Ad & Da Converters

SEE 3263: ELECTRONIC SYSTEMS

Chapter 6: Chapter 6: A/D And D/A Converters A/D And D/A Converters

11

SEE 3263 A/D & D/A CONVERTERS

INTRODUCTIONINTRODUCTIONIn real world, most signal processing involves analog quantity

INTRODUCTION INTRODUCTION

quantity.Analog quantity can take on any value over a continuous range of values and most important its exact value is significantsignificant. A digital quantity will have a value that is specified as one of two possibilities such as 0 or 1, LOW or HIGH, TRUE or FALSE and so onTRUE or FALSE and so on.Actual value is not important but must falls within the specified ranges. For example:

0 V to 0.8 V ⇒ logic 02 V to 5 V ⇒ logic 1

2


INTRODUCTIONINTRODUCTIONDigital logic circuits require special interfacing techniques to input and output analog data

INTRODUCTION INTRODUCTION

techniques to input and output analog data.Physical quantities with an infinite range of values, such as temperature, pressure, fluid flow, velocity, p p yacceleration and voltage are analog quantities.Analog-to-digital (A/D) conversion is the process of converting analog values to digital codes representingconverting analog values to digital codes representing the analog value.Digital-to-analog (D/A) conversion is the process of g g ( ) pconverting digital codes to proportional analog values.Digital audio, digital sampling and music synthesis equipment are some exciting examples of A/D and D/Aequipment are some exciting examples of A/D and D/A applications.

3


The diagram below shows the elements used in the digital technique to monitor and control the analog physical variable.

Transducer : A device used to convert the physical variable to an l t i l i bl F l th i t h t ll delectrical variable. For example a thermistor, photocell and

tachometer.Analog to digital converter : To convert an analog input to equivalent

digital output.g pDigital System : The digital information is process according to a program

instructions.Digital to analog converter : To convert a digital information to a

proportional analog quantity (voltage orproportional analog quantity (voltage orcurrent).

Actuator : A device that control the physical variable. 4


DigitalDigital--ToTo--Analog ConversionAnalog ConversionDigitalDigital--ToTo--Analog ConversionAnalog ConversionIs the process of taking a value represented in digital code (such as straight binary or BCD) and converting it to a voltage or current which is proportional to the digital value.

Fro the diagram, there are 4 digital inputs means that it is a 4-bit DAC. D3 is the MSB and D0 is the LSB. it is a 4 bit DAC. D3 is the MSB and D0 is the LSB. Analog output voltage VO is proportional to the input value. The digital input D3 to D0 will produce 24 = 16 of 4 bit binary number of 4-bit binary number.

5


Graph plot of VGraph plot of VOO(analog) versus (analog) versus OO( g)( g)VVinin(digital) for 4 bit DAC (digital) for 4 bit DAC

age.

VO

12

13

14

15

og O

utpu

t Vol

ta

8

9

10

11

Ana

l

5

6

7

8

1

2

3

4

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

Digital Input

6


In generalIn general,

Analog output = R x digital inputAnalog output = R x digital input

where R is the resolution.Analog output can be voltage or current Analog output can be voltage or current. Therefore R can either be in unit volt or ampere.

If R = 0.25 V,Then VO = (0.25V) x digital input

For digital input of 10002 = 810VO = 0.25 V x 8 = 2.0 VO

7

SEE 3263 A/D & D/A CONVERTERSA 4-bit DAC produce an output current. For a di it l i t f 1010 th t t t idigital input of 10102, the output current is 5mA. What is the value of IO for a digital input 01012 ?

A5I mA 0.5 10mA5

Input DigitalIR O ===

For a digital input of 01012 = 510

IO = R x digital input= 0.5mA x 5 = 2 5mA= 2.5mA

8


What is the maximum output voltage for the 8 bit DAC th t d 1V t t f di it l8-bit DAC that produce 1V output for digital input of 001100102?

