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Lecture 4 PN Junction and MOS Electrostatics(I) Electrostatics in Thermal Equilibrium

Outline • Non-uniformly doped semiconductor in thermal equilibrium • Relationships between potential, φ(x) and

equilibrium carrier concentrations, po(x), no(x) –Boltzmann relations & “60 mV Rule” • Quasi-neutral situation

Reading Assignment: Howe and Sodini; Chapter 3, Sections 3.1-3.2

6.012 Spring 2007 Lecture 4 1 1. Non-uniformly doped semiconductor in thermal equilibrium Consider a piece of n-type Si in thermal equilibrium with non-uniform distribution:

Nd Nd(x)

x

What is the resulting concentration in thermal equilibrium?

n-type ⇒ lots of , few holes ⇒ focus on electrons

6.012 Spring 2007 Lecture 4 2 OPTION 1: electron concentration follows doping concentration EXACTLY ⇒

no(x) = N d(x)

no, Nd Nd(x) no(x)=Nd(x)?

x

Gradient of electron concentration ⇒ net electron ⇒ not in thermal equilibrium!

6.012 Spring 2007 Lecture 4 3 OPTION 2: electron concentration uniform in space

no(x) = nave ≠ f(x)

no, Nd Nd(x)

no = f(x)?

x

Think about space charge density:

ρ(x) ≈ qN[ d(x) − no(x)]

If Nd(x) ≠ no(x) ⇒ρ(x) ≠ 0 ⇒ electric field ⇒ net electron drift ⇒ not in thermal equilibrium!

6.012 Spring 2007 Lecture 4 4 OPTION 3: Demand that Jn = 0 in thermal equilibrium at every x (Jp = 0 too)

Diffusion precisely balances Drift

drift diff J n(x) = J n (x) + Jn (x) = 0

What is no(x) that satisfies this condition?

partially uncompensated charge no, Nd Nd(x) + no(x) - net electron charge x

Let us examine the electrostatics implications of

no(x) ≠ Nd(x)

6.012 Spring 2007 Lecture 4 5 Space charge density

ρ(x) = qN[ d(x) − no(x)]

partially uncompensated donor charge no, Nd Nd(x) + no(x) - net electron charge x ρ

+ − x

6.012 Spring 2007 Lecture 4 6 Electric Field

Poisson’s equation: dE ρ(x) = dx εs

Integrate from x = 0: x 1 E(x)− E(0) = ρ(x ′ )dx ′ εs ∫ 0

no, Nd Nd(x) + no(x) -

x ρ

+ − x

E x

6.012 Spring 2007 Lecture 4 7 Electrostatic Potential dφ =−E(x) dx Integrate from x=0: x φ(x)− φ(0) =−∫ E(x ′ )dx′ 0 Need to select reference (physics is in the potential difference, not in absolute value!);

Select φ(x = 0) = φref

no, Nd Nd(x) + no(x) -

x ρ

+ − x

E x

φ

φ ref

x

6.012 Spring 2007 Lecture 4 8 2. Relationships between potential, φ(x) and

equilibrium carrier concentrations, po(x), no(x) (Boltzmann relations) dn J = 0 = qn µ E + qD o n o n n dx µ dφ 1 dn n • = • o Dn dx no dx Using Einstein relation: q dφ d(ln n ) • = o kT dx dx

Integrate:

q no ()φ − φref = ln no − ln no,ref = ln kT no,ref

Then: ⎡ ⎤ q(φ − φref ) n = n exp⎢ ⎥ o o,ref ⎢ kT ⎥ ⎣ ⎦

6.012 Spring 2007 Lecture 4 9 Any reference is good

In 6.012, φref = 0 at no,ref = ni

Then: qφ kT no = nie

If we do same with holes (starting with Jh = 0 in thermal 2 equilibrium, or simply using nopo = ni ); −qφ kT po = nie

We can re-write as: kT n φ = • ln o q ni and

kT p φ =− • ln o q ni

6.012 Spring 2007 Lecture 4 10 “60 mV” Rule

At room temperature for Si: n n φ = ()25mV • ln o = ()25mV • ln() 10 •log o ni ni

Or n φ ≈ ()60mV • log o ni

EXAMPLE 1:

18 −3 no = 10 cm ⇒ φ = (60mV)× 8 = 480 mV

6.012 Spring 2007 Lecture 4 11 “60 mV” Rule: contd.

With holes: p p φ =−()25mV • ln o =−()25mV • ln() 10 • log o ni ni

Or p φ ≈−()60 mV •log o ni

EXAMPLE 2:

18 −3 2 −3 no = 10 cm ⇒ po =10 cm ⇒ φ =−()60mV ×−8 = 480mV

6.012 Spring 2007 Lecture 4 12 Relationship between φ, no and po:

−3 po, equilibrium hole concentration (cm )

p-type intrinsic n-type

10191018 1016 1014 1012 1010 108 106 104 102 101

φ φ φ p+ (mV) n+ −550 −480 −360 −240 −120 0 120 240 360 480 550

−550 −480 −360 −240 −120 0 120 240 360 480 550 φ φ p+ φ (mV) n+

101 102 104 106 108 1010 1012 1014 1016 1018 1019

p-typeintrinsic n-type

−3 no, equilibrium electron concentration (cm )

Note: φ cannot exceed 550 mV or be smaller than -550 mV. (Beyond this point different physics come into play.)

6.012 Spring 2007 Lecture 4 13 Example 3: Compute potential difference in thermal 17 -3 equilibrium between region where no = 10 cm and no = 1015 cm-3.

17 −3 φ(no = 10 cm ) = 60 × 7 = 420 mV

15 −3 φ(no = 10 cm ) = 60 × 5 = 300 mV

17 −3 15 −3 φ(no = 10 cm ) − φ(no = 10 cm ) = 120 mV

Example 4: Compute potential difference in thermal 20 -3 equilibrium between region where po = 10 cm and po = 1016 cm-3.

20 −3 φ( po = 10 cm ) = φmax =−550mV

16 −3 φ( po = 10 cm ) =−60 × 6 =−360mV

20 −3 16 −3 φ( po = 10 cm ) − φ(po = 10 cm ) =−190 mV

6.012 Spring 2007 Lecture 4 14 3. Quasi-neutral situation

If Nd(x) changes slowly with x ⇒ no(x) also changes slowly with x. WHY?

Small dno/dx implies a small diffusion current. We do not need a large drift current to balance it.

Small drift current implies a small electric field and therefore a small space charge

Then: no(x) ≈ Nd (x)

no(x) tracks Nd(x) well ⇒ minimum space charge ⇒ semiconductor is quasi-neutral

no, Nd Nd(x) no(x) = Nd(x)

x

6.012 Spring 2007 Lecture 4 15 What did we learn today?

Summary of Key Concepts

• It is possible to have an electric field inside a semiconductor in thermal equilibrium – ⇒ Non-uniform doping distribution. • In thermal equilibrium, there is a fundamental relationship between the φ(x) and the equilibrium

carrier concentrations no(x) & po(x) – Boltzmann relations (or “60 mV Rule”). • In a slowly varying doping profile, majority carrier concentration tracks well the doping concentration.

6.012 Spring 2007 Lecture 4 16