Transcript Block D: Semiconductor Electronics

EE2301: Basic Electronic Circuit
Recap in Unit 2
EE2301: Block B Unit 2
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The Transistor


A transistor is a 3-terminal semiconductor device (cf
Diode is a 2-terminal device)
Performs 2 main functions fundamental to electronic
circuits:
1) Amplification – magnifying a signal
2) Switching – controlling a large current or voltage across 2
terminals (on/off)

2 major families of transistors:
1) Field Effect Transistors (FETs) – Unit 2
2) Bipolar Junction Transistors (BJTs) – Unit 3
Unit 8
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Unit 8
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EE2301: Basic Electronic Circuit
A quick revision on
Basic Structure of a MOSFET
EE2301: Block B Unit 2
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N-channel MOSFET
G
S
IG=0
D
D
Gate
Oxide
n+
p
Bulk (substrate)
n+
Substrate
G
IG=0
S
Gate current sees a capacitor formed by the gate, oxide
(which is insulating), and substrate (conducting)
EE2301: Block C Unit 2
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NMOS is normally off
When VGS > VT
EE2301: Block C Unit 2
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Operating Regions
When
VGS - VT = VDS
ID = K VDS2
SATURATION
CUTOFF
EE2301: Block C Unit 2
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EE2301: Basic Electronic Circuit
Let’s con’t Unit 3
Bipolar Junction Transistor
EE2301: Block B Unit 2
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Block C Unit 3 Outline
 The bipolar transistor (BJT) operation
> Construction of the BJT
> Basic working principle of the BJT
> Operation modes of the BJT
 Transistor amplifier circuits
> Biasing the transistor in a circuit
> Small signal equivalent circuit (hybrid model)
G. Rizzoni, “Fundamentals of EE” Sections 10.1 – 10.4
EE2301: Block C Unit 3
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Summary
Construction
of a BJT
1. A BJT has an emitter, collector and base, while a MOSFET has a gate,
source and drain.
 The
Formed
by joining
3 sectionsdepends
of semiconducting
materialatofthe
alternating
2.
operation
of MOSFET
on the voltage
oxide- doping
nature gate electrode, while the operation of BJT is dependent on the
insulated
 This can
be base.
either:
current
at the
3. BJTs
are preferred
for low current
applications,
while(called
MOSFETs
are for
> Thin
n region sandwiched
between
p+ and p layers
pnp-transistor)
high >
power
Thin functions.
p region sandwiched between n+ and n layers (called npn-transistor)
4. In digital and analog circuits, MOSFETs are considered to be more
 Comprises three terminals (each one associated with one of the three sections)
commonly used than BJTs these days.
known as the emitter (E), base (B) and collector (C)
IE = IB + IC
EE2301: Block C Unit 3
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EE2301: Basic Electronic Circuit
How does it work?
Let’s have a quick revision on p-n
junction first
EE2301: Block B Unit 2
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p-n Junction
The pn junction forms the basis of the semiconductor diode
Within the depletion region, no free carriers exist since the holes and
electrons at the interface between the p-type and n-type recombine.
EE2301: Block C Unit 1
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Diode Physics
iD
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Forward Biased:
Diode conducts
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--- +++ - - - --- +++
- - --- +++ - - - -
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Semiconductor Electronics - Unit 1: Diodes
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Diode Physics
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Reverse Biased:
Little or no current
----+ + + --------+ + +
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Website: http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/diode.html
Semiconductor Electronics - Unit 1: Diodes
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EE2301: Basic Electronic Circuit
3 different operation regions
compared to MOSFET
EE2301: Block B Unit 2
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Large Signal Model
Like the MOS, the BJT is also characterized by 3 operating regions, but differ in
what each region means. The operation principle is also obviously different between
the two.
For a BJT, it is easier and typical to think of the emitter current (IE) to be controlled
by the base current (IB) and the voltage across the collector and emitter (VCE)
 Cut-off region
> BE junction is reverse-biased
 Linear active region
> BE junction is forward-biased
(VBE = Vγ typically around 0.7V)
> BC junction is reverse-biased
 Saturation region
> BOTH BE and BC junctions are
forward-biased (VBE = Vγ)
EE2301: Block C Unit 3
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Cutoff mode
In the cutoff mode, the pn junctions at both the emitter-base and collector-base
interfaces are reverse biased. This is similar to having two back to back reverse
biased diodes, where it can be seen that no current can flow between the emitter
and collector terminals. As such, the transistor is simply off. There is also no
current flowing into the base in this mode.