5000110010

V0.02 50V 1

I tDi it lVR

5000110010

O

102

===

=

50InputDigital

For digital input of 111111112 = 255100

VO = R x digital input= 0.02 V x 255 = 5.1 V

9


RESOLUTION (Step Size)RESOLUTION (Step Size)RESOLUTION (Step Size) RESOLUTION (Step Size)

The Resolution of a DAC is defined as the smallestThe Resolution of a DAC is defined as the smallest change that can occur in the analog output as a result of a change in the digital input.

The resolution is always equal to the weight of the LSB and is also referred to as the step size since it is the amount that output will change as the digital input value is changed from one step to nextinput value is changed from one step to next.

10


For the given 4For the given 4--bit DAC, each digital input depends on bit DAC, each digital input depends on its weight. Therefore the its weight. Therefore the resolution = LSB = 0.5 Vresolution = LSB = 0.5 V

DD CC BB AA VVOO(V)(V)

00 00 00 11 0.50.5

00 00 11 00 1100 00 11 00 11

00 11 00 00 2.02.0

11 00 00 00 4 04 011 00 00 00 4.04.0

Note that there are 16 l l i l t t 16

Generally for N-bit DAC, levels equivalent to 16 input state but there are 15 steps between level 0 and

No of levels = 2N

No of steps = 2N -1pthe full scale.

11


PERCENTAGE RESOLUTIONPERCENTAGE RESOLUTIONPERCENTAGE RESOLUTIONPERCENTAGE RESOLUTION

R l ti l b d fi d thResolution can also be defined as the percentage of the full-scale(F.S) output.

% 100 x (F.S) scale full

size step resolution % =

• Or it can also be calculated from:

% 100x steps ofnumber total

1resolution % =

12


An 8-bit DAC has a step size of 10 mV. Determine the full scale output voltage and the percentage resolution.

No of steps = 28 – 1 = 256 –1 = 255

Full-scale voltage = 10 mV x 255 = 2.55 V

% 0.39 100% x V 2.55

mV 10resolution % ==

This shows that the percentage resolution becomes smaller as the number of input bits is increased.

13


WHAT DOES RESOLUTION MEAN?WHAT DOES RESOLUTION MEAN?WHAT DOES RESOLUTION MEAN?WHAT DOES RESOLUTION MEAN?A DAC cannot produce a continuous range of outputvalues, and so its output is not truly analog., p y gA DAC produces a finite set of output values.The DAC’s resolution (number of bits) determines howmany possible voltage values.many possible voltage values.If a 6-bit DAC is used, there will be 63 possible steps of0.159V between 0 and 10V.When an 8-bit DAC is used there will be 255 possibleWhen an 8 bit DAC is used, there will be 255 possiblesteps of 0.039 V between 0 and 10V.The greater the number of bits, the finer the resolution(the smaller the step size)(the smaller the step size).The resolution limits how close the DAC output cancome to a given analog value.Generally the cost of DACs increases with the numberGenerally, the cost of DACs increases with the numberof bits, and so the designer will use as few bits asnecessary. 14


BCD INPUT CODEBCD INPUT CODEBCD INPUT CODEBCD INPUT CODEThe DACs we have considered thus far have used a binary input codebinary input code.Many DACs use a BCD input code where 4-bit code groups are used for each decimal digit.g p g

DAC10204080D1

C1B1A1

BCD for MSD

Vout 100 possible values since input ranges from 00 to 99

1248D0

C0B0A0

BCD for LSD

ranges from 00 to 99

Step size = weight of A0

15


If weight of AO is 0.2 V, determine the following:(a) Step size.(b) Full-Scale output and percentage resolution.(c) Vout for D1C1B1A1 = 01012 and D0C0B0A0 =

00112.

(a) Step size = weight of A = 0 2 V(a) Step size = weight of AO = 0.2 V(b) There are 99 steps from 00 to 99.

