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1 . Lecture #19
ANNOUNCEMENT
• Quiz #4 (Thursday 4/3) to cover Chapters 10 & 11
OUTLINE
• BJT transient response
• BJT small-signal model, fT
Reading: Chapter 12
Spring 2003 EE130 Lecture 19, Slide 1
BJT Switching - Qualitative
Spring 2003 EE130 Lecture 19, Slide 2
1
2 . Turn-on transient
• We know: dQB = I BB − QB where IBB=VS/RS
dt τB
• The general solution is:
QB (t ) = I BBτ B + Ae − t / τ B
• Initial condition: QB(0)=0.
since transistor is in cutoff
QB (t ) = I BBτ B (1 − e − t / τ B ) t r = τ B ln
1
VCC / RL
QB (t ) I BBτ B + Ae − t / τ B 1− I τ
τ = τt
L0 ≤ t ≤ tr BB B
iC (t ) = t
VCC
Lt ≥ t r
RL
Spring 2003 EE130 Lecture 19, Slide 3
Turn-off transient
• We know: dQB = −ξI BB − QB
dt τB
• The general solution is:
QB (t ) = −ξI BBτ B + Ae − t / τ B
• Initial condition: QB(0)=IBBτB
[
QB (t ) = I BBτ B (1 + ξ )e − t / τ B − ξ ]
1+ ξ
I CC L0 ≤ t ≤ t sd t sd ≅ τ B ln
I CCτ t
iC (t ) = QB (t ) I BBτ B [(1+ξ )e −t / τ B −ξ ] I τ +ξ
τ = Lt ≥t BB B
t τt sd
Spring 2003 EE130 Lecture 19, Slide 4
2
3 . Small-Signal Model
B C
Forward-active mode,
Common-emitter configuration:
+
Cπ vbe rπ gm vbe
I C = α F I F e qVBE / kT
−
E E
transconductance:
dI C d
gm ≡ = (α F I F e qVBE / kT ) At 300 K, for example,
dVBE dVBE
gm=IC /26mV.
q
= α F I F e qVBE / kT = I C /(kT / q )
kT
g m = I C /(kT / q )
Spring 2003 EE130 Lecture 19, Slide 5
Small-Signal Model (cont.)
1 dI B 1 dI C g
= = = m
rπ dVBE β dc dVBE β dc
β dc β dc
rπ = rπ =
gm gm
dQF d (τ F I C )
Cπ = = = τ F gm
dVBE dVBE
This is the minority-carrier charge-storage capacitance, better
known as the diffusion capacitance.
Add the depletion-layer capacitance, CJBE :
Cπ = τ F g m + CdBE
Spring 2003 EE130 Lecture 19, Slide 6
3
4 . Forward Transit Time τF
QF
τ F = τ E + τ BE + τ t + τ BC =
IC
where
τ E = emitter delay time
τ BE = emitter-base depletion region transit time
τ t = base transit time
τ BC = base-collector depletion-region transit time
• To reduce the forward transit time, the emitter as well
as the depletion layers must be kept thin.
Spring 2003 EE130 Lecture 19, Slide 7
Example: Small-Signal Model Parameters
A BJT is biased at IC = 1 mA and VCE = 3 V. βdc=90, τF=5 ps,
and T = 300 K. Find (a) gm , (b) rπ , (c) Cπ .
Solution:
1 mA mA
(a) g m = I C /(kT / q) = = 39 = 39 mS (milli siemens)
26 mV V
(b) rπ = βdc / gm = 90/0.039 = 2.3 kΩ
c) Cπ = τ F g m = 5 ×10 −12 × 0.039 ≈ 1.9 ×10 −14 F = 19 fF(femto farad)
Spring 2003 EE130 Lecture 19, Slide 8
4
5 . Application of Small-Signal Model
Once the model parameters have been determined, one can
analyze circuits with arbitrary source and load impedance.
B C
The parameters are routinely
+
Signal C
vbe rπ gm vbe
Load determined through comprehensive
source π
-
measurement of the BJT AC
E E
and DC characteristics.
rb Cµ rc
B C
+
Cπ rπ C dBC
Full BJT equivalent circuit: vbe gm vbe ro
−
re
E
Spring 2003 EE130 Lecture 19, Slide 9
Cutoff Frequency fT
B C
+
Signal Load 1
C
source π
vbe rπ gm vbe
β ac = 1 at fT =
- 2π (τ F + C JBE kT / qI C )
E E
The load is a short circuit, and the signal source is a current
source, ib , at frequency, f. At what frequency does the a.c.
current gain fall to unity?
ib ib
vbe = =
input admittance 1 / rπ + jωCπ
ic = g m vbe
ic gm 1
β (ω ) = = =
ib 1 / rπ + jωCπ 1 / β F + jωτ F + jωCdBE kT / qI C
Spring 2003 EE130 Lecture 19, Slide 10
5
6 .For the full BJT 1
fT =
equivalent circuit: 2π (τ F + (CdBE + CdBC )kT / (qI C ) + CdBC (re + rc ))
SiGe HBT by IBM
fT is commonly used as a metric for the speed of a transistor.
Spring 2003 EE130 Lecture 19, Slide 11
Cutoff Frequency fT
1
fT =
2π (τ F + (CdBE + CdBC )kT / (qI C ) + CdBC (re + rc ))
• To maximize ft:
– Increase IC
– Minimize CdBE, CdBC
– Minimize re, rc
– Minimize τF
Spring 2003 EE130 Lecture 19, Slide 12
6
7 .Base Widening at High IC: the Kirk Effect
• At very high current densities (> 0.5 mA/µm2), base widening
occurs*, so QB increases.
→ tt and τBC increase, so τF increases and fT decreases.
*For an NPN BJT, the electron
Top to bottom : density in the collector (n = NC)
VCE = 0.5V, 0.8V, becomes insufficient to support
1.5V, 3V. the collector current even if the
electrons move at the saturation
velocity.
I C = qAnvsat
IC
ρ dep ,C = qN C − qn = qN C −
Avsat
Eventually, ρ changes sign as IC
increases (for fixed VBC), and the
base width is effectively widened.
Spring 2003 EE130 Lecture 19, Slide 13
BJT Structure for High Speed
B E C
P+polySi N+ polySi P+polySi
p+ P base p+
Shallow N+
N collector trench
Deep Deep
trench N+ subcollector trench
• Narrow base P− substrate
• n+ poly-Si emitter
• Self-aligned p+ poly-Si base contacts
• Lightly-doped collector
• Heavily-doped epitaxial subcollector
• Shallow trenches and deep trenches filled with SiO2
for electrical isolation
Spring 2003 EE130 Lecture 19, Slide 14
7
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