Conversion Devices

‫ محمد محمد حيمة‬.‫د‬

m_heima@yahoo.com
 ‫بسم هللا الرحمن الرحيم‬
‫نحمده ونصلي على رسوله الكريم‬
 Data Acquisition System Overview
Conversion Devices
 In the last few years, industrial PC I/O interface products have become
increasingly reliable, accurate and affordable. PC-based data
acquisition and control systems are widely used in industrial and
laboratory applications like monitoring, control, data acquisition and
automated testing.
 Selecting and building a DA&C (Data Acquisition and Control)
system that actually does what you want it to do requires some
knowledge of electrical and computer engineering.
 Transducers and actuators
 Signal conditioning
 Data acquisition and control hardware
 Computer systems software
 Data Acquisition System Introduction I

 A data acquisition system consists of many

components that are integrated to:
 Sense physical variables (use of transducers)
 Condition the electrical signal to make it readable
by an A/D board
 Convert the signal into a digital format acceptable
by a computer
 Process, analyze, store, and display the acquired
data with the help of software
 Data Acquisition System Introduction II

 A data acquisition system II consists of many

components that are integrated to:
 Convert the output signal of computer into an analog
format this is done by the use of D/A board.
 Condition the electrical signal to make it readable by
the transducers
 Transfer the signal by the use of transducers into real
physical form.
Block Diagram of Data Acquisition System I
Block Diagram of Data Acquisition System II
 Transducers

Sense physical phenomena and translate it into electric signal.

 Temperature  Displacement
 Pressure  Level
 Light  Electric
signals
 Force  ON/OFF switch
 Transducers and Actuators
‫محوالت الطاقة والمشغالت الميكانيكية‬

 A transducer converts temperature, pressure, level,

length, position, etc. into voltage, current,
frequency, pulses or other signals.
 An actuator is a device that activates process
control equipment by using pneumatic, hydraulic
or electrical power. For example, a valve actuator
opens and closes a valve to control fluid rate.
 Signal Conditioning

 Signal conditioning circuits improve the quality of

signals generated by transducers before they are
converted into digital signals by the PC's data-
acquisition hardware.
 Examples of signal conditioning are signal scaling,
amplification, linearization, cold-junction
compensation, filtering, attenuation, excitation,
common-mode rejection, and so on.
 Signal Conditioning

 One of the most common signal

conditioning functions is amplification.
 For maximum resolution, the voltage range
of the input signals should be approximately
equal to the maximum input range of the
A/D converter. Amplification expands the
range of the transducer signals so that they
match the input range of the A/D converter.
For example, a x10 amplifier maps
transducer signals which range from 0 to 1 V
into the range 0 to 10 V before they go into
the A/D converter.
 Signal Conditioning

Electrical signals are conditioned so they

can be used by an analog input board. The
following features may be available:
Amplification
Isolation
Filtering
Linearization
 Data Acquisition

 Data acquisition and control hardware

generally performs one or more of the
following functions:
 analog input,
 analog output,
 digital input,
 digital output and
 counter/timer functions.

1st Conversion Devices

 1st Voltage to Current Conversion V/I
ICVS (Transconductance Amplifier)

Vin
 Neglecting VIO, the (-) +
and (+) terminals are at _
the same voltage, Vin. IL RL
Therefore, VR1 = Vin. I=0
 Since I = 0, I1 R1
 IL = I1 = Vin/R1
 Current independent
of RL
 Floating load
 1st Voltage to Current Conversion V/I
ICVS (Transconductance Amplifier)

R1 R2
Vin + Vo
_
 For grounded load R3 R4
 If R2/R1 = R4/R3
RL IL
V-IN
 IL  Ii 
R3
 Constant-Current Source
IL

Ri RL
 For the basic constant- _
current circuit, the op- Ii
amp has a very high VIN +
Zin, thus, IL = Ii.
 If RL changes, IL
remains constant as VIN
IL  Ii 
long as VIN and Ri are Ri
held constant.
 ICVS Example
IL
Load Given: Vin = 2 V, R1 = 2 k
I1 R1 Vo(max) =  10 V
_
+ Find: iL , gm and RL(max)
Vin +
_ Solution:
IL = I1 = Vin/R1 = 2 / 2000 = 1 mA
Note: gm = Io/Vin = 1/R1 = 1 / 2000
 If RL > RL(max) the op amp = 0.5 mS
will saturate
 The output current, iL is
RL(max) = Vo(max)/iL = 10 V / 1 mA
independent of the load = 10 k 
resistance.
 2nd Current to Voltage Conversion I/V
VCIS (Transresistance Amplifier)
 V0=I R

I should not be too large otherwise offset voltage will be too high.

IF
RF
General Equations: Iin
IF = Iin
_
A V0
Vo = -IFRF +
Rm = Vo/Iin = RF
VCIS (Transresistance Amplifier) Summary

 Transresistance Amplifiers are used for low-power applications

to produce an output voltage proportional to the input current.

