total text of second sem record

7/22/2019 Total Text of Second Sem Record

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ENGINEERING PHYSICS LABORATORYII

1 G.V.P. COLLEGE OF ENGINEERING FOR WOMEN

THERMISTORTHEORY:

The name thermistor comes from thermally

sensitive resistor. They are basically

semiconducting materials and are of two distinct

classes:

1. METAL OXIDES: They are made from finepowders that are compressed and sintered at high

temperature. Mn2O3 (manganese oxide), Ni O

(nickel oxide), Co O3 (cobalt oxide), Cu2O3

(copper oxide), Fe2O3 (iron oxide), TiO3

(titanium oxide) U2O3 (uranium oxide) etc, are

the few examples. They are suitable for

temperatures 200-700 K. If the temperature is

higher than this range then Al2O3, Be O, Mg O,

ZrO2 Y2O3 and Dy2O3 (Dy :dysprosium) are

used.

2.SINGLE CRYSTAL SEMICONDUCTORS:They are usually Germanium and Silicon doped

with 1016 to 1017 dopant atoms/cm3. Ge

thermistors are suitable for cryogenic range 1-

100 K. Si thermistors are suitable for 100-250 K.

After 250 K the Silicon thermistors will become

PTC (positive temperature coefficient) from

NTC.

The resistivity and the conductivity of thethermistor are related to the concentration of

electrons and holes n andp of the semiconductor

though the relation, = = + ... (1)The concentrations n and p are strongly

dependent on temperature Tin Kelvin.

Where Ea is called activation energy which is

related to the energy band gap of that

semiconductor. Hence, As temperature

increases, the resistance R(T) changes according

to the relation,

() = .(2)Where RO is the resistance of the thermistor at

absolute temperature To. B is a characteristic

temperature that lies between 2000K to 5000K.

The temperature coefficient of resistance isdefined as the ratio of fractional change in

resistance to the infinitesimal change intemperature .

= = = ..(3)The typical value of is about 0.05/K. It isalmost 10 times more sensitive compared with

ordinary metals. Thermistors are available from

1K to 1M.

Advantages:They are low cost, compact and highly

temperature sensitive devices. Hence are more

useful than conventional thermometric devices.

Using eq. (2) at some constant reference

temperature, say 0, the resistance will be =

Where, A represents the characteristic resistance

of the thermistor at reference (zero Kelvin). To

make the expression to look like a linear relation

to determine the values of A and B constants,

take natural logarithm on both sides of the above

expression,

ln = ln + .. (4)The exponential curve now became linear. If we

plot the variable ln 1, we will get A andB constants from the intercept and slope of the

straight line.

DESIGN OF EXPERIMENT:

PRINCIPLE: If we measure the resistance I of

thermistor at various temperatures (T), we can

plot the graph and obtain thevalues ofA and B.

How to vary the temperature T?Using an electric heater we can change the

temperature roughly from 30 to 60.How to measure the resistance R?Using Wheatstones bridge.

Wheatstones bridge principle:The circuit shown here is a Wheatstones bridge

and it consists of four resistors R1, R2, R3 and R4,

a galvanometer

(G) and a Battery(V). Suppose the

resistance R4 be

unknown. The

voltage applied

to this circuit by

the battery is

only to set up

some current and

its magnitude has

no importance,

i.e. whether or 2V or 5V it does not matter at all.Wheatstone bridge gets balanced, i.e. the

Galvanometer shows a zero deflection when,

G

R3 R4

V

R1 R2


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12 =

34 Or 4 = 231

If the resistances R1 and R2 are equal, then the

bridge will be balanced, i.e. the null deflection in

Galvanometer, when R4 = R3. If we choose R3 as

a variable resistor, like a decade resistance box,

the unknown resistance R4 will be equal to the

resistance maintained in the box.

Measurement of resistance of thermistor:Here in this experiment we employ a

1K (at room temperature) thermistor. We form

a wheatstones bridge with two fixed value

resistors each of 1K resistance along with a

variable decade resistance box. Two arms of the

bridge are occupied by 1K resistors and the

other two arms, one with thermistor and the

other with decade resistance box. The reason forchoosing a 1K fixed resistor. The sensitivity of

measurement of resistance will be better when

all the four resistors here are of same

(comparable) magnitude hence the remaining

Rs are 1K each.

