A/D And D/A Converters -.:: GEOCITIES.ws ::.

10/28/2008

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SEE 3263: ELECTRONIC SYSTEMS

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

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SEE 3263 A/D & D/A CONVERTERS

In real world, most signal processing involves analog quantity.A l tit t k l ti

INTRODUCTION INTRODUCTION

Analog quantity can take on any value over a continuous range of values and most important its exact value is significant. 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 on.Actual value is not important but must falls within the specified ranges For example:specified ranges. For example:

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

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

INTRODUCTION INTRODUCTION

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

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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 electrical variable. For example a thermistor, photocell andtachometer.

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

Digital System : The digital information is process according to a programinstructions.

Digital to analog converter : To convert a digital information to a proportional analog quantity (voltage orcurrent).

Actuator : A device that control the physical variable. 4

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DigitalDigital--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 g g pit 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.

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

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Ana

log

Out

put V

olta

ge. V

O

6

7

8

9

10

11

12

13

14

1

2

3

4

5

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

Digital Input

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In general,

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

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

If R = 0.25 V,Then V = (0 25V) x digital inputThen VO = (0.25V) x digital input

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

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SEE 3263 A/D & D/A CONVERTERSA 4-bit DAC produce an output current. For a digital input of 10102, the output current is 5mA. What is the value of IO for a digital input 01012 ?

mA 0.5 10mA 5

Input DigitalIR O ===

For a digital input of 01012 = 510g p 2 10

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

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What is the maximum output voltage for the 8-bit DAC that produce 1V output for digital input of 001100102?

V 0.02 50V 1

Input DigitalVR

50 00110010

O

102

===

=

di i l i f 2For digital input of 111111112 = 25510

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

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RESOLUTION (Step Size) RESOLUTION (Step Size)

The Resolution of a DAC is defined as the smallest change that can occur in the analog output as achange 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 digitalis the amount that output will change as the digital input value is changed from one step to next.

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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 11

00 11 00 00 2.02.0

11 00 00 00 4.04.0

Note that there are 16 Generally for N-bit DAC, Note that there are 16 levels equivalent to 16 input state but there are 15 steps between level 0 and the full scale.

Generally for N bit DAC,

No of levels = 2N

No of steps = 2N -1

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PERCENTAGE RESOLUTIONPERCENTAGE RESOLUTION

Resolution 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 % =

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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 V2 55

mV 10resolution % ==V2.55

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

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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.A DAC produces a finite set of output values.A DAC produces a finite set of output values.The DAC’s resolution (number of bits) determines howmany 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 possiblesteps of 0.039 V between 0 and 10V.The greater the number of bits the finer the resolutionThe greater the number of bits, the finer the resolution(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 numberof bits, and so the designer will use as few bits asnecessary. 14

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

204080D1

C1B1

BCD for MSD

DAC10

1248

A1

D0C0B0A0

BCD for LSD

Vout 100 possible values since input ranges from 00 to 99

Step size = weight of A0

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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 AO = 0.2 V(b) There are 99 steps from 00 to 99.

FS = 99 x 0 2 = 19 8 V thus % resolution = %1100%x.2V0=FS = 99 x 0.2 = 19.8 V thus % resolution =

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

= 0.2 V x 53 = 10.6 V

% 1 100%x V19.8

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DIGITALDIGITAL--TOTO-- ANALOG CONVERTERANALOG CONVERTER

There are 2 types of typical DAC converterThere are 2 types of typical DAC converter circuit:

DAC binary weighted

DAC R – 2R ladder network

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Binary Weighted DAC Binary Weighted DAC This is the basic circuit for one type of 4-bit DAC. The inputs are bi i t hi h d t 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 these input voltages.

The output is evaluated for any RV I ,

2RV I ,

4RV I ,

8RV I D

3C

2B

1A

O ====p y

input condition by setting the appropriate inputs to either 0V or 5V. For example, if the digital input is 10102, then VD = VB = 5V and VC = VA = 0V. Thus VOUT = -(V+0+1/4V+0) = 6.25V ⎟

⎠⎞

⎜⎝⎛ +++−=

−=

DF

CF

BF

AF

FFO

VRR V

2RR V

4RR V

8RRRIV

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

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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 levelsThe precision of the input voltage levels.

