and
1.25 kohms, RF = 2kohm.,
Vin=5V
 dv= 15/(24-1)=1V
• 0101 ==> -(0*2k/1.25k+
1*2k/2.5k+0*2k/5k+1*2k/10k)*5
= -5V
• 1001==> -(1*2k/1.25k+
0*2k/2.5k+0*2k/5k+1*2k/10k)*5
= -9V
Example
• Determine the smallest • Resolution is dv,
step size (or
smallest non-zero
resolution) of an 8 bit
voltage value.
DAC for a range of 0-  dv = Voltage
12 V.
range/(2n-1) = 12/(281) = 12/255= 47.1milli
Volt
Example
• Find the minimum
number of bits
required in a DAC if
the range is 10 V and a
resolution of 10mV is
required.
 dv = (voltage
range)/(2n-1)
• 10(10-3) = 10/ (2n-1)
• (2n-1) = 10/0.01
• n (log2) = log(1000+1)
• n = log(1001)/log2 = 9.97
• minimum bits req. = 10