19 The Bipolar Junction Transistor n
Physical Structure: oxide-isolated, low-voltage, high-frequency design ... typical of the bipolar transistor found in a BiCMOS process, such as the MicroLinear tile array chips used in the laboratory experiments
metal contact to base
, , ,
n+ polysilicon contact to n+ emitter region
p+
n+ buried
field oxide
n
layer
p-type base
,, ,,, ,
,,, ,, A
metal contact to collector
n+
n+ buried
n+ - p - n sandwich (intrinsic npn transistor)
A'
layer
p-type substrate
(a)
,,,,,,,,, ,,,,,,,,, ,,,, , , , , ,,,, ,,,, , , ,,,,,,, ,,,,,,,, , , , (base)
A
p+
p
n + emitter area, AE (intrinsic npn transistor)
(emitter)
edge of n + buried layer
field oxide
A'
n+
(collector)
(b)
EECS 105 Fall 1998 Lecture 19
Circuit Symbol and Terminal Characteristics n
As with MOSFETs, we have two devices that have complementary characterisitcs, in this case the npn transistor and the pnp transistor The direction of the diode arrow indicates whether the central layer (the base) is n or p C B IB
+ VBE
− E
E +
IC + VCE − −IE
(a)
npn normal operation: VCE positive IC positive VBE = 0.7 V IB positive -IE positive n
VEB −
B −IB
C
IE + VEC − −IC
(b)
pnp normal operation: VEC positive -IC positive VEB = 0.7 V -IB positive IE positive
The pnp usually has a very different physical structure ... we will concentrate on the npn and then consider the pnp briefly
EECS 105 Fall 1998 Lecture 19
npn BJT Collector Characteristics n
Similar test circuit as for n-channel MOSFET ... except IB is controlled instead of VBE (for convenience) IC = IC(IB, VCE)
+ V − CE IB
(a) IC (µA) 300
IB = 2.5 µA IB = 2 µA
250 200
IB = 1.5 µA
(saturation)
150
IB = 1 µA
(forward active)
100 IB = 500 nA 50 −3
−2
IB = 0 (cutoff)
−1 1
IB = 1 µA IB = 2 µA
2
3
4
5
6
VCE (V)
−4 (reverse active)
−8
(b)
EECS 105 Fall 1998 Lecture 19
Regions of Operation n
Constant-current region is called forward active ... corresponds to MOSFET saturation region (!?!) IC = βF IB = IS exp[VBE/Vth](1 + VCE / VA) ... (VA)-1 is like λ for MOSFET
n
Constant-voltage region is called saturation ... corresponds to MOSFET triode region V CE ≈ V CE ( sat ) = 0.1V or 0.2 V
n
Cutoff ... corresponds to MOSFET cutoff region
n
Reverse active ... terminal voltages for npn sandwich are flipped so that VCE is negative and VBC = 0.7 V. Only occasionally useful.
Boundary between saturation and forward-active regions: V CE > V CE ( sat )
and
IB > 0
... much easier to apply this test than VDS > VDS(sat)
EECS 105 Fall 1998 Lecture 19
Small-Signal Model of the Forward-Active npn BJT n
Transconductance (same concept as for MOSFET):
gm =
Ebers-Moll (forward-active):
∂i C ∂ v BE
iC = I S e
Q
v BE ⁄ V th
iC
iC
IC + ic IC
Q
slope = gm
IC Q VBE
0.2
0.4
0.6 V
BE
VBE + vbe
vBE
vBE
Evaluating the derivative, we find that
IC I S V BE ⁄ V th ------------gm = e = V V th th
EECS 105 Fall 1998 Lecture 19
Input Resistance n
The collector current is a function of the base current in the forward-active region (recall IC = βFIB). At the operating point Q, we define ∂i C βo = ∂ iB Q
and so ic = βo ib. (Note that the “DC beta” βF and the small-signal βo are both highly variable from device to device) n
Since the base current is therefore a function of the base-emitter voltage, we define the input resistance rπ as:
–1
rπ =
n
∂i B ∂ v BE
= Q
∂i B ∂ iC
∂i C Q
∂ v BE
Q
1 = ------ g m β o
Solving for the input resistance β o V th βo kTβ o r π = ------ = -------------- = -----------gm IC qI C
n
For a high input resistance (often desirable), we need a high current gain or a low DC bias current.