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PRECISION REFERENCE PRECISION REFERENCE SUPPLYSUPPLY

VRef

IO IO/2 IO/4 IO/8Use semiconductor

-

+

R 2R 4R 8RRF

IF= IO

IO VOSwitch closed when input bit = 1

switch like the CMOS transmission gate

B3 B2 B1 B0

MSB LSB

RVWhere I

8IB

4IB

2IBIBI

REFO

O0

O1

O2O3O

=

×+×+×+×=

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RR--2R LADDER NETWORK2R LADDER NETWORK

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

4 current switches ill b ti t d will be activated

depends on the digital input.

input)D(digital )21)(

RV(

16I

)input digitalD(Bcurrent LSI

4REFO

out

×==

×=

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

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RVI n

REF)LSB(O ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

V I R ( t LSB) R DVO = -IoutRF = -(current LSB) x RF x D

Voltage Resolution = voltage LSB =

Therefore VO = -(voltage resolution) x D

FnREF R R

V⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

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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

A0 06251V101VIl i REF ⎞⎜⎛⎞

⎜⎛⎞

⎜⎛⎞

⎜⎛

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

mA 0.062521

k10V10

21

RVIresolution 4n

REF)LSB(O =

⎠⎞

⎜⎝⎛⎠⎞

⎜⎝⎛

Ω=

⎠⎞

⎜⎝⎛⎠⎞

⎜⎝⎛==

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

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DAC 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 voltages.

outI

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It provides two output current terminals that can be used to increase the capability of the DAC-08.If the internal switch is at position ‘1’, the ladder current

ill fl h h b I d if h i h i i i ‘0’will flow through bus Iout and if the switch is at position ‘0’, the ladder current will flow through busResolution (current )= LSB =Iout = (LSB) x DIFS = (LSB) x (2n–1) = (LSB) x 255Total branch current in DAC 08 = IFS

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RV

nREF ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

OUTI

Total branch current in DAC 08 IFS

OUTFS II −=outI

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R0 1 F

-15 V +15 V

UNIPOLAR ANALOG OUTPUT UNIPOLAR ANALOG OUTPUT VOLTAGEVOLTAGE

-

+Vref

Rref

Iref

IOUT

outI

RF

5

133

21121110976 8

4

0.1 F0.1 F0.1 F

1614

15

DAC -082

3 6

-15 V

+15 V

+10 V

5 K

5 K

5 K

MSB LSB

Vo= IoutRF

D1D2D3 D0D4D5D6D7

D sulotion voltage reV

R21

RVsolutionVoltage re

O

FnREF

×=

×⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

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-

+Vref

Rref

Iref

IOUT

outI

RF

5

133

21121110976 8

4

0.1 F0.1 F0.1 F

1614

15

DAC -082

3 6

-15 V +15 V

15 V

+15 V

+10 V

5 K

5 K

MSB LSB

Vo= IoutRF

For unipolar DAC-08, determine VO for the following inputs:(a) 00000001 (b) 11111111

D1D2D3 D0D4D5D6D7

-15 V5 K

(a) 000000012 (b) 111111112

mV 39kΩ521

kΩ5V10R

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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

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BIPOLAR ANALOG OUTPUT BIPOLAR ANALOG OUTPUT VOLTAGEVOLTAGE

outI FoutO R)I(V outI−=

FoutO R)I(V outI−= )II( outFS −=outI

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

A 821

kΩ5V24.10

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RVsolutionCurrent re 8n

REF µ=⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

IFS = 8µA x 255 = 2.04mAFS µIout = 8µA x 12710 = 1.016mA ,

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

mA024.1mA016.1mA04.2Iout

=−=

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SEE 3263 A/D & D/A CONVERTERSoutI

For bipolar DAC-08 determine Vo for the following inputs: (a) 000000002 (b) 011111112

(c) 100000002 (d) 111111112

Current resolution = = = 8µAIFS = 8µA X 255 = 2.04mAIout = 8µA x 0 = 0, Vo = (0 - 2.04mA)5KΩ = -10.2 V

N t th t th f ll l lt h h th i t i 0 d

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RV

nREF ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

6521

5K10.24

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

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

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

-ve full-scaleNegative zeroPositive zero+ve full-scale

0 0 0 0 0 0 0 00 1 1 1 1 1 1 11 0 0 0 0 0 0 01 1 1 1 1 1 1 1

01.016 1.024 2.04

2.041.0241.016

0

-10.2-0.0400.04010.2

outI


An 8-bit DAC has a full scale output of 2 mA and a full scale error ± 0.5 %. What are the possible output range for an input of 100000002.2

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

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

An ADC takes an analog input voltage and after a t i t f ti d di it l t t dcertain amount of time produces a digital output code

which represents the analog input.