FS = 99 x 0.2 = 19.8 V thus % resolution = % 1 100% x V19.8

.2V0=

(c) 01012 = 510 and 00112 = 310, then Vout = step size x digital input

= 0.2 V x 53 = 10.6 V

16


DIGITALDIGITAL TOTO ANALOG CONVERTERANALOG CONVERTERDIGITALDIGITAL--TOTO-- ANALOG CONVERTERANALOG CONVERTER

There are 2 types of typical DAC converter circuit:

DAC binary weightedDAC binary weighted

DAC R – 2R ladder networkDAC R 2R ladder network

17


Binary Weighted DACBinary Weighted DACBinary Weighted DAC Binary Weighted DAC This is the basic circuit for one t f 4 bit DAC Th i t type of 4-bit DAC. The inputs are binary input which are assumed to have values of either 0V or 5V.

The op-amp is employed as a summing amplifier, which produces the weighted sum of

i VVVVthese input voltages.

The output is evaluated for any input condition by setting the

RV I ,

2RV I ,

4RV I ,

8RV I D

3C

2B

1A

O ====

R 2R 4R 8R f t th i ht d f 23 p y g

appropriate inputs to either 0V or 5V. For example, if the digital input is 10102, then VD = VB = 5V

⎞⎛

−= FFO RIV

R, 2R, 4R, 8R refer to the weighted of 23, 22, 21 and 20. Thus

and VC = VA = 0V. Thus VOUT = -(V+0+1/4V+0) = 6.25V ⎟

⎠⎞

⎜⎝⎛ +++−= D

FC

FB

FA

F VRR V

2RR V

4RR V

8RR

18


How close the circuit comes to producing an accurate values depends primarily on two factors:

The precision of the input and feedback resistors.

The precision of the input voltage levels.

19


PRECISION REFERENCEPRECISION REFERENCEPRECISION REFERENCE PRECISION REFERENCE SUPPLYSUPPLY

VRef

R 2R 4R 8RRF

IO IO/2 IO/4 IO/8

IF= IOUse semiconductor switch like the CMOS transmission

t

-IO VOSwitch closed

h i t bit 1

gate

+when input bit = 1

B3 B2 B1 B0

MSB LSB

IBIBIBIBI O0

O1

O2O3O ×+×+×+×=

RVWhere I

8B

4B

2BIBI

REFO

012O3O

=

+++

20


RR--2R LADDER NETWORK2R LADDER NETWORKRR--2R LADDER NETWORK2R LADDER NETWORK

4-bit R-2R is 4 bit R 2R is constructed with 3 resistors R and 5 resistors 2R resistors 2R. Normally R = 10 kΩand 2R = 20 kΩ.

4 current switches will be activated d d th depends on the digital input.

1VI)input digitalD(Bcurrent LSIout ×=

input)D(digital )21)(

RV(

16I

4REFO ×==

21


DAC OUTPUT VOLTAGEDAC OUTPUT VOLTAGEDAC OUTPUT VOLTAGEDAC OUTPUT VOLTAGEIn general, for n-bit,

21

RVI n

REF)LSB(O ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

2R ⎠⎝⎠⎝

VO = -IoutRF = -(current LSB) x RF x D

Voltage Resolution = voltage LSB = FnREF R R

V⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

21

Therefore VO = -(voltage resolution) x D22


Assume VREF = 10 V for 4-bit R-2R withR = 10kΩ. Determine:(a) Current resolution.(b) IO when the digital input is 11112

(a) n = 4(a) n = 4

mA 0.062521

k10V10

21

RVIresolution 4n

REF)LSB(O =⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

Ω=⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛==

(b) IO = 1O(LSB) x D = (0.0625 mA) (15) = 0.9375 mA

2k102R)( ⎠⎝⎠⎝ Ω⎠⎝⎠⎝

23


DAC INTEGRATED CIRCUITDAC INTEGRATED CIRCUIT(DAC 0808/ MC1408)(DAC 0808/ MC1408)

24


DAC 0808/ MC 1408 is an 8-bit DACDAC 0808/ MC 1408 is an 8 bit DAC.Pin 13 and 3 are the power supply terminal +ve and –ve respectively.Pin 14 and 15 allow the +ve and ve reference voltagesPin 14 and 15 allow the +ve and –ve reference voltages.

