unit 1 transistors, ujts and thyristors. unit 1: transistors, ujts and thyristors objectives:...

Unit 1

Transistors, UJTs and Thyristors

Unit 1: Transistors, UJTs and Thyristors

• Objectives:• Operating Point• CE configuration• Thermal runaway• UJT• SCR

1.1 Operating (Q) Point

To design an amplifier:1. DC & AC analysis

2. Operating Point in active region by biasing circuits

1. Fixed bias2. Emitter bias3. Collector-to-base bias4. Voltage divider bias with emitter bias

…selecting suitable Q point

…selecting suitable Q point

• Point A no biasCut-off region

• Point B near to knee portion Do not allow more output swing

• Point C close to PD(max) curve Output’s +ve swing is limited

• Point D middle of active regionAllows +ve & -ve excursions of output

1.2 CE configuration• Popular as provides considerable AV & AI

To learn:• Biasing Circuits:

1. Fixed bias2. Emitter bias (self bias)3. Voltage divider bias with emitter bias

• Analysis: DC, load-line

• merits & demerits

05/06/23 10:33 AM Unit 1: Transistors, UJTs and SCR

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fixed bias circuit• Simplest

• Biasing components:– 2 resistors (RB & RC)– Vcc

• BE junction FB by Vcc through RB (100s of kΩ)

• CB junction RB by Vcc through Rc (few k Ω)

05/06/23 10:33 AM Unit 1: Transistors, UJTs and SCR 8

DC analysis

To get DC equivalent circuit:

Open Circuit all capacitors, and redraw the circuit


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…From fundamentals• VBE = VB – VE

Similarly• VCE = VC – VE

Next, by KVL to BE loop

VCC – IBRB – VBE = 0

VCC – VBE IB = ----------------- RB


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…But IC = β IB where β is transistor gain

VCC - VBE

IC = β ( ---------------- ) RB

VCC

IC = β ( ---------- ) -----(1) RB


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…

β VCC

IC ≈ ------------ RB

IC

Where β is transistor current gain = ------ IB


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…

Next, apply KVL to output loop,

VCC = ICRC + VCE

VCE = VCC – ICRC ----(2)

Put Ic = 0 in eqn (2)∴ VCE = Vcc

Put VCE = 0 in eqn (2)∴ Ic = Vcc / Ic



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…

• Q point is influenced by

– IB – base current– RC – collector resistance– VCC – supply voltage


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Numerical example:

Find values ofICQ & VCQ

Find values ofIC & VCE,

forcing each value to zero,

each time


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Solution:


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• Thus,

ICQ = 1.43 mA & VCEQ = 9.28 V

Which are the coordinates of Q point


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Emitter bias / self bias configuration• Additional resistor RE, improves stability,as it produces negative feedback.


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DC analysis

• By KVL to input base loop,

VCC – IB RB – VBE – IE RE= 0

VCC – VBE

IB = -------------------- ----(1) RB + (β+1)RE

…

neglecting VBE VCC

IB = -------------------- RB + (β+1)RE

Vin

Compare this equation with Iin = -------- Rin

…

We get input resistance = RB + (β+1)RE

Next, applying KVL to output collector loop,

VCC – IC RC – VCE – IE RE = 0

VCE = VCC – IC (RC + RE) ------(2)

From basics, IC = IB + IEIB is << IE = IC

…Also VE = IE RE = IC RE

From eqn. (1) & (2)

VCC – VBE

ICQ = β -------------------- RB + (β+1)RE

VCEQ = VCC – IC (RC + RE)

The coordinates of Q point

Load line analysis

We have

VCE = VCC – IC (RC + RE)

Put IC = 0

VCE = VCC│IC = 0 is point of load line on x – axis (VCC, 0)

…

For second point,

Put VCE = 0

VCC

IC = ---------------(RC + RE) VCE = 0

…thus we have

on y- axis

(0, VCC/(RC + RE))

as another

point on load line.

How stability increased:

VCC – VBE

ICQ = β -------------------- RB + (β+1)RE

VCEQ = VCC – IC (RC + RE)

RE introduces negative (ac emitter) feedback, due to which• Stability of Av increases

How stability is improved by RE:

IEVE

VBE

IB

IE ≈ Ic IE = VE / RE

VCC = IB RB + VBE + VE

VBE = IB RBIB = β IB

Due to temperature or else, ICIC

Initial increase and later decrease proves that RE improves stability

Small Signal Operation:

AC and DC Currents in an Amplifier:

IBQ

ib

IB = IBQ + ib

ieIEQ

IE = IEQ + ie

ICQ

ic

IC = ICQ + ic

Numerical example:• Refer to figure. Find the values of RB, RC &

RB, given that IB=40µA, VE=2V, IC=4mA, VCE=12V & supply voltage VCC=15V. Assume silicon transistor.

