Devices Assignment 2

8/8/2019 Devices Assignment 2

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1. Sys tem De s ign

This section is divided into two main parts; design of the remote -controlledsystem and design of the power supply. From the specification, I mapped out ablock diagram showing my proposed design:

Since the power supply had to be designed around the requirements of theremote-controlled system then I decided to design the remote-controlled systemfirst.

1.1. R emote-Controlled Sys tem De sign

As you can see from the previous section, the main components making upthis system are the encoder, transmitter, receiver, decoder, toggle,transistor and relay coil.

To begin the design process I first began with the encoder configuration.

1.1.1. Encoder & Tran smitter Configuration

I decided to use the H T12 E RF Encoder for the design as I found that itwas slightly cheaper than the HT1 2A IR Encoder; however either onecould be used. On the next page ( figure 2 ) you can see the PINschematic of the HT1 2E encoder taken from the data sheet.

Encoder Transmitter

Battery

Power

Transformer &Rectifier

Regulator 1(Remote Control

System )

Receiver

Regulator 2 (Lighting Load )

DecoderToggle

Transistor &Relay Coil

Relay Contact LightingLoad

F igure 1



3

With thi s parti c

¡

ar model it i s po ssible to u se PINS 10 ¢ 13 a s data line s or addre ss line s £ for thi s parti cular u se ( s ¤ itching a light on or off ¥ itwa s only ne cessar y to u se the data from DOUT therefore I de cided touse PINS 10 ¢ 13 a s additional addre ss line s ¦ With the en coder set-up inthi s configuration it wa s po ssible to set 4 096 (212

¥ differentcombination s for the addre ss line; thi s make s the addre ss harder tocra ck compared to the HT 12A en coder.

It is stated in the data sheet that the input voltage (VDD ¥ can operateat a minimum of 2.4 Volt s DC, typically at 5 Volt s DC and at amaximum of 12 Volt s DC. Since I wanted thi s en coder to be part of abatter y powered hand-held devi ce then I cho se to u se a small batter y re commended for garage door opener s, ke yless entr y and car alarm s1;a lithium CR3 032 3V batter y.

Lastly, I had to choo se a frequen cy for the o scillator to o scillate at. I

cho se to set the frequen cy to 3 kHz whil st con sidering the tran smitterdata frequen cy range of 2 00 Hz to 4 000 Hz. To find the corre ctre sistan ce value needed to set the o scillator frequen cy I used thefollowing Oscillator Frequen cy vs. Suppl y Voltage table provided inthe data sheet ( see fi gu § e 3 o n t he next pag e ).

1 Everead y Batter y Compan y Inc, Electroni c Batterie s Cross Referen ced b y Device Type,http://www.energizer. com/produ cts/ spe cialt yphotobatterie s/wat ch-button/Page s/PDA-alarm-remote.a spx, a ccessed on 05/ 04/ 10 @ 15:4 0

¨ i gu § e 2



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As you can see, with a 3V power suppl y and a 3 kHz o scillatorfrequen cy, an 82 0 k re sistor mu st be u sed.

For the tran smitter I cho se the QAM-TX 1-433 RF module. I cho se thi s parti cular model since it ha s a low operating voltage between 1.5Volt s DC and 5 Volt s DC which make s it suitable with the 3 Voltbatter y. Also, it ha s a ver y stable operating frequen cy, a suffi cienttran smit range (5 0 meter s) and i s low cost.

Below i s a diagram showing m y final de sign for the en coder andtran smitter:

F i gu © e 3

F i gu © e 4





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As you can see, with a 5V power suppl y and a 150 kHz oscillatorfrequen cy, a 5 1 k re sistor mu st be u sed.

The QAM-RX2-433 i s the re commended re ceiver to mat ch the QAM-TX1-433 tran smitter. In the data sheet it i s re commended that it i s typically supplied with 5 Volt s DC.

Below i s a diagram showing m y final de sign for the de coder andre ceiver:

When the de coder find s a mat ching signal from the en coder, it fir stche cks the signal 3 time s to confirm that it i s definitel y the corre ctsignal and then a high pul se will be sent out of VT to the toggleswit ch .

F i gu e 5

F i gu e 6

To To gg e

S itch

Tr an sistor &

Rel a Coil Circu it



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1.1. 3 . Toggle Sw itch, Tran sis tor & R ela y Coil Circuit

The signal from the output of the decoder cannot be fed straight intothe lighting load; it must be used to activate a switch between thelighting load and its power source. To do this, I used the following

circuit containing a D-type flip flop toggle switch, Bipolar JunctionTransistor and a Relay Coil:

Since the output of the D-type flip flop was equal to 5V (Vcc) then Iused a pair of resistors to create a potential divider in order to lower

the voltage at the transistor; I did this to match the BC 547 C transistor

on voltage which had to lie between 0. 55 V and 0. 7 V asrecommended in the data sheet. To work out the resistor values Iused the potential divider rule:

This could be re-arranged to:

I chose Vout to be 0.6 V so that it would meet the transistor on voltagerequirement and I picked a random E1 2 value for R 2 (15 k ) so that I

could find the appropriate value for R1. Vin was set by the output of the d-type flip flop ( 5V).

