ECE305#Homework#SOLUTIONS:#Week#8# - nanoHUB/uploads/Week8HW_Solutions5.pdf · Mark%Lundstrom% %...

transcript

Mark Lundstrom 10/17/2014

ECE-‐305 Fall 2014 1

ECE 305 Homework SOLUTIONS: Week 8

Mark Lundstrom Purdue University

(10.31.14) 1) A silicon diode is asymmetrically doped at N D = 10

19 cm-‐3 and N A = 1016 cm-‐3. (Note

that at N D = 1019 the semiconductor is on the edge of degeneracy, but we can assume

that non-‐degenerate carrier statistics are close enough for this problem.) Answer the following questions assuming room temperature. Assume that the minority electron and hole lifetimes are

! n = ! p = 10

"6 s. The lengths of the N and P regions are

L = 500 µm and L >> xp ,xn . Assume an “ideal diode” and answer the following questions.

1a) Compute JD = ID A , the diode current density at a forward bias of VA = 0.5 V.

Solution: Since this in a one-‐sided junction with N D >> N A , essentially all of the current is due to electrons injected into the P-‐region.

JD = q

ni2

N A

DnLn

eqVA kBT !1( ) = J0 eqVA kBT !1( ) (*) From Fig. 3.5 on p. 80 of SDF, µn = 1248 cm

2 V-s at N A = 1016 .

Using the Einstein relation, we find

Dn =

kBTq

µn = 0.026!1248 = 32.4 cm2 /s

The diffusion length is:

Ln = Dn! n = 32.4"10#6 = 57 µm


ECE-‐305 Fall 2014 2

HW8 solutions (continued): Since Ln


ECE-‐305 Fall 2014 3

HW8 solutions (continued): 2) Consider the diode in problem 1) above and answer the following questions.

2a) If the diode is biased such that JD = 10!6 A/cm2 , what is VA ?

Solution:

Begin with JD = q

ni2

N A

DnLn

eqVA kBT !1( ) = J0 eqVA kBT !1( ) In moderate forward bias, we can drop the -‐1:

JD = J0eqVA kBT

so

VA =

kBTq

lnJDJ0

!"#

$%&

Putting in numbers:

VA = 0.026ln

10!6

9.1"10!12#$%

&'(= 0.302 V

VA = 0.302 V

2b) If the temperature changes from 300 K to 301 K, how much does VA change? Solution: Differentiate the expression that we obtained above:

dVAdT

= ddT

kBTq

lnJDJ0

!"#

$%&

'()

*)

+,)

-)=

kBq

lnJDJ0

!"#

$%&+

kBTq

J0JD

ddT

JD J0`.1( )

dVAdT

=kBq

lnJDJ0

!"#

$%&'

kBTq

1J0

dJ0dT

!"#

$%&

dJ0dT

= ddT

qni

2

N A

DnLn

!

"#$

%&

The strongest part of the temperature-‐dependence comes from the exponential factor in ni

2 , so we can ignore the temperature dependence of the diffusion coefficient, the diffusion length, and the effective densities-‐of-‐states and write:


ECE-‐305 Fall 2014 4

HW8 solutions (continued):

J0 = Ke!EG kBT

then

1J0

dJ0dT

= 1Ke!EG kBT

Ke!EG kBTEG

kBT2

"

#$%

&'=

EGkBT

2

"

#$%

&',

which can be used to find

dVAdT

=kBq

lnJDJ0

!"#

$%&'

kBTq

1J0

dJ0dT

!"#

$%&=

kBq

lnJDJ0

!"#

$%&'

kBTq

EGkBT

2

!

"#$

%&

so

dVAdT

=kBq

lnJDJ0

!"#

$%&'

EG qT

!"#

$%&.

