ee2301: basic electronic circuit recap in unit 2 ee2301: block b unit 21

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


A quick revision on

Basic Structure of a MOSFET


EE2301: Block C Unit 2 5

N-channel MOSFET

Bulk (substrate)

GS D

p n+n+G

D

S

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

IG=0

Gate

Oxide

Substrate

IG=0


NMOS is normally off

When VGS > VT


Operating Regions

CUTOFF

ID = K VDS2

TRIO

DE/O

HM

IC

SATURATION

BR

EA

KD

OW

N

When

VGS - VT = VDS


Let’s con’t Unit 3

Bipolar Junction Transistor



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


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

Summary1. 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.


How does it work?

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



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 - Unit 1: Diodes 13

Diode Physics

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Semiconductor Electronics - Unit 1: Diodes 14

Diode Physics

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

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Reverse Biased:

Little or no current


3 different operation regions compared to MOSFET



Large Signal Model

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)

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)


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


Linear active mode

n+

np C

B

E

Vγ VCB > 0

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

IB

IE

IC

Electrons in the emitter are injected into the base

Electron flowSome 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 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


Saturation mode

n+ np C

B

E

Vγ VCB < 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.

IE IC

Electron flow

IB

Holes in the base are also injected into the collector now

Potential gradient between collector and emitter is greatly reduced since VCE < Vγ (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

IBB injects current to forward-bias the BE junction

When Collector (C) is open

Unit 8 21


Collector Characteristic: Family of curves (for each value of IB, an IC-VCE curve can be generated)

Unit 8 22


Both junctions are forward-biased

VCE is small

Both junctions are reverse-biased

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

Breakdown region: Physical limit of operation of the device


Recap in last lecture

Bipolar Junction Transistor



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

Unit 8 25


Both junctions are forward-biased

VCE is small

Both junctions are reverse-biased

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

Breakdown region: Physical limit of operation of the device


Large Signal Model

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)

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)


Basic Techniques to determine operation region


Block C Unit 3 P.1


C

B

E

1. Identify the terminals of the transistor2. 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 regiona. Cutoff if both are reversed biasedb. Active linear if VBE is forward and VBC

is reversedc. Saturation if both are forward biased

VBC = -0.7 (reversed)

VBE = VBC –VEC = -0.7 – (-0.2) = -0.5 (reversed)

Both reversed biased cutoff

Block C Unit 3 P.1


C

B

E

VBE = 0.7 (forward)

VBC = VBE –VCE = 0.7 – (0.3) = 0.4 (forward)

Both reversed biased saturation

Block C Unit 3 P.1


C

B

E

PNP type

VBE = -0.6 (forward)

VBC = VBE –VCE = -0.6 – (-4.0) = 3.4V (reversed)

VBE is forward and VBC is reversed active linear

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)

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

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 = 375ΩLoad 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

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)


Some More Examples


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

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

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)


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


Mathematics will not covered in this course

Vi=VBE; i1=IB;V0=VCE; i0=IC

±2.5V

±50uA

h11 (input resistance): hi; h12 (reverse transfer voltage ratio): hrh21 (forward transfer current ratio): hf h22 (output conductance): ho


End of Block C Unit 3


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: basic electronic circuit recap in unit 2 ee2301: block b unit 21

Documents

block c unit

block b unit

cont unit

mosfet ee2301

pn junction

idid slide

v ds slide

n layers