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For 1-4 below, your explanation of why should include an appropriate mathematical
equation substantiating your answer.
- If you want to build a diode that has an ideality factor of 1 at room temperature,
would you choose Si, Ge, or GaAs? Why?
Ge.
The forward bias current can be written as the sum of the diffusion current and the
recombination current.
qV kT i qV kT i n A
n
p D
p For
A (^) e A
qnW n e L N
D
L N
D
J q
2 / / 2
2
τ
The diffusion current is proportional to
2 n (^) i and the recombination current is proportional
to n (^) i. If we want the diffusion current to dominate (since it has an ideality factor of one)
we want the material with the largest ni so that the
2 n i term dominates. Since
E kT i C V
G n N N e
2 − / = , we want the material with the smallest bandgap, Ge.
- For a diode with n=1 biased under forward bias at a fixed voltage, V (^) A > 3kT, will the
current increase or decrease as the temperature is increased? Write down an equation
that gives the dominant temperature dependence.
For n=1, the current is given by the diffusion current.
E kT qV kT C V n A
n
p D
p For
qV kT i n A
n
p D
p For
G A
A
N N e e L N
D
L N
D
J q
n e L N
D
L N
D
J q
/ /
2 /
−
3 N (^) C NV ∝ T , but the dominant temperature dependence is in the exponential.
( E qV) kT For
G A J e
− − / ∝. Therefore, since EG > qVA, as T increases, the current increases for
a fixed VA.
- Under low forward bias the ideality factor of a Si diode is n=2. Explain why.
Under low forward bias,the current is dominated by the recombination current, the
second term on the right of Eq. (1) above, which has an ideality factor of 2. The
exponential dependence of
qV (^) A kT e
/ 2 comes from the expression for Rmax in the depletion
region which occurs when n=p. When n=p,
( ) (^ E F)^ kT
i
F E kT i
n i i p n ne ne
− / − / = = (3)
Multiplying the 2 terms on the right together and taking the square root gives
qV kT i n = p =ne A/^2 (4)
using Fn -Fp = qVA. The expression for the recombination rate
( )
( ) ( )
( ) ( )
E E kT n i
E E kT p i
qV kT i T i i T
A
n ne p ne
n e R / /
2 / 1
− − τ + +τ +
becomes, assuming ET=Ei, VA > 3kT/q, and τn = τp,
i qVA^ kT e
n R
/ 2 max 2 τ
When then approximate the recombination current as
∫ τ
W i qV kT rec e A
qWn J dxqR qR W
0
/ 2 max 2
giving the ideality factor of 2.
- Under very high forward bias the ideality factor of a Si diode is also n=2. Explain
why.
Under high level injection, one can reach a condition where
n (^) p ( xp) = pp( xp) (8)
Then, from
qV kT i np =n^2 e A/
( )
qV kT p p i n x =n e A/^2
resulting in the ideality factor of 2.
( )
p n p
n
p
n p p B
p
p p
p
p
p
W
t
p p
p p
n W J t W
qD
t W
D
n W t W
q N q
n W t W
t W
t W
n W x t t W
n W N dx
p
∫
−
−
2
2
(f) For V (^) A = 0.7V, ND = 10
18 /cm
3 , NA = 10
16 /cm
3 , t (^) p -x (^) p =0.1 μm, T=300K, and μn = 1500
cm
2 /Vs, calculate J (^) n
( )
(e ) J
n J q
q
k T D
i n
n
B n
^ ≈
− 2
- 7 / 0. 026 16
2
5
2
cm
kA 1 6. 47 10 10
μ 0. 026 ( 1500 ) 39 cm /s
for an n
p junction. n (^) i = 1.45e10/cm
3 was used.
(g) Compare this value to the current for the above diode with t (^) p -x (^) p >> Ln and τn = 1ns.
Replace t (^) p -W with Ln in (e).
- 975 10 (cm)
− 4
Ln = Dn τ n= ×.
So
J (^) n = Jn (short base) / 19.75 = 328 A/cm
2 .
(h) Considering the linear expression for ∆np (x) from part (b), derive, as we did in class,
an expression for the ac conductance and diffusion capacitance under forward bias. What
is the frequency dependence?
Using standard linear response approach, we let all physical quantities take the form
j t
j t
jt
nt n ne
J t J Je
V t V Ve
ω
ω
ω
0 1
0 1
0 1
etc.
where the amplitudes V 1 , J 1 , and ∆n1 are complex quantities. We write down the minority
carrier diffusion equation, group all of the terms proportional to e
jωt together and cancel
the e
jωt terms leaving the ac minority carrier diffusion equation for ∆n 1.
n
n
n
x
n j n D
1 2
1 1
∆ = or
n
n
n
x
n D
1 2
= where n
n n
j ωτ
Now, the solution for ∆n 1 is the usual expression for ∆n with the replacement
everywhere of τn
for τ n. However, in the linear approximation of (b), the Ln and thus
the τn
cancels leaving
( p p)
p p x t
x t n x n x − +
( )
( )/ 0 1 / 0 / 1 / ∆ − = − = − = −
qV Ve kT po
qVt kT p po
B
jt B B
jt n x n e B^ n e n e e
ω ω
Under forward bias,
V kBT e 0 /
1, so we ignore the 1 in the parenthesis. Also, since this is
linear response, qV 1 /kB T << 1, so we have
j t
B
p DC DC
j t
B
qV kT po
qV kT po
qV kT qVe kT p po
e k T q
V
n x n n
e k T q
V
n x n e e n e n e B B B
jt B
ω
ω ω
1
0 /^1 / 0 / 0 /^1
From this we see that
k T q
V
n n B
DC /
1 ∆ 1 =∆
The ac current amplitude is thus
k T q
V
n x J t x
qD n x dx
d J qD
B
p DC p p
n n xp /
1 1 1 ∆ 1 − = −
The ac admittance is J 1 / V 1 = JDC/(kBT/q).This quantity is purely real. It is the ac
conductance. The susceptance is zero and therefore the diffusion capacitance is zero (at
least due to the electrons – there will be some contribution due to the holes which we are
ignoring in the n
p juncition). There is no frequency dependence in this approximation.
