Friday, 26 April 2013
TYPES OF FEEDBACK
1) VOLTAGE – SHUNT or SHUNT- SHUNT FEEDBACK
2) VOLTAGE – SERIES or SERIES – SHUNT FEEDBACK
3) CURRENT – SERIES or SERIES – SERIES FEEDBACK
4) CURRENT – SHUNT or SHUNT –SERIES FEEDBACK
|
INPUT
|
OUTPUT
|
|
SHUNT
|
SHUNT
|
|
SERIES
|
SHUNT
|
|
SERIES
|
SERIES
|
|
SHUNT
|
SERIES
|
NOTE: ASSUME SHUNT AS DECREASING AND
INPUT AND OUTPUT IMPEDANCE
|
TYPE OF FEEDBACK
|
INPUT
|
OUTPUT
|
|
VOLTAGE – SHUNT or
SHUNT- SHUNT FEEDBACK
|
DECREASING
|
DECREASING
|
|
VOLTAGE – SERIES or
SERIES – SHUNT FEEDBACK
|
INCREASING
|
DECREASING
|
|
CURRENT – SERIES or
SERIES – SERIES FEEDBACK
|
INCREASING
|
INCREASING
|
|
CURRENT – SHUNT or
SHUNT –SERIES FEEDBACK
|
DECREASING
|
INCREASING
|
By how much factor it is going to increase or decrease
The factor by how much it is going to increase or decrease
is ( 1+ Aβ
)
If RIN and R0 are input impedance and
output impedance before feedback, then after feedback
|
TYPE OF FEEDBACK
|
INPUT
|
OUTPUT
|
|
VOLTAGE – SHUNT or
SHUNT- SHUNT FEEDBACK
|
RIN /( 1+ Aβ
)
|
R0 / ( 1+ Aβ
)
|
|
VOLTAGE – SERIES or
SERIES – SHUNT FEEDBACK
|
RIN ( 1+ Aβ
)
|
R0 / ( 1+ Aβ
)
|
|
CURRENT – SERIES or
SERIES – SERIES FEEDBACK
|
RIN ( 1+ Aβ
)
|
R0 ( 1+ Aβ
)
|
|
CURRENT – SHUNT or
SHUNT –SERIES FEEDBACK
|
RIN / ( 1+ Aβ
)
|
R0 ( 1+ Aβ
)
|
Friday, 26 April 2013 0
Thursday, 25 April 2013
Consider
VBE1 = VBE2 = 0.7 V,β1 = 100,β2=50.
Find IB1,IB2,I1,I2,IC2,IE1,V01,V02
SOLUTION
By applying thevenin’s theorem at
input
VTH = VCC R2
/ (R1 + R2 )
= 24*10 / (10 + 82)
= 2.61V
RTH = 82 || 10 = 8.91 KΩ
Applying KVL at the input
12 – (8.91 KΩ + 100 KΩ )IB2
– VBE2 – VBE1 – (IE1 * 100Ω )= 0
IE2= ( β2 + 1 )IB2
= 51 IB2
IB1 = IE2
IE1 = ( β1 + 1 )
IB1 = ( β1 + 1 ) ( β2 + 1 )IB2 =101
* 51 * IB2
= 5151 * IB2
= 5151 * IB2
2.61 – ( 108.91 KΩ * IB2 )
– 0.7 – 0.7 – ( 5151 * IB2 * 100Ω ) = 0
620.1K* IB2 = 1.21
IB2 = 1.21 / 624.01K
= 1.94 µA
Thursday, 25 April 2013 0
Tuesday, 23 April 2013
|
ASIA PARIS
|
LAHORE
|
|
BENGAL’S SORROW
|
RIVER DAMODAR
|
|
BLUE MOUNTAINS
|
NILGIRI HILLS
|
|
BLUE RIVER
|
YANGTZE KIANG,CHINA
|
|
BRITAIN OF SOUTH
|
NEW ZEALAND
|
|
BACKBONE OF ENGLAND
|
PENNINE MOUNTAINS
|
|
BACKBONE OF ITALY
|
APENNINE MOUNTAINS
|
|
BIG APPLE
|
NEWYORK
|
|
CITY OF GOLDEN TEMPLE
|
AMRITSAR
|
|
CITY OF MAGNIFICENT DISTANCES
|
WASHINGTON D.C
|
|
CITY OF MAGNIFICENT DISTANCES(INDIA)
|
CHENNAI
|
|
CITY OF GOLDEN GATE
|
SAN FRANCISCO,U.S.A
|
|
CITY OF FACTORIES
|
KANPUR,U.P
|
|
CITY OF FESTIVALS
|
MADURAI,TAMIL NADU
|
|
CITY OF DREAMING SPIRES
|
OXFORD,ENGLAND
|
|
CITY OF ETERNAL SPRINGS
|
QUITO,SOUTH AMERICA
|
|
CITY OF PALACES
|
KOLKATA
|
|
CITY OF SKYSCRAPERS
|
NEWYORK
|
|
CITY OF POPES
|
ROME
|
|
CITY OF SEVEN HILLS
|
ROME
|
|
COCKPIT OF EUROPE
|
BELGIUM
|
|
CITY OF SEVEN ISLANDS
|
MUMBAI
|
|
CITY OF SNOW
|
CANADA
|
|
CITY OF ARABIAN NIGHTS
|
BAGHDAD,IRAQ
|
|
CITY OF MOTOR CARS
|
DETROIT,MICHIGAN,U.S.A
|
|
CITY OF ROSES
|
CHANDIGARH
|
|
CITY OF SMOKE
|
OSAKA,JAPAN
|
|
STEEL CITY OF INDIA
|
JAMSHEDPUR
|
|
CITY OF TEMPLES
|
VARANASI
|
|
COSTLIEST COAST
|
COSTA RICA
|
Tuesday, 23 April 2013 0
Let the input terminal to which source is connected is
considered as shunt
The terminal connected to ground at the input is considered
as series
The output
terminal from which output is taken is considered as voltage
The other end of
the load resistor is considered to be current
NOTE :
Naming the feedback is done from the output to input.