outI

25


It provides two output current terminals that can be used to increase the capability of the DAC-08.If th i t l it h i t iti ‘1’ th l dd tIf the internal switch is at position ‘1’, the ladder current will flow through bus Iout and if the switch is at position ‘0’, the ladder current will flow through bus

⎞⎛⎞⎛OUTI

Resolution (current )= LSB =Iout = (LSB) x D

21

RV

nREF ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

OUT

IFS = (LSB) x (2n–1) = (LSB) x 255Total branch current in DAC 08 = IFS

OUTFS II −=outI

26


UNIPOLAR ANALOG OUTPUT UNIPOLAR ANALOG OUTPUT VOLTAGEVOLTAGE

RF0.1 F0.1 F0.1 F

-15 V +15 V

5 K

-

+Vref

Rref

Iref

IOUT

outI

133

2

416

14

15

DAC -082

3 6

+15 V

5 K

MSB LSB

Vo= IoutRF

+Vref

D1D2D3 D0D4D5D6D7

521121110976 8

15

-15 V+10 V

5 K

MSB LSB

R1VsolutionVoltage re REF ×⎞⎜⎛⎞

⎜⎛

D sulotion voltage reV

R2R

solutionVoltage re

O

FnREF

×=

×⎠

⎜⎝⎠

⎜⎝

=

27


RF0.1 F0.1 F0.1 F

-15 V +15 V

-Rref

Iref

IOUT

I

1334

1614

DAC -082

6

+15 V

5 K

5 K

Vo= IoutRF

+Vref

D1D2D3 D0D4D5D6D7

outI

521121110976 8

15 3 6

-15 V+10 V

5 K

MSB LSB

For unipolar DAC-08, determine VO for the Ofollowing inputs:(a) 000000012 (b) 111111112

mV 39kΩ521

kΩ5V10R

21

RVV 8Fn

REFLSB =×⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=×⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

(a) VO = VLSB x D = 39 mV x 1 = 39 mV(b) VO = VLSB x D = 39 mV x 255 = 9.961 V 28


BIPOLAR ANALOG OUTPUT BIPOLAR ANALOG OUTPUT VOLTAGEVOLTAGE

outI FoutO R)I(V outI−=

FoutO R)I(V outI−= )II( outFS −=outIFoutO )( out )( outFSout

29


For bipolar DAC-08, determine VO forFor bipolar DAC 08, determine VO for an input of 011111112

A 821

kΩ5V24.10

21

RVsolutionCurrent re 8n

REF µ=⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

2kΩ52R ⎠⎝⎠⎝⎠⎝⎠⎝

IFS = 8µA x 255 = 2.04mAIout = 8µA x 12710 = 1.016mA , mA016.1mA04.2Iout −=Iout 8µA x 12710 1.016mA ,

V = (1 016mA 1 024mA)x5kΩ = 0 04 V

mA024.1mA016.1mA04.2Iout

=

VO = (1.016mA - 1.024mA)x5kΩ = - 0.04 V

30

SEE 3263 A/D & D/A CONVERTERSoutI

For bipolar DAC-08 determine Vo for the f ll i i tfollowing inputs: (a) 000000002 (b) 011111112

(c) 10000000 (d) 11111111

Current resolution = = = 8µA 21

RV

nREF ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

6521

5K10.24

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

(c) 100000002 (d) 111111112

µIFS = 8µA X 255 = 2.04mAIout = 8µA x 0 = 0, V = (0 - 2 04mA)5KΩ = -10 2 V

2R ⎠⎝⎠⎝ 6525K ⎠⎝⎠⎝

Vo = (0 - 2.04mA)5KΩ = -10.2 V

Note that the –ve full-scale voltage happen when the input is 0 and the +ve full scale voltage happen when all inputs are 1the +ve full-scale voltage happen when all inputs are 1.

