Chapter 1 Basic Principles of Digital Systems outlin e 1



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Solution

a. Va__(0/256) Vref _ 0 V

b. Va__(128/256) Vref__(1/2) Vref__5 V

c. Va__(64/256) Vref__(1/4) Vref__2.5 V

d. Va__(192/256) Vref__(3/4) Vref__7.5 V

In general, a DAC input code consisting of 1 followed by all 0s generates an output

value of 1⁄2 full scale. A code of 01 followed by all 0s yields an output of 1⁄4 full scale. An

output of 11 followed by all 0s generates an output of 3⁄4 full scale.

❘❙❚

❘❙❚ SECTION 12.2C REVIEW PROBLEM



12.4 Calculate Va for an 8-bit R-2R ladder DAC when the input code is 10100001. Assume

that Vref is 10 V.

MC1408 Integrated Circuit D/A Converter

Multiplying DAC A DAC whose output changes linearly with a change in DAC

reference voltage.

K E Y T E R M

12.2 • Digital-to-Analog Conversion 579

A common and inexpensive DAC is the MC1408 8-bit multiplying digital-to-analog converter.

This device also goes by the designation DAC0808. A logic symbol for this DAC is

shown in Figure 12.11.



FIGURE 12.11

MC1408 DAC

The output current, Io, flows into pin 4. Io is a binary fraction of the current flowing

into pin 14, as specified by the states of the digital inputs. Other inputs select the range of

output voltage and allow for phase compensation.

Figure 12.12 shows the MC1408 in a simple D/A configuration. R14 and R15 are approximately

equal. Pin 14 is approximately at ground potential. This implies:

1. That the DAC reference current can be calculated using only Vref (_) and R14 (Iref _



Vref (_)/R14)

2. That R15 is not strictly necessary in the circuit. (It is used primarily to stabilize the circuit

against temperature drift.)

The reference voltage must be set up so that current flows into pin 14 and out of pin 15.

Thus, Vref (_) must be positive with respect to Vref (_). (It is permissible to ground pin 14

if pin 15 is at a negative voltage.)



Io is given by:

Since the output is proportional to Vref (_), we refer to the MC1408 as a multiplying



DAC.

Io should not exceed 2 mA. We calculate the output voltage by Ohm’s law: Vo _

_Io RL. The output voltage is negative because current flows from ground into pin 4.

The open pin on the Range input allows the output voltage dropped across RL to

range from _0.4 V to _5.0 V without damaging the output circuit of the DAC. If the

Range input is grounded, the output can range from _0.4 to _0.55 V. The lower voltage

range allows the output to switch about four times faster than it can in the higher

range.

I

b b b b b b b b V

oR







7 6 5 4 3 2 1 0 

2 4 8 16 32 64 128 256 14

ref ( )

580 C H A P T E R 1 2 • Interfacing Analog and Digital Circuits

❘❙❚ EXAMPLE 12.5 The DAC circuit in Figure 12.12 has the following component values: R14 _ R15 _

5.6 k_; RL _ 3.3 k_. Vref (_) is _8 V, and Vref (_) is grounded.

Calculate the value of Vo for each of the following input codes: b7b6b5b4b3b2b1b0 _

00000000, 00000001, 10000000, 10100000, 11111111.

What is the resolution of this DAC?



Solution First, calculate the value of Iref.

Iref _ Vref (_)/R14

_ _8 V/5.6 k_ _ 1.43 mA

Calculate the output current by using the binary fraction for each code. Multiply _Io

by RL to get the output voltage.



b7b6b5b4b3b2b1b0 _ 00000000

Io _ 0, Vo _ 0

b7b6b5b4b3b2b1b0 _ 00000001

Io _ (1/256) (1.43 mA) _ 5.58 _A

Vo__(5.58 _A)(3.3 k_) __18.4 mV

b7b6b5b4b3b2b1b0 _ 10000000

Io _ (1/2) (1.43 mA) _ 714 _A

Vo__(714 _A)(3.3 k_) __2.36 V

FIGURE 12.12

MC1408 Configured for

Unbuffered Analog Output

12.2 • Digital-to-Analog Conversion 581

b7b6b5b4b3b2b1b0 _ 10100000

Io__(1/2 _ 1/8)(1.43 mA) _ (5/8)(1.43 mA) _ 893 _A

Vo__(893 _A)(3.3 k_) __2.95 V

b7b6b5b4b3b2b1b0 _ 11111111

Io _ (255/256) (1.43 mA) _ 1.42 mA

Vo__(1.42 mA)(3.3 k_) __4.70 V

Resolution is the same as the output resulting from the LSB: 18.4 mV/step

❘❙❚

❘❙❚ SECTION 12.2D REVIEW PROBLEM



12.5 The output voltage range of an MC1408 DAC can be limited by grounding the Range

pin. Why would we choose to do this?



