Steady-State Illumination
•
x
The rate equation which results from the uniform
illumination condition:
∂ (∆n p )
∂t
= DN
∂ 2 (∆n p )
∂x 2
−
( ∆n p )
τn
+ GL
GaAs
∆n p = GLτ n
•
•
•
•
This is also just follows from our original
recombination analysis
Illuminate a p-type GaAs (direct bandgap) sample
using a low-power semiconductor laser which emits
10mW assume the spot diameter at the sample is
0.1mm2 and that the minority carrier lifetime τn = 10ns
What is the steady-state carrier density?
NOTE: Ephoton(eV) =1.24/[wavelength in µm]
Center for Compound Semiconductors
Dr. S.-C. Shen, ECE3040B
Photogeneration
•
•
•
First find the Generation Rate G—assume all photons create EHPs
– 10mW at 850nm (GaAs typical emission wavelength)
– The number of photons emitted per second in this laser source
– Photons/s = Joules/s x photons/Joule
– Input = (10x10-3 Joules/s) x (1.24eV/0.850) x (1/q Joules/eV) =
12.4/[(0.850 x 1.6)] x 1016 photons/s ~ x1017 photons/s
– Volume = Area x depth ~ 0.001cm2 (0.5x10-4 cm) = 5x10-8cm3
• Depth is approx the absorption depth of 0.5micron
Then G = 1017EHP/s /[ ~10-7 cm-3] ~ 1024 EHP/s-cm3
∆np0 = Gτn = 1024EHP/s-cm3 x10ns = 1016 cm-3
Note: silicon, having an indirect band structure, typically has a excess
carrier lifetime of 10’s of ms
– Therefore it is relatively easy to create a large steady-state excess
carrier concentration in Si due to light absorption
Center for Compound Semiconductors
Dr. S.-C. Shen, ECE3040B
Quasi-Fermi Levels in Steady State
∆np(x)
∆np0=1013cm-3
0
∆ n p ( x ) = 10 13 e (− x / LN )cm −3
x
0
Light
Light absorbed
in a thin layer
0
5Ln
p-type silicon
No light excitation here
Center for Compound Semiconductors
x
Dr. S.-C. Shen, ECE3040B
Example: Quasi Fermi
Level
E
Ec
FN(x)
NA = 1x1015 cm-3
Ei
no = 1x105 cm-3, ni = 1x1010 cm-3
Excess carriers ∆np0 = 1013 cm-3
EF , Fp
p ≅ p0, → F p ≅ E F ≅ E i − kT ln( N A / ni )
n ≅ no + ∆n p 0 e
− x / LN
Ev
→
∆np0=1013cm-3
E i + kT ln[ no / ni + ( ∆n p 0 / ni )e − x / LN ]
Where K ∆n p ( x ) >> n0 →
Fn ≅ E i + kT ln[( ∆n p 0 / ni )e − x / LN ] →
∆np(x)
Fn = E i + kT ln( n / n i ) =
∆ n p ( x ) = 10 13 e (− x / LN )cm −3
Fn = E i + kT ln( ∆n p 0 / ni ) − kT ( x / LN )
Fn = E i + kT ln(10 3 ) − kT ( x / LN )
Center for Compound Semiconductors
0
0
5LN
x
Dr. S.-C. Shen, ECE3040B
Illumination and Diffusion—Another Example
•
•
A steady-state laser beam illuminates a narrow region of a uniformly doped p-type
sample of silicon maintained at 300K and results in a steady state excess of ∆np0 =
1011 /cm3 electrons at x = 0. Photogeneration only occurs at x = 0.
The bar extends from x = -L to x = +L
∆np(-L) = ∆np(+L) = 0. NA = 1016/cm3 and there is no electric field E = 0 inside the
bar.
Laser Illumination
∆np(-L)=0
silicon
-L
∆np(L)=0
at 300K
0
L
x
1.