Applications of thermistors:1. They are used as temperature sensing

elements in microwave ovens, heaters

and also in some electronic

thermometers.

2. Used as sensor in cryogenic liquidstorage flasks.

3. Used as compensator for providingthermal stability to transistor based

circuits.

4. Used in fire alarms, Infrared detectors assensor.

THERMISTOR EXPERIMENTAim:

1.To study the variation of resistance of a thermistor with temperature.2.To find the temperature (thermoelectric) coefficient of resistance () of the thermistor.3.To determine I and B coefficients.

Apparatus:Thermistor(1 K), electric heater (max 700 C), 1.5 volt battery or a D.C. power supply, mercury

or benzene thermometer (0 110), test tube containing insulating oil (edible oil / castor oil),resistors (1k - 2 No.s), Galvanometer (30 0 30), resistance box (1 to 1000 range),

connecting wires.

Formulae:

= =

=

Procedure:1.Construct the bridge according to the

circuit diagram (Maintain at least 1000 resistance in the Resistance box beforeconnecting the circuit, i.e. remove the 1000

plug key).

2.The 1 K resistors are already connected

on the back panel of the board. Hence no

need to connect them again.

3. If a variable D.C. source is given instead

of a battery, set the voltage to 1.5 or 2 Volt

with the help of a multimeter.

4.The bridge gets balanced (Galvanometer

shows 0 deflection) when the resistance of

thermistor gets equal to that of the resistance

box. Remove the plug keys of resistance box

and find out the null point resistance.

T

G

RB RT

1.5 V

R1=1K R2=1K

Electric heater

Test tube withCoconut oil

T

G

1.5 V

R2=1KR1=1K

RB


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5.Start heating the thermistor by turning on the heater switch on the board.

6. Measure the resistance of thermistor for every two degrees centigrade rise in temperature. Note

the readings up to 600 C in steps of 20 C.

7. At each temperature bridge is not balanced initially and it shows some deflection. It can be made

zero by adjusting the resistances in the variable resistance box. Tabulate the readings.

8. Remove the power supply or battery, soon after you complete the experiment . If you forget

doing this, it will cause the galvanometer to deflect more causing damage to its restore spring.

Graph:A graph is plotted by taking R versus T(0 C). This graph gives the value of.

Another graph is plotted between ln R and (1/T(K)). The slope of this graph gives B and its

intercept on y (ln R) axis gives ln A from which A can be calculated. But it is not possible to find

out the intercept from the graph. It can be done with the help of least square fit method as

described in the Appendix.

Use this method to compute both slope (B) and intercept (ln A) of the straight line. Here assume X as

(1/T) and Y as lnR. The intercept Cgives the value of ln A and the slope will give B (in K). From the

intercept find out the value of A (in ).

Precautions:1. Temperature of the thermistor should be

less than 700 C.

2. Thermistor must be immersed completelyinside the hot oil bath.

3. Readings of thermometer must be notedwithout parallax.

4. Connections should be made properlywithout any loose contact.5. Resistance must be varied quickly in the

resistance box to get the null point within

the 20C intervals.

6. Battery must be disconnected immediatelyafter completion of the experiment.

Viva-Voce Questions:

1. Where do you find applications of

thermistor? Name a few of them.

They are useful in temperature sensing andcontrolling equipments. Ex. Microwave

ovens, Infrared heat sensors, Liquefied gas

temperature sensors in cryogenics.

2. Explain the principle of Wheatstonesbridge.

In the bridge circuit, the potential at the two

nodes across which the galvanometer is

connected will be same when the four

resistors R1 to R4 satisfy the relation12

=34

3. After obtaining the data from this

experiment, you will have the values of A

and B coefficients. Can you determine

the temperature of your body? I will

provide you only a thermistor and a

multimeter. If yes, describe the method.

