Recitation 16
Small Signal Model of p-n Diode
6.012 Spring 2009
Recitation 16: Small Signal Model of p-n Diode
Review
Last week we learned about the IV characteristic of p-n diode:
I = Io (eqVo /kT − 1)
Dp
1
1
Dn
2
+
where Io = qAni
Na wp − xp Nd wn − xn
If we plot,
Yesterday, we discussed the small signal model for p-n diode. Under forward bias, the small
signal (ss) model of a p-n diode is:
1
Recitation 16
Small Signal Model of p-n Diode
6.012 Spring 2009
Cj is the junction capacitance as we talked about before, Cd is called “diffusion capacitance”,
this is new
Small signal circuit model of p-n diode: (two terminal device)
In a small signal model, we have linearized conductance (resistance) and capacitances.
2
Recitation 16
Small Signal Model of p-n Diode
6.012 Spring 2009
Linearized Conductance (Resistance)
1
,
gd
γd =
=⇒ γd
gd =
δIo · (eqVD /kT − 1) δiD =
δVD VD
δVD
VD
q qVD /kT
= Io ·
e
kT
q
=
· (ID + Io )
kT
kT
q
= Vth
=
ID ,
q
kT
Vth
=
ID
Vth constant, the larger operating current ID , the smaller is γd
Depletion Capacitance (due to p-n junction)
Cjo =
s
=
xdo
Cj (VD ) =
qs Na Nd
2(Na + Nd )φB
Cjo
For forward bias limit VD to
VD
φB
1−
φB
2
capacitance/unit area!
Diffusion Capacitance
For this, we need to look at the majority carrier concentration as well. To keep quasi-neutral,
nn (x) = Nd + pn (x)
pp (x) = Na + np (x)
3
Recitation 16
Small Signal Model of p-n Diode
6.012 Spring 2009
It is a capacitor without two parallel plates! And the “+” and “-” charges are just mixed
with each other! Isn’t that amazing!
charge stored on the n-side:
q pn
Cdn
1. Cdn =
1
= −qNn = qA (wn − xn ) ·
2
dqpn wn − xn
=
·
= qA
dVD VD
2
n2i qVD /kT
(e
− 1)
Nd
ni2 q qVD /kT
e
Nd kT
qA
qA
(wn − xn ) · pno eqVD /kT =
(wn − xn ) · pn (xn )
2Vth
2Vth
2. Write in terms of ID
Cdn
=
Cdn
=
Define :
T Tp
=
Cdn
Cdn
=
Similarly, Cdp
Dp
n2i
wn − xn wn − xn q
·
· eqVD /kT ·
·
·
Nd wn − xn
2
Dp
kT
2
q (wn − xn )
·
· IDp
kT
2Dp
transit time of holes through n-QNR:
wn − xn
length
(wn − xn )2
=
=
2Dp
2Dp /(wn − xn )
velocity
q
· TTp · IDp
kT
IDp · TTp
Vth
q
· TTn · IDn
kT
qA
Charges on both side are added together. The two capacitors are in parallel.
q
(TTn · IDn + TTp · IDp )
kT
qA
=
((wp − xp )npo + (wn − xn )pno )eqVD /kT
2Vth
Cd = Cdn + Cdp =
Cd = Cdn + Cdp
Discussions:
1. Where do the extra majority carriers come from?
2. Majority current: -diffusion + drift
3. Cd ∝ eqVD /kT : for reverse bias, depletion capacitance dominate. For forward bias,
diffusion capacitance dominate
4
Recitation 16
Small Signal Model of p-n Diode
6.012 Spring 2009
Exercise: p-n Diode
wp = 0.5 μm
Na = 2.5 × 1017 cm−3
Dn = 14 cm2 /s
wn = 1.0 μm
Nd = 4.0 × 1016 cm−3
Dp = 10 cm2 /s
Vd = 720 mV
ID = 50 μA
Find Vd , Cj , and Cd . For this calculation, ignore xn , xp .
Jo =
qni2
Dp
Dn
+
Na wp Nd wn
= 5.79 × 10−11 A/cm2
ID = Io (eqVo /kT − 1) Io · eqVo /kT = Io · 10720/60 = 50 μA
=⇒ Io = 5 × 10−17 A =⇒ Jo = Io /A =⇒ A = 8.64 × 10−7 cm2
Vth
0.025 V
kT Na Nd
γd =
=
= 500 ω φB =
ln
= 840 mV
ID
50 μA
q
n2i
qs Nd Na
Cjo = A ·
2(Na + Nd )φB
1.6 × 10−19 C × 8.85 × 10−14 F/cm · 11.9 × 4 × 1016 × 2.5 × 1017 cm−6
= A
2(4 × 1016 + 2.5 × 1017 )0.84 V · cm−3
= 8.64 × 10−7 cm2 × 5.88 × 10−8 F/cm2
Cj
= 50.8 fF = 50.8 × 10−15 F
Cjo
50.8 fF
= =
= 71.8 fF
√1
1 − VφDB
2
Cd =
=
=
=
qA
· (wp · np (−xp ) + wn · pn (xn ))
2Vth
1.6 × 10−19 C × 8.64 × 10−7 cm2
1020
1020
−4
(0.5 × 10−4 ×
+
1
×
10
×
) eqVo /kT
4 × 1016 2 × 0.025
2.5 × 1017
1.6 ×
10−19 C
10−7 cm2
1.6 ×
10−19 C
10−7
× 8.64 ×
2 × 0.025 V
× 8.64 ×
2 × 0.025 V
1012
−4
14
15
(0.5
10−4 × 2.5
×10 × 4 ×10 + 1 ×
×
10 )
cm
cm−3
× 2.7 × 1011 = 746 fF
5
cm
cm−3
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6.012 Microelectronic Devices and Circuits
Spring 2009
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