Chapter 2. - DC Biasing - BJTs

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1 Chapter 2. - DC Biasing - BJTs Objectives To Understand : Concept of Operating point and stability Analyzing Various biasing circuits and their comparison with respect to stability BJT A Review Invented in 1948 by Bardeen, Brattain and Shockley Contains three adjoining, alternately doped semiconductor regions: Emitter (E), Base (B), and Collector (C) The middle region, base, is very thin Emitter is heavily doped compared to collector. So, emitter and collector are not interchangeable. Three operating regions Linear region operation: Base emitter junction forward biased Base collector junction reverse biased Cutoff region operation: Base emitter junction reverse biased Base collector junction reverse biased Saturation region operation: Base emitter junction forward biased Base collector junction forward biased Three operating regions of BJT Cut off: V CE = V CC, I C 0 Active or linear : V CE V CC /2, I C I C max /2 Saturation: V CE 0, I C I C max Q-Point (Static Operation Point) 1

2 The values of the parameters I B, I C and V CE together are termed as operating point or Q ( Quiescent) point of the transistor. Q-Point The intersection of the dc bias value of I B with the dc load line determines the Q- point. It is desirable to have the Q-point centered on the load line. Why? When a circuit is designed to have a centered Q-point, the amplifier is said to be midpoint biased. Midpoint biasing allows optimum ac operation of the amplifier. Introduction - Biasing The analysis or design of a transistor amplifier requires knowledge of both the dc and ac response of the system.in fact, the amplifier increases the strength of a weak signal by transferring the energy from the applied DC source to the weak input ac signal The analysis or design of any electronic amplifier therefore has two components: The dc portion and The ac portion During the design stage, the choice of parameters for the required dc levels will affect the ac response. What is biasing circuit? Once the desired dc current and voltage levels have been identified, a network must be constructed that will establish the desired values of I B, I C and V CE, Such a network is known as biasing circuit. A biasing network has to preferably make use of one power supply to bias both the junctions of the transistor. Purpose of the DC biasing circuit To turn the device ON To place it in operation in the region of its characteristic where the device operates most linearly, i.e. to set up the initial dc values of I B, I C, and V CE Important basic relationship V BE = 0.7V I E = ( + 1) I B I C I C = I B 2

3 Biasing circuits: Fixed bias circuit Emitter bias Voltage divider bias DC bias with voltage feedback Miscellaneous bias Fixed bias The simplest transistor dc bias configuration. For dc analysis, open all the capacitance. DC Analysis Applying KVL to the input loop: V CC = I B R B + V BE From the above equation, deriving for IB, we get, I B = [V CC V BE ] / R B The selection of R B sets the level of base current for the operating point. Applying KVL for the output loop: V CC = I C R C + V CE Thus, V CE = V CC I C R C 3

4 In circuits where emitter is grounded, V CE = V E V BE = V B Design and Analysis Design: Given I B, I C, V CE and V CC, or I C, V CE and, design the values of R B, R C using the equations obtained by applying KVL to input and output loops. Analysis: Given the circuit values (V CC, R B and R C ), determine the values of I B, I C, V CE using the equations obtained by applying KVL to input and output loops. Problem Analysis Given the fixed bias circuit with V CC = 12V, R B = 240 k, R C = 2.2 k and = 75. Determine the values of operating point. Equation for the input loop is: I B = [V CC V BE ] / R B where V BE = 0.7V, thus substituting the other given values in the equation, we get I B = 47.08uA I C = I B = 3.53mA V CE = V CC I C R C = 4.23V When the transistor is biased such that I B is very high so as to make I C very high such that I C R C drop is almost V CC and V CE is almost 0, the transistor is said to be in saturation. I C sat = V CC / R C in a fixed bias circuit. Verification Whenever a fixed bias circuit is analyzed, the value of I CQ obtained could be verified with the value of I CSat ( = V CC / R C ) to understand whether the transistor is in active region. In active region, I CQ = ( I CSat /2) Load line analysis A fixed bias circuit with given values of V CC, R C and R B can be analyzed ( means, determining the values of I BQ, I CQ and V CEQ ) using the concept of load line also. Here the input loop KVL equation is not used for the purpose of analysis, instead, the output characteristics of the transistor used in the given circuit and output loop KVL equation are made use of. 4

