Electrónica Geral
Transparências de apoio às aulas
Mestrado Integrado em Engenharia Biomédica
Mestrado Bioengenharia e Nanossistemas
2º semestre 2011/2012
João Costa Freire
Instituto Superior Técnico
Fevereiro de 2012
1
Revisões e Problemas
Based on PowerPoint Overheads for
Sedra/Smith
Microelectronic Circuits 5/e
Chap 5. – BJTs
Revisões do BJT – No parágrafo 5.1 apenas se
introduz a equação exponencial da corrente iC
(expressão 5.3) e a relação com iB (expressão
5.10 e 5.11) e não as suas deduções
(expressões 5.1, 5.2 e 5.6 a 5.9) e não se dá
5.1.4. Nos parágrafos 5.2 a 5.10 não é
necessário saber as fórmulas de 5.8.1 a 3 mas
apenas o modelo dado em 5.8.4 e exemplos de
aplicação dados de seguida.
Problemas: Rever Example 5.1 a 5.19 do Sedra e BJT1 a BJT3
©2004 Oxford University Press.
Bipolar Junction Transistors (BJTs)
Teóricas EG 2010/2011
Problemas EG 2011/2012
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Figure 5.13 Circuit symbols for BJTs:
npn (a) and pnp (b).
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wB
For having a high
current gain
β = iC/ iB ⇒ iC ≈ iE
Figure 5.3 Current flow in an npn transistor biased to operate
in the active mode.
Figure 5.6 Crosssection of an npn
BJT.
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1.
nE > nC
E with more
electrons than C
2.
Small Base width wB
3.
Emitter surrounded
by Collector
⇓
Typically
β ≈ 100 – 300
iC = α iE ⇒ α>0.99≈
≈1
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Figure 5.20 Large-signal equivalent-circuit models of an npn BJT operating in the active mode
in the common-emitter configuration.
Both model are equivalent since the saturation current of theBaseEmitter junction is defined as Is BE = Is / β
and
iC = Is exp (vBE / VT) (it is assumed n=1 – abrupt pn junction)
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Table 5.3
npn and pnp BJT ´symbols and equivalent simple models indicating the positive
current flow and voltages when they are biased at forward active region.
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3 operation regions
1. BE and BC OFF
vBE, vBC < VDON=0.5V
BJT OFF (iC,B,E≈0)
2. BE and BC ON
vBE ≈ vBC > VDON ⇒
vCE = vCB - vBE ≈ 0V
(VCEsat ≈ 0.2-0.3V)
BJT ON (saturation)
3. BE ON and BC OFF
vBE > VDON and
Figure 5.23 An expanded view of the common-emitter
characteristics in the saturation region.
Zone 1 and 2 – Digital applications (ON or OFF)
vCE>0.3V ⇒ vBC<VDON
Zone 3 – Analog applications (active with gain)
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BJT active
(with gain)
8
ICQ
oQ
VCEQ
Figure 5.19 (a) Conceptual circuit for measuring the iC –vCE characteristics
of the BJT. (b) The iC –vCE characteristics of a practical BJT.
From fig.5.19 – iC ≈ Is evBE/nVT (1+vCE/VA) @ active region
From Early Voltage VA we obtain the output resistance
rCE = ro = (VCEQ + VA) / ICQ ≈ VA / ICQ
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if
VCEQ << VA
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Figure 5.21 Common-emitter characteristics. Note that the horizontal scale
is expanded around the origin to show the saturation region in some detail.
Usually BVCEO (breakdown CE voltage for base open – IB=0) is at least
several Volt but can be tenths or hundred Volt ⇒ BVCEO >> VCEsat
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Figure 5.26 (a) Basic common-emitter amplifier circuit. (b) Transfer characteristic of the
circuit in (a). The amplifier is biased at a point Q, and a small voltage signal vi is superimposed
on the dc bias voltage VBE. The resulting output signal vo appears superimposed on the dc
collector voltage VCE. The amplitude of vo is larger than that of vi by the voltage gain Av.
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Figure 5.28 Graphical construction for the
determination of the dc base current in the
circuit of Fig. 5.27.
Figure 5.27 Circuit whose operation
is to be analyzed graphically.