Numerical:

3. Voltage-divider-bias with emitter-bias configuration:

• Stability further improved, ( but gain decreases)

• Most commonly used configuration

• Gain can be improved by bypass capacitor, later studied.

DC equivalent circuit:

To get DC equivalent circuit:

Open Circuit all capacitors,

and redraw the circuit

DC analysis2 methods:

1. Accurate method (uses Thevenin’s theorem)

2. Approximate method

1. Accurate Method:Using Thevenin’s Theorem:

DC Equivalent Circuit

Thevenin’s Equivalent Circuit

Redrawn showing Vcc

Calculating RTH & VTH:RTH:

Identify the 2 points short circuit all DC voltages

RTH

Little consideration will show that –RTH is = parallel combination of RB1 & RB2

RTH = RB1 ║ RB2

VTH:Consider the circuit, again –

VTH is nothing but – the voltage across the 2 points OR across RB2

RB2 VCC

VTH = ---------------- (RB1 + RB2)

VTH

Now replace the base circuit with Thevenin’s equivalent circuit -

• Applying KVL to the base loop –

VTH – IB RTH – VBE – IE RE = 0

Substituting IE = (β+1)IB

VTH - VBE

IB = ---------------------- RTH + (β+1)RE

IE = IC + IBDividing by IC, as it is to be eliminated,

IE IC + IB---- = ----- -----IC IC IC

1 = 1 + ----

β

β + 1 = ---------- β

IC (β + 1) = -------------- IC / IB

IE = IB (β + 1)

Proof

… VTH - VBE

IB = β --------------------- -----(1) RTH + (β+1)RE

By KVL to output collector loop,

VCC –IC RC – VCE – IE VE = 0

VCE = VCC – IC RC – IC RE as IC = IE

VCE = VCC – ICQ (RC + RE) -----(2)

Hence equations (1) & (2) represent Q point.

Q

Q

Load line analysis:

• Same as that of emitter bias:

Advantages & disadvantages of VDB:

• Excellent stability (against temperature & β), because of –ve feedback introduced by RE.

• But RE reduces gain AV.

• Again CE bypasses ac signal, increasing AV, also maintaining stability.

IE ie

DC do not pass through CE asf=0 in XC = 1 / (2πfC)= resistance

Numerical:

Solution:

4. Collector-to-base bias configuration:Little bit looking back:We have studied –1. base bias2. emitter (self) bias3. voltage-divider-bias with emitter-biaswith,• DC analysis (to know Q point/ICQ & VCEQ) • load-line analysis (points A & B of load-

line)

• RB provides negative feedback, voltage-shunt feedback

• Better (not excellent, as VDB with EB) stability

DC analysis:

IB + IC

IB IC

• KVL to base-emitter loop,VCC – (IB + IC) RC – IB RB – VBE = 0

• Substitute IC = β IB

VCC – VBE

IB = --------------------- RB + (β + 1) RC

VCC – VBE

ICQ = β --------------------- -----(1) RB + (β + 1) RC

• KVL to collector-emitter loop, VCC – (IB + IC) RC –VCE = 0

VCE = VCC – (IB + IC) RC

• Ignoring IB, VCE = VCC – IC RC

At quiescent conditions, VCEQ = VCC – ICQ RC -----(2)

Advantages & disadvantages:• Provides stability

• Reduced gain, due to –ve feedback, can be reduced by bypassing capacitor, as shown below,

We are not studying …

• 4.3 Common base circuit• 4.4 Common collector circuit• 4.5 Bias stabilization• 4.6 Bias compensation

But let us now study …

• 4.7 Thermal Runaway

4.7 Thermal Runaway

• Power dissipation depends on• Physical size• Its construction• Mounting arrangement

• Maximum power rating is limited by• Temperature that CB junction can withstand• Ambient temperature

• When ambient temperature , power rating , is known as “power derating”

…• Power dissipation capability range – 100s

milliwatts to 250 Watts

• Max. CB temperature – Si – 150 °C to 225 °C – Ge – 60 °C to 100

°C

Case temperature °C

Power dissipation PD

PD(MAX)

Typical power derating curve100 200

Maximum operating temperature

Defn.:when the transistor is in operation it

dissipates power & its junction temperature rises, which in turn causes collector current to increase. This may lead to more power dissipation & further increase in temperature & subsequent increase in collector current. If this cycle continues, it may result in permanent damage to the transistor. This phenomenon is known as “Thermal Runaway”.