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Digital Ramp ADC Digital Ramp ADC

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

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

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Total possible steps = 210 – 1 = 102310

Step size = mV101023

23.10=p

Since VA = 3.728 V and VT = 0.1 mVVAX must reach 3.7281 VThis needs

1023

steps 373 372.81 mV10

V 7281.3==

37310 = 01011101012

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

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

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START

Clear all bits

Start at

Assume a 4-bit SAC with a step size of 1V.Let assume the

l i t MSB

Set bit = 1

Clear bit back to 0

ISVAX > VA ?

Have

Yes

No

No

analog input, VA=10.4V

all bits beenchecked?

Go to nextlowest bit

Conversion is complete and result is in REGISTER

Yes

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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 =

Step 108 would produce VAX = 2.16 V

5.108 mV20

V17.2=

Step 108 would produce VAX 2.16 VStep 109 would produce VAX = 2.18 VThe SAC always produces a final VAX that is at the step below VAThus, for VA = 2.17 V, the digital result would be 10810 = 011011002

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CONVERSION TIMECONVERSION TIMETC for SAC = N x 1 clock cycle.This conversion time will be the same regardless of the value of V because the control logic has to processvalue 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 both utilizes a 500 kHz clock frequency.

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

For SAC, tC = 10 x 2µs = 20µs.40

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THE ADC0804 INTEGRATED THE ADC0804 INTEGRATED CIRCUIT CIRCUIT

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

li d t V ( ) hil V ( ) i t d t lapplied to VIN(+) while VIN(-) is connected to analog ground.During normal operation, the converter uses VCC = + 5V as its reference voltage, and the analog input can range from 0 to 5V full scale.

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With 8-bits, the resolution is =

It has an internal clock generator circuit that produces a frequency of 1

mV 19.6 255

V 5=

frequency of

where R and C are values of externally connected components.A typical clock frequency is 606 kHz using R = 10 kΩand C = 150 pF. If desired, an external clock frequency

RC1.11f =

can be used by connecting it to the CLK IN pin.With 606kHz clock frequency, TC = 13.2µs.It has separate ground connections for digital and analog voltages at pin 10 and pin 8 respectively.

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TYPICAL CONNECTION OF TYPICAL CONNECTION OF ADC0804 ADC0804

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APPLICATION EXAMPLEAPPLICATION EXAMPLE

+ 5V

+ 5V

LED 0VCCVIN(+)

VIN(-)

A.GND

Vref/2

CLK R

CLK in

Vin

10 K

150 pF

1K

1K

1K

1K

1K

1K

1K

ADC 08041K

5V

D0

D1

D2

D3

D4

D5

D6

D7

LED 1

+-

2.5 K

VZ(2.5V)10 K

RP

LED2

LED 3

LED 4

LED 5

LED6

LED7

CS

RD

D.GND

150 pF + 5V

INTR

WR 10 K

3.3 F

START

74HCT14

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Referring to the figure above, RP is the 10 kΩpotentiometer. If RP is set so that V+ = 1.28 V, determine:

The input voltage range Vin

The voltage resolutionThe conversion timeThe LED that will light up when Vin = 2.26 VThe input voltage when the digital output is 10101112

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V56.2VV28.1V2

Vref

ref =⇒== +

mV0410V56.2resolutionvoltage

Input voltage range Vin = 0 V hingga 2.56 V

mV04.10255

resolution voltage ==

1.22526.2stepsTotal ==

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

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1.225mV04.10

steps Total

Thus total steps = 22510 = 111000012=D7D6D5D4D3D2D1D0LED that will light up: LED4, LED3, LED2, LED1Vin = 8710 x 10.04mV = 0.8735V although actual Vinshould be slightly greater than 0.8735V.

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THE FLASH ADCTHE FLASH ADCFlash ADC is the highest-speed ADC, butA l i C t O t t Di it l O t thighest speed ADC, but it requires much more circuitry.For example, a 6-bit flash ADC requires 63 analog comparators, while an 8-bit unit requires 255

Analog in Comparator Outputs Digital Outputs

VA C1 C2 C3 C4 C5 C6 C7 A B C

0 – 1 V

1 – 2 V

2 – 3 V

3 – 4 V

4 – 5 V

5 – 6 V

1 1 1 1 1 1 1

0 1 1 1 1 1 1

0 0 1 1 1 1 1

0 0 0 1 1 1 1

0 0 0 0 1 1 1

0 0 0 0 0 1 1

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1requires 255 comparators, and a 10-bit converter requires 1023 comparators.

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6 – 7 V

>7 V

0 0 0 0 0 0 1

0 0 0 0 0 0 0

1 1 0

1 1 1


T H E T H E EE N DN D

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A/D And D/A Converters -.:: GEOCITIES.ws ::.

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