5 The method of load line analysis is as below: 1. Consider the equation V CE = V CC I C R C This relates V CE and I C for the given I B and R C 2. Also, we know that, V CE and I C are related through output characteristics We know that the equation, V CE = V CC I C R C represents a straight line which can be plotted on the output characteristics of the transistor. Such line drawn as per the above equation is known as load line, the slope of which is decided by the value of R C ( the load). Load line The two extreme points on the load line can be calculated and by joining which the load line can be drawn. To find extreme points, first, Ic is made 0 in the equation: V CE = V CC I C R C. This gives the coordinates (V CC,0) on the x axis of the output characteristics. The other extreme point is on the y-axis and can be calculated by making V CE = 0 in the equation V CE = V CC I C R C which gives I C( max) = V CC / R C thus giving the coordinates of the point as (0, V CC / R C ). The two extreme points so obtained are joined to form the load line. The load line intersects the output characteristics at various points corresponding to different I B s. The actual operating point is established for the given I B. Q point variation As I B is varied, the Q point shifts accordingly on the load line either up or down depending on I B increased or decreased respectively. As R C is varied, the Q point shifts to left or right along the same I B line since the slope of the line varies. As R C increases, slope reduces ( slope is -1/R C ) which results in shift of Q point to the left meaning no variation in I C and reduction in V CE. Thus if the output characteristics is known, the analysis of the given fixed bias circuit or designing a fixed bias circuit is possible using load line analysis as mentioned above. 5

6 Emitter Bias It can be shown that, including an emitter resistor in the fixed bias circuit improves the stability of Q point. Thus emitter bias is a biasing circuit very similar to fixed bias circuit with an emitter resistor added to it. Input loop Writing KVL around the input loop we get, V CC = I B R B + V BE + I E R E (1) We know that, I E = ( +1)I B (2) Substituting this in (1), we get, V CC = I B R B + V BE + ( +1)I B R E V CC V BE = I B (R B + ( +1) R E ) Solving for I B : I B = (V CC V BE ) /[(R B + ( +1) R E )] 6

7 The expression for I B in a fixed bias circuit was, Equivalent input loop: I B = (V CC V BE ) /R B R EI in the above circuit is ( +1)R E which means that, the emitter resistance that is common to both the loops appears as such a high resistance in the input loop. Thus Ri = ( +1)R E ( more about this when we take up ac analysis) Output loop Collector emitter loop Applying KVL, V CC = I C R C + V CE + I E R E I C is almost same as I E 7

8 Thus, V CC = I C R C + V CE + I C R E = I C (R C + R E ) +V CE V CE = V CC - I C (R C + R E ) Since emitter is not connected directly to ground, it is at a potential V E, given by, V E = I E R E V C = V CE + V E OR V C = V CC I C R C Also, V B = V CC I B R B OR V B = V BE + V E Problem: Analyze the following circuit: given = 75, V CC = 16V, R B = 430k, R C = 2k and R E = 1k Solution: I B = (V CC V BE ) /[(R B + ( +1) R E )] = ( ) / [ 430k + (76) 1k] = A I C = ( 75) (30.24 A) = 2.27mA V CE = V CC - I C (R C + R E ) = 9.19V V C = V CC I C R C = 11.46V V E = V C V CE = 2.27V V B = V BE + V E = 2.97V V BC = V B V C = = V 8

9 Improved bias stability Addition of emitter resistance makes the dc bias currents and voltages remain closer to their set value even with variation in transistor beta temperature Stability In a fixed bias circuit, I B does not vary with and therefore whenever there is an increase in, I C increases proportionately, and thus V CE reduces making the Q point to drift towards saturation.in an emitter bias circuit, As increases, I B reduces, maintaining almost same I C and V CE thus stabilizing the Q point against variations. Saturation current In saturation V CE is almost 0V, thus Thus, saturation current V CC = I C ( R C + R E ) I C,sat = V CC / ( R C + R E ) Load line analysis The two extreme points on the load line of an emitter bias circuit are, (0, V CC / [ R C + R E ]) on the Y axis, and ( V CC, 0) on the X axis. Voltage divider bias +V CC R 1 R C C C 2 1 v in v out R 2 R E C 3 9

10 This is the biasing circuit wherein, I CQ and V CEQ are almost independent of. The level of I BQ will change with so as to maintain the values of I CQ and V CEQ almost same, thus maintaining the stability of Q point. Two methods of analyzing a voltage divider bias circuit are: Exact method can be applied to any voltage divider circuit Approximate method direct method, saves time and energy, can be applied in most of the circuits. Exact method In this method, the Thevenin equivalent network for the network to the left of the base terminal to be found. To find Rth: From the above circuit, R th = R1 R2 = R1 R2 / (R1 + R2) 10

