Problems on Electrical Properties Solutions

Problems on electrical properties

1. A pure pure alum alumin inum um inter interco conn nnect ect with with a semic semicirc ircul ular ar cros cross-s s-sec ectio tion n runs runs betw betwee een n two two transistors transistors on an integrated circuit. circuit. The interconnect interconnect is 10 microns long and has a radius of  200 nm. Across its length is a voltage of 5 V. The room-temperature conductivity of pure Al is .! "10# $% m& -1.  a& 'etermine 'etermine the current density density flowing flowing through through the interconnec interconnect. t.  b& The operating operating tempera temperature ture of the the device is is (0 o). *ould *ould you e"pect the current densit y flowing through the interconnect to be higher+ lower+ or the same as that at room temperature, riefly ustify your answer.  c& /uppose an Al - 5 wt )u alloy was used in place of pure pure aluminum. *ould *ould you e"pect the current density flowing through the interconnect to be higher+ lower+ or the same as that for a pure Al interconnect, riefly ustify your answer. ans wer. a& −(

L 1  L 1 10 · 10 = 2.! Ω  R=ρ = σ = # − 1 ·  A  A .! · 10 (Ω · m ) π · ( 200 · 10− m )2

 J  =

5V  I  = V  = = 1.# · 101  A / m 2 −  2  A  R · A 2.! Ω · π · ( 200 · 10 m )

 b&  increases with T+ T+ because the increased atomic vibration amplitude increases the  probability of electron scattering. 1

 ∝ ρ ⇒  J decreasesasT increases  J  ∝ c&  increases as the the Al is alloyed alloyed with with copper+ copper+ because because the substitu substitutional tional copper copper atoms atoms introduce some amount of local strain in the Al 3)) lattice and this local strain increases the  probability of electron scattering. Therefore+ 4 decreases when )u is added.

2. A resistiv resistivee heating heating element element is to be built built from from tungst tungsten en wire 0.05 0.05 mm in diamet diameter er.. The element should dissipate 50 *atts *atts of power. The resitivity of tungsten is  m6cm.  a& 7stimate the length of the wire re8uired for such a device operating at 110 110 V. V.  b& /uppose the temperature temperature dependence of the resistivity is included included in the calculation. *ould the design now call for a longer or shorter wire relative rela tive to when the Tdependence was ignored, a&

2

 2

 2

 2

V  V  · A V  · A  P = I · V  = = ⇒ L = =  R ρ· L ρ · P 

( 110 V  ) · ⟦ π · ( (  · 10

−5

5 · 10

−5

2

m

2

)⟧

Ω · m ) · ( 50 W )

= 5. · 10− m =5. mm

 b&  increases with increasing T+ T+ so a slightly shorter wire could be used.



 . A 120 120 V source is placed directly directly across a tungsten tungsten filament 10 m in length and 50 microns microns -5 in diameter. $'ata9 resistivity of * is 610  6m+ heat capacity of * is c: 25 4;mol6<+ mass density 1.25 g;cm &  a& *hat *hat current current is passed passed initia initially lly,,  b& *hat is the power power in *atts,  c& =f the filament filament were perfectly perfectly insulated insulated thermally thermally and allowed allowed to heat heat due to the resistance+ apro"imately for how much will the temperature change if the initial current flowed for 1 second, d& *hat would would the resistance resistance be at this temperature temperature if the temperature temperature coefficient coefficient of o resistance of tungsten can be appro"imated as >:1;$250 )& , a&

V  V · A  I = = = ρ l   R

 b&

120

V · π· (

 · 10

 =( 2(2 · 10 P = I · V  =(

−(

−5

50 · 10

−(

m

2

2

)

Ω · m · 10 m

= 2(2 μ  A

 )= 1.!! mW   A ) · ( 120 V  )=

c& Q = m · c · Δ T = P · t = I · V · t  −  ) · ( 1 s ) Δ T  =  = I ·V ·t = ( 2(2 · 10  A ) · ( 110 V  ) (

m· c

 Densidad · Volumen Volum en · c

( 2(2 · 10−  A ) · ( 110 V ) · ( 1 s ) (

Δ T = 1.25



2

g / cm · ⟦( 10 · 10 cm ) · (

50 · 10

−!

