ECE 474: Principles of Electronic Devices · V.M. Ayres, ECE474, Spring 2011 New: Doped Si @ 300K + laser light Given: 1017 cm-3 e- hole pairs (EHPs) are generated every microsecond

ECE 474:Principles of Electronic Devices

Prof. Virginia AyresElectrical & Computer EngineeringMichigan State [email protected]

V.M. Ayres, ECE474, Spring 2011

Lecture 23:

Chp. 04

Photo-generated carriers – IIIDiffusion current

Examples of each


Lecture 23:

Chp. 04


Examples of each


Photo-generated carriers:How to deal with the electron and hole increases that are due to light

Total electrons: n = n0 + δn

Total holes: p = p0 + δp


New:Doped Si @ 300K + laser light

Given: 1017 cm-3 e- hole pairs (EHPs) are generated every microsecond by laser light on the n-doped Si. τn = τp = 1 μsec. Find the increases in the e-and hole concentrations δn and δp:

goptical = 1017 cm-3 EHP10-6 sec

= 1025 cm-3 s-1

δn = gop τn= (1025 cm-3 s-1)(1 x 10-6 s)= 1019 cm-3

= 100 x 1017 cm-3

δp = δn = 100 x 1017 cm-3

Usual: Doped Si @ 300K

Nd = 1017 cm-3 P atomsFully ionised: Nd = Nd

+

Therefore:

Majority carrier concentration:n0 = 1017 cm-3

Minority carrier concentration:p0 = 2.25 x 103 cm-3


Comparison:Doped Si @ 300K versus Doped Si @ 300K + laser light

n0 = 1x1017 cm-3

n = 101 x 1017 cm-3

=> increase factor: 102

p0 = 2.25x103 cm-3

p ≈ 100 x 1017 cm-3


Some increase in majority carrier concentration and a HUGE increase in minority carrier concentration!


Fermi energy level => 2 Quasi-Fermi energy levels Fn and Fp .(They’re called “quasi” because they only exist as long as the light is on. Especially the one for the minority carriers.)

EF

Ei

EV

EC

Egap=

1.1 eV

n = ni exp(Fn-Ei)/kT

p = ni exp(Ei-Fp)/kT

(note: np = ni2 does not work)

0.407 eV

For this example:

Fn

Ei

EV

EC

Egap=

1.1 eV

Fp

0.526 eV

0.526 eVFn

Ei

EV

EC

Egap=

1.1 eV

Fp

0.526 eV

0.526 eV


Time dependence is possible when EHPs are created by an ON/OFF flash of light rather than a continuous application of light. Look for clue in the problem wording: “ created at t = 0”.

Fig. 4-7 is a flash situation.The x-axis is time (ns).


Aside: note that this Streetman example is for a p-type semiconductor.


New:Doped Si @ 300K + laser light

Given: 1017 cm-3 e- hole pairs (EHPs) are generated every microsecond by laser light on the n-doped Si. τn = τp = 1 μsec. Find the increases in the e-and hole concentrations δn and δp:

goptical = 1017 cm-3 EHP10-6 sec

= 1025 cm-3 s-1

δn = gop τn= (1025 cm-3 s-1)(1 x 10-6 s)= 1019 cm-3

= 100 x 1017 cm-3

δp = δn = 100 x 1017 cm-3

Usual: Doped Si @ 300K


+

Therefore:



This example assumes that gop never changes its value


Example 01 (time):Doped Si @ 300K + laser light flash

Given: optical generation rate goptical(t=0) = 1025 cm-3 s-1

δn(t=0) = gop (t=0) τn= (1025 cm-3 s-1)(1 x 10-6 s)= 1019 cm-3

= 100 x 1017 cm-3

δp(t=0) = δn(t=0) = 100 x 1017 cm-3

n(t=0) = Δn = 101 x 1017 cm-3


p(t=0) = Δp ≈ 100 x 1017 cm-3


Doped Si @ 300K


+

Therefore:



Even if gop(t) does change,its value at t=0 is calculated exactly the same:




δn(t=0) = gop (t=0) τn= (1025 cm-3 s-1)(1 x 10-6 s)= 1019 cm-3

= 100 x 1017 cm-3

δp(t=0) = δn(t=0) = 100 x 1017 cm-3

n(t=0) = Δn = 101 x 1017 cm-3


p(t=0) = Δp ≈ 100 x 1017 cm-3


Doped Si @ 300K


+

Therefore:



Note: same n and p, butspecial new names at t = 0:

V.M. Ayres, ECE474, Spring 2011 Time (ns)

1020

105

100

1010

1015

Example 01:Put a dashed line across for the n0 and p0 values

Put a dot at t = 0 for then(t=0) ≡ Δnp(t=0) ≡ Δp

values

0

n0

p0

Δn

Δp


1020

105

100

1010

1015



values.

Over time, the light generated carriers are not maintained. So the concentration drop back down to just the doped + temp concentrations

0

n0

p0

Δn

Δp




δn(t=0) = gop (t=0) τn= 109 cm-3

δp(t=0) = δn(t=0) = 109 cm-3

n(t=0) ≈ 1017 cm-3

p(t=0) ≈ 109 cm-3

Doped Si @ 300K


+

Therefore:




1020

105

100

1010

1015

0

n0

p0

Δn

Δp



values


1020

105

100

1010

1015

0

n0Δn

Δp

p0



values.



