Bipolar Junction Transistor (BJT) - Introduction

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1 Bipolar Junction Transistor (BJT) - Introduction It was found in 1948 at the Bell Telephone Laboratories. It is a three terminal device and has three semiconductor regions. It can be used in signal amplification and digital logic circuits as well. Its current conduction process is due to both holes and electrons. That is why the name bipolar. S. Sivasubramani EE101 - BJT 1/ 60

2 NPN Transistor E n type p type n type Emitter region Base region Collector region C EB Junction B CB Junction C B E Figure: Symbol S. Sivasubramani EE101 - BJT 2/ 60

3 PNP Transistor E p type n type p type Emitter region Base region Collector region C EB Junction B CB Junction E B C Figure: Symbol S. Sivasubramani EE101 - BJT 3/ 60

4 BJT - Modes of Operation Mode EB Junction CB Junction Active Forward Reverse Saturation Forward Forward Cutoff Reverse Reverse Active mode - Amplification Saturation and Cutoff - Digital logic circuits S. Sivasubramani EE101 - BJT 4/ 60

5 Active Mode - NPN Transistor E I E N P N Emitter Base Collector I C C V BE I B B i C = I S exp( v BE V t ) V CB i B = i C β where β is common-emitter current gain. Its typical value is in the range of 50 to 200. S. Sivasubramani EE101 - BJT 5/ 60

6 Active Mode By KCL i E = i C i B i E = i C β 1 β i C = αi E where α is a constant. α = β β 1 i E = I S α exp(v BE V t ) If β = 100, then α α is called as common-base current gain. S. Sivasubramani EE101 - BJT 6/ 60

7 i v Characteristics - CE i B I B I C V BE I E V CE v BE (V) S. Sivasubramani EE101 - BJT 7/ 60

8 i v Characteristics - CE Saturation Region i C Active Region v BE = v BE = v BE = v BE = Cutoff Region v CE As long as v CE > v CE sat, BJT is in active region. v CE sat = 0.2 V. If v CE falls below v CE sat, BJT will enter into saturation region. S. Sivasubramani EE101 - BJT 8/ 60

9 BJT Model - Active Mode B i B i C C v BE βi B i E E This model is nonlinear. (large signal) As E is common to both input and output, it is called as common emitter configuration. S. Sivasubramani EE101 - BJT 9/ 60

10 i v Characteristics - Practical BJT * C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\BJT_Characteristics.sch Date/Time run: 09/16/14 17:20:34 (A) BJT_Characteristics (active) 1.0A Temperature: A 0A 0V 5V 10V IC(Q1) V_V2 Date: September 16, 2014 Page 1 Time: 17:22:04 S. Sivasubramani EE101 - BJT 10/ 60

11 Early Voltage As Practical BJTs have finite slope in the i C Vs v CE characteristics while acting in active mode. This is accounted for by the following relation. i C = I S exp( v BE V t )(1 v CE V A ) where V A is the Early voltage whose value is in the range of 10 V to 100 V. The nonzero slope indicates the output resistance is finite. S. Sivasubramani EE101 - BJT 11/ 60

12 Large Signal Model Including Early Voltage B i B i C C v BE βi B r o i E E It is the large signal model of common emitter NPN transistor in active region. As there is no restriction in the signal range, it is called as a large signal model. S. Sivasubramani EE101 - BJT 12/ 60

13 Summary - Active Mode i C = I S exp( v BE V t ) i B = i C β = I S β exp(v BE V t ) i E = i C α = I S α exp(v BE V t ) i C = βi B ; i C = αi E α = β β 1 β = α α 1 S. Sivasubramani EE101 - BJT 13/ 60

14 BJT - Various Modes of Operation I B V BE I C V BC V CE V BE = V BC V CE V BC = V BE V CE 1 Active mode : V BE = 0.7V, V CE > 0.2V. 2 Saturation mode : V BE = 0.7V, V CE 0.2V. 3 Cutoff mode: V BE < 0.5V, V BC < 0.5V. S. Sivasubramani EE101 - BJT 14/ 60

15 BJT Models - Active Mode B I B > 0 i C C V BE =0.7 V βi B V CE >0.2 V E EBJ : Forward biased CBJ: Reverse biased S. Sivasubramani EE101 - BJT 15/ 60

16 BJT Models - Saturation Mode B I B > 0 V BE =0.7 V i C V CE C =0.2 V E EBJ : Forward biased CBJ: Forward biased S. Sivasubramani EE101 - BJT 16/ 60