KVL – input loop
VBB + vi = RB iB + vBE
input load line
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KVL – output loop
VCC = RC iC + vCE
output load line
DC design goals
Figure 5.29 Graphical construction for determining
the dc collector current IC and the collector-to-emitter
voltage VCE in the circuit of Fig. 5.27.
VCE ≈ 3.2 V; IC ≈ 1 mA
input data
BJT β = 100
DBE: IS = 1 fA; n=1.2; VT
= 25 mV.
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Figure 5.48 (a) Conceptual
circuit to illustrate the
operation of the transistor as
an amplifier. (b) The circuit of
(a) with the signal source vbe
eliminated for dc (bias)
analysis.
DC circuit
Total circuit
AC sources eliminated
Capacitors Open Circuit
Inductors Short Circuit
Figure 5.50 The amplifier circuit of Fig. 5.48(a) with the dc
sources (VBE and VCC) eliminated (short circuited). Thus only the
signal components are present. Note that this is a representation of
the signal operation of the BJT and not an actual amplifier circuit.
AC circuit
DC sources eliminated & BJT incremental linear model
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Figure 5.51 Two slightly different versions of the simplified hybrid-π model for the
small-signal operation of the BJT. The equivalent circuit in (a) represents the BJT
as a voltage-controlled current source (a transconductance amplifier), and that in
(b) represents the BJT as a current-controlled current source (a current amplifier).
AC BJT incremental linear model
•BE junction: active region (ICQ>>Is) ⇒ iB ≈ (Is / β) exp (vBE / VT)
⇒ gbe = ∂iB / ∂vBE |VBEQ = ICQ / (β
β VT) ⇒ rπ = rbe = β (VT / ICQ)
•Collector current source: ic = β ib = β vbe / rπ = gm vbe
with gm = β / rπ = ICQ / VT (= ∂iC / ∂vBE |VBEQ – iC controlled by vBE)
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Polarização e Modelo dinâmico: EXERCÍCIO BJT1 (1de2)
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Polarização e Modelo dinâmico: EXERCÍCIO BJT1 (2de2)
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VBB + vi = RB iB + vBE
input load line
VCC = RC iC + vCE
output load line
AC (slope rbe) DC (slope -RB)
Slope -RC
Figure 5.30 Graphical determination of the signal components vbe, ib, ic, and vce when a signal
component vi is superimposed on the dc voltage VBB (see Fig. 5.27).
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Figure 5.40 Circuits for Example 5.10.
VBB = RBB iB + vBE + RE iE
BJT DC Bias with
For VBB>>∆
∆VBE
RE negative feedback V = R i + v + R i
CC
C C
CE
E E
and RE>>RBB/β
β
iC↑ ⇒ vRE↑ ⇒ vBE↓
iE ≈ iC ⇓ β>>1
⇓
⇒ iC↓
iC ≈ (VBB - vBE)/(RE+RBB/β
β) iC ≈ constant
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BJT DC Bias
RB negative feedback
iC↑ ⇒ vCE↓ & iBRB↑
⇒ vBE↓ ⇒ iC↓
Figure 5.46 (a) A common-emitter transistor amplifier biased
by a feedback resistor RB. (b) Analysis of the circuit in (a).
iC = (VCC - vBE) / (RC + RB / β)
BJT DC Bias
vCE = RB iB + vBE
VCC = RC iC + vCE
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⇒
⇓
iC ≈ constant for
∆vBE<<VCC and RC >> RB/β
β
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Bias Network Capacitors
DC decoupling Ci & Co
AC bypass CE
Co
Ci
CE
For each capacitor a
transfer function have 1
zero ωz and 1 pole ωp
Figure 5.71 (a) Capacitively coupled common-emitter amplifier.
ωz = 0 for Ci & Co
and ωz < ωp for CE
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⇒
have high-pass behaviour
(ideal C → ∞)
⇓
Ci, Co & CE define circuit low
frequency behaviour
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EB and CB junctions have charge storage
⇒ capacitors on the AC BJT model
rx – access to base:
metal strip resistance
Due to rx the circuit
topology has a Π
shape but
asymmetrical “hybrid-π
π model”
µ
x
Figure 5.67 The high-frequency hybrid-π
π model.
