Bipolar Junction
Transistors (BJTs)
Course: ELEC231
Prof. Farid Touati
1
5.1 DEVICE STRUCTUE & MODES OF OPERATION
Figure 5.1 A simplified structure of the npn transistor.
Figure 5.2 A simplified structure of the pnp transistor.
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Active Mode for an NPN BJT
Compare with MOSFET….
Figure 5.3 Current flow in an npn transistor biased to operate in the active mode. (Reverse current components due to drift of thermally
generated minority carriers are not shown.)
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Modes of Operation
MODE
EBJ
CBJ
Cutoff
Reverse
Reverse
Active
Forward
Reverse
Saturation
Forward
Forward
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Large-Signal Equivalent Circuit in the NPN Active Mode
Current Equations
iB =
iC
β
, iC = I S e vbe / VT
iE = iC + iB → iE = αiC
Figure 5.5 Large-signal equivalent-circuit models of the npn BJT operating in the forward active mode.
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Large-Signal Equivalent Circuit in the NPN Reverse Mode
Figure 5.7 Model for the npn transistor when operated in the reverse active mode (i.e., with the CBJ forward biased and the EBJ reverse biased).
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Active Mode for an PNP BJT
Figure 5.11 Current flow in a pnp transistor biased to operate in the active mode.
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Large-Signal Equivalent Circuit in the PNP Active Mode
Figure 5.12 Large-signal model for the pnp transistor operating in the active mode.
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CIRCUIT SYMBOLS AND CONVENTIONS
Figure 5.13 Circuit symbols for BJTs.
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VOLTAGE POLARITY AND CURRENT FLOW
Figure 5.14 Voltage polarities and current flow in transistors biased in the active mode.
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NPN BJT large-signal models
Active
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Saturation
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Cutoff
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GRAPHICAL REPRESENTATION OF BJT
CHARACTERISITCS
iC-vBE curve compare with iD-vGS
See Example 5.10 on this!
Figure 5.16 The iC –vBE characteristic for an npn transistor.
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Figure 5.17 Effect of temperature on the iC–vBE characteristic. At a constant emitter
current (broken line), vBE changes by –2 mV/°C.
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EXAMPLE 5.1
VBE=0.7V @ iC=1mA, β=100. Design the circuit so that iC=2mA
and VC=5V.
Figure 5.15 Circuit for Example 5.1.
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Solve graphically….
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BJT biased in Active Region
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BJT biased in Saturation or
Cutoff Region
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EXAMPLE 5.4
Determine all nodes
voltage, β=100.
Check for active
mode
ONBOARD….
Figure 5.34 Analysis of the circuit for Example 5.4: (a) circuit; (b) circuit redrawn to remind the reader of the convention used in this book to
show connections to the power supply; (c) analysis with the steps numbered.
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EXAMPLE 5.5
Determine all nodes
voltage, β=100.
Check for active
mode
DO IT YOURSELF!
Figure 5.35 Analysis of the circuit for Example 5.5. Note that the circled numbers indicate the order of the analysis steps.
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EXAMPLE 5.6
Determine all nodes
voltage, β=100.
Check for active
mode
DO IT YOURSELF!
Figure 5.36 Example 5.6: (a) circuit; (b) analysis with the order of the analysis steps indicated by circled numbers.
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EXAMPLE 5.8
Determine all nodes
voltage, β=100.
Check for active
mode
DO IT YOURSELF!
T rises à Ib & Ic increase à Q unstable à solution??
Figure 5.38 Example 5.8: (a) circuit; (b) analysis with the steps indicated by the circled numbers.
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EXAMPLE 5.10
Determine all nodes
voltage, β=100.
Check for active
mode
ONBOARD….
Stable BIASING!
Q is independent of β
Q is stable also!
Figure 5.40 Circuits for Example 5.10.
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BIASING SCHEMES
Figure 5.43 Two obvious schemes for biasing the BJT: (a) by fixing VBE; (b) by fixing IB. Both result in wide variations in
IC and hence in VCE and therefore are considered to be “bad.” Neither scheme is recommended.
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BIASING USING TWO POWER SUPPLIES
Figure 5.45 Biasing the BJT using two power supplies. Resistor RB is needed only if the signal is to be capacitively
coupled to the base. Otherwise, the base can be connected directly to ground, or to a grounded signal source, resulting in
almost total β-independence of the bias current.
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BIASING USING FEEDBACK R
Figure 5.46 (a) A common-emitter transistor amplifier biased by a feedback resistor RB. (b) Analysis of the circuit in (a).
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BIASING USING A CONSTANT CURRENT SOURCE
Figure 5.47 (a) A BJT biased using a constant-current source I. (b) Circuit for implementing the current source I.
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SMALL-SIGNAL OPERATION AND MODELS
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.
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CE Amplifier Transfer Characteristic
Logic Inverter….. Can you see!
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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BJT transconductance gm
∂iC
gm =
∂vBE
iC = I C
IC
=
VT
Figure 5.49 Linear operation of the transistor under the small-signal condition: A small signal vbe with a triangular
waveform is superimposed on the dc voltage VBE. It gives rise to a collector signal current ic, also of triangular waveform,
superimposed on the dc current IC. Here, ic = gmvbe, where gm is the slope of the iC–vBE curve at the bias point Q.
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The Hybrid-π Model
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.
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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 currentcontrolled current source (a current amplifier).
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The T Model
Figure 5.52 Two slightly different versions of what is known as the T model of the BJT. The circuit in (a) is a voltage-controlled
current source representation and that in (b) is a current-controlled current source representation. These models explicitly show
the emitter resistance re rather than the base resistance rπ featured in the hybrid-π model.
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Application of the Small-Signal Equivalent Circuits
The process of analysis of BJT amps consists of:
1. Determine the Bias conditions (e.g. IC)
2. Calculate the values of the small-signal model
parameters: gm=IC/VT, rπ=β/gm, and re=VT/IE=α/
gm
3. Short all dc voltage sources and open all dc
current sources.
4. Replace the transistor by one of its models.
5. Analyze the resulting circuit to find the required
quantities (voltage/current gain, i/p and o/p
resistance, etc…
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Example 5.14
Analyze the circuit below to find its Av, Avo, Rin, Ro, Ai.
Assume β=100 and VT=25mV
On Board…..
Figure 5.53 Example 5.14: (a) circuit; (b) dc analysis; (c) small-signal model.
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Constraint: vbe < 10mV à vi<0.91V àvC=vO< 2.77V
We took vipeak=0.8V àibpeak=0.008mA, vbe,peak= 8.6mV à
icpeak=8mA, vcpeak=2.43V.
Figure 5.54 Signal waveforms in the circuit of Fig. 5.53.
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Augmenting the Small-Signal Models to account
for the Early effect.
The iC-vCE Curve and the Early Voltage VA
−1
⎡ ∂i
⎤
ro = ⎢ C vBE = cons tan t ⎥ ,
⎣ ∂vCE
⎦
ro ≅
VA
IC
Figure 5.58 The hybrid-π small-signal model, in its two versions, with the resistance ro included.
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The Common-Emitter (CE) BJT Amplifier
Figure 5.60 (a) A common-emitter amplifier using the structure of Fig. 5.59. (b) Equivalent circuit obtained by replacing the
transistor with its hybrid-π model.
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The CE BJT Amplifier with Emitter R
Figure 5.61 (a) A common-emitter amplifier with an emitter resistance Re. (b) Equivalent circuit obtained by replacing the transistor with its T
model.
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The Common-Base (CB) BJT Amplifier
Figure 5.62 (a) A common-base amplifier using the structure of Fig. 5.59. (b) Equivalent circuit obtained by replacing the transistor with its T
model.
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The CC or Emitter-Follower BJT Amplifier
Figure 5.63 (a) An emitter-follower circuit based on the structure of Fig. 5.59. (b) Small-signal equivalent circuit of the emitter follower with the
transistor replaced by its T model augmented with ro. (c) The circuit in (b) redrawn to emphasize that ro is in parallel with RL. This simplifies the
analysis considerably.
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Figure 5.64 (a) An equivalent circuit of the emitter follower obtained from the circuit in Fig. 5.63(c) by reflecting all resistances in the emitter to
the base side. (b) The circuit in (a) after application of Thévenin theorem to the input circuit composed of vsig, Rsig, and RB.
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Figure 5.65 (a) An alternate equivalent circuit of the emitter follower obtained by reflecting all base-circuit resistances to the emitter side. (b)
The circuit in (a) after application of Thévenin theorem to the input circuit composed of vsig, Rsig / (β 1 1), and RB / (β 1 1).
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Figure 5.66 Thévenin equivalent circuit of the output of the emitter follower of Fig. 5.63(a). This circuit can be used to find vo and hence the
overall voltage gain vo/vsig for any desired RL.
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