17 BJT Models - Cutoff Mode B I B = 0 I C = 0 C V BE < 0.5 V E EBJ : Reverse biased CBJ: Reverse biased S. Sivasubramani EE101 - BJT 17/ 60

18 BJT - DC Analysis Whenever we want to analyze BJT circuits only with DC voltages, use the following steps. 1 Assume that the transistor is operating in active mode. 2 Determine I C, I B, V CE and V BE using the active mode model. 3 Check for consistency of results with active-mode operation such as V CE > V CE sat. 4 If it is satisfied, analysis is over. 5 If not, assume saturation mode and repeat the analysis like active mode. This analysis is mainly used to identify the operating point. S. Sivasubramani EE101 - BJT 18/ 60

19 Check yourself Which mode is transistor operating? β=100 and V BE = 0.7 V. 10 V By KVL in Base Emitter loop, 4 = V BE 3.3I E 4.7 k Ω I B = I E = 1 ma I E β 1 = = 9.9µA I C = I E I B = 0.99 ma 4 V 3.3 kω By KVL in CE loop, V CE = I C 3.3 I E = V Since V CE > V CE sat, it is operating in active mode. S. Sivasubramani EE101 - BJT 19/ 60

20 Amplifier - An Introduction v I (t) i I i o R L v O (t) v I (t) v O (t) S. Sivasubramani EE101 - BJT 20/ 60

21 Amplifier - An Introduction v O (t) = Av I (t) As v O and v I are linearly related, it is called as a linear amplifier. v O Slope= A v I S. Sivasubramani EE101 - BJT 21/ 60

22 Amplifier - An Introduction v I (t) i I i o R L v O (t) An amplifier takes power from battery to amplify the signal given at its input. This is how it amplifies the input signal. S. Sivasubramani EE101 - BJT 22/ 60

23 BJT- Amplifier V CC R C i C v BE is the input voltage. R C is the load resistance. The output v CE is taken between the collector and ground. v CE = V CC i C R C v BE v CE v CE = V CC I S exp( v BE V t )R C S. Sivasubramani EE101 - BJT 23/ 60

24 BJT - Voltage Transfer Characteristics (VTC) v CE Active Cutoff Saturation 0.2 V 0.5 V v BE S. Sivasubramani EE101 - BJT 24/ 60

25 BJT - Voltage Transfer Characteristics * C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\BJT_Characteristics.sch Date/Time run: 09/20/14 10:49:03 (A) BJT_Characteristics (active) 5.0V Temperature: V 0V 0V 0.5V 1.0V 1.5V V(Q1:c) V_V1 Date: September 20, 2014 Page 1 Time: 10:50:58 S. Sivasubramani EE101 - BJT 25/ 60

26 BJT - Biasing V CC v CE I C R C V CE Q Quiescent point V BE V CE V BE V CE = V CC I S exp( V BE V t )R C v BE S. Sivasubramani EE101 - BJT 26/ 60

27 BJT- Amplifier V CC i C R C v be v CE V BE v BE S. Sivasubramani EE101 - BJT 27/ 60

28 BJT-Amplifier v CE Active Cutoff Y Saturation V CE Q 0.2 V Z 0.5 V V BE v BE S. Sivasubramani EE101 - BJT 28/ 60

29 Small Signal Voltage Gain If the input signal v be is very small, the output signal v ce will be v ce = A v v be where A v is the voltage gain that is determined by evaluating the slope of VTC at Q point. A v = dv ce dv be VBE A v = d(v CC I S exp( v BE )R C ) V t dv be VBE = I S exp( V BE )R C V t V t A v = I C R C V t S. Sivasubramani EE101 - BJT 29/ 60

30 Graphical Analysis - To determine Bias point v BE R C V CC i C v CE i C V CC R C I C i C = V CC v CE R C Z Q V CE v BE = v BE = v BE = Y v BE = V CC v CE S. Sivasubramani EE101 - BJT 30/ 60

31 How to choose a bias point There are mainly two factors that determine the location of Q or bias or Operating point. 1 Gain 2 Swing limit at the output It is clear from graphical analysis, there are three operating points( Z, Q & Y). If we choose Z, gain will be more but distorted negative portion of signal. If we choose Y, reduction in gain and distortion in positive portion of signal. If Q is chosen, gain as well as swing in the output signal are achieved. Choose an operating point that should give gain and maximum swing S. Sivasubramani EE101 - BJT 31/ 60