VOLTAGE-SERIES FEEDBACK OR SERIES-SHUNT FEEDBACK
STEP 1: identify the feedback resistor
STEP 2: The terminal connected to ground at the input is
considered as series
STEP 3: The output terminal from which output is taken is
considered as voltage
STEP 4: Name the feedback is considered from the output to
input
As the feedback resistor is connected from output voltage ( VOLTAGE
) to input connected to ground (SERIES)
So,this circuit is called voltage series feedback
How to convert the name VOLTAGE- SERIES to SERIES-SHUNT FEEDBACK
Rename voltage as shunt and interchange the
names as
VOLTAGE- SHUNT or SHUNT-SHUNT FEEDBACK
STEP 1: identify the feedback resistor
STEP 2: Let the input terminal to which source is connected
is considered as shunt
STEP 3: The output terminal from which output is taken is considered as voltage
STEP 3: The output terminal from which output is taken is considered as voltage
STEP 4: Name the feedback is considered from the output to
input
As the feedback resistor is connected from output voltage ( VOLTAGE
) to input (SHUNT)
So,this circuit is called voltage shunt feedback
How to convert the name VOLTAGE- SHUNT to SHUNT-SHUNT
FEEDBACK
Rename voltage as shunt and interchange the names
as
CURRENT-SERIES or SERIES-SERIES FEEDBACK
STEP 1: identify the feedback resistor
STEP 2: The terminal connected to ground at the input is
considered as series
STEP 3: The other end of the load resistor is considered to
be current
STEP 4: Name the feedback is considered from the output to
input
As the feedback resistor is connected from other end of the
load resistor ( CURRENT ) to input terminal connected to ground (SERIES)
So,this circuit is called current series feedback
How to convert the name CURRENT- SERIES to SERIES-SERIES
Rename current as series and interchange the
names as
CURRENT- SHUNT or SHUNT-SERIES FEEDBACK
STEP 1: identify the feedback resistor
STEP 2: Let the input terminal to which source is connected
is considered as shunt
STEP 3: The other end of the load resistor is considered to
be current
STEP 4: Name the feedback is considered from the output to
input
As the feedback resistor is connected from other end of the
load resistor ( CURRENT ) to the input terminal to which source is connected (
SHUNT )
So, this circuit is called current shunt feedback
How to convert the name CURRENT- SHUNT to SHUNT-SERIES
Rename current as series and interchange the names as
Sunday, 14 April 2013
a)if V1 = 1V find V0
b)if ICBO = 10nA at 250C.find maximum temperature
at which the transistor is in cut-off.
SOLUTION
a)Given V1 = 1V
Solving above equation we get VTH = -0.695V
From the circuit VTH = VBE
As VBE ≤ 0.6V
the transistor is in cutoff.
So,there is no IC ie.,IC = 0
Therefore V0 = 12V
b)
IB = I2 – I1
As the transistor is in cut-off IB is
approximately equal to -ICBO
-ICBO = I2 – I1
ICBO = I1 – I2
I1 = 0.6 – (-12 ) / 100K = 0.126mA
I2 = 1 – 0.6 / 15k = 0.4 /15k =0.026mA
ICBO = 0.126mA – 0.026mA
=0.1mA
ICBO (T) = ICBO (250C ) * 2 (
T – 25 )/10
0.1mA = 10nA * 2 ( T – 25 )/10
Applying logarithm
( T – 25 ) /10 = 13.28
T –25 =132.8
T = 157.80C
Sunday, 14 April 2013 0
Subscribe to:
Posts (Atom)