If No, justify your answer.Yes, it is possible. Suppose that you want

to measure your body temperature. Just

keep it in tight contact with your body

(cover it tightly with skin). Use the

multimeter to measure the resistance of this

thermistor. After few seconds of contact

with body, thermistor attains constant

resistance. With the known A and B

coefficients, we can measure the body

temperature by substituting in

= =

T1 T2T in0 C

R1

R2

1

in K-1

Slope = B

ln R


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REFERENCES:

1. Physics of semiconductor devices, S. M. Sze, 3rd ed, John Wiley publications, chapter 14, sensors,Thermal sensors, p.744-746.

2. The art of Electronics, Paul Horowitz, 2nd ed, Cambridge university press, chapter 15,Measurement transducers, Thermistors, p.992-993

3. Electronic devices and circuit theory, R. Boylestad, 7th ed, Prentice hall publications, Art. 20.11Thermistors, p.837-838

4. Electronic sensor circuits and projects, Forrest MimsIII, Master publishing, p.13, 46-47.

BAND GAP OF SEMICONDUCTOR USING PN JUNCTION DIODE

THEORY:PN junction diode is an example for extrinsic

semiconductor. It can be biased in both

forward and reverse directions. The current

that flow through the diode when its junction

is biased with a voltage Vwill be

= 1With = +

.

Where,

V = applied voltage across junction

Is = Reverse saturation current, a constant

dependent on temperature of junction

= a constant equal to 1 for Ge (high

rated currents) and 2 for Si (low rated

currents)

VT = Volt equivalent of temperature

=

11600, T = Temperature of junction in

Kelvin

A = area of crosssection of junction

e = elementary charge =1.6 1016CDp(n)= Diffusion constant for holes

(electrons)

= for holes and for electrons = mobility of holesLp(n)= Diffusion length for holes (electrons)

pno = equilibrium concentration of holes (p)

in the ntype material

=2

npo = equilibrium concentration of

electrons (n) in ptype material

=2

ni = intrinsic carrier concentration (/cm3)

ni2 = 3

B = a constant independent of T

EG = Energy band gap of semiconductor

(in Joule)

NA = Acceptor ion concentration (/m3)

ND = Donor ion concentration (/m3)

The term Is is highly temperature dependent.

The expression for it can be written as

= +

=

+

=

2

+

2

=2 +

= 3

+

=

11600

3 +

= 11600 2 +

Experimentally it was observed that themobility term in the bracket varies as2.Hence,

= (1) is a constant whose magnitude is in nano or

pico ampere.

Under reverse biased condition applied

voltage V will be negative and hence the

expression for current through diode will be,

=

1

=

0

1

=

Diode will have only the reverse saturation

current flowing through it. The negative sign

indicates that the current is flowing in opposite


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direction to that of forward bias. Hence the

current ID through diode in reverse bias will be

= = 0 exp . (2)Applying natural logarithms on both sides

implies,

ln (

) = l n (

0) + l n

exp

ln () = ln (0) 1T . (3)This is the equation of the straight line with

ln(ID) as ordinate(yaxis) and 1/T as abscissa

(x axis). Ln(I0) is the y intercept of thegraph. If we plot 1/T versus ln(ID) graph, its

slope with x axis gives the value of .By knowing the Boltzmann constant kB we

can evaluate the energy band gap of the

semiconductor, similarly we can estimate thevalue of Boltzmann constant if we know the

energy band gap of the given semiconductor.

Applications:

1. We can use this to make a diodethermometer.

DESIGN OF EXPERIMENT:

PRINCIPLE: If we measure the reverse

saturation current through the diode by

varying its temperature, we can plot the

graph and obtain slope .Which diode is suitable for this?

OA79, Germanium diode, used as envelope

detector in amplitude demodulation circuits.

Why this OA 79? Why not any other?Because reverse current variation is more in

the case of Germanium than with silicon.

Hence for a small temperature range of

variation (300 to 600C), it is better to choose

Ge diode than any other silicon diodes. If we

want to do this experiment with silicon diodes,we must have an electric heater capable of

giving temperatures up to 1500C.

How to vary the temperature T?Using an electric heater we can change the

temperature roughly from 30 to 60.How to measure the reverse current?

Using a moving coil micro ammeter.