2

cm

=1.? K 

2

) ⟧ · 0.1! J / g · K 

d&  R =ρ ·

l  =ρ 0 ( 1+α · Δ T ) · l   A  A

 R =  · 10

−5

Ω · m · ( 1+

1 250

 · 1.? ºC ) · ºC 

10

[ π· (

m

50 · 10 2

−(

m

2

)]

=0.! M  Ω  Ω

 !. 3or intrinsic intrinsic gallium arsenide+ arsenide+ the room-tempera room-temperature ture electrical conducti conductivity vity is 10 -( $6m&-1@ the electro electron n and hole hole mobilit mobilities ies are+ respecti respectively vely++ 0.?5 0.?5 and 0.0! 0.0! m 2;V6s. )ompute the intrinsic carrier concentration ni at room temperature. ecause the material is intrinsic+ carrier concentration may be computed as9 − (Ω · m )− 10 σ = n i= ∣e∣· (μ e +μ h) ( 1.( · 10− C ) ·[( + (

1

0.?5

1

0.0! )  m

= # · 10 m− 12

2

/V · s ]



 5. To high-puri high-purity ty silicon silicon is added added 10 10 2 m- arsenic atoms.  a& =s this this material material n-type n-type or p-type, p-type,  b& )alculate )alculate the room-temperature room-temperature electrical electrical conductivity conductivity of this material. material. c& )ompute )ompute the the conductivi conductivity ty at # <. $'ata9 e$2# <&:0.0# m 2;V s+  e$# <&:0.0! m 2;V s a& Arsenic Arsenic is a Broup VA VA element element and+ therefore therefore++ will act as a donor in silicon+ silicon+ which means that this material is n-type.  b& At room temperature $2? <& we are within the e"trinsic temperature region+ which means that virtually all of the arsenic atoms have donated electrones $i.e. n : 10 2 m-&. 3urthermore+ inasmuch as this material is e"tinsic n-type+ conductivity may be computed using 78uation 1. )onse8uently+ it is necessary to determine the electron mobility for a donor concentration of 102 m-. Thus+ conductivity is ust9

σ =n ·∣e∣· μ e ( !uation 1 ) σ =( 10 m − ) · ( 1.( · 10− C ) · ( 0.0# m / V · s )= 1120  (Ω  ( Ω · m )− 2



1

2

1

c&

σ =( 10 m − ) · ( 1.( · 10 − C ) · ( 0.0! m / V · s )= (!0  (Ω  ( Ω · m )− 2



1

2

1

 (. =f conductivi conductivity ty of semicond semiconducting ucting material material 250 250 $ m&-1 at 20 o) and 1100 $ m& -1 at 100 o)+ *hat is the magnitude of its band gap, $C :1.? 10 -2 4;<+ 1 eV:1.( 10 -1 4&   g 

ln ( σ T  )= ln (σ 0)−

2C  " T 1

1

ln ( σT  )= ln (σ 0)− 2

  g  2C  " T 2

( a)

( #)

/ubtracting $b& from $a&+ thus

−   g  2<  "

σ(T  ) ( ln ( σ ))

( ln (

1

=

(T 2)

(

1

T 1

−

1

T 2

)

−(

⇒   g =( 2 · ?(.2 · 10 eV  / / K )

(

1 #

1100

−1

 K 

250

−

))

1 2

=0.! eV  −1

 K 

)

 #. )onsider )onsider a capacitor capacitor with a plate plate area of 1 mm " 1 mm and and a plate separation separation of 100 100 mm. =f  the gap is filled with a titanate whose dielectric constant is 50+ what is the ma"imum charge that can be stored at when for an applied voltage of 5 V. =f the titanate is removed and replaced replaced by air+ air+ what plate area will be re8uired to store this same amount amount of charge, charge, $data -12 D0:?.?5!610  3;m& −(

Q  = ε A ⇒ Q =ε 0 · εr  ( V · A ) = ?.?5! · 10−12 $  · 50 ( 5 V · 10 m C =  =ε ( 0.1 m ) V  l  l  m

2

)=

2.21 · 10

−2

 %C 

=f the titanate is removed and replaced by air+ the area re8uire to store the same amount of charge is9

 ( Q · l ) ( 2.21 · 10− C · 0.1 m )  A & = = =5 · 10− m $  (V · ε ) (5 V · ?.?5! · 10− ) 1!

5

0

12

m

2

 ?. The polariEation polariEation F of of a dielectric material material positioned positioned within a parallel-plate parallel-plate capacitor capacitor is 10-( );m2. $data D0:?.?5!610-12 3;m&  a& *hat is the the dielectric dielectric constant constant if an electric electric field field of 5610 5610! V;m is applied,  b& *hat will be the dielectric displacement ', a& −(

εr =

2

( 10 C / m ) P  +1= +1 =.2( (ε 0 ·  ) ( ?.?5! · 10−12 $ / m · 5 · 10 ! V  / / m )

 b& −12

 D=ε0 ·  + P =?.?5! · 10

!

−(

 $  / m · 5 · 10 V  / m+10

2

−(

C  / m =1.!! · 10

C / m

2