Transistor fundamentals Nafees Ahamad Asstt. Prof., EECE Deptt, DIT University, Dehradun Website: www.eedofdit.weebly.com
Transistor A transistor consists of two PN junctions formed by sandwiching either p-type or n-type semiconductor between a pair of opposite types. Accordingly ; there are two types of transistors, namely; (i) n-p-n transistor (ii) p-n-p transistor Nafees Ahamad 2
Why name Transistor? Transfer + Resistor Transistor. It is transferring the signal from low resistance to high resistance. It amplifies the signal if it is properly biased. Nafees Ahamad 3
Transistor: Actual look Nafees Ahamad 4
Naming the Transistor Terminals Nafees Ahamad 5
Naming the Transistor Terminals 1. Emitter: The section on one side that supplies charge carriers (electrons or holes) is called the emitter. 2. Base: The emitter is always forward biased w.r.t. base. The middle section between the emitter and collector is called the base. 3. Collector: The section on the other side that collects the charges is called the collector. The collector is always reverse biased. Nafees Ahamad 6
Biasing Transistor Terminals The base-emitter junction is forward biased, allowing low resistance for the emitter circuit. The base-collector junction is reverse biased and provides high resistance in the collector circuit. Nafees Ahamad 7
Transistor Action (i) Working of npn transistor: The forward bias causes the electrons in the n-type emitter to flow towards the base. This constitutes the emitter current I E. As these electrons flow through the p-type base, they tend to combine with holes. As the base is lightly doped and very thin, therefore, only a few electrons (less than 5%) combine with holes to constitute base current I B. The remainder (more than 95%) cross over into the collector region to constitute collector current I C. Nafees Ahamad 8
Transistor Action Nafees Ahamad 9
Transistor Action (ii) Working of pnp transistor: The forward bias causes the holes in the p-type emitter to flow towards the base. This constitutes the emitter current I E. As these holes cross into n-type base, they tend to combine with the electrons. As the base is lightly doped and very thin, therefore, only a few holes (less than 5%) combine with the electrons. The remainder (more than 95%) cross into the collector region to constitute collector current I C. Nafees Ahamad 10
Transistor Action Note: Direction of current is same as the direction of flow of Nafees Ahamad electrons and opposite to that of holes. 11
Transistor Symbols: NPN transistor: Nafees Ahamad 12
PNP transistor: Nafees Ahamad 13
Transistor Connections There are three methods (i) Common Base connection (CB) (ii) Common Emitter connection (CE) (iii) Common Collector connection (CC) Note: Each circuit connection has specific advantages and disadvantages. Nafees Ahamad 14
Transistor Connections CB Common Base Connection (CB Connection): Base is common Nafees Ahamad 15
Transistor Connections CB 1. Current amplification factor (α): Output Current α = (when only DC ) Input Curretn VCB =Constamt Change in Output Current α = Change in Input Curretn VCB =Constamt α = I C I E VCB =Constamt < 1 Commercial transistors 0.9 < α < 0.99 Nafees Ahamad 16
Transistor Connections CB 2. Expression for collector current: Total collector current, I C = α I E + I leakage Where α I E = That part of emitter current which reaches the collector terminal (by using α = I C I E ) I leakage = Current due to the movement of minority carriers across base-collector junction (Very less) I leakage = I CBO = Collector-base current with emitter open (See diagram on next slide) Nafees Ahamad 17
Transistor Connections CB I leakage = I CBO Nafees Ahamad 18
Transistor Connections CB So I C = αi E + I CBO Put I E = I C + I B and solve Nafees Ahamad 19
Transistor Connections CB Various currents are as shown in following figure Nafees Ahamad 20
Question Q1. In a common base connection, α = 0.95. The voltage drop across 2 kω resistance which is connected in the collector is 2V. Find the base current. Solution: Nafees Ahamad 21
Question Q2. For the common base circuit shown in Fig., determine I C and V CB. Assume the transistor to be of silicon. Nafees Ahamad 22
Question Solution: Above circuit can be redrawn as Nafees Ahamad 23
Question Since the transistor is of silicon, So V BE = 0.7V. Applying KVL to the emitter-side loop, we get, Neglecting leakage current I leakage = I CBO Nafees Ahamad 24
Question Applying KVL to the collector-side loop, we have, Nafees Ahamad 25
