session one notes1
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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
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