How to simplify and apply the "Minority Carrier Continuity Equations"
Semiconductor Devices - Hour 15
How to simplify and apply the "Minority Carrier Continuity Equations"
Last time: Equations governing population of m...
Where use: δn ( x , t) = n ( x , t) − no δp ( x , t) = p ( x , t) − po Semiconductor_Devices_15.mcd
Deviations from equilibrium or "excess" concentrations
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ASSUMPTIONS / LIMITATIONS: 1) Only apply to a material's MINORITY CARRIERS
2) Only apply with LOW-LEVEL INJECTION Assumption that whatever we did to change the minority carrier populations the alteration was not so great that majority carrier populations changed That is: if have P-type material, require that δp
to jump up and undo the last step:
Best compromise giving good electron capture AND then hole capture is trap level ~ middle of bandgap Conduction Band
The "ideal" trap
For silicon crystal this is the case for metal atoms such as gold and silver
Valence Band
Net recombination rate for an indirect bandgap semiconductor then becomes: Semiconductor_Devices_15.mcd
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Net recombination rate = δcarrier / τdue to traps
More traps => faster recombination => shorter τ
Minority carrier lifetime in indirect bandgap silicon is not a fundamental property! Varies piece of silicon to piece of silicon:
Silicon minority carrier τ vs. Concentration of gold
−8
−9
− 10
10
14
10
15
10
16
10
17
10
18
Metal Concentration (1/cc) Semiconductor_Devices_15.mcd
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In general (but not always) it is good to have minimal recombination: Minority carriers live longer, go further
Would then want highest purity Si that the product could afford: Integrated Circuits: Want τ = 10 -100 µsec => metals 1013 -1014 / cc ~ 1 part per billion Solar Cells:
Want τ = 10 - 100 msec
=> metals 1010 - 1011 /cc ~ 1 metal per 1012 Si atoms !!
Why workers in Si technology are PARANOID about gold contamination! Also why Si technology waited until 1960's: Took that long to figure out how to purify Si!
Will use this input to get a handle on the tau's appearing in the minority carrier continuity equations
How to simplify and apply the "Minority Carrier Continuity Equations"
Semiconductor_Devices_15.mcd
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Simplification #1: "Steady-State"
Everything has settled in and is now ~ constant in time
Consider P-silicon where the minority carrier = n
d dt
δn = 0 = μn ⋅ ξ ⋅
d dx
δn + Dn ⋅
d2 dx
2
δn −
δn + Gnon_thermal τn
Assume I now shine high energy (UV) light onto one end of a P-silicon bar:
UV
P - silicon
Chose UV light because its energy is much greater than Si's bandgap => is absorbed very quickly
Gnon_thermal
new minority electrons created by light energy x
Approximate as a delta function of added carriers ~ at left end of Si bar Semiconductor_Devices_15.mcd
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Gnon_thermal
delta function blip of extra electrons ~ at left end surface
x Solve for region to deeper in the bar:
Simplification #2: Gnon thermal = 0
0 = μn ⋅ ξ ⋅
d dx
δn + Dn ⋅
AWAY from surface, light already absorbed => no non-thermal generation
d2 dx
2
δn −
Simplification #3: ξ electric field ~ 0
0 = Dn ⋅
0=
d2 dx
2
δn −
δn τn
⎛ 1 ⎞ ⋅ δn δn − ⎜ 2 dx ⎝ D n ⋅ τn ⎠ d2
Semiconductor_Devices_15.mcd
~ left end to deep into bar
δn τn
There is no applied field! Only small field from imbalance of carriers
Reduced original 5 terms down to 2 terms: Equation now solvable! ELIMINATION OF TERMS = GENERAL SOLUTION STRATEGY
Define "Diffusion Length" =
10
D n ⋅ τn
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0=
d2 dx
2
1
δn −
δn = A ⋅ e
Ln −x Ln
2
+ B⋅ e
⋅ δn
x Ln
Look past strange δn - You know the solution to this PDE!
Where A and B are as yet arbitrary constants
If the bar is VERY long (on scale of diffusion length), second term would blow up at large x So for long bar must have B = 0
δn ( x) = A ⋅ e
−x Ln
then δn( x=0 ) => A so:
δn ( x) = δn ( 0) ⋅ e
−x Ln
From intensity of UV light I cold figure out what the concentration of excess electrons was in that surface blip δn ( x) δn ( 0) Putting it all together, expect:
δn ( x) = δn ( 0) ⋅ e
−x Ln
x
Semiconductor_Devices_15.mcd
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Let's work this case with some real numbers to see values and test assumptions
Assume P-type silicon doped with
Na := 1 × 10
18 1
⋅
cc
(and negligible Nd) at
cc := cm
3
T := 300K
Say that absorption of UV light a surface adds: 9 1
δnat_0 := 10 ⋅
Because light promotes electrons from valence band to conduction band
cc
it creates new electrons and holes in pairs
9 1
δpat_0 := 10 ⋅
cc
First question: Do we indeed have the "low level" injection we assumed? Doping is modest so we can assume all of the acceptors ionize creating holes po := Na
po = 1 × 10
18 1
cc
then using Si value of:
ni := 1.5 ⋅ 10
10 1
⋅
cc
2
ni no := po
Semiconductor_Devices_15.mcd
1 no = 225 cc
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Low level injection requires that change in MAJORITY carrier be small: po = 1 × 10