To turn the transistor on, we must forward bias the base-emitter junction, VBE>0
EE2301: Block C Unit 3
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Linear active mode
The active region is characterized by a small base current greatly amplified to
give a much larger collector current, such that IB << IC
It is common to expression their relationship by a parameter symbolized by β:
IC = βIB, where β is typically large in the active region (from 50 to above 400)
As a result of this large current gain, we find that IC ~ IE
Holes in the base are
injected into the emitter
Vγ
VCB > 0
B
IB
E
IE
n+
Electrons in the emitter
are injected into the base
n
p
IC
C
Electron flow
Some of the emitter electrons recombine
in the base but most it is swept into the
collector
Large electric field across the junction sweeps electrons in the base into the collector
with minimal recombination occurring in the thin base layer.
EE2301: Block C Unit 3
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Saturation mode
The saturation region is characterized by low current gain between the base and
collector.
Base current is now larger
due to hole injection into
the collector from the base
Vγ
B
VCB < 0
Holes in the base are
also injected into the
collector now
IB
E
n+
IE
n
p
IC
C
Electron flow
Potential gradient between collector and
emitter is greatly reduced since VCE < Vγ
(but still positive), having an effect on the
collector current
Accelerating electric field across the junction is now switched off. Electrons
diffuse into collector.
EE2301: Block C Unit 3
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BJT I-V Characteristic (1)
 BE junction similar to
diode
 IBB injects current to
forward-bias the BE
junction
Unit 8
When Collector (C) is open
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BJT I-V Characteristic (2)
Collector Characteristic: Family of curves (for each value of
IB, an IC-VCE curve can be generated)
Unit 8
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BJT I-V Characteristic (3)
Both junctions are forward-biased
BE forward-biased, CB reversebiased (amplifier operation)
VCE is small
Both junctions are reverse-biased
Unit 8
Breakdown region: Physical limit of
operation of the device
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EE2301: Basic Electronic Circuit
Recap in last lecture
Bipolar Junction Transistor
EE2301: Block B Unit 2
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Construction of a BJT
 Formed by joining 3 sections of semiconducting material of alternating doping
nature
 This can be either:
> Thin n region sandwiched between p+ and p layers (called pnp-transistor)
> Thin p region sandwiched between n+ and n layers (called npn-transistor)
 Comprises three terminals (each one associated with one of the three sections)
known as the emitter (E), base (B) and collector (C)
IE = IB + IC
EE2301: Block C Unit 3
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BJT I-V Characteristic (3)
Both junctions are forward-biased
BE forward-biased, CB reversebiased (amplifier operation)
VCE is small
Both junctions are reverse-biased
Unit 8
Breakdown region: Physical limit of
operation of the device
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Large Signal Model
Like the MOS, the BJT is also characterized by 3 operating regions, but differ in
what each region means. The operation principle is also obviously different between
the two.
For a BJT, it is easier and typical to think of the emitter current (IE) to be controlled
by the base current (IB) and the voltage across the collector and emitter (VCE)
 Cut-off region
> BE junction is reverse-biased
 Linear active region
> BE junction is forward-biased
(VBE = Vγ typically around 0.7V)
> BC junction is reverse-biased
 Saturation region
> BOTH BE and BC junctions are
forward-biased (VBE = Vγ)
EE2301: Block C Unit 3
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EE2301: Basic Electronic Circuit
Basic Techniques to determine
operation region
EE2301: Block B Unit 2
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Block C Unit 3 P.1
C
B
E
VBC = -0.7 (reversed)
VBE = VBC –VEC
= -0.7 – (-0.2)
= -0.5 (reversed)
1. Identify the terminals of the transistor
2. Determine whether it is PNP or NPN
a. PNP : VBE and VBC < 0  forward
biased
b. NPN : VBE and VBC > 0  forward
biased
3. Determine VBE and VBC to see if they are
forward or reversed biased
4. Determine the operation region
a. Cutoff if both are reversed biased
b. Active linear if VBE is forward and VBC
is reversed
c. Saturation if both are forward biased
Both reversed biased  cutoff
EE2301: Block C Unit 3
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Block C Unit 3 P.1
C
VBE = 0.7 (forward)
VBC = VBE –VCE
= 0.7 – (0.3)
= 0.4 (forward)
B
E
EE2301: Block C Unit 3
Both reversed biased  saturation
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Block C Unit 3 P.1
C
PNP type
VBE = -0.6 (forward)
B
E
VBC = VBE –VCE
= -0.6 – (-4.0)
= 3.4V (reversed)
VBE is forward and VBC is reversed  active linear
EE2301: Block C Unit 3
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Finding the operating mode 1
Given following voltmeter readings, find the operating mode of the BJT:
V1 = 2V; V2 = 1.3V; V3 = 8V
The first question to ask is whether the BJT is on/off. This
depends on VBE and if there is a base current. So we try to
find VBE first:
VBE = V1 – V2 = 0.7V (this indicates the BJT is on)
Furthermore, IB = (4 – V1)/40 = 50µA (current flows into the
base as if the BJT is on)
Now we need to check if it is in active and saturation mode.
We know that if the BJT is in active mode, then β will be
large. Therefore, we need to find the ratio of IC over IB.