32 Check yourself Find the operating point. Assume V BE = 0.7 V. V CC = 5V I C (ma) 10 I B = 50µA 500 Ω 8 I B = 40µA 215 kω 6 I B = 30µA 4 I B = 20µA v i v o 2 I B = 10µA V CE (V) S. Sivasubramani EE101 - BJT 32/ 60

33 Solution By KVL in BE loop, 5 V 215 kω 500 Ω 5 V 5 = I B V BE I B = = 20 µa KVL in CE loop, (Load line) 5 = I C 500 V CE I C = 5 V CE 500 S. Sivasubramani EE101 - BJT 33/ 60

34 Solution - contd... I C (ma) I CQ 4 2 Q I B = 50µA I B = 40µA I B = 30µA I B = 20µA I B = 10µA V CE Q V CE (V) S. Sivasubramani EE101 - BJT 34/ 60

35 Transistor Biasing Biasing means establishing a constant current in the collector of BJT. The Q point or bias or Operating point should not be sensitive to temperature variations and variations in β. The location of DC bias point should allow for maximum output signal swing. Transistor is operated in active region by biasing. To calculate I C and V CE corresponding to DC bias, DC analysis should be carried out. (Instead of using two different batteries, one DC battery V CC can be used for both EB and CB junctions.) S. Sivasubramani EE101 - BJT 35/ 60

36 Two Basic Schemes - Not to be used V CC V CC V CC V CC I C I C R B1 R C R B R C V CE V CE I B I B R B2 V BE V BE (a) (b) Figure: Bad Schemes S. Sivasubramani EE101 - BJT 36/ 60

37 Bad Schemes 1 First scheme (Figure a) It applies constant V BE through voltage divider action. For a small change in V BE due to temperature variation, I B changes drastically because of exponential relationship between them. It will change I C as I C = βi B. Hence, It is not a good approach. 2 Second scheme (Figure b) Figure b applies constant I B. If there is a change in β, I C will change. Hence, It is also not a good approach As these two schemes are sensitive to variation in temperature and variation in β, they should not be used for biasing a transistor. S. Sivasubramani EE101 - BJT 37/ 60

38 Classical Biasing Scheme V CC V CC I C R 1 I B R C V CE In order to make biasing scheme insensitive to temperature and β variations, R E is included in emitter circuit. R 2 V BE R E S. Sivasubramani EE101 - BJT 38/ 60

39 Classical Bias Scheme - Thevenin Equivalent V BB = V CC R 2 R 1 R 2 V CC I C I B R C V CE R B = R 1 R 2 V BE R E S. Sivasubramani EE101 - BJT 39/ 60

40 Classical Bias Scheme The current I E can be determined by writing KVL in base-emitter loop. V BB = I B R B V BE I E R E Substituting I B = I E β 1 As I C = αi E and α = V BB = I E I E = β β 1 R B β 1 V BE I E R E V BB V BE R E R B /(β 1) I C = β(v BB V BE ) R E (β 1) R B S. Sivasubramani EE101 - BJT 40/ 60

41 Contd... If we choose R E (β 1) >> R B, I C β(v BB V BE ) R E (β 1) Normally β lies in the range of 10 to 100. I C can be written as β 1 β I C (V BB V BE ) R E Now I C is independent of β. As V BB >> V BE, change in V BE will also not affect I C much. In this bias scheme, Q point is insensitive to variation in β and variation in temperature. S. Sivasubramani EE101 - BJT 41/ 60

42 Transistor - Small Signal Analysis To perform small signal analysis, small signal equivalent circuit should be drawn. 1 Calculate the operating point using DC analysis Click here to refer. 2 Draw the small signal equivalent circuit at this operating point. 3 As small signal circuit is linear, calculating the voltage, current would be easy. Small signal analysis is also called as ac analysis. S. Sivasubramani EE101 - BJT 42/ 60

43 Circuit with small signal Let us consider this circuit to draw small signal equivalent and analyze. i C R C v be v BE v CE V CC V BE S. Sivasubramani EE101 - BJT 43/ 60

44 Circuit with small signal - DC Analysis To find the operating point by DC analysis, ac signal should be removed. R C I C Calculate I C, V CE as follows. I C = I S exp( V BE V t ) I B V CC V CE = V CC I C R C Mostly I C is determined as V BE I E V CE I C = βi B where I B is found using KVL in base emitter loop. S. Sivasubramani EE101 - BJT 44/ 60