Digital Input Analog OutputD7 D6 D5 D4 D3 D2 D1 D0 Iout(mA) (mA) Vo(V)

-ve full-scale 0 0 0 0 0 0 0 0 0 2.04 -10.2outI

31

Negative zeroPositive zero+ve full-scale

0 1 1 1 1 1 1 11 0 0 0 0 0 0 01 1 1 1 1 1 1 1

1.016 1.024 2.04

1.0241.016

0

-0.0400.04010.2


An 8-bit DAC has a full scale output of p2 mA and a full scale error ± 0.5 %. What are the possible output range for

i t f 10000000an input of 100000002.

Step size= 2 mA/255 = 7.84 µAInput 100000002 = 12810Ideal output current = ILSB x D = 7.84 µA x 128 = 1004 µAMaximum error = ± 0.5 % x 2 mA = ± 10 µAµThus an ideal output current range = 1004 µA ± 10 µA = 994 µA to 1014 µA

32


ANALOGANALOG--TOTO--DIGITAL DIGITAL CONVERTER (ADC)CONVERTER (ADC)

A ADC t k l i t lt d ftAn ADC takes an analog input voltage and after a certain amount of time produces a digital output code which represents the analog input.

33


Digital Ramp ADCDigital Ramp ADCDigital Ramp ADC Digital Ramp ADC

34


For a digital ramp ADC, if fclk = 1 MHz, VT = 0.1 mV, full-scale output = 10.23 V and a 10-, pbit input, determine:

The digital equivalent obtained for VA =The digital equivalent obtained for VA 3.728 V.The conversion timeThe conversion time.The resolution of this converter.

35


Total possible steps = 210 – 1 = 1023102310Step size =

Si V 3 728 V d V 0 1 V

mV101023

23.10=

Since VA = 3.728 V and VT = 0.1 mVVAX must reach 3.7281 VThis needs steps373372 81V7281.3This needs

37310 = 01011101012

steps 373372.81mV 10

==

10 2

Require 373 steps to complete the conversion, so need 373 clock pulses = 373 µs = tcresolution = step size = 10 mV

36


SUCCESSIVE APPROXIMATIONSUCCESSIVE APPROXIMATIONSUCCESSIVE APPROXIMATION SUCCESSIVE APPROXIMATION ADC (SAC)ADC (SAC)

37


STARTAssume a 4-bit Clear all

bits

Assume a 4 bit SAC with a step size of 1V.L t th

Start at MSB

Let assume the analog input, VA=10.4V

Set bit = 1

Clear bit back to 0

ISVAX > VA ?

Yes

Have all bits been checked?

Go to next lowest bit

No

No

Conversion is complete and result is in REGISTER

Yes

END38


An 8-bit SAC has a resolution of 20mV. What will its digital output be for an analog input of 2.17 V.

No of Steps = 5.108 V20V17.2

=

Step 108 would produce VAX = 2.16 VStep 109 would produce VAX = 2 18 V

mV20

Step 109 would produce VAX 2.18 VThe SAC always produces a final VAX that is at the step below VATh f V 2 17 V th di it l lt ld b Thus, for VA = 2.17 V, the digital result would be 10810 = 011011002

39


CONVERSION TIMECONVERSION TIMECONVERSION TIMECONVERSION TIMETC for SAC = N x 1 clock cycle.This conversion time will be the same regardless of theThis conversion time will be the same regardless of the value of VA because the control logic has to process each bit to see whether a 1 is needed or not.

Compare the maximum conversion times of a 10-bit digital-ramp ADC and a 10-bit SAC if a 10-bit digital-ramp ADC and a 10-bit SAC if both utilizes a 500 kHz clock frequency.

For digital-ramp ADC, tC =1023 x 2µs = 2046µs.

For SAC, tC = 10 x 2µs = 20µs.40


THE ADC0804 INTEGRATEDTHE ADC0804 INTEGRATEDTHE ADC0804 INTEGRATED THE ADC0804 INTEGRATED CIRCUIT CIRCUIT

It is an 8-bit ADC that performs A/D conversion using the successive-approximation method.ppIt has two analog inputs: VIN(+) and VIN(-) to allow differential inputs.The actual analog input V = V (+) V ( )The actual analog input, VIN = VIN(+) - VIN(-).In single-ended measurements, the analog input is applied to VIN(+) while VIN(-) is connected to analog

dground.During normal operation, the converter uses VCC = + 5V as its reference voltage, and the analog input can range g g gfrom 0 to 5V full scale.