Characteristics of CB Connection Two types 1.Input characteristics: I E Vs V EB at constant V CB 2.Output characteristics: I c Vs V CB at constant I E Nafees Ahamad 26
Characteristics of CB Connection 1. Input characteristics: (I E Vs V EB ) Input resistance (r i ): (r i -low value ) Nafees Ahamad 27
Characteristics of CB Connection 2. Output characteristics: (I c Vs V CB ) Output resistance(r o ): (r o -High value ) Nafees Ahamad 28
Transistor Connections CE Here, emitter of the transistor is common to both input and output circuits Most commonly used (90% to 95% transistor applications) Nafees Ahamad 29
Transistor Connections CE 1. Base Current amplification factor (β): (Similar to CB) Output Current β = Input Curretn = I C (when only DC ) I B Change in Output Current β = Change in Input Curretn β = I C I E > 20 For commercial transistors 20 < β < 500. Nafees Ahamad 30
Transistor Connections CE Relation between β and α. Nafees Ahamad 31
Transistor Connections CE Nafees Ahamad 32
Transistor Connections CE 2. Expression for collector current: Where I CEO = Collector-Emitter current when base is open (leakage current) Nafees Ahamad 33
Transistor Connections CE Various currents are as Nafees Ahamad 34
Transistor Connections CE Various currents are as shown in following figure Nafees Ahamad 35
Characteristics of CE Connection Two types (Similar to CB connection) 1.Input characteristics: I B Vs V BE at constant V CE 2.Output characteristics: I c Vs V CE at constant I B Nafees Ahamad 36
Characteristics of CE Connection Nafees Ahamad 37
Characteristics of CE Connection 1. Input characteristic: Input resistance(r i ): r i few hundred ohms Nafees Ahamad 38
Characteristics of CE Connection 2. Output characteristic: Output resistance(r o ): Nafees Ahamad 39
Transistor Connections CC I/p- between Base & Collector, O/p between Emitter & Collector Nafees Ahamad 40
Transistor Connections CC 1. γ = γ = Current amplification factor γ: (Similar to CB) Output Current Input Curretn Change in Output Current Change in Input Curretn = I E I B (when only DC ) γ = I E I B > 20 Nafees Ahamad 41
Transistor Connections CC Relation between γ and α: Expression for collector current: Nafees Ahamad 42
Comparison of Transistor Connections SN Characteristic CB CE CC 1 I/p Resistance Low ( 100 Ω) Low ( 750 Ω) Very high ( 750 KΩ) 2 O/p Resistance Very high ( 450 KΩ) High( 45 KΩ) Low ( 3 Voltage Gain 4 Applications High frequency applications 150 >500 <1 Audio frequency applications 50 Ω) Impedance matching 5 Current Gain No (<1) High(β) Applicable Nafees Ahamad 43
Transistor Load Line Analysis Load line analysis is used to calculate I C. DC load line: Consider a common emitter npn transistor where no signal is applied. Therefore, d.c. conditions prevail in the circuit. Nafees Ahamad 44
Transistor Load Line Analysis Nafees Ahamad 45
Transistor Load Line Analysis The value of V CE is given by V CE = V CC I C R C (Known as DC load line) ---(1) Above can be represented by a straight line on the output characteristics with following two end points. Put I C = 0 in equ(1) V CE = V CC I C R C => V CE = V CC Point B Put V CE = 0 in equ(1) 0 = V CC I C R C => I C = V CC /R C Point A Nafees Ahamad 46
Transistor Load Line Analysis Operating Point: The zero signal values of I C and V CE are known as the operating point(points B & A). It is also called quiescent (silent) point or Q-point because it is in the absence of the signal. Nafees Ahamad 47
Transistor Load Line Analysis Draw V CE Vs I C characteristics when I B = 5 μa as shown in previous slide Then intersection point Q, of load line and V CE Vs I C characteristics will represent Q-point as shown in above graph with V CE = OC Volts I C = OD Amp Nafees Ahamad 48
Questions on Load Line Q. In the circuit diagram shown in Fig., if V CC = 12V and R C = 6 kω, draw the d.c. load line. What will be the Q point if zero signal base current is 20μA and β = 50? Nafees Ahamad 49
Questions on Load Line Solution We know V CE = V CC I C R C ---(1) Put I C = 0 in equ(1) V CE = V CC I C R C => V CE = V CC =12 V Point B Put V CE = 0 in equ(1) 0 = V CC I C R C => I C = V CC /R C =12/6kΩ = 2 ma Point A Now draw the dc load line joining points A & B Nafees Ahamad 50
Questions on Load Line Zero signal base current, I B = 20 μa = 0.02 ma Current amplification factor, β = 50 Zero signal collector current, I C = β I B = 50 0.02 = 1 ma V CE = V CC I C R C = 12 1 ma 6 k Ω = 6 V Operating point is 6 V, 1 ma. Nafees Ahamad 51
Thank you Nafees Ahamad 52