Directsensing.ppt

Direct sensing

(Electronics)

Cambridge IA2 Physics

Applications Section – Paper 4

Syllabus Section VII.28

Candidates should be able to:

(a) show an understanding that an electronic sensor consists of a sensing device and a circuit that provides an output voltage.

(b) show an understanding of the change in resistance with light intensity of a lightdependent resistor (LDR).

(c) sketch the temperature characteristic of a negative temperature coefficient thermistor.

(d) show an understanding of the action of a piezo-electric transducer and its application in a simple microphone.

(e) describe the structure of a metal wire strain gauge.

(f) relate extension of a strain gauge to change in resistance of the gauge. (g) show an understanding that the output from sensing devices can be

registered as a voltage.

(n) show an understanding of the use of relays in electronic circuits.

(o) show an understanding of the use of light-emitting diodes (LEDs) as devices to indicate the state of the output of electronic circuits.

(p) show an understanding of the need for calibration where digital or analogue meters are used as output devices.

Potential Divider

Potential difference is shared between

resistors in series

V

_input

V

_output

R



R



V

_output

LDR sensors

V

_input

V

output

High light intensity

LDR resistance is low



V

_output

= LOW

Low light intensity

LDR resistance is high



V

_output

= HIGH

10

0

Transistors

• Base

• Collector

• Emitter

Small base current controls a larger collector

current.

LDR sensors

High R



High V

LDR sensors

• LED lights when

LDR is in light

High V

Low R



Thermistor circuits

Potential difference is shared between

resistors in series

V

_input

V

_output

10

0



Low temperature

Thermistor resistance is

high



V

_output

= HIGH

High temperature

LM35 Temperature Sensor

• LM35 solid-state temperature sensor

LM35 Temperature Sensor

• Output voltage proportional to the

C temperature.

• The scale factor is .01V/

C

• The LM35 does not require external calibration

• Maintains an accuracy of

+/-0.4

C at room

temperature and +/- 0.8

C in range 0

C to +100

C.

• Draws only 60 micro-amps and so possesses a low

self-heating capability. This self-heating causes

less than 0.1

C temperature rise in still air.

Piezo-electric Effect

piezocompressMS.avi

Piezo-electric Effect

Change in the dimensions of a

crystal when a p.d. is applied

Piezo-electric ‘Poling’

An electric field of 2000Vmm

-1

is applied to a

heated crystal (above the Curie temperature)

Piezo-electric Effect

Ultrasound

- transducer

Frequency range <600MHz

Earthed case

Co-axial cable

Piezo-electric crystal

Plastic

cover

Silver electrodes

Backing material

Input/output

p.d.

We will return to the

piezo-electric effect

in the section called

Strain Gauge

Q22.3

Practice in

Physics

(PiP)

Strain Gauge

Strain =



L/L

Resistance,

R =



L/A

(R+



R) =



(L+



L)/A





R =



L/A



R



strain

Assumes that the

change in A is

The LED

• When current flows emits light

• Indicates the state of a circuit

The LED

Electrons in a

semi-conductor

diode fall into

‘holes’ and emit

light of a specific

colour.

Calibration

•

_{Linear output}

Scales are based on two fixed points. Regular

calibration may be required to ensure these fixed

points (and all between) have not changed value.

•

_{Non-linear output}

Need to verify the nature of the variation and the

positions of any fixed points.

e.g. Using a thermistor as a thermometer

Electric Relay

Electric Relay

Relays allow one circuit to

switch a second completely

separate circuit.

e.g. a low voltage battery

circuit can use a relay to

switch a 230V AC mains

circuit.

There is no

electrical

Electric Relay

relay.swf

Electric Relay

Used to maintain the strength of a signal over

very long distances

i.e. in telegraphy

Fig 1.5 (

Page 4, Telecommunications

)

Relay station

every few kms

Press here

Small p.d.

activates the circuit

driven by a

Syllabus Section VII.28

Candidates should be able to:

(h)

recall the main properties of the

ideal operational

amplifier

(op-amp).

(i) deduce, from the properties of an ideal operational amplifier,

the use of an operational amplifier as a

comparator

.

(j) show an understanding of the effects of

negative feedback

on the gain of an operational amplifier.

(k) recall the circuit diagrams for both the

inverting

and the

non-inverting

amplifier for single signal input.

(l) show an understanding of the

virtual earth approximation

and derive an expression for the gain of

inverting amplifiers

.

(m) recall and use expressions for the voltage gain of

invertin

g

and of

non-inverting amplifiers

.

–

Operational Amplifier

• An integrated circuit (IC) using silicon

semi-conductor technology

• Nearly

ideal

characteristics

•

_{Differential amplifier}

_{– it amplifies the}

difference between two input potentials

• Versatile – can be made to perform a

number of mathematical

operations

Load

Operational Amplifier

The op-amp circuit symbol

Inverting

input

V

Non-Inverting

input

V

Positive input potential

+V

e.g. +15V

Negative input potential

–V

e.g.

–

15V

Output potential

V

–

What’s Inside an Op-Amp?

• On this course we do not care what is inside

the op-amp chip or how it works!!

• We need to remember only

the two

ideal op-amp

relations

to use the

op-amp in practical

circuits (slides 37 & 38)

–

Operational Amplifier

Single Chip Op-Amp (8 pin)

–

+

7

6

5

8

2

3

4

1

PIN

1

2 V

_–

3 V

₊

4 –V

_s

(–15V)

5

6 V

_out

Operational Amplifier

• High-gain differential amplifier

• A

₀

= ‘open-loop’ gain = V

_out

/V

_in

(

Typically 10

5

for DC and low-frequency inputs)

Operational Amplifier

Example 1



V

_s

=



15V

V

₊

= +3V

V

_–

= +2V

A

₀

= 10

5

V

_out

= A

₀

(V

₊

– V

_–

)

V

_out

= 10

5

x (3 – 2 )

V

_out

= 100,000V!