LightingLoad

Regulator2

F rom Decoder O utput (VT)

R2

R1

CLK

D

GND

V cc

Q

PRE

V LR Q

SN74LV C2 G74D-type Flip Flop

5V

BC5 47C

Tyco1 393 793 -6Relay Coil

5V

110k

15k R2

R1

F igure 7



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k to the nearest E1 2 value.

As you can see, when the voltage goes high on the decoder output,the toggle (D-type flip flop ) switches and provides a voltage to the

transistor. With 0.6 Volts present at the BJT base pin, the transistorswitches allowing the regulator voltage to flow through to ground andthe relay coil then becomes magnetically charged and closes the relaycontact making the connection between the lighting load and itspower source.

I chose the Tyco 1 393 7 93 -6 relay coil as it was non -latching and

operated at typical voltage of 5 volts.

With the remote-control system design complete, I now had to design

an appropriate power supply to meet the specifications of the abovecomponents.

1.2. Po w er Suppl y Design

From my block diagram on page 2 ( figure 1 ) you can see that the lightingload power supply and receiver sub-system are powered from the same

mains connection but separate voltage regulators as specified in the brief.

The first regulator (Regulator 1 ) will supply the receiver sub-system whichrequires around 5 Volts DC ( see decoder, receiver, toggle and relay coil

voltage requirements ) and the second regulator (Regulator 2) will need to

provide 1 2 Volts (lighting load requires 1 2V s 2% as specified in brief ).Before choosing the regulators, I first had to choose a suitable Transformerand design a Rectifier.

1.2.1. Tran sformer, R ectifier & Smoothing Capacitor

The first consideration I made when choosing a Transformer was thatit is powered by the mains ( 24 0 Volts @ 50 Hz). Before I could beginthe calculations, I needed to know an approximate step -down voltage.Since I knew that voltage would be lost across the diodes and thelighting load regulator had to be at least 2.5V above the output ( at



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least 1 4.5 V) then I estimated that 1 8 Volts would be an appropriatesupply voltage.

I decided to choose an RS10- 5860 Transformer as it is suitable with24 0 Volts input, p rovides 1 8 Volts output and 1.1 Amps of current.

Also, it was the cheapest I could find that matched my requiredspecification.

The current specification was based on the current being drawn by

the load and the receiver system circuit. To work out the currentdrawn by the load (I Load) I used the following formula:

The values for W and V were taken from the lighting load

specification; 12 V @ 10 W.

To find the current being drawn by the receiver system circuit, Isummed the typical current values of the receiver, decoder, toggleand relay coil as specified in the datasheets:

y Receiver - 3.5 mAy Decoder - 0. 2 mAy Toggle - 100 mAy Relay Coil 80 mA

Total = 0.1 8 Amps

Therefore an approximate current amount being drawn by the systemwas equal to 1.01 Amp s . This confirmed that my chosen transformerwas able to provide enough current.

Below is a circuit representation of the transformer:

18 V

F use

240 V @ 50Hz

125 mA

F igure 8



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To find an appropriate value for the fuse I used the following formula:

I choose to use a 1 25 mA fuse since it was the closest value I could

find above 75 mA.

The next part of the power supply chain was the rectifier. I chose touse a common bridge rectifier with four silicon diodes; I chose 1 N41 48silicon diode s as they have a reasonably small voltage drop of 0. 7 V.

See the rectifier circuit diagram below.

Before I could continue any calculations I first had to convert the RMSvoltage to a peak voltage:

= 25.46 V peak

The job of the rectifier is to convert the A.C voltage into a D.C voltage.With the bridge rectifier, only two diodes allow a voltage to passthrough at one time. Since there is a voltage drop ( Vf ) of 0. 7 V acrosseach diode and only two diodes active at o ne instant, then the total

voltage drop across the rectifier is 1. 4 V.

25 .46 1. 4 = 2 4.0 6 V olt s peak

With the wave fully rectified, the signal then had to be passed througha capacitor to smooth out the ripple; this is known as the smoothingcapacitor. To work out a value for the capacitor, I wanted to make

18 V RMS fromTransformer

Smoothing

C apacitor C

F igure 9



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sure that ripple at the input of Regulator 1 was sm all enough so thatthe output would not exceed 100 m V peak to peak as specified in thebrief. To do this, I used the ripple rejection information f rom the

A7805 C datasheet ( see page 1 3 for Regulator 1 choice ).