Putting numbers in:

dVAdT

= 1.38!10"23

1.6!10"19ln 10

"6

9.1!10"12#$%

&'(" 1.12

300#$%

&'(

dVAdT

= 1.00!10"3 " 3.73!10"3 = "2.7 !10"3

dVAdT

! "3 mV/K

So we need to lower the applied bias about 3 millivolts to keep the current constant as the temperature increases 1 K or 1 degree C. PN junctions can be used as thermometers because the diode current depends sensitively on temperature. A more careful treatment would include the temperature dependencies of the bandgap, effective densities-‐of-‐states, diffusion coefficient, etc., but the result would be close to the value obtained here.


ECE-‐305 Fall 2014 5

HW8 solutions (continued): 3) The sketch below shows the carrier concentrations in a PN junction at room

temperature. Answer the following questions.

3a) Is the diode forward or reverse biased? Explain your answer.

Solution: Forward biased because there are excess electrons on the P-‐side and excess holes on the N-‐side.

3b) What is the acceptor concentration on the P-‐side?

Solution: NA = 1016 cm-3

3c) What is the donor concentration on the N-‐side?

Solution: ND = 1014 cm-3

3d) What is the intrinsic carrier concentration?

Solution:

n0 p0 = ni2

On the P-‐side: n0 p0 = 1016 !107 = 1023 ni = 10

23 = 3.16!1011 cm-3

On the N-‐side: n0 p0 = 1014 !109 = 1023 ni = 10

23 = 3.16!1011 cm-3

ni = 3.16!10

11 cm-3


ECE-‐305 Fall 2014 6

HW8 solutions (continued): 3e) Do low level injection conditions apply?

Solution: YES.

On the P-‐side: !n "xp( ) = 1010


ECE-‐305 Fall 2014 7


4a) Is the diode forward or reverse biased?

Solution: Forward biased because Fn > Fp .

4b) What is the value of the applied bias?

Solution:

qVA = Fn ! Fp

VA = +0.5 V

4c) What is the bandgap of the semiconductor?

Solution: Reading from the graph:

EC ! EV = 1.25 eV

4d) What is the built-‐in potential of the junction.

Solution: From the plot:

Vj =Vbi !VA = 0.25 V

Since: VA = +0.5 V

Vbi =Vj +VA = 0.75 V

Vbi = 0.75 V

5) Consider the Si diode of prob. 1) and make one change. The minority carrier lifetime

increases by a factor of 1000, so ! n = ! p = 10

"3 s. The lengths of the N and P regions

are still L = 500 µm and L >> xp ,xn . Assume an “ideal diode” and compute the

forward-‐biased current density at VA = 0.6 V.

Solution: We recognize that the minority carrier diffusion length will change.

Ln = Dn! n = 32.4"10#3 = 1800 µm

Now Ln >> L , so the long base diode of prob. 1) has become a short base diode. Instead of eqn. (*) in prob. 1), we have to replace the minority carrier diffusion length by the length of the P-‐region:


ECE-‐305 Fall 2014 8


JD = q

ni2

N A

DnL

eqVA kBT !1( ) = J0 eqVA kBT !1( )

Write the saturation current density as

J0 = q

ni2

N A

DnL

= qni

2

N A

DnLn

!"#

$#

%&#

'#

LnL

The term in brackets was computed in prob. 1a), and so was the diffusion length. Using these results:

J0 = q

ni2

N A

DnLn

!"#

$#

%&#

'#

LnL

= 9.1(10)12{ }( 57500 = 1.04(10)12 A/cm2

The current becomes:

JD = J0 e

qVA kBT !1( ) = 1.04"10!12 e0.6/0.026 !1( ) = 1.04"10!12 1.05"1010 !1( ) = 1.1"10!2 JD 0.6 V( ) = 1.1!10"2 A/cm2 Note: The lifetime is longer, so the current density must be smaller – this is a sanity check. Key point to remember: Long base diode: L >> Ln

JD = q

ni2

N A

DnLn

eqVA kBT !1( )

Short base diode, L

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