Biasing the diode:Use a constant voltage D.C. power supply or a

battery to bias it in reverse direction. The

voltage applied must be very low, 2 Volt. Incase of an ideal diode the reverse current does

not vary with applied reverse voltage. But in

practical diode case, it increases with increase

in reverse voltage. This is due to the increase

of leakage currents across the junction with

applied voltage. At room temperature, the

reverse current may be small and different for

same type of diodes, but it follows the

equation (2). The values ofIo may vary from

diode to diode.

Description of heater:The heater contains an electric heating

element attached to a stainless steel container

holding some cold water. A test tube

containing oil is immersed in the water bath.

Oil is an insulator of electricity and hence it

is used for heating the diode. This also

provides uniform heating of diode. The diode

with properly insulated connecting wires is

immersed in the oil bath. Thermometer is also

kept inside the oil bath to measure its

temperature. We cannot directly insert the

diode inside the water bath as tap water

contains lots of minerals dissolved in it and

acts like conductor. This will short circuit the

diode.

Useful data:From the data sheet of the OA 79 diode:

Material of the diode is Germanium. Maximum surrounding temperature is

600C.

Maximum allowed reverse currentthrough the diode is 60A.

BAND GAP EXPERIMENT

Aim:To determine the energy band gap of the

material of the semiconductor by studying the

variation of reverse saturation current through

given PN junction diode with temperature.

Apparatus:OA 79 Ge diode, heater (max 600 C),

thermometer, test tube containing insulating

oil (edible oil or castor oil), power supply (2V

D.C.), connecting wires, micro ammeter (0 -

50 A) and a voltmeter or multimeter.

Formula:Reverse current through diode is given by

= Where, EG is the energy band gap of the

material of the semi conductor diode, T is theabsolute temperature of the diode junction and

kB = 1.38 x 10-23 J/K is Boltzmann constant.


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VD

ID

T

Stainless steel container water bath

Test tube containing oil

A2 V

++

__

+

Electric Heater

=

Circuit diagram:

Caution: Set the applied reverse bias voltageat 2 Volt. Do not increase this value more. Do

not heat the diode beyond 600C.

Procedure:

1. Build the circuit as shown in the circuitdiagram.

2. Observe the initial temperature of thethermometer. If it is high (>300 C) then

replace the water in the heater jar with

some cold water and try to reduce the

temperature below 300 C.

3. Apply the reverse voltage (2 Volt) byadjusting the potentiometer (if a battery isgiven, then there is no need of doing this

adjustment).

4. Switch on the heater. Note down thereverse current in the micro ammeter for

say, every 20C rise, in temperature of the

diode (if micro ammeter is not available,

you can use a multimeter in D.C. current

mode under 200 A ranges).

5. Tabulate the readings.6. Complete the calculations relevant to the

tabular form and get the answer for slope.

7. Plot a graph between lnI and 1/Tto obtainits slope.

8. Calculate the EG from both slopes obtainedfrom graph and table.

Precautions:

1. Readings of thermometer must be notedwithout any parallax error.

2. Reverse bias voltage must be regulated at 2Volt throughout the experiment.

3. Diode should be completely immersed insidethe oil bath.

GRAPH:Plot a graph by taking the values of ln I vs

1/T. Find out the slope of the curve. Do not

consider the origin of this graph.

Usually we start at 300K and go up to 333K,

hence 1/T varies roughly from 2.99

103

1 to 3.33 103

1. So start at 2.98

and go up to 3.34 by choosing the scaleOn 1/T axis as

1 = 0.02 1031Usually ID varies from 2 A to 60 A. So ln

ID varies roughly from to13.2. So start at

9.7 and go up to 13.2 by choosing the

scale

On lnIaxis as

1 = 0.1Slope (EG/kB) can be calculated from both

straight line data fit as well as from thegraph.

Viva-Voce questions:

1.Distinguish intrinsic and extrinsic

semiconductor.If the semiconductor material consists of no

impurities (dopants), then it will BE intrinsic

(pure) semiconductor. If it contains dopants

{acceptor type [p-type] III group elements

or Donor type [n-type]IV group elements}then it will be an extrinsic semiconductor.

2.What are the band gaps of Silicon and

Germanium?