45 Small signal model - BJT (refer lecture notes on two port network) Let us consider BJT as a two port network with emitter as a common terminal. v BE B i B i C C v CE E i C = I S exp( v BE V t ) i B = i C β S. Sivasubramani EE101 - BJT 45/ 60

46 The small signal voltages and currents are related by linear equations. i b = y bb v be y bc v ce i c = y cb v be y cc v ce As i B and i C are only function of v BE, y bc and y cc will be zero. y bb = i B v BE VBE = I S βv t exp( V BE V t ) = I C βv t y cb = i C v BE VBE = I C V t S. Sivasubramani EE101 - BJT 46/ 60

47 B i b i c C v be r π g m v be v ce E E where r π = 1 y bb = βv t I C is the resistance between base and emitter and g m = y cb = I C V t is the transconductance. S. Sivasubramani EE101 - BJT 47/ 60

48 Small signal model - circuit i C R C v be v BE v CE V CC B v be E i b r π C g m v be R C E v ce V BE S. Sivasubramani EE101 - BJT 48/ 60

49 Transistor Small Signal Analysis - Steps to be followed 1 Eliminate the signal source and determine the DC operating point in particular I C and V CE. 2 Calculate the value of small signal model parameters r π and g m at the operating point. 3 Draw the small signal equivalent circuit by short circuiting the DC sources and open circuiting the DC current sources. 4 Analyze this circuit to determine voltage gain, input resistance and output resistance. S. Sivasubramani EE101 - BJT 49/ 60

50 Example - Small signal analysis Calculate small signal gain v o v i. Assume β = kω V O v o v i 100 kω 10 V 3 V S. Sivasubramani EE101 - BJT 50/ 60

51 Example - Solution DC Analysis: 3 kω 100 kω 10 V 3 V Assume V BE =0.7 V. I B = I B = ma I C = βi B = 2.3 ma V CE = V CC I C R C = 3.1 V S. Sivasubramani EE101 - BJT 51/ 60

52 Solution - Contd Small signal analysis (AC analysis): 100 kω B C v i r π v be g m v be 3 kω v o E E g m = I C = 2.3 = V t 25 r π = β g m = = Ω r π v be = v i = v i 100kΩ r π v o = g m v be 3000 = 2.96v i v o v i = 2.96 S. Sivasubramani EE101 - BJT 52/ 60

53 AC Coupling It is used to add input signal to the amplifier and output signal to the load without affecting bias point. It is done with the help of large capacitance. Large capacitance will become an open circuit in DC analysis. Large capacitance will behave like a short circuit at signal frequency in ac analysis. In addition to two capacitances, one more capacitance is connected across R E. It is called as a bypass capacitor which will act as a short circuit and ground emitter without R E in ac analysis. S. Sivasubramani EE101 - BJT 53/ 60

54 CE Amplifier - Example V CC R 1 R C C 2 R S C 1 v i R 2 R L v o R E C E S. Sivasubramani EE101 - BJT 54/ 60

55 CE Amplifier - DC Analysis To find the operating point, the circuit is redrawn like this. V CC I C R 1 R C C R 2 R E E V CE S. Sivasubramani EE101 - BJT 55/ 60

56 Small Signal Equivalent Circuit R S B i b C i c v i R 2 R 1 v be r π g m v be R C R L v ce E S. Sivasubramani EE101 - BJT 56/ 60

57 BJT - Digital Logic 1 Saturation : When EB junction and CB junction are forward biased, BJT will act as a closed switch between collector and emitter. 2 Cutoff : When EB junction and CB junction are reverse biased, BJT will act as an open circuit between collector and emitter. The above two modes can be used to implement logic functions. S. Sivasubramani EE101 - BJT 57/ 60

58 BJT - NOT Gate V CC =5 V R C V out When V in < 0.7 V (Low), EB and CB junctions are reverse biased. I B = 0, I C = 0. V out = V CC = 5V (High) V in R B When V in = 5 V (High), EB and CB Junctions are forward biased. V out = V CEsat = 0.2V (Low) S. Sivasubramani EE101 - BJT 58/ 60

59 Check yourself V CC =5 V R C C Positive Logic : 0 V - Low 0 5 V - High 1 B A R B R B NAND Gate A B C S. Sivasubramani EE101 - BJT 59/ 60

60 Check yourself B A R B R B V CC =5 V C R E Positive Logic : 0 V - Low 0 5 V - High 1 AND Gate A B C S. Sivasubramani EE101 - BJT 60/ 60

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