41


With 8 bit th l ti i V19 6V5With 8-bits, the resolution is =

It has an internal clock generator circuit that produces a

mV19.6 255

=

It has an internal clock generator circuit that produces a frequency of

RC1.11f =

where R and C are values of externally connected components.A t i l l k f i 606 kH i R 10 kΩ

C

A typical clock frequency is 606 kHz using R = 10 kΩand C = 150 pF. If desired, an external clock frequency can be used by connecting it to the CLK IN pin.y g pWith 606kHz clock frequency, TC = 13.2µs.It has separate ground connections for digital and analog

lt t i 10 d i 8 ti lvoltages at pin 10 and pin 8 respectively.

42


TYPICAL CONNECTION OF TYPICAL CONNECTION OF ADC0804 ADC0804

43


APPLICATION EXAMPLEAPPLICATION EXAMPLEAPPLICATION EXAMPLEAPPLICATION EXAMPLE+ 5V

LED 0

+ 5V

VCCVIN(+)

VIN(-)

A.GNDVin

1K

1K

1K

D0

D1LED 12.5 K

LED2A.GND

Vref/2

CLK R

Vin 1K

1K

1K

1K

1K

ADC 0804

D2

D3

D4

D5

+-VZ(2.5V)

10 K

RP

LED2

LED 3

LED 4

LED 5

CLK in

CS

RD

10 K

150 pF

1KADC 08041K

+ 5V

D6

D7

WR 10 K

LED6

LED7

RD

D.GND INTR

WR 10 K

3.3 F

START

74HCT14

44


Referring to the fig re abo e R is the 10 kΩReferring to the figure above, RP is the 10 kΩpotentiometer. If RP is set so that V+ = 1.28 V, determine:determine:

The input voltage range Vin

The voltage resolutionThe voltage resolutionThe conversion timeThe LED that will light up when V = 2 26 VThe LED that will light up when Vin = 2.26 VThe input voltage when the digital output is 101011110101112

45


V562VV281VV

fref =⇒== + V56.2VV28.1V2 ref⇒

Input voltage range Vin = 0 V hingga 2.56 V

mV04.10255

V56.2 resolution voltage ==

T = 1.1 RC = 1.1 x 10kΩ x 150pF = 1.65 µsTherefore tc = N x T = 8 x 1.65 µs = 13.2 µsWhen Vin = 2.26 V

1.225mV04.10

26.2 steps Total ==

W e Vin . 6 V

Thus total steps = 22510 = 111000012=D7D6D5D4D3D2D1D0LED that will light up: LED4, LED3, LED2, LED1V 87 10 04 V 0 8735V lth h t l V

46

Vin = 8710 x 10.04mV = 0.8735V although actual Vinshould be slightly greater than 0.8735V.


THE FLASH ADCTHE FLASH ADCTHE FLASH ADCTHE FLASH ADCFlash ADC is theFlash ADC is the highest-speed ADC, but it requires much more circuitry

Analog in Comparator Outputs Digital Outputs

VA C1 C2 C3 C4 C5 C6 C7 A B C

0 – 1 V 1 1 1 1 1 1 1 0 0 0circuitry.For example, a 6-bit flash ADC requires 63 analog comparators

1 – 2 V

2 – 3 V

3 – 4 V

0 1 1 1 1 1 1

0 0 1 1 1 1 1

0 0 0 1 1 1 1

0 0 1

0 1 0

0 1 1analog comparators, while an 8-bit unit requires 255 comparators and a 10

4 – 5 V

5 – 6 V

6 – 7 V

0 0 0 0 1 1 1

0 0 0 0 0 1 1

0 0 0 0 0 0 1

1 0 0

1 0 1

1 1 0

comparators, and a 10-bit converter requires 1023 comparators.

>7 V 0 0 0 0 0 0 0 1 1 1

47


T H E T H E EE N DN D

48

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