 Photodiodes and Phototransistors, which are used in the

production of solar power are commonly modeled as current
sources.

 Current to Voltage Converters can be used to convert these

current sources to more commonly used voltage sources.
 Current to Voltage Converter Example
IF RF

Given: Iin = 10 mA
_
RF = 200 
+
Iin + VO Find: IF , Vo and Rm
-
Solution:
IF = Iin = 10 mA

Vo = -IFRF = 10 mA * 200 
=2V
Rm = Vo/Iin = RF = 200 
 Current to Voltage Converter I/V
Rf
Ii
 Since the inverting Vin If
_
terminal is at virtual  Vout
ground, +
 Vout = -If Rf = -Ii Rf
 As the amount of light
changes, the current
through the photocell Circuit for sensing light level
changes; thus and converting it to a
proportional output voltage
 Vout = |  Ii| Rf
 3rd Digital to Analog converter (D/A)

 The opposite of analog to digital conversion is digital to

analog (D/A) conversion. This operation converts digital
information into analog voltage or current. D/A devices
allow the computer to control real-world events.
 Analog output signals may directly control process
equipment. The process can give feedback in the form of
analog input signals. This is referred to as a closed loop
control system with PID control.
 Analog outputs can also be used to generate waveforms. In
this case, the device behaves as a function generator.
Analog Outputs (D/A)
 What is a DAC ?

A digital-to-analog converter (DAC) is a circuit

that produces an analog current or voltage that
is proportional to an analog reference (voltage or
current) and an N-bit binary word.

Vout = k x Vref x (Binary Word)

 In English

 DACs generate piecewise continuous signals

from digital code.
 OR
 DAC converters are devices that receive a
binary word from the microprocessor and
convert it to a scaled analog voltage (or current).
 DAC Configuration

N-Bit Binary Word

Analog Reference Digital to Analog Analog

Voltage (Vref) Converter (D/A) Output (Vout)
 DAC Types

Multiplying DAC
- reference source is external to the DAC
package
Non-multiplying DAC
- reference source is inside the DAC package
 DAC Circuit Types

Two types of DAC Circuits:

1. Binary weighted

2. R-2R ladder
 The binary weighted resistor network

 Comprises of a register and resistor network

 Output of each bit of the register will depend on whether a 1 or a
0 is stored in that position
 e.g. for a 0 then 0V output
 for a 1 then 5V output
 Resistance R is inversely proportional to binary weight of each
digit

R
MSB
2R RL VL
4-bit
registe 4R
r
8R
LSB
 Buffering the resistor network

 Best solution is to follow the resistor network with a

buffer amplifier
 Has high impedance, practically no current flows
 All input currents sum at S and go through Rf
 Vo = -IfRf
MSB R I1 Rf

2R I2 If
4-bit register

4R I3 -
S I

LSB 8R I4 +
Vo

Vo = -If  Rf = -( I1+ I2+ I3+ I4 )  Rf

N-Bit Binary Weighted DAC
 Binary Weighted Principles

N
bi
I 0  VR   i 1
i 12 R
I 0  sum of currents leaving junction
R  resistance corresponding to MSB
N  number of input bits
b1  MSB
 Principles Cont’d

V0 = -Rf I0
V0 = voltage output from amplifier
Rf = feedback resistance
Resolution= VR / 2N

Note: For a gain of 1, R = 2Rf

 Example

Find output voltage and current for a binary

weighted resistor DAC of 4 bits where :

R = 10 kΩ, Rf = 5 kΩ and VR = 10 Volts. Applied

binary word is 1001.
 Solution
Rf = (R/2)

I i

Vo
8R 4R 2R R

4-bit 3-bit 2-bit 1-bit

MSB

VR
 Solution Cont’d

10 V  1 0 0 1 
Io     
  2 *10 2 *10 2 *10 23 *104 
0 4 1 4 2 4

I 0  - 0.001125 A
V0  - R f I 0
V0  (5 )(0.001125 A)  5.625 V
3
 Solution Cont’d

Binary input = 1001 = 9

From example, V0 = 5.625V

V0/VR = 5.625V/10V = 9/16

 D/A Example 2

 Calculate the output voltage for an input

code word 0110 if a logic 1 is 10V and a
logic 0 is 0V, and R = RF=1k
 I1 = I 4 = 0
 I2 = 10 V / 2R = 10 / 2k = 5 mA
 I3 = 10 V / 4R = 10 / 4k = 0.25 mA
 Vo = -If x Rf = -(0.0075) x 1000 = -7.5 volts
Vo = -If  Rf = -( I1+ I2+ I3+ I4 )  Rf

 The weighted resistor network

 Seldom used when more than 6 bits in the code word

 to illustrate the problem consider the design of an 8-bit
DAC if the smallest resistor has resistance R
 what would be the value of the largest resistor?

 what would be the tolerance of the smallest resistor?