Op Amp Buffering of MC1408

The MC1408 DAC will not drive much of a load on its own, particularly when the Range

input is grounded. We can use an operational amplifier to increase the output voltage and

current. This allows us to select the lower voltage range for faster switching while retaining

the ability to drive a reasonable load. The output voltage is limited only by the op amp

supply voltages. We use a 34071 high slew rate op amp for fast switching.

Figure 12.13 shows such a circuit. The 0.1-_F capacitor decouples the _5-V supply.

(The manufacturer actually recommends that the _5-V logic supply not be used as a reference

voltage. It doesn’t matter for a demonstration circuit, but may introduce noise that is

unacceptable in a commercial design.) The 75-pF capacitor is for phase compensation.

Vref(_)

Range


Ground

Comp


b3

b6

b7



(5)

R15


b5

b4

b2



b1

b0

VCC VEE



_12 V

_12 V


_12 V

75 pF


_5 V

Vref(_)


0.1 _F

R14B


5 k_

RFB


10 k_

RFA


4.7 k_

R14A


2.7 k_

1 k_


Vref _ _ 5 V

I0

I0



I0

Va

MC1408



MSB

LSB


_

_

_



_

(6)


(8)

(7)


(9)

(10)


(11)

(12)


(15)

(2)


(1)

(14) (4)


(13)

(16)


(3)

Va

_



_

FIGURE 12.13

DAC With Op Amp Buffering



582 C H A P T E R 1 2 • Interfacing Analog and Digital Circuits

Va is positive because the voltage drop across RF is positive with respect to the virtual

ground at the op amp (_) input. This feedback voltage is in parallel with (i.e., the same as)

the output voltage, since both are measured from output to ground.

We can develop the formula for the analog voltage, Va, in three stages:

1. Calculate the reference current:

Iref _ Vref(_)/R14

2. Determine the binary-weighted fraction of reference current to get DAC output current:

3. Use Ohm’s law to calculate the op amp output voltage:

The resistor values in the above formulae are the total resistances for the corresponding

part of the circuit. That is, R14 _ R14A _ R14B and RF _ RFA _ RFB. These both consist

of a fixed and a variable resistor, which has two advantages: (a) The reference current

and output voltage can be independently adjusted within a specified range by the variable

resistors. (b) The resistances defining the reference and feedback currents cannot go below

a specified minimum value, determined by the fixed resistance, ensuring that excessive

current does not flow into the reference input or the DAC output terminal.



Va can, in theory, be any positive value less than the op amp positive supply (_12 V in

this case). Any attempt to exceed this voltage makes the op amp saturate. The actual maximum

value, if not the same as the op amp’s saturation voltage, depends on the values of RF

and R14.

❘❙❚ EXAMPLE 12.6 Describe a step-by-step procedure that calibrates the DAC circuit in Figure 12.13 so that it

has a reference current of 1 mA and a full-scale analog output voltage of 10 V, using only

a series of measurements of the analog output voltage. When the procedure is complete,

what are the resistance values in the circuit? What is the range of the DAC?



Solution Since the maximum output of the DAC is 1 LSB less than full scale, we must

indirectly measure the full scale value. We can do so by setting the digital input code to

10000000, which exactly represents the half-scale value of output current, and making appropriate

adjustments.

Set the variable feedback resistor to zero so that the output voltage is due only to the

fixed feedback resistor and the feedback current. Measure the output voltage of the circuit

and adjust R14B so that Va _ 2.35 volts. Ohm’s law tells us that this sets the feedback current

to IF _ 2.35 V/4.7 k_ _ 0.5 mA. Since the digital code is set for half scale, Iref _ 2 IF

_ 1 mA.

Adjust RFB so that the half-scale output voltage is 5.00 V.



After adjustment, R14 _ 2.7 k_ _ 2.3 k_ _ 5 k_ and RF _ 4.7 k_ _ 4.3 k_ _ 10

k_. In both cases the variable resistors were selected so that their final values are about

half-way through their respective ranges.

The range of the DAC is 0 V to 9.961 V.