For silicon;in electron volts = 1.21 3.60 104; For Ge, = 0.785 2.23 104. T is the temperature of thesample in Kelvin. At 300K, EG = 0.72 eV for

Ge; EG= 1.1 eV for Si.

3.How do you test the diode for its polarity

using a multimeter?

There will be a symbol of diode on themultimeters mode changing dial. Turn the

dial to diode testing mode. Connect the two

leads of the multimeter to the two leads of


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the diode. If the multimeter shows infinite

resistance (it shows a 1 Or OL means

out of range, very large), then it is reverse

biased and the terminal of diode that is

connected to positive (red probe) of

multimeter will be the cathode of the diode

and the other one will obviously be the

anode. Similarly, if the meter shows somefinite resistance like few hundreds (150, 540

etc), then it is forward biased, i.e. the

terminal of diode that is connected to positive

(red probe) of multimeter will be the Anode

of the diode and the other one will be the

Cathode. During this process, multimeter

applies some known voltage across its leads

and measures its resistance.

4. If I reveal the material of the diode used,

can you estimate the Boltzmann constant

from this experiment? If yes, describe how

do you do it, if no, say why?

(Think and answer)

5.Why do we observe small current (of the

order of Micro amp) in this experiment?

What are responsible for this small

current?

Because reverse current is due to the

minority carries only. As their number is

very small the current is also small.

6. In which biasing of diode are you doingthis experiment?Reverse bias.

7.Can you determine the band gap by

changing the bias of the diode? If yes,

describe how you do it. If no, explain

why. (think and answer)

8. If I give you a silicon diode and the same

experimental set up (micro ammeter 0-

50range), can you find out its band gap?

Justify your answer.No, the reverse current variation is very

small of the order of few nano amperes per

degree centigrade and hence it not possible

to observe the variation in reverse current

with the micro ammeter for a temperature

range of 30-600 C

9.What is the magnitude of reverse current

in silicon?Few nano amperes.

10. Can you make a diode thermometer

using this setup? If yes, say how? If no,

say why?

Yes, once if we know the value of I0

(antilog of intercept of lnI vs 1/T graph)

from the experiment, we can measure the

T. Just bring the diode in contact with the

body whose temperature is to be measured

and measure the reverse current (ID)

accurately. As we know the I0 and ID we

can determine the T in Kelvin for that body

using the relation = 0exp .

References:

1. Electronic devices and circuits, Millman and Halkias, McGraw hill student editionp.126-132.

2. Semiconductor device physics and technology, SM Sze, M K Lee, 3 rd Ed, John wiley,P.107


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RC CIRCUITTIME CONSTANT

THEORY:Resistor and capacitor combination circuits

have got great importance in the field of both

electrical and electronics engineering. This is

the fundamental circuit to understand theworking of many complex electronic circuits.

Consider a resistor of resistance R and

a capacitor of capacitance C. They can be

biased with an external D.C. source to start the

process of charging.

Suppose that switch S is closed.

Apply Kirchhoff voltage law (KVL) to the

above circuit of Charging. () () = 0Where, the sub scripts R and C denote the

voltages across the resistor and capacitor. If

we assume a current of i(t), a function of time,then from Ohms law, = .If the instantaneous charge on the plates of the

capacitor is, then = () , Where, Cis the capacitance of the capacitor.

Substituting them in above eqn. implies,

() = 0;But, is equal to the rate of change ofcharge in the circuit, i.e.

= ()

() = 0Differentiate this expression with respect to

time t.

22

1

= 0Voltage V across battery does not change with

time,

0 2()

2

1

()

= 0

22 +1

= 0

Put trial solution, = 0 exp = 0 exp =

2

2 = 20 exp = 2 Going back to differential equation,

22 +1

= 0

2 + 1 = 0 = 1 (Assuming 0)This implies

=

0 exp

1

Voltage across capacitor will be,

= = exp

1

= , when capacitor charges to maximumvalue q0, the voltage across it will be V.