 Very difficult to manufacture very accurate resistors over

this range
Limitations of the Binary Weighted D/C

Has problems if bit length is longer than 8 bits

For example, if R = 10 k Ohms
R8 = 28-1(10 k Ohms) = 1280 kΩ
If VR = 10 Volts,
I8 = 10V/1280 k Ohms = 7.8 A

Op-amps to handle those currents are

expensive because this is usually below
the current noise threshold.
 Limitations Cont’d

If R = 10 Ohms and Vref = 10 V

I = VR/R = 10 V / 10 Ω = 1 A

This current is more than a typical op-amp can handle.

Intuitively, the resistance values must be accurate to less

than one part in 2N for the RN input to be meaningful. This
is difficult to do, especially in IC’s.
The R-2R Ladder Resistor Network

 Has a resistor network which requires resistance values

that differ 2:1 for any sized code word
 The principle of the network is based on Kirchhoff's
current rule
 The current entering N must leave by way of the two
resistors R1 and R2
I N R2

R1
The R-2R Ladder Resistor Network

 Works on a current dividing network

A B
I R I1

2R 2R 2R

I1 I2

 Resistance to right of B = 1/(1/2R + 1/2R)

 Resistance to right of A = R +2R/2 = 2R
 Current divides I1 = I/2 I2 = I/4 divides again
 The R-2R Ladder Resistor Network

 The network of resistors to the right of A have an

equivalent resistance of 2R, and so the right hand
resistance can be replaced by a copy of the network
I I/2 R I/4 R I/8
Bit Current
3 I/2
2 I/4
2R 2R 2R 2R
1 I/8
0 I/16
I1=I/2 I2=I/4 I3=I/8

bit 3 bit 2 bit 1

bit 0
The R-2R Ladder Resistor Network
I R R R The state of the bits
is used to switch a
voltage source
Vs 2R 2R 2R 2R 2R

I/2 I/4 I/8 I/16

Rf

+
Vo
4-bit register
MSB

LSB

Vo  -R f (b3 I 2  b2 I 4  b I 8 b0 I 16)

1
 Example

I R R R

Vs 2R 2R 2R 2R 2R

I/2 I/4 I/8 I/16

Rf

+
Vo
4-bit register
MSB

LSB

Vo  -R f (b3 I 2  b2 I 4  b I 8 b0 I 16)

1

 For the circuit shown above with I = 10 mA and Rf = 2k,

calculate the output voltage V0 for an input code word 1110.
 Example
 I = 10mA
 Rf = 2k
 input code word 1110
 Vo = -2000( 0.01/2 + 0.01/4 + 0.01/8 + (0 x 0.1)/8 )
 = - 2000 * (0.04 + 0.02 + 0.01) / 8
 = 17.5 volts
R/2R ladder DAC

 Most popular single

package DAC
 Resolves BWL problems
 Only two resistor values
 Equations governing R/2R
N
bi
Vo  VR  i ; where b1 is MSB
i 1 2

N
VR bi
 Ii  2R 
i 1 2
 i 1 ; where b1 is MSB

VR
Resolution  N ; where N is number of bits
2
 1 
Vo  fs   VR 1  N ; where N is number of bits
 2 
 Principles of Operation

 Binary Switch  true  Inverting Op-amp used

ground w/ LOW input to generate analog
 Binary Switch  op- output voltage
amp virtual ground w/ HI  Performed many times
input per second  semi-
 Splits current at each bit continuous DAC
 After multiplication of
binary word  Io
 DAC Errors

 Resolution:
 Increases (improves) as number of bits increases
 Most microcontrollers use 8 bit DAC
 Some 12 bit DAC used in high end applications
 Overshoot & Settling Time:
 Time for DAC to come w/in 0.5 LSB {Vo ± 0.5*(VR /2N)} of new voltage
after binary change
 Typical current output of DAC’s conversion times is (10 ns to 1 μs)
 Absolute Accuracy Error:
 Difference between theoretical and actual output
 Conversion Speed:
 Rapidly fluctuating inputs require high conversion speed to be interpreted
accurately
 DAC Errors Cont’d

 Saturation:
Use of op-amps requires that input voltage and scaling voltages be bounded to the
specifications of the op-amp.
Non-Monotonicity:

Certain conditions where increased input results in decrease V o
 Differential Non-Linearity:
 Deviation of actual converter step size from the ideal predicted wave step
 Gain Error:
 Gain too low = same analog output; gain too high = too large an output
Offset Error:
 Constant error of DAC
Reference Voltage

 Internal / external
Applications of DAC

 Control Systems
 Digital Audio
 Digital Telephones
 Cruise Control
 Waveform Generation
‫وآخر دعوانا ان الحمد‬
‫هلل رب العالمين‬