session one notes1

8/9/2019 Session One Notes1

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ECTL La bs

Electronics and Computers T raining Labs=========================================================

//) ) ) ) ))

) ) ) @_@) Core Electronics Course Day 1 Notes( ( ( = ) [July 2010]

) ) ) - (_ __

/ `-'\\ /,\\\`

=========================================================Core Electronics CourseResistor Applications: 1-L-Section AttenuatorThe schematic:


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The at t enuat ors propert ies:

1-Input Resist ance [ Equivalent Resist ance bet ween t he t wo

t erminals of t he at t enuat or when connect ed t o t he load must =R

Where R is t he Source Resist ance= L oad Resist ance

Why?t o make t he [ L -sect ion + The L oad] r eceives t he maximum

power f rom t he sour ce + minimize t he ref lect ions of t he

radiat ion come out f rom t he source [ Ref lect ions in t he light

occur s due t o passing t hr ough d if f erent mat erials , in

elect romaget ic c ircuit s ;it is occur red when passing t hr ough

d if f erent resist ances[ Impedance in general]

2-The ratio between the output voltage to the input voltage is the

attenuation ratio =1/k

K =Inverse of t he at t enuat ion rat io.


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2in Sourcein

2out out Load

Source Load

In dB [Decibel] notation

dB=10*log (Power Ratio)

or

dB=20*log (Voltage Ratio)

V /RP[As 10*log 10*log

P V /R

{Here R =R and generally we do

Resistance normalization (Assume tha

=>

=

2in in in

2out outout

t R=1) }

P V VSo 10*log 10*log 20*log ]

P VV= =


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Applying KCL @the input node:

Rg R1

Rg

R1

g in in out

1

g in

g in in in in

in

I =I

I =current through the source resistance

I =current through the first attenuator resistance

V -V V -VSo :

R Ras V =2V so the numenator in

the first fraction =V -V 2V -V =V

VSo

R

=

=

in out in out in1

1 in out

out

1

V -V V -V VSo R =R* But K=

R V V

So dividing both numenator and denumenator by Vk-1

R R*k

=

=


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Applying KCL @the output node:

R1 R2 Rout

R1

R2

Rout

in out out out

1 2

in

out

I I +I

I =current through the first attenuator resistance

I =current through the Second attenuator resistance

I =current through the load resistance

V -V V VSo :

R R R

VBut K=

V

So

=

= +

out

1 2

1

dividing both sides by V

k-1 1 1

R R R

k-1But R =R* So

k

k-1

= +

k-1R* 2

22 2

1 1 thenR R

kK 1 1 K-1 1 R

So = So R =R R R R R K-1

= +

= +

The f inal deduc t ion : Given values of k[ 1/A t t enuat ionf ac t or] and R[ sour ce=load resist ance]

Then

1 2

K-1 RR =R * and R =

K K-1


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The homework[for those who dont get the hardcopy material]Prove for a [L-Section Type B] attenuator as shown in the following figure;

Values of R1and R2 are ;represectively ; equal =>

1

2

KR =R ( )K-1

R =R (K-1)

where K =I nverse of t he at t enuat ion rat io.

in

out

VK=

V=10 dB = 3.16 A bsolu t e.

and R= Load Impedance = Source Impedance=50 Ohm.

Then verify for attenuation value=10dBR1= 73.12 OhmR2=108.11Ohm

session one notes1

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