Ripple Rejection = - 78 dB

Therefore:

Now by re-arranging this formula, I could find the maximum inputripple voltage:

Vout ripple is the desired ripple as specified in the brief; 100 mV peak

to peak (0.1 Volts).

From this you can see that for Regulator 1 to provide an output rippleno bigger than 100 m V peak to peak, the input ripple could be as bigas 794 Volts peak to peak. Since I knew that whatever capacitor Ichose would satisfy this ripple amount, I decided to base thecalculation of the smoothing capacitor around the voltagerequirements of Regulator 2.

Regulator 2 was required to provide 1 2 Volts to the lighting load;therefore I chose the A7812 C regulator ( see page 1 3). According tothe A7812 C regulator datasheet, the voltage input had to be at least2.5 Volts above the output voltage; therefore the voltage input of Regulator 2 had to be 1 4.5 Volts or above.





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1.2.2. R egulator 1

I chose to use the A7805 C regulator for Regulator 1 since it has atypical output voltage of 5 V to match the voltage needs of theReceiver, Decoder, Toggle and Relay Coil. According to the datasheet,

for the output to be 5 Volts then the input voltage had to be between7 V and 25 V RMS (at least 2 Volts bigger than the output ); the inputvoltage was set as 1 7 Volts DC.

Below is a circuit diagram showing the Regulator with noise rejectioncapacitors as recommended in the datasheet:

1.2. 3 . R egulator 2

I chose Regulator 2 as the regulator to feed the lighting load. Since Iwanted the output to provide 1 2V ± 2% then I choose the A7812 Cregulator.

According to the datasheet, for the output to be 1 2 Volts then the

input voltage had to be between 1 4.5 V and 30 V (at least 2.5 Voltsbigger than the output ); the input voltage was set as 1 7 Volts DC.

One the next page ( figure 11 ) is a circuit diagram showing theRegulator with noise rejection capacitors as recommended in the

datasheet.

Vin = 17 V Vout = 5 V

A7805 CR egulator

Common

IN OUT

0.33 F 0.1 F

F igure 10





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The heat dissipation system can be thought of as the equivalent circuitbelow:

y TJ is taken as the maximum junction temperature taken from the

datasheet (1 50oC).y JC is taken as the thermal resistance from junction to case and is

given in the datasheet as 4oC/W.y CS is taken as the thermal resistance from case to sink and refers

to the thermal grease; this was taken as an average of 1 oC/W.y SAis taken as the thermal resistance from the sink to ambient; this

is the unknown that must be found in order to find an appropriate

heat-sink.y TA is taken as the maximum ambient temperature; I based this on

an assumption that the system would not be used in temperaturesexceeding 70oC.

To find the unknown ( SA) I used the following formula:

This could be re -arranged to:

TJ

TA

JC

Cs

SA

4oC/W

1oC/W

70oC

150oC

F igure 12



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Therefore, to find SA for Regulator 1:

And for Regulator 2:

With the values for SA found, I chose the following heat -sinks:

y Regulator 1 PF 720 2 8 oC/W H eat- sink y Regulator 2 SW252 1 3 oC/W H eat- s ink

I choose heat-sinks with a slightly better temperature per Wattdifference so as to exceed the requirements.

2. V irtual Simulation

Since there are many considerations to make when designing the power supply, I

decided to perform a virtual simulation in order to confirm that the powerrequirements for the system were being met.

It was stated in the assignment brief that is was not required to simulate the

mains transformer and that it could be assumed that the transformer could berepresented as a Thevenin equivalent a.c generator of 22 volts peak amplitudewith an output resistance of 0.9 ohms.

To perform the simulation I used National Instruments Multisim 10. On the nextpage is a screenshot of the power supply circuit ( figure 1 3).



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Unfortunatel y, the software did not have the exa ct regulator model t ype s, so Iused equivalent regulator s; a TA7805F (5 Volt s) and a TA 7812F (12 Volt s).

As you can see from the multi-meter re sult s, 5. 00 4 Volt s wa s being provided b y Regulator 1 and 11 .973 Volt s wa s being provided b y Regulator 2, thi s mat chedthe de sired value s of 5 Volt s and 12 Volt s.

To confirm that m y calculation for the capa citor en sured a rea sonabl y minimumripple I de cided to mea sure it u sing an o scilloscope:

As you can see ( circled above ), the ripple i s approximatel y 10 3.55 mV peak topeak whi ch meant that the ripple requirement s were ex ceeded for bothRegulator s.