EECS 105 Fall 1998 Lecture 19
Output Resistance n
The Ebers-Moll model has perfect current source behavior in the forward-active region -- actual characteristics show some increase: IC
−VAn
n
VCE
Why? Base width shrinks due to encroachment by base-collector depletion region Approximate model: introduce Early voltage VAn to model increase in iC
Model:
n
iC = IS e
v BE ⁄ Vth
v CE 1 + --------- V An
Output resistance: ∂i C –1 ro = ∂ v CE
Q
IC --------≅ V An
EECS 105 Fall 1998 Lecture 19
Complete Small-Signal Model n
Add the depletion capacitance from the base-emitter junction to find the total base-emitter capacitance: Cπ = CjE + Cb C jE =
2C jEo
CjEo is proportional to the emitter-base junction area (AE) n
Depletion capacitance from the base-collector junction: Cµ C µo C µ = ------------------------------------1 + V CB ⁄ φ Bc Cµo is proportional to the base-collector junction area (AC)
n
Depletion capacitance from collector (n+ buried layer) to bulk: Ccs C cso C cs = -----------------------------------1 + V CS ⁄ φ Bs Ccso is proportional to the collector-substrate junction area (AS) base ib +
rb
rc
+ Cπ
vbe
collector ic
Cµ
vπ
rπ
gmvπ
−
ro
+
Ccs substrate
vce
rex −
− emitter
EECS 105 Fall 1998 Lecture 19
Numerical Values of Small-Signal Elements
ib +
base
vbe −
ic collector +
+ vπ
gmvπ
rπ
ro
−
vce −
emitter n
Transconductance:
IC = 100 µA, Vth = 25 mV -->
gm = 4 mS = 4 x 10-3 S
Note: gm varies linearly with collector current and is independent of device geometry, in contrast to the MOSFET
n
Input resistance:
βo = 100, IC = 100 µA, Vth = 25 mV -->
n
rπ = 25 kΩ
Output resistance:
IC = 100 µA, VAn = 35 V -->
ro = 350 kΩ
VAn = Early voltage increases with increasing base width and decreases with decreasing base doping.
EECS 105 Fall 1998 Lecture 19
npn BJT SPICE model Close correspondence to Ebers-Moll and small-signal models Name
Parameter Description
Units
IS
transport saturation current [IS]
Amps
BF
ideal maximum forward beta [βF]
None
VAF
forward Early voltage [VAn]
Volts
BR
ideal maximum reverse beta [βR]
None
RB
zero bias base resistance [rb]
Ohms
RE
emitter resistance [rex]
Ohms
RC
collector resistance [rc]
Ohms
CJE
B-E zero-bias depletion capacitance [CjEo]
Farads
VJE
B-E built-in potential [φBe]
Volts
MJE
B-E junction exponential factor
None
CJC
B-C zero-bias depletion capacitance [Cµo]
Farads
VJC
B-C built-in potential [φBc]
Volts
MJC
B-C junction exponential factor
None
CJS
substrate zero-bias depletion capacitance [Ccso]
Farads
VJS
substrate built-in potential [φBs]
Volts
MJS
substrate junction exponential factor
None
TF
ideal forward transit time [τF]
Seconds
.MODEL MODQN NPN IS=1E-17 BF=100 VAF=25 TF=50P + CJE=8E-15 VJE=0.95 MJE=0.5 CJC=22E-15 VJC=0.79 MJC=0.5 + CJS=41E-15 VJS=0.71 MJS=0.5 RB=250 RC=200 RE=5
EECS 105 Fall 1998 Lecture 19