(FS _ 1 LSB _ 10 V _ (10 V/256) _ 9.961 V)

V IR

R

R

o F V

F

a ref


digital code

256














14

I

b b b b b b b b

I

b b b b b b b b V

R

V

R

o



























7 6 5 4 3 2 1 0

7 6 5 4 3 2 1 0

14

2 4 8 16 32 64 128 256



2 4 8 16 32 64 128 256

ref


ref

14

digital code ref



256

12.2 • Digital-to-Analog Conversion 583

❘❙❚ EXAMPLE 12.7 Figure 12.14 shows the circuit of an analog ramp (sawtooth) generator built from an

MC1408 DAC, an op amp, and an 8-bit synchronous counter. (A ramp generator has numerous

analog applications, such as sweep generation in an oscilloscope and frequency

sweep in a spectrum analyzer.)

CTR DIV 256

Q3

Q6

Q7



Q5

Q4

Q2



Q1

Q0

CLK



Vref(_)

Range


Ground

b3

b6



b7

b5

b4



b2

b1

b0



VCC VEE

_5 V _12 V

Vref(_)

0.1 _F


5 k_

10 k_ 4.7 k_

2.7 k_

Vref _ _ 5 V



I0

Va

MC1408



_

_

_



_

75 pF


FIGURE 12.14

Example 16.5

DAC Ramp Generator

Briefly explain the operation of the circuit and sketch the output waveform. Calculate

the step size between analog outputs resulting from adjacent codes. Assume that the DAC

is set for _6-V output when the input code is 10000000.

Calculate the output sawtooth frequency when the clock is running at 1 MHz.

Solution The 8-bit counter cycles from 00000000 to 11111111 and repeats continuously.

This is a total of 256 states.

The DAC output is 0 V for an input code of 00000000 and (12 V _ 1 LSB) for a

code of 11111111. We know this because a code of 10000000 always gives an output

voltage of half the full-scale value (6 V _ 12 V/2), and the maximum code gives an

output that is one step less than the full-scale voltage. The step size is 12 V/256 steps _

46.9 mV/step. The DAC output advances linearly from 0 to (12 V _ 1 LSB) in 256

clock cycles.

Figure 12.15 shows the analog output plotted against the number of input clock cycles.

The ramp looks smooth at the scale shown. A section enlarged 32 times shows the analog

steps resulting from eight clock pulses.

One complete cycle of the sawtooth waveform requires 256 clock pulses. Thus, if



fCLK _ 1 MHz, fo _ 1 MHz/256 _ 3.9 kHz.

(Note that if we do not use a high slew rate op amp, the sawtooth waveform will not

have vertical sides.)

584 C H A P T E R 1 2 • Interfacing Analog and Digital Circuits

❘❙❚


Bipolar Operation of MC1408

Many analog signals are bipolar, that is, they have both positive and negative values. We

can configure the MC1408 to produce a bipolar output voltage. Such a circuit is shown in

Figure 12.16.

We can model the bipolar DAC as shown in Figure 12.16b. The amplitude of the

constant-current sink, Io, is set by Vref (_), R14, and the binary value of the digital inputs.



Is is determined by Ohm’s law: Is _ Vref (_)/R4.

The output voltage is set by the value of IF:



Va _ IF RF _ IF (RFA _ RFB)

By Kirchhoff’s current law:



Is _ IF _ Io _ 0

or

IF _ Io _ Is

Thus, output voltage is given by:

Va _ (Io _ Is)RF _ Io RF _ Is RF

_ __


b

2

7_



_ _

b

4

6_



_ _

b

8

5_



_ _

1

b

6

4_

_ _



3

b

2

3_



_ _

6

b

4

2_

_ _



1

b

2

1



8

_ _ _


2

b

5

0



6 _

_ _


R

R

1

F

4

_ Vref _ _



R

R

F

4

_ Vref



_ __

digit


2

a

5



l

6

code



____

R

R

1

F

4

__ Vref _ _



R

R

F

4

_ Vref



How do we understand the circuit operation from this mathematical analysis?

FIGURE 12.15

Example 12.7

Sawtooth Waveform Output of Circuit in Figure 12.14

12.2 • Digital-to-Analog Conversion 585

The current sink, Io, is a variable element. The voltage source, Vref (_), remains constant.

To satisfy Kirchhoff’s current law, the feedback current, IF, must vary to the same degree

as Io. Depending on the value of Io with respect to Is, IF can be positive or negative.

We can get some intuitive understanding of the circuit operation by examining several

cases of the equation Va.