Regulator 1

Regulator 2

F i gu re 13

F i gu re 14



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3 . Component s Cos ting

With the design and simulation complete, I decided to find an approximate costfor the system. Below is a table showing each component and its associatedprice:

Component Model/T ype Co s t Each Qt y Web s ite TotalEncoder HT1 2E £0.57 1 Rapidonline.com £0. 57

Transmitter QAM-TX1- 433 £2.29 1 Onecall.farnell.com £ 2.29

Decoder HT1 2F £0.72 1 Rapidonline.com £0. 72

Receiver QAM-RX 2-433 £2.76 1 Onecall.farnell.com £ 2.76

Push Button Switch R1 2-81 RED £0.37 1 Onecall.farnell.com £0. 37

Lithium Battery CR 3032 3V £0.6 8 1 Onecall.farnell.com £0.6 8

12 bit DIL Switch MCEI-1 2 £0. 41 2 Onecall.farnell.com £0. 82

D-Type Flip-Flop SN74LVC2G74DCTR £0. 48 1 Onecall.farnell.com £0. 48

BJT Transistor BC 54 7C £0.0 4 1 Onecall.farnell.com £0.0 4

Relay Coil Tyco 1 393793-6 £1.06 1 Onecall.farnell.com £1.06

Resistor 82 0 k £0.0 2 1 Onecall.farnell.com £0.0 2

Resistor 51 k £0.0 2 1 Onecall.farnell.com £0.0 2

Resistor 1 5 k £0.0 2 1 Onecall.farnell.com £0.0 2

Resistor 110 k £0.0 2 1 Onecall.farnell.com £0.0 2

Transformer RS10- 58 60 £ 5.55 1 rs-online.com £ 5.55

Diode 1N 4148 £0.0 3 4 Onecall.farnell.com £0.1 2

Smoothing Cap Electrolit 6 80 F £0.16 1 Onecall.farnell.com £0.16

Regulator 1 A 7805CKTE £0.31 1 Onecall.farnell.com £0. 31

Regulator 2 A7812CKTE £0.38 1 Onecall.farnell.com £0. 38

Heatsink PF 720 £0. 23 1 Onecall.farnell.com £0. 23

Heatsink SW 252 £0.66 1 Onecall.farnell.com £0.66

Fuse 1 25 mA £0.1 3 1 Rapidonline.com £0.1 3

Plug BS 1363 £1.0 4 1 Onecall.farnell.com £1.0 4

Capacitor 0. 33 F £0.0 4 2 Onecall.farnell.com £0.0 8

Capacitor 0.1 F £0.0 4 2 Onecall.farnell.com £0.0 8

TOTAL £1 8 . 61

This is only an approximate cost and does not include components such as the printed circuit board, cables and enclosures.

As you can see, the total cost of the system is £1 8.56. To see how reasonable thisprice was, I decided to compare the cost with 3 other remotely-controlledlighting systems of a similar specification ( see figure 1 6 on the next page ).

F igure 15



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Description Web s ite PriceWireless Light & Appliance Remote Control Overstock.com £1 8.17

Remote Control & Receiver Light Pack U-diy.co.uk £1 8.99Infrared Remote Control Light Switch Maplin.co.uk £ 24 .99

Comparing the cost of my system to already existing systems, I w ould say thatthe total price was fairly reasonable. If I was to set up a company then I would

most likely be able to get a trade price on the components which would lowerthe cost. Also, the prices given for my components was the individual price percomponent, however if the components are bought in bulk then the individualprice per component works out cheaper.

F igure 1 6



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Conclu sion

One assumption that I made in my design was that all the components were workingat 100 % efficiency; therefore there may have been slight voltage and currentvariables not taken into consideration. However, I made sure that all voltage and

current values were in reasonable limits. For example, I made sure not to set avoltage of 2.9 when the maximum voltage allowed was 3 Volts. All of the voltage andcurrent values were in reasonable limits so that if variable component tolerancescaused a slightly higher or lower output than expected, the system would not beaffected.

After I had finished my design, I worried that even though the RF encoder anddecoder were cheaper than the IR versions , the price of the RF transmitter andreceiver were more expensive than the IR emitter and receiver; to confirm this I

decided to find the total cost of the IR components. What I found was that even

though the price of an IR emitter was a lot cheaper than the RF transmitter, the IRreceiver circuit cost approximately the same price as the RF transmitter and receivertogether, therefore my initial choice proved to be the wisest.



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Bibliograph y

Book s:

y Electronic & Electical Engineering: Principles and Practice (Third Edition 2002) Warnes, L

Web s ite s:

y Onecall.farnell.com

y www.electronicsandyou.com

y www.energizer.com

y www.maplin.co.uk

y www.overstock.com

y www.rapidonline.com

y www.rs-online.com

y www.u-diy.co.uk

y www.wikipedia.com

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Devices Assignment 2