11 To find Eth From the above circuit, E th = V R2 = R 2 V CC / (R1 + R2) In the above network, applying KVL ( E th V BE ) = I B [ R th +( + 1) R E ] I B = ( E th V BE ) / [ R th +( + 1) R E ] Analysis of Output loop KVL to the output loop: V CC = I C R C + V CE + I E R E I E I C Thus, V CE = V CC I C (R C + R E ) Note that this is similar to emitter bias circuit. 11

12 Problem For the circuit given below, find I C and V CE. Given the values of R 1, R 2, R C, R E and = 140 and V CC = 18V. For the purpose of DC analysis, all the capacitors in the amplifier circuit are opened. Solution Considering exact analysis: 1. Let us find R th = R1 R2 = R1 R2 / (R1 + R2) = 3.55K 2. Then find E th = V R2 = R2V CC / (R1 + R2) 3. Then find IB = 1.64V I B = ( Eth V BE ) / [ Rth +( + 1) R E ] = 4.37 A 4. Then find I C = I B = 0.612mA 5. Then find V CE = V CC I C (RC + RE) = 12.63V 12

13 Approximate analysis: The input section of the voltage divider configuration can be represented by the network shown in the next slide. Input Network The emitter resistance R E is seen as ( +1)R E at the input loop. If this resistance is much higher compared to R 2, then the current I B is much smaller than I 2 through R 2. This means, Ri >> R2 OR ( +1)R E 10R2 OR R E 10R2 This makes I B to be negligible. Thus I 1 through R 1 is almost same as the current I 2 through R 2. Thus R 1 and R 2 can be considered as in series. Voltage divider can be applied to find the voltage across R 2 ( V B ) V B = V CC R 2 / ( R 1 + R 2 ) Once V B is determined, V E is calculated as, V E = V B V BE After finding V E, I E is calculated as, I E = V E / R E I E I C V CE = V CC I C ( R C + R E ) 13

14 Problem Given: V CC = 18V, R 1 = 39k, R 2 = 3.9k, R C = 4k, R E = 1.5k and = 140. Analyse the circuit using approximate technique. In order to check whether approximate technique can be used, we need to verify the condition, R E 10R 2 Here, Thus the condition Solution R E = 210 k and 10R 2 = 39 k R E 10R 2 satisfied Thus approximate technique can be applied. 1. Find V B = V CC R 2 / ( R 1 + R 2 ) = 1.64V 2. Find V E = V B 0.7 = 0.94V 3. Find I E = V E / R E = 0.63mA = I C 4. Find V CE = V CC I C (R C + R E ) = 12.55V Comparison Exact Analysis Approximate Analysis I C = 0.612mA I C = 0.63mA V CE = 12.63V V CE = 12.55V Both the methods result in the same values for I C and V CE since the condition R E 10R 2 is satisfied. It can be shown that the results due to exact analysis and approximate analysis have more deviation if the above mentioned condition is not satisfied. For load line analysis of voltage divider network, Ic,max = V CC / ( R C +R E ) when V CE = 0V and V CE max = V CC when I C = 0. 14

15 DC bias with voltage feedback Input loop Applying KVL for Input Loop: V CC = I C1 R C + I B R B + V BE + I E R E Substituting for I E as ( +1)I B and solving for I B, I B = ( V CC V BE ) / [ R B + ( R C + R E )] Output loop 15

16 Neglecting the base current, KVL to the output loop results in, V CE = V CC I C ( R C + R E ) DC bias with voltage feedback Input loop Applying KVL to input loop: V CC = I C R C + I B R B + V BE + I E R E I C I C and I C I E Substituting for I E as ( +1)I B [ or as I B ] and solving for I B, I B = ( V CC V BE ) / [ R B + ( R C + R E )] Output loop 16

17 Neglecting the base current, and applying KVL to the output loop results in, V CE = V CC I C ( R C + R E ) In this circuit, improved stability is obtained by introducing a feedback path from collector to base. Sensitivity of Q point to changes in beta or temperature variations is normally less than that encountered for the fixed bias or emitter biased configurations. Problem: Given: V CC = 10V, R C = 4.