ωz = ∞ for Cπ & Cµ
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⇒
π
o
have low-pass behaviour
(ideal C → 0)
⇓
Cπ & Cµ define circuit high
frequency behaviour
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Ci, Co & CE define low
frequency behaviour
Cπ & Cµ define high
frequency behaviour
Ci, Co & CE → ∞ : short-circuit
medium frequency 
⇒
Cπ & Cµ → 0 : open-circuit
behaviour (constant) 
Figure 5.71 (b) Sketch of the magnitude of the gain of the CE amplifier versus frequency.
The graph delineates the three frequency bands relevant to frequency-response determination.
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


iµ
V’sig
R’sig
zπ
Z’be
Figure 5.72 Determining the high-frequency response of the CE amplifier: (a)
equivalent circuit.
Vπ = Iµ (1/sCµ) + (Iµ - gmVπ)R’L ⇒ Vπ = Iµ (1/sCµ) + (Iµ - gmVπ)R’L
⇒ Z’be = Vπ / Iµ = [(1/sCµ) + R’L] / (1 + gmR’L)
Thévenin left of rπ: R’sig = rx+RB//Rsig and V’sig = Vsig[RB/(Rsig+RB)]
Voltage divider: Vπ = V’sig (zπ//Z’
// be) / [ R’sig + (zπ//Z’
// be) ]
Output node: Vo = ( -gmVπ + Iµ ) R’L = Vπ ( Y’be – gm ) R’L
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Gv = Vo / Vsig =
[RB/(Rsig+RB)] { (zπ//Z’
// be) / [ R’sig + (zπ//Z’
// be)] } ( Y’be – gm ) R’L
where zπ = rx // Cπ = 1 / ( gπ + s Cπ ) and
Z’be = [(1/sCµ) + R’L] / (1 + gmR’L) = 1 / Y’be
Gv = Gv0 (1 + s / sz ) / [ (1 + s / sp1) (1 + s / sp2) ]
Low-pass circuit with two poles |Gv|
ω1,2 = |sp1,2| and one zero ωz = |sz|
exercise: calculate the
expressions of ω1,2 and ωz
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


CE low frequency gain






with Gv0 = Gv (s=0) = – gm R’L [ RB / ( Rsig + RB ) ] [ rπ / ( R’sig + rπ )]
input voltage divider
Gv0
-20dB/dec
-40dB/dec
ω1 ω2 ωz
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-20dB/dec
logω
ω
25
Polarização e Modelo dinâmico: EXERCÍCIO BJT2 (1de3)
1.Example 5.14: (a)
Em regime contínuo (vulgo DC)
calcule as correntes e tensões em todos os ramos do circuito
sabendo que o BJT tem β=100. Enuncie as Leis da Teoria de
Circuitos e as relações do BJT que utilizar.
KVL malha de entrada V = R i + v
BB
BB B
BE
+
Relação da junção BE vBE ≈ 0,7 V
⇓
para V~Volt e R~kΩ
Ω ⇒ I~mA
iB = (VBB - vBE)/RBB = 23µ
µA
KVL malha de saída VCC = RC iC + vCE + RE iE
+
Relação de correntes no BJT iC = β iB
Figure 5.53 Example 5.14: (a) circuit;
(b) dc analysis; (c) small-signal model.
β>>1 ⇓ iE ≈ iC
iC ≈ (VBB - vBE)/(RE+RBB/β
β) = 2,3 mA
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VCE ≈ VCC – RC iC = 3,1 V
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Polarização e Modelo dinâmico: EXERCÍCIO BJT2 (2de3)
1.Example 5.14:
(b)Se o parâmetro b duplicar qual é o novo valor das
tensões e correntes no circuito?
(c)E se a resistência RC aumentar para 4kΩ
Ω?
(a) Não se altera a malha de entrada logo iB = 23µA
iC = 200 × 23µ
µA = 4,6 mA duplica também ⇒
polarização não estabilizada
vCE = 10 – 3k × 4,6m = -3,8 V ⇒ impossível porque BJT saturou
VCE < VCEsat ~ 0,2 a 0,3 V para IC ~ mA ⇒ iC ≠ β iB
KVL malha de saída
iC = (VCC - VCEsat) / RC ≈ 3,2 mA
(b) Não se altera a malha de entrada logo iB = 23µA e iC = 2,3 mA
vCE = 10 – 4k × 2,3m = 0,8 V ⇒ BJT ainda está na zona activa
(VCE > VCEsat) logo iC = β iB e os cálculos está correctos
RC não afecta iC desde que BJT continue na zona activa
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Polarização e Modelo dinâmico: EXERCÍCIO BJT1 (3de3)
1.Example 5.14: (d) Em regime dinâmico (vulgo AC) com
os dados fornecidos e calculados na alínea anterior calcule
o ganho de tensão Av a impedância de entrada Zi e a
impedância de saída Zo do circuito da figura 5.53.