What are the dominant physical processes that determine the steady-state excess
electron concentration [∆np(x)] in the regions of the bar removed x=0?
2.
Sketch the expected general form of ∆np(x) inside the bar (-L ~ x ~ L) under steady
state conditions
Center for Compound Semiconductors
Dr. S.-C. Shen, ECE3040B
Diffusion and Recombination
1.
2.
Dominant processes: Diffusion and recombination
Sketch the distribution
∆np(x)
Shape depends upon Ln and L
If L>>Ln
0
5Ln
3.
4.
5Ln
x
Low injection satisfied? YES! ∆np0 << NA
What is the differential equation that describes the diffusion?
∂ (∆n p )
∂t
= DN
∂ 2 (∆n p )
∂x
Center for Compound Semiconductors
2
−
( ∆n p )
τn
=0
Dr. S.-C. Shen, ECE3040B
Recombination in p-n Junctions—
Similar Solutions are Expected
In a biased p-n junction diode, excess minority carriers are injected at the
junction at a steady rate.
– What’s the distribution of the diffusion currents in space?
∆ pn ( x ) = ∆ pn 0 e
Lp =
D pτ p
( − x / LP )
Hole injection
•Because of the hole gradient, a hole current
flows
J p ( x ) = −q
Dp
τp
∆pn0
∆ pn ( x )
Center for Compound Semiconductors
∆pn(x)
•
0
x
5Lp
Dr. S.-C. Shen, ECE3040B
ECE 3040 - Microelectronic Circuits
Lecture 5
2-Terminal Semiconductor Devices:
PN Junction, Diodes
Instructor: Dr. Shyh-Chiang Shen
Lecture Outline
•
•
•
•
•
•
Qualitative description of PN junction
Quantitative analysis of PN junction
I-V characterisitics
Small signal model
Frequency response
Circuits using PN junctions
Center for Compound Semiconductors
Dr. S.-C. Shen, ECE3040B
Formation of PN junction
•
Upon the formation of p-n junction
– Electrons diffuse from n-side to pside
– Holes diffuse from p- to n-side
– Internal electric field built-up
(d(Ei)/dx )
• Electrons tend to drift back to nside
• Holes tend to drift back to p-side
•
Equilibrium state:
– The vicinity of the junction presents a
built-in field that e & h are depleted,
leaving fixed space charges near the
junction (Depletion region)
– Electrons from n-side see a potential
barrier in conduction band
– Holes from p-side see a potential
barrier in valence band
Ec
EF1
N-type
f(E)|EF1
EV
Ec
p-type
f(E)|EF2
EF2
EV
Center for Compound Semiconductors
Dr. S.-C. Shen, ECE3040B
What is a PN Junction for???
• PN junction forms a
“diode” that conducts
current only under
forward bias
– Switches
– Rectifier
Diode conduct current
under forward bias
I (A)
• Other important devices
made of diodes
–
–
–
–
–
LEDs
LASERs
Optical sensor
Optical modulator
Solar Cells
When a diode turns
on, it behaves just
like a resistor
Diode conducts ~0 current
under reverse bias
Center for Compound Semiconductors
VA (V)
Dr. S.-C. Shen, ECE3040B
Energy Bandgap Diagram v.s. VA
•
Reverse Bias: the barrier
height of the junction is
further increased, the e’s
and h’s have close-to-zero
probability to run across
the junction. The major
contributing current is the
“leakage current” from RG centers
•
I (A)
Forward Bias: the barrier
height of the junction is
reduced, the e’s and h’s have
higher probability to be swept
across the junction and a current
is flowing from P to N
Ec
e-
-
-
-
Ei
-
+ + +
+
EV
+
+
•
h+
-
+
VA (V)
Zero Bias: e and h are blocked by the
built-in potential, the probability of travel
across the junction is very low.
Net current =0 under no light
illumination
Center for Compound Semiconductors
Dr. S.-C. Shen, ECE3040B