PD Tj IC

…• Steady state junction temperature,

Tj – TA = θ PD

Where Tj – junction temperature TA – ambient temperaturePD – power dissipated in transistor

θ – thermal resistance °C/W

Thermal Resistance:ratio of the rise in transistor junction

temperature to the amount of power dissipated.

Which depends on –

• Transistor size• Size of heat sink• Other cooling method, such as forced air

Thermal resistance (θ):

Tj – TA dTj

θ = ----------- = -------- PD dPD

dPD 1 ----------- = -------- dTj θ

Operating Point considerations against thermal run-away:

At Q2:IC ↑ Q2 to move on

100mW curve PD ↓ thermally stable

At Q3:IC ↑ Q3 to move on

200mW curve PD ↑ thermally instable checked for thermal runaway

4.8 Transistor switch

• Another major application of transistor is Switch

…While designing,Transistor is heavily saturatedIC(sat) = VCC / RC

IB(sat) = IC(sat)/ βUsually IB(max) = IB(sat) x 0.25The minimum i/p voltage to drive

transistor in saturation isVIH = IB(max) RB + VBE

Rsat = VCE(sat) / IC(sat) ≈ few tens of Ohms

Transistor Switching Delay:

During ON:tD=delay time=to 10%tr=rise time=10% to

90%turn-ON time=tD+tr

During OFF:ts=storage time=fall to

10%tf=fall time=90% to

10%toff=turn-OFF time =ts+tf

Numerical:

An input pulse is applied to the transistor switch shown in figure. What is the minimum input voltage required to make the LED glow? It is given that the minimum current required by the LED to glow is 10mA, voltage drop across LED is 1.5V, BE voltage 0.7V, CE voltage at saturation is 0.5V.

IE=IC=10mA 1.5V

VBE = 0.7V

IB=IC / β =10mA / 100 = 1µA

VCE(sat) = 0.5V

1..Vp ?

2..IC(sat) ?

3..IB(sat) ?

=β x IC(sat)

4..IB(max) ?

=1.25 x IB(sat)

UJT (unijunction transistor)

• Only 1 pn junction, unlike 2 (BE & CB)• Current controlled• Negative resistance exists• May be used as switch

Construction:

Equivalent Circuit:

Intrinsic stand-off ratio:

= RBB1 / (RBB1 + RBB2)

VI characteristics:VP = VBB + V

Operation:when VE ↑ -IEO reaches 0, then ↑ to IP at VP, IE ↑ with VE ↓ up to IV & VV known as negative resistance, further behaves as a normal resistance.

UJT Relaxation Oscillator:

SCR (silicon controlled rectifier):Operation:A momentary pulse

applied to the gate increases the base current of npn transistor initiating regenerative feedback action;. This action ultimately drives both transistors to saturation causing switching-ON by conducting heavily.

Characteristics:

CB

BEFB

BE FB

CB

Questions:

1. Explain selecting a suitable operating point (104)

2. Explain DC & load-line analysis of fixed bias (105)

3. What are the parameters that vary the Q –point (ans:- IB, RC, VCC -108)

4. Problem 4.1 (109)5. Explain DC & load-line analysis of emitter /

self bias (110)6. Problem 4.2 & 4.3 (113)

7. Explain DC & load-line analysis of VDB with EB circuit (115)

8. Problem 4.4 (119)9. Explain DC & load-line analysis of collector-to-

base bias (124)10. Problem 4.7 (127)11. Define thermal runaway. Explain operating point

considerations in thermal runaway (147)12. Explain transistor as a switch. Brief about switching

delays (152)13. Explain UJT, relaxation oscillator (215)14. Explain SCR (227)

End

ofUnit 1: Transistors, UJT & Thyristors

IC (Integrated Circuit)

Overview


Vin = HIGH (≈5Volts)Saturation region, IC(sat)=VCC / RC

VCE = VCC - IC(sat) RC

= 0 Volt = logic LOW

Cut-off region, IC ≈ 0VCE = VCC - IC RC

= logic LOW

unit 1 transistors, ujts and thyristors. unit 1: transistors, ujts and thyristors objectives:...

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