Enuncie as Leis da Teoria de Circuitos e as relações do
BJT que utilizar.
+
Zo
Zi
vo
−
modelo dinâmico do BJT
gm = IC / VT ≈ 2,3 mA / 25 mV = 92 mA/V (mS)
rπ = β / gm ≈ 100 / 92 = 1,09 kΩ
Ω
KVL malha de saída vi = RBB ib + rπ i b
= (RRB + rπ) vbe / rπ
+
KVL malha de saída v = - g R v
o
m C be
Figure 5.53 Example 5.14: (a)
KCL nó de saída
circuit; (c) small-signal model.
⇓ ⇒ Av = vo/vi = - gm RC [r /(RBB + r )] = - 3,04
π
π
Zi = Vi/Ii = RBB + rπ = 101kΩ
Ω ; Zo = Vo/Io = RC = 3kΩ
Ω
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Polarização e Modelo dinâmico: EXERCÍCIO BJT3 (1de3)
2. Example 5.13: (a) Dimensione a rede de polarização do BJT da figura para
se ter IC=1mA para que seja pouco sensível às variações tèrmicas de -25ºC a
125ºC sabendo que β = 100 @ 25ºC e vBE(25ºC,mA) ≈ 0,7 V.
Relações do BJT
vBE ≈ 0,7 V
para V~Volt e R~kΩ
Ω ⇒ I~mA
iC = β iB & β>>1 ⇒ iE ≈ iC
KVL malha de entrada
VBB = RBB iB + vBE + RE iE
= [(RBB/β
β)+RE] iC + vBE
Figure 5.44 Classical biasing for BJTs using a single
power supply: (a) circuit; (b) circuit with the voltage
divider supplying the base replaced with its Thévenin
equivalent.
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iC = (VBB - vBE)/ [(RBB/β
β)+RE]
KVL malha de saída
VCC = RC iC + vCE + RE iE
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Polarização e Modelo dinâmico: EXERCÍCIO BJT3 (2de3)
∆vBE (T) ≈ - 2 mV / ºC
(a)
Variações típicas de BJT com T ∆T = 125 – 25 = 150ºC ⇒ ∆V = 0,3 V
BE
Para IC pouco sensível
VBB >> ∆VBE e RBB<< β RE
Se VBB = 4V ⇒ RE ≈ (VBB – vBE)/IC = 3,3kΩ
Ω
& RBB << 330 kW ⇒ RBB < 33 kΩ
Ω
Para VB insensível a IB
VBB = VCC R2 / (R1 + R2) ⇒ R1 = 2 R2
Se RBB = 10 kΩ
R1 R2 / (R1 + R2) = 10 kΩ
Ω ⇒ R2 = 15 kΩ
Ω & R1 = 30 kΩ
Ω
IC RC = VCC – VRE - VCE ⇒ RC ≈ 4,35 kΩ
Ω
Se ICRC = VCE
VCE ≈ 4,35 V
Se R1 + R2 ↓ ⇒ R2 e R1 ↓ ⇒ RBB ↓ ⇒ mais válidas as aproximações
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Polarização e Modelo dinâmico: EXERCÍCIO BJT3 (3de3)
2. Example 5.13: (b) Quais sâo os valores de IC e VCE se β = 500 a 25ºC?
Não faça aproximações.
(c) Quais os valores de IC e VCE se RC=0 e RC=6 kΩ?
Ω?
(b) IC aumenta e VCE baixa ligeiramente
IC=(VBB-vBE)/[(RBB/β
β)+RE(β
β+1)/β
β] = 0.998mA
VCE = - RC IC + VCC - RE IE = 4,36 V
(c) IC mantem-se e VCE baixa (cuidado)
IC = (VBB - vBE)/ [(RBB/β
β)+RE] = 0,990 mA
VCE = - RC IC + VCC - RE IE = 4,39 V
Colector com polarização da base fixa é uma fonte de corrente
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