Recap in Unit 2

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EE2301:

Basic Electronic Circuit

Recap in Unit 2

EE2301: Block B Unit 2

1

Unit 8

2

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

3

EE2301:

Basic Electronic Circuit

A quick revision on

Basic Structure of a MOSFET

EE2301: Block B Unit 2

4

EE2301: Block C Unit 2

5

N
-
channel MOSFET

Bulk (substrate)

G

S

D

p

n
+

n
+

G

D

S

Gate current sees a capacitor formed by the gate, oxide
(which is insulating), and substrate (conducting)

I
G
=0

Gate

Oxide

Substrate

I
G
=0

EE2301: Block C Unit 2

6

NMOS is normally off

When V
GS

> V
T

EE2301: Block C Unit 2

7

Operating Regions

CUTOFF


I
D

= K V
DS
2

SATURATION

When

V
GS
-

V
T

= V
DS

EE2301:

Basic Electronic Circuit

Let’s con’t Unit 3

Bipolar Junction Transistor

EE2301: Block B Unit 2

8

EE2301: Block C Unit 3

9

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

10

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)

I
E

= I
B

+ I
C

Summary

1. A BJT has an emitter, collector and base, while a MOSFET has a gate,
source and drain.

2. The operation of MOSFET depends on
the
voltage

at the oxide
-
insulated gate electrode, while the operation of BJT is dependent on the
current
at the base.

3. BJTs are preferred for
low current applications
, while MOSFETs are for
high power functions
.

4. In digital and analog circuits, MOSFETs are considered to be
more
commonly

used than BJTs these days.

EE2301:

Basic Electronic Circuit

How does it work?

Let’s have a quick revision on p
-
n
junction first

EE2301: Block B Unit 2

11

EE2301: Block C Unit 1

12

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.

Semiconductor Electronics
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Unit 1: Diodes

13

Diode Physics

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Diode conducts

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Semiconductor Electronics
-

Unit 1: Diodes

14

Diode Physics

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Reverse Biased:

Little or no current

EE2301:

Basic Electronic Circuit

3 different operation regions
compared to MOSFET

EE2301: Block B Unit 2

15

EE2301: Block C Unit 3

16

Large Signal Model


Cut
-
off region

>
BE junction is reverse
-
biased


Linear active region

>
BE junction is forward
-
biased


(V
BE

= V
γ

typically around 0.7V)

>
BC junction is reverse
-
biased


Saturation region

>
BOTH BE and BC junctions are
forward
-
biased (V
BE

= V
γ
)

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 (I
E
) to be controlled
by the base current (I
B
) and the voltage across the collector and emitter (V
CE
)

EE2301: Block C Unit 3

17

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, V
BE
>0

EE2301: Block C Unit 3

18

Linear active mode

n+

n

p

C

B

E

V
γ

V
CB

> 0

Large electric field across the junction sweeps electrons in the base into the collector
with minimal recombination occurring in the thin base layer.

I
B

I
E

I
C

Electrons in the emitter
are injected into the base

Electron flow

Some of the emitter electrons recombine
in the base but most it is swept into the
collector

Holes in the base are
injected into the emitter

The active region is characterized by a small base current greatly amplified to
give a much larger collector current, such that I
B

<< I
C


It is common to expression their relationship by a parameter symbolized by
β
:
I
C

=
β
I
B
, 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
I
C

~ I
E


EE2301: Block C Unit 3

19

Saturation mode

n+

n

p

C

B

E

V
γ

V
CB

< 0

Accelerating electric field across the junction is now switched off. Electrons
diffuse into collector.

The saturation region is characterized by low current gain between the base and
collector.

I
E

I
C

Electron flow

I
B

Holes in the base are
also injected into the
collector now

Potential gradient between collector and
emitter is greatly reduced since V
CE

< V
γ

(but still positive), having an effect on the
collector current

Base current is now larger
due to hole injection into
the collector from the base

Unit 8

20

BJT I
-
V Characteristic (1)


BE junction similar to
diode


I
BB

injects current to
forward
-
bias the BE
junction

When Collector (C) is open

Unit 8

21

BJT I
-
V Characteristic (2)

Collector Characteristic: Family of curves (for each value of
I
B
, an I
C
-
V
CE

curve can be generated)

Unit 8

22

BJT I
-
V Characteristic (3)

Both junctions are forward
-
biased

V
CE

is small

Both junctions are reverse
-
biased

BE forward
-
biased, CB reverse
-
biased (amplifier operation)

Breakdown region: Physical limit of
operation of the device

EE2301:

Basic Electronic Circuit

Recap in last lecture

Bipolar Junction Transistor

EE2301: Block B Unit 2

23

EE2301: Block C Unit 3

24

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)

I
E

= I
B

+ I
C

Unit 8

25

BJT I
-
V Characteristic (3)

Both junctions are forward
-
biased

V
CE

is small

Both junctions are reverse
-
biased

BE forward
-
biased, CB reverse
-
biased (amplifier operation)

Breakdown region: Physical limit of
operation of the device

EE2301: Block C Unit 3

26

Large Signal Model


Cut
-
off region

>
BE junction is reverse
-
biased


Linear active region

>
BE junction is forward
-
biased


(V
BE

= V
γ

typically around 0.7V)

>
BC junction is reverse
-
biased


Saturation region

>
BOTH BE and BC junctions are
forward
-
biased (V
BE

= V
γ
)

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 (I
E
) to be controlled
by the base current (I
B
) and the voltage across the collector and emitter (V
CE
)

EE2301:

Basic Electronic Circuit

Basic Techniques to determine
operation region

EE2301: Block B Unit 2

27

Block C Unit 3 P.1

EE2301: Block C Unit 3

28

C

B

E

1.
Identify the terminals of the transistor

2.
Determine whether it is PNP or NPN

a.
PNP

: V
BE

and V
BC

< 0


forward
biased

b.
NPN

: V
BE

and V
BC

> 0


forward
biased

3.
Determine V
BE

and V
BC

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 V
BE

is forward and V
BC

is reversed

c.
Saturation if both are forward biased


V
BC

=
-
0.7 (reversed)


V
BE

= V
BC


V
EC



=
-
0.7


(
-
0.2)


=
-
0.5 (reversed)


Both reversed biased


cutoff

Block C Unit 3 P.1

EE2301: Block C Unit 3

29

C

B

E

V
BE

= 0.7 (forward)


V
BC

= V
BE


V
CE



= 0.7


(0.3)


= 0.4 (forward)


Both reversed biased


saturation

Block C Unit 3 P.1

EE2301: Block C Unit 3

30

C

B

E

PNP type


V
BE

=
-
0.6 (forward)


V
BC

= V
BE


V
CE



=
-
0.6


(
-
4.0)


= 3.4V (reversed)


V
BE

is forward and V
BC

is reversed


active linear

Finding the operating mode 1

EE2301: Block C Unit 3

31

Given following voltmeter readings, find the operating mode of the BJT:

V
1

= 2V; V
2

= 1.3V; V
3

= 8V

The first question to ask is whether the BJT is on/off. This
depends on V
BE

and if there is a base current. So we try to
find V
BE

first:

V
BE

= V
1



V
2

= 0.7V (this indicates the BJT is on)

Furthermore, I
B

= (4


V
1
)/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 I
C

over I
B
.

V
CC

= I
C
R
C

+ V
3


12 = I
C
*1 + 8

I
C

= 4mA


β

= 4/0.05 = 80 (BJT is in active mode)

To further verify this, we can check whether I
E

≈ I
C
:

I
E

= V
2
/321 = 4.05mA (hence equal to I
C

+ I
B
)



Finding the operating mode 2

EE2301: Block C Unit 3

32

Part II: Find the state of the BJT once again if V
BB

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 V
BE

or the direction of the base
current.

If V
BB

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 I
B

= 0

Option 2: If we assume the base is at a positive voltage, then
I
B

flows out of the base

Which ever option you choose, the conclusion is still the
same


the BJT is in cutoff

Finding the Q
-
point (by graph)

EE2301: Block C Unit 3

33

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.

I
BB

= 0.15mA; V
CC

= 15V; R
C

= 375
Ω

Load line: Apply KVL around the right side mesh,

V
CC

= V
CE

+ I
C
R
C

Since we want to draw this on the I
-
V curve, we should express this as I
C

against V
CE
:

I
C

= V
CC
/R
C



V
CE
/R
C

Finding Q
-
point (by calculation)

EE2301: Block C Unit 3

34

Given: R
B

= 62.7k
Ω
; R
C

= 375
Ω
; V
BB

= 10V; V
CC

= 15V; V
γ

= 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

V
BE

= V
γ
.


I
B

= (V
BB



V
γ
)/R
B

= 0.15mA

I
C

=
β
I
B

= 21.75mA

V
CE

= V
CC



I
C
R
C

= 15


21.75*0.375 = 6.84V

How do we know the transistor is operated in active linear region?

We must test whether
V
BE
is forward biased
and
V
BC

is reversed biased

V
BE

= V
BB
-

I
B
R
B

= 10


(62.7)(0.15)

= 0.595V>0 (forward baised);


V
BC
=V
BE



V
CE
= 0.595
-
6.84 =
-
6.245 <0 (reversed biased)

EE2301:

Basic Electronic Circuit

Some More Examples

EE2301: Block B Unit 2

35

Block C Unit 3 Q.3

EE2301: Block C Unit 3

36

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:

I
C
=1.25mA;
V
CE
=8.101V; active region

Block C Unit 3 Q.6

EE2301: Block C Unit 3

37

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:

I
C
=2.25mA;
V
CE
=7.857V; active region

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

38

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

EE2301: Block C Unit 3

39

Mathematics will not
covered in this course

V
i
=V
BE
; i
1
=I
B
;V
0
=V
CE
; i
0
=I
C

±
2.5V

±
50uA

h
11

(input resistance):
h
i;

h
12

(reverse transfer voltage ratio):
h
r

h
21

(forward transfer current ratio):
h
f

h
22

(output conductance):
h
o

EE2301:

Basic Electronic Circuit

End of Block C Unit 3

EE2301: Block B Unit 2

40

Finding the operating region 3

EE2301: Block C Unit 3

41

Given following voltmeter readings, find the operating mode of the BJT:

V
1

= 2.7V; V
2

= 2V; V
3

= 2.3V

Once again, we start with the base of the BJT.

V
BE

= V
1



V
2

= 0.7V (hence BJT is on)

This time V
BB

is unknown, so we cannot find I
B

directly.
But we follow the same path as previous, it looks like we
can still find I
B

through I
C

and I
E
.

I
E

= V
2
/R
E

and
I
C

= (V
CC
-
V
3
)/R
C


But R
E

and R
C

are both unknown, so it seems we cannot
find any of the currents.

However, we can find V
BC

or V
CE
:

V
BC

= V
1



V
3

= 0.4V (This is sufficient to forward
-
bias the
CE junction

BJT is in saturation)

V
CE

= V
3



V
2

= 0.3V (low value indicative of saturation)