FIGURE 12.16

MC1408 as a Bipolar D/A Converter



586 C H A P T E R 1 2 • Interfacing Analog and Digital Circuits

Case 1: Io _ 0. This corresponds to the digital input b7b6b5b4b3b2b1b0 _ 00000000. The

output voltage is:

This is the maximum negative output voltage.

Case 2: 0 _ Io _ Is. The term (Io _ Is) is negative, so output voltage is also a negative

value.


Case 3: Io _ Is. The output is given by:

Va _ (Io _ Is)RF _ 0

The digital code for this case could be any value, depending on the setting of R14. To set the

zero-crossing to half-scale, set the digital input to 10000000 and adjust R14 for 0 V.

Case 4: Io Is. Since the term (Io _ Is) is positive, output voltage is positive. The largest

value of Io (and thus the maximum positive output voltage) corresponds to the input code



b7b6b5b4b3b2b1b0 _ 11111111.

The magnitude of the maximum positive output voltage of this particular circuit is 2

LSB less than the magnitude of the maximum negative voltage. Specifically, Va _

(127/128)(RF/R4)(Vref) if R4 _ 2R14.

To summarize:

Input Code Output Voltage

00000000 Maximum negative*

10000000 0 V**

11111111 Maximum positive

*As adjusted by RFB

**As adjusted by R14B



Negative Range:

00000000 to 01111111 (128 codes)



Positive Range:

10000001 to 11111111 (127 codes)



Zero:

00000000 (001 code)

256 codes

❘❙❚ EXAMPLE 12.8 Calculate the values to which R14 and RF must be set to make the output of the bipolar

DAC in Figure 12.16 range from _12 V to (_12 V _ 2 LSB). Describe the procedure you

would use to set the circuit output as specified.

Confirm that the calculated resistor settings generate the correct values of maximum

and minimum output.



Solution Set R14 so that the DAC circuit has an output of 0 V when input code is

b7b6b5b4b3b2b1b0 _ 10000000. We can calculate the value of R14 as follows:

R

R

V

R

R

F FV

2

0



14 4

ref−ref



V I I R I R

R

R

a o s F s F V

( −) −−F ref

4

12.2 • Digital-to-Analog Conversion 587

The first term is set by the value of the input code. Solving for R14, we get:

To set the maximum negative value, set the input code to 00000000 and adjust RFB for

_12 V. RFB _ RF _ RFA. Solve the following equation for RF:

_ _

R

R

F

4

_ Vref _ _12 V



__

10

R

k

F

_

(5 V) _ _12 V



RF _ (12 V)(10 k)/5 V _ 24 k

RFB _ 24 k _ 18 k _ 6 k

Settings

R14 _ R4/2 _ 5 k_ for zero-crossing at half-scale.

RF _ 24 k_ for output of _12 V.

Check Output Range

For b7b6b5b4b3b2b1b0 _ 00000000:



Va _ __

R

0

14



_ _ _

R

1

4



__ RF Vref _ __

(24


1

k

0


k

)
(5 V)

__ _12 V

For b7b6b5b4b3b2b1b0 _ 11111111:



Va _ __

25

2



6

5

R

5

14

__ _



R

1

4



__ RF Vref

_ __


(256

2

)(



5

5

5



k)

_ __


10

1

k



__ (24 k)(5 V) _ 11.906 V

(Note: 12 V _ 2 LSB _ 12 V _ (12 V/128) _ 12 V _ 94 mV _ 11.906 V.)

❘❙❚ SECTION 12.2E REVIEW PROBLEM

12.6 Why is the actual maximum value of an 8-bit DAC less than its reference (i.e., its apparent

maximum) voltage?

DAC Performance Specifications

A number of factors affect the performance of a digital-to-analog converter. The major factors

are briefly described below.

1

2

1



0

1

2



1

0

1



2

1

2



2 10

14 4


14 4

14 4


14 4

14 4


R R

R V

R R

R R

R R

R R

−F







−






k /2 = 5 k

ref


/ 

588 C H A P T E R 1 2 • Interfacing Analog and Digital Circuits

Monotonicity. The output of a DAC is monotonic if the magnitude of the output voltage

increases every time the input code increases. Figure 12.17 shows the output of a DAC that

increases monotonically and the output of a DAC that does not.

We show the output response of a DAC as a series of data points joined by a straightline

approximation. One input code produces one voltage, so there is no value that corresponds

to anything in between codes, but the straight-line approximation allows us to see a

trend over the whole range of input codes.

Absolute accuracy. This is a measure of DAC output voltage with respect to its expected

value.



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