Streetman Example:


Streetman Example:

Majority carrier concentration:p0 = 1.0 x 1015 cm-3

Minority carrier concentration:n0 = (2 x 106 cm-3)2/1015 cm-3

= 4 x 10-3 cm-3

δn(t=0) = Δn = gop(t=0) τn= (gop cm-3 s-1)(1 x 10-8 s)= 1014 cm-3

δp(t=0) = Δp = gop τn = 1014 cm-3

p(t=0) = 1.0 x 1015 cm-3 + 1014 cm-3

= 1.1 x 1015 cm-3

n(t=0) = 4 x 10-3 cm-3 + 1014 cm-3

≈ 1014 cm-3


1015

100

10-5

105

1010

0

n0

n

p

Δn

Δp p0

Streetman Example 03:Put a dashed line across for the n0 and p0 values


values


1015

100

10-5

105

1010

0

n0

p0

np

Streetman Example 03:Put a dashed line across for the n0 and p0 values


values.


Δn

Δp


1015

100

10-5

105

1010

All that remains is to work out an expression for the dramatic drop of the minority carrier back to its doped + temp concentration.

Follow the derivation in Streetman pp. 125-126.

For the “low level injection”conditions, it is:

If Holes p(t) = MajorityThen Electrons n(t) = Minority

0

n0

p0

np

n

t

ntn τ−

Δ= exp)(

n

t

ntn τ−

Δ= exp)(

Δn

Δp


1020

105

100

1010

1015

0

n0n

p

p0

In our Example 02:

If Electrons n(t) = MajorityThen Holes p(t) = Minority

p

t

ptp τ−

Δ= exp)(

p

t

ptp τ−

Δ= exp)(

Δn

Δp

Note: usual in Streetman:τp = τn


Lecture 23:

Chp. 04


Examples of each


High power n-channel field effect transistor:Note Idrift ____ and Idiffusion ____ regions.

n np

Wilkipedia

OFF


High power n-channel field effect transistor:Note Idrift ____ and Idiffusion ____ regions.

n np

Wilkipedia

ON


Si

np

n

Si

np

n

Si

np

n

OFF

I

V

I

V

Transistor: a device for binary logic (Chp. 06)

ONSaturation regime

Typically not used:Linear (ohmic) regime

I

V


Si

np

n Getting a good OFF is Key

OFF is achieved by this 2 pnjunctions + depletion region energy barrier design(Chp. 05)

OFF is NOT achieved by just using the bandgap of Si

= 1.11 eV


ON and Diffusion Current (Chp. 04):

VDS = IRWhere is the line of charge that creates the potential drop?

Si

np

nn

VDS

Si

np

nn

VDS


ON and Diffusion Current:

VDS = IRWhere is the line of charge that creates the potential drop? At the Ohmic contacts (Chp. 05)

Si

np

nn

VDS

Si

np

nn

VDS


ON and Diffusion Current:

VDS = IRThe rest (blue) is just ordinary n-type material.Carriers move by diffusion, not VDS = IR: no VDS.

Si

np

nn

VDS

Si

np

nn

VDS


Idrift = q(n0 <velocityelectron>+ p0<velocityhole>) Area

Idrift = q(n0 μn + p0 μp ) E Area

Jdrift = Idrift/Area = q(n0 μn + p0 μp ) E = σ E

From Lecture 21:

There’s more:


Diffusion Current:(Si @300K, Doping yes, laser light no)

Consider putting 2 different n-type doping concentrations together in the same block of Si. What happens?

n0

= 1017 cm-3

n0

= 1013 cm-3

x

z

y


Consider putting 2 different n-type doping concentrations together in the same block of Si. What happens?

n0

= 1017 cm-3

n0

= 1013 cm-3

Lots of e- Less e-

x

z

y

Diffusion Current:


n0

= 1017 cm-3

n0

= 1013 cm-3

Let’s go! e- Less e-

A current Idiff starts to flow based on the concentration imbalance

x

z

y

Diffusion Current:


n0

= 1017 cm-3

n0

= 1013 cm-3

NOTE: now you are piling up P+

P+ e-x

z

y

Diffusion Current:


n0

=

1017

cm-3

P+ e-

P+ e-

P+ e-

P+ e-

n0

=

1013

cm-3

An internal E-field E develops that hangs on to the e- , so the diffusion comes to a halt: Idiff = Jdiff = 0

+ E -

Diffusion Current:

x

z

y


n0

=

1017

cm-3

P+ e-

P+ e-

P+ e-

P+ e-

n0

=

1013

cm-3

Another way to say that an internal E-field E develops: an energy barrier is created.

+ E -

Diffusion Current:

x

z

y


Example 01: describe the energy barrier: individually:

EF

Ei

EV

EC

Egap=

1.1 eV

EF

Ei

EV

EC

Egap=

1.1 eV

0.407 eV0.2 eV

n0= 1017 cm-3 n0= 1013 cm-3


Line the Fermi energy levels up

EF

Ei

EV

EC

Egap=

1.1 eV

EF

Ei

EV

EC

Egap=

1.1 eV

0.407 eV 0.2 eV

n0= 1017 cm-3 n0= 1013 cm-3


EF straight across

EF

Ei

EC

EV

Egap=

1.1 eV

EF

Ei

EV

EC

Egap=

1.1 eV

0.407 eV 0.2 eV

n0= 1017 cm-3 n0= 1013 cm-3


0.2 eV

Connect: far left to far right:

EV

EC

Egap=

1.1 eV

EF

Ei

EV

EC

Egap=

1.1 eV

0.407 eV

n0= 1017 cm-3 n0= 1013 cm-3

Ei


0.2 eV

Finished diagram:

EV

EC

Egap=

1.1 eV

EF

Ei

EV

EC

Egap=

1.1 eV

0.407 eV

n0= 1017 cm-3 n0= 1013 cm-3

Ei

Documents

ECE 474: Principles of Electronic Devices · V.M. Ayres, ECE474, Spring 2011 New: Doped Si @ 300K + laser light Given: 1017 cm-3 e- hole pairs (EHPs) are generated every microsecond