V

_out

= +15V

(actually



14V)

This condition is known as

positive saturation

.

The actual output for a



15V supply cannot exceed +15V

Operational Amplifier

Example 2



V

_s

=



15V

V

₊

= – 4.3V

V

_–

= – 4.2V

A

₀

= 10

5

V

_out

= A

₀

(V

₊

– V

_–

)

V

_out

= 10

5

x (

_–

4.3 – (

_–

4.2))

V

_out

=

–

10,000V!

V

_out

= –15V

(actually



–14V)

This condition is known as

negative saturation

.

The actual output for a



15V supply cannot exceed –15V

Operational Amplifier

Example 3



V

_s

=



15V

V

₊

= +5.8733V

V

_–

= +5.8732V

A

₀

= 10

5

V

_out

= A

₀

(V

₊

– V

_–

)

V

_out

= 10

5

x (5.8733 – 5.8732)

V

_out

= +10V

This illustrates that if the output voltage V

is not saturated

Then the inputs must be nearly equal.

Op-Amp Output Saturation

Typical gain 10

Linear region very limited (due to fixed power supply)

Ideal Op-Amp Relations

Golden Rule No. 1

V

₊

= V

_–

“If the output is not saturated, then the two

inputs must have nearly the same potential”

Ideal Op-Amp Relations

Golden Rule No. 2

I

₊

= I

_–

“If the output is not saturated, then no current

flows into or out of the two inputs”

(i.e. the resistance between them is



)

V

V

₊

= V

_–

I

₊

= I

_–

= 0

opamp.swf

i.e. R





Summary

Property

Ideal

Op-Amp

Real

Op-Amp

Input resistance



no current is drained

10

6

to 10

12



Output resistance

Zero

can drive any load

About 100



Voltage Gain



at any frequency

10

5

– 10

6

dependent on frequency

Bandwidth & Slew Rate

• Bandwidth

The gain of a real op-amp is not constant for all frequencies

• Infinite Bandwidth

For the

ideal op-amp

the gain is assumed to be constant for

all frequencies – i.e. Infinite bandwidth

• Slew rate

Questions

SAQ 12.1

(

Page 106, Telecommunications)

Calculate output voltage

SAQ 12.2

The comparator (switch)

•

_{Open-loop mode}

_{(no feedback)}

• When one input changes with respect to the

other the output

switches

from one

saturation level to the other

6V

0.06A

–9V

+9V

The Comparator

(Fig12.4,

page 107, Telecommunications

)

Output circuit

Comparator circuit

(Op-Amp)

LDR

Sensor circuit

Power

22k



10k



LDR 3.9k



0V

0V

Comparator – LDR in daylight

Inverting input

(V

)

V

= +9 x (3.9/13.9) = +2.5V

Non-inverting input in daylight

(V

)

R

< 500





_V

_{< +9 x (0.5/22.5)}



_V

_{< +0.2V}



_V

_{> 10}

x (0.2 – 2.5) = –230,000V

Comparator – LDR in darkness

Inverting input

(V

)

V

= +9 x (3.9/13.9) = +2.5V (does not change)

Non-inverting input in darkness

(V

)

R

> 100k



V

> +9 x (100/120)



V

> +7.5V



V

> 10

x (7.5 – 2.5) = +500,000V



_{Vout = +8V (positive saturation)}

+8V

+0V

Voltage follower (

‘buffer’

)

•

Negative feedback

• Unity gain (output follows input)

•

Buffers

or isolates one circuit from another

Negative feedback loop

Note:

Here V

represents

the input voltage

(not the supply)

Inverting Amplifier

• Negative gain (inverter)

• Two resistors, Z

₁

(input) and Z

₂

(feedback)

Note:

Here the symbol

Z is used to

represent

resistance

“Virtual Earth Approximation”

For the Inverting Amplifier:

• V

₊

(non-inverting input) is

earthed

= 0V

• V

₊

= V

_–

(ideal op-amp Golden Rule No.1)

The inverting input (V

_–

) is referred to as the “

virtual

earth”

input

• Input resistance = R

₁

(or ‘Z

₁

’)

Inverting Amplifier

V

_out

= –V

_–



R

₂

/R

₁

V

_–

=

inverting

input potential

V

₊

= 0V (earthed)

Non-Inverting Amplifier

• Positive gain

• Two resistors, Z

₁

and Z

₂

Note:

Again the symbol

Z is used here

to represent

resistance

Non-Inverting Amplifier

V

_out

= V

_–



(1

+

R

₂

/R

₁

)

V

_–

=

inverting

input

V

₊

=

non-inverting

input

Questions

SAQ 12.4

(

page 110, Telecommunications)

Design an op-amp with gain = 4

SAQ 12.5 (

page 111

)

Design an op-amp with gain = –12

and input resistance = 68k



Questions (

page 115

)

Further Operations

•

_{The following operations are not on the}

syllabus

• They are explained fully in

opamp.swf

• They are included, because it is hoped that

A2-level students will be interested in how

a simple device can be used to perform

Differentiator

• Op-amp

differentiates

the input signal

i.e. output depends on the

rate of change

of input

Differentiator

Integrator

• Op-amp

integrates

the input signal

i.e. output depends on the

sum of inputs

over time

Op-Amp Applet