EE2301:

Basic Electronic Circuit

Recap in Unit 2

EE2301: Block B Unit 2 1

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 2

Unit 8 3

EE2301:

Basic Electronic Circuit

A quick revision on

Basic Structure of a MOSFET EE2301: Block B Unit 2 4

S

n +

N-channel MOSFET

p

G

I G =0

D

n +

Gate Oxide Substrate

I G =0

Bulk (substrate) G D S Gate current sees a capacitor formed by the gate, oxide (which is insulating), and substrate (conducting) EE2301: Block C Unit 2 5

When V GS > V T EE2301: Block C Unit 2

NMOS is normally off

6

When V GS - V T = V DS

Operating Regions

I D = K V DS 2 SATURATION CUTOFF

EE2301: Block C Unit 2 7

EE2301:

Basic Electronic Circuit

Let’s con’t Unit 3

Bipolar Junction Transistor EE2301: Block B Unit 2 8

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 9

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

  nature the

voltage

at the oxide insulated gate electrode, while the operation of BJT is dependent on the Thin

n

region sandwiched between p+ and p layers (called pnp-transistor)  > Thin

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)

I E = I B + I C

EE2301: Block C Unit 3 10

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

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 12

i D

+ Forward Biased:

Diode conducts

Diode Physics

+ + + + + + + + + + +

- - - - - - -

+ + + + + + + + +

-

+ + + + + + + + + + +

-

+ + +

+ -

Semiconductor Electronics - Unit 1: Diodes 13

Diode Physics

+ Reverse Biased:

Little or no current + + + + + + + + + + +

- - - - - - -

+ + + + + + + + +

-

+ + + + + + + + +

- - - - - - - - - - - - -

+ + + + + + + + + + + + + + +

+

Website: http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/diode.html

Semiconductor Electronics - Unit 1: Diodes 14

EE2301:

Basic Electronic Circuit

3 different operation regions compared to MOSFET

EE2301: Block B Unit 2 15

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

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 ) EE2301: Block C Unit 3 16

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 17

Linear active mode

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 = βI 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 Holes in the base are

V γ

B

V CB > 0

injected into the emitter

I B

E n+

I E

p

I C

n 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 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 18

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

V γ

B

V CB < 0

Holes in the base are also injected into the collector now

I B

E n+

I E

p n

I C

C Electron flow Potential gradient between collector and emitter is greatly reduced since V CE < V γ (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 19

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 20

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 21

BJT I-V Characteristic (3)

Both junctions are forward-biased V CE is small BE forward-biased, CB reverse biased (amplifier operation) Both junctions are reverse-biased Unit 8 Breakdown region: Physical limit of operation of the device 22

EE2301:

Basic Electronic Circuit

Recap in last lecture

Bipolar Junction Transistor EE2301: Block B Unit 2 23

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

EE2301: Block C Unit 3 24

BJT I-V Characteristic (3)

Both junctions are forward-biased V CE is small BE forward-biased, CB reverse biased (amplifier operation) Both junctions are reverse-biased Unit 8 Breakdown region: Physical limit of operation of the device 25

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

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 ) EE2301: Block C Unit 3 26

EE2301:

Basic Electronic Circuit

Basic Techniques to determine operation region

EE2301: Block B Unit 2 27

Block C Unit 3 P.1

B C E

V BC = -0.7 (reversed) V BE = V BC –V EC = -0.7 – (-0.2) = -0.5 (reversed) Both reversed biased 

cutoff

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

biased : V BE and V BC > 0  forward 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 reversed is forward and V BC c. Saturation if both are forward biased EE2301: Block C Unit 3 28

B E C

Block C Unit 3 P.1

V BE = 0.7 (forward) V BC = V BE –V CE = 0.7 – (0.3) = 0.4 (forward) Both reversed biased 

saturation

EE2301: Block C Unit 3 29

B E C

Block C Unit 3 P.1

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

EE2301: Block C Unit 3 30

Finding the operating mode 1

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 find V BE first: and if there is a base current. So we try to 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) EE2301: Block C Unit 3 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 ) 31

Finding the operating mode 2

Part II: Find the state of the BJT once again if V BB the voltmeter readings will change as a result. is now short-circuited. Note that all Once again we start with the base to check if the BJT is on or off first. We can check for V BE current. or the direction of the base 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 EE2301: Block C Unit 3 32

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. I BB = 0.15mA; V CC = 15V; R C = 375Ω Load line: Apply KVL around the right side mesh, I 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 C = V CC /R C – V CE /R C against V CE : EE2301: Block C Unit 3 33

Finding Q-point (by calculation)

Given: R B = 62.7kΩ; R C = 375Ω; 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 = βI 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: Block C Unit 3 34

EE2301:

Basic Electronic Circuit

Some More Examples

EE2301: Block B Unit 2 35

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?

EE2301: Block C Unit 3

Answer:

I C

=1.25mA;

V CE

=8.101V; active region 36

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?

EE2301: Block C Unit 3

Answers:

I C

=2.25mA;

V CE

=7.857V; active region 37

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

h 11 h 21

(input resistance):

h i; h 12

(forward transfer current ratio):

h f

(reverse transfer voltage ratio):

h r h 22

(output conductance):

h o

Mathematics will not covered in this course ± 50uA ± 2.5V

V i =V BE ; i 1 =I B ;V 0 =V CE ; i 0 =I C EE2301: Block C Unit 3 39

EE2301:

Basic Electronic Circuit

End of Block C Unit 3

EE2301: Block B Unit 2 40

Finding the operating region 3

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.

EE2301: Block C Unit 3 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) 41