Hence,

= exp 1 Current will be,

= = 0 exp 1 = exp 1 ; 0 = 0

= 0 exp Hence,

At t = 0, is , the maximum currentflowing through the circuit. Using, = 0

= = 0 exp Where, i0 is the current in the circuit at t=0,

i0R represents the initial voltage across the

resistor, i.e. V;

= 1 exp = = 0 exp

= exp

S

V

R

C VC

VR


9/11



DISCHARGING OF CAPACITOR: Similarly for discharging of capacitor, KVL

gives + () = 0Proceeding in the same manner as above, we

can show that voltage across the capacitor

during discharging will be

() = exp

GRAPHS:

Aim:

To study the Resistor and Capacitor series (R-C) circuit and hence to determine the time

constant of the circuit from the charging and discharging curves.

Apparatus:

Resistors, capacitors, Toggle switch and a D.C. power source.

Miscellaneous: Multimeter, bread board, connecting wires and stop watch.

Circuit

diagram

TimeTime

DURING

VMaxVMa

0.63

DURING

VMaxV

TimeTime

0.37

R

ES

I

S

T

O

R

DISCHARGINGDURING

x

VMax0.63

CHARGINGDURING

Max

VMax0.37

C

A

P

A

CI

T

O

R

S

V

R

CV

C

VR


10/11



s:

Charging:

Arrows in the circuit show the direction of flow of current during the process.Voltage across capacitor during CHARGINGat any time t from the start of the process is given by

Voltage across capacitor during DISCHARGING at any time t from the start of the process is

given by

Formulae:

= 1 Or = 1 .. While charging and = Or = .. While discharging

Components used in this experiment are:

RESISTANCE R = .. ; CAPACITANCE = .. FARADTime constant of the circuit is = RC (theoretical) =.. Seconds

Procedure:Charging process:1. Connect the circuit on the bread board as shown in the above figure (charging).2. Calculate the theoretical value of the time constant using the above formula.

__

VC

+RR

VVCharging of capacitorDischarging

+

00.0Volt DC

D.C.Ran e: 20 V

-+_

+

VC

of capacitor

+

11.92Volt

D.CRange: 20 V

-+

Discharging:


11/11



3. Switch on the circuit and simultaneously switch on the stop watch and start counting of time.Sometimes it is difficult to switch on the stop watch simultaneously while closing the key wire.

So, join the key wire to power bus without any hesitation to start the stop-watch. If we join the

ends of the capacitor with a wire, it gets discharged and the voltage across it becomes zero. As

the multimeter leads are connected to the ends of the capacitor,join the leads of the multimeter

to discharge the capacitor. This will bring the voltage across C to zero. Now hold the two

leads of multimeter together until you switch on the stop watch and thereafter leave them

separate.4. Measure the voltage across the capacitor with multimeter in regular intervals of time (say 10 or

20 sec) and tabulate it. {it can be decided on the basis of time constant, if time constant is about

220 sec, then we can go in steps of 20 sec; if it is about 100 sec or less, we can go in steps of 5 or

10 sec}.

5. Take the readings until the capacitor charges to maximum voltage (of power supply).Discharging process:6. Repeat the same process by connecting the

switch

wire to the ground and simultaneously switch on the stop watch and start counting of time. There

will be a problem similar to the one which you have encountered while charging the

capacitor. If we connect the key wire to ground, instantaneously the capacitor discharges

and the voltage start falling down. Dont be in a hurry to switch on the stop watch. Justtake the positive lead of the multimeter which is already connected to the positive plate of

the capacitor and join it directly to the power bus and hold it. This will charge the capacitor

to the power supply voltage. Now switch on the stop watch

7. and simultaneously remove themultimeter lead from the power bus.

Dont forget to remember the valueshown by the multimeter while

switching on the stop watch. This will be

the reading at t=0 sec, whiledischarging. Again note down the values

of the voltage across capacitor in the same

intervals of time as above until some

minimum value of voltage appear across it.

(Say about one volt)

Graph: Plot a graph by taking the voltage

across capacitor versus time both for charging

and discharging cases.

From the graph calculate the timeconstant of the circuit. During charging, the

time taken for the capacitor to charge to 63%

of the maximum value is its time constant.

Similarly during discharging, time taken by

the capacitor to discharge to 37% of its

maximum value is also its time constant.

Time

Discharging of capacitor

Voltageacro

ssC VMax

0.63 VMax

Charging of capacitor

VMax

Time

0.37 VMax

total text of second sem record

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