VCC = ICRC + V3 12 = IC*1 + 8 IC = 4mA
β = 4/0.05 = 80 (BJT is in active mode)
To further verify this, we can check whether IE ≈ IC:
IE = V2/321 = 4.05mA (hence equal to IC + IB)
EE2301: Block C Unit 3
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Finding the operating mode 2
Part II: Find the state of the BJT once again if VBB is now short-circuited. Note that all
the voltmeter readings will change as a result.
Once again we start with the base to check if the BJT is on
or off first. We can check for VBE or the direction of the base
current.
If VBB is short circuited, what then are the options we can
choose from?
Option 1: If we assume (or guess) the base is also grounded
as a result, then IB = 0
Option 2: If we assume the base is at a positive voltage, then
IB flows out of the base
Which ever option you choose, the conclusion is still the
same  the BJT is in cutoff
EE2301: Block C Unit 3
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Finding the Q-point (by graph)
Just like the MOS, the approach (and motivation) for finding the Q-point is the same for the
BJT. Once again, the Q-point is found at the intersection of the load line and the I-V curves.
IBB = 0.15mA; VCC = 15V; RC = 375Ω
Load line: Apply KVL around the right side mesh,
VCC = VCE + ICRC
Since we want to draw this on the I-V curve, we should express this as IC against VCE:
IC = VCC/RC – VCE/RC
EE2301: Block C Unit 3
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Finding Q-point (by calculation)
Given: RB = 62.7kΩ; RC = 375Ω; VBB = 10V; VCC = 15V; Vγ = 0.6V; β = 145
Find the Q-point , assuming that the transistor is operated in active region
We start by assuming the BJT is on VBE = Vγ.
IB = (VBB – Vγ)/RB = 0.15mA
IC = βIB = 21.75mA
VCE = VCC – ICRC = 15 – 21.75*0.375 = 6.84V
How do we know the transistor is operated in active linear region?
We must test whether VBE is forward biased and VBC is reversed biased
VBE = VBB- IBRB = 10 – (62.7)(0.15) = 0.595V>0 (forward baised);
VBC=VBE – VCE = 0.595 -6.84 = -6.245 <0 (reversed biased)
EE2301: Block C Unit 3
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EE2301: Basic Electronic Circuit
Some More Examples
EE2301: Block B Unit 2
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Block C Unit 3 Q.3
Given the circuit of Figure P10.3, determine the operation point of the silicon
device with β = 100. Assume a 0.6-V offset voltage In what region is the
transistor?
Answer:
IC=1.25mA; VCE=8.101V; active region
EE2301: Block C Unit 3
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Block C Unit 3 Q.6
Given the circuit of Figure below, determine the operating point of the
transistor. Assume a 0.6-V offset voltage and β = 150. In what region is
the transistor?
Answers:
IC=2.25mA; VCE=7.857V; active region
EE2301: Block C Unit 3
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Hybrid model equivalent circuit
 The BJT is inherently a non-linear device (basic circuit theory
is based on linear problems)
 We can instead define the linear characteristics of the BJT for
a given Q-point for small variations about the Q-point
 We can do so using a small signal equivalent circuit that will
now allow us to linearize the behavior of the BJT for a given
operating point
 Now we can use linear circuit theory analysis applicable for
the conditions are the parameters of the model are valid
 For the BJT, one popular model is the h-model (h for “hybrid”
as in h-parameters)
EE2301: Block C Unit 3
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h parameter model
 Linearizes the characteristics of a BJTT (which is otherwise nonlinear) for
a given operating point
 Only valid for a particular operating point
 Only valid for small varying signals
h11 (input resistance): hi;
h12 (reverse transfer voltage ratio): hr
h21 (forward transfer current ratio): hf
h22 (output conductance): ho
Mathematics will not
covered in this course
±50uA
±2.5V
Vi=VBE; i1=IB;V0=VCE; i0=IC
EE2301: Block C Unit 3
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EE2301: Basic Electronic Circuit
End of Block C Unit 3
EE2301: Block B Unit 2
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Finding the operating region 3
Given following voltmeter readings, find the operating mode of the BJT:
V1 = 2.7V; V2 = 2V; V3 = 2.3V
Once again, we start with the base of the BJT.
VBE = V1 – V2 = 0.7V (hence BJT is on)
This time VBB is unknown, so we cannot find IB directly.
But we follow the same path as previous, it looks like we
can still find IB through IC and IE.
IE = V2/RE and IC = (VCC-V3)/RC
But RE and RC are both unknown, so it seems we cannot
find any of the currents.
However, we can find VBC or VCE:
VBC = V1 – V3 = 0.4V (This is sufficient to forward-bias the
CE junction BJT is in saturation)
VCE = V3 – V2 = 0.3V (low value indicative of saturation)
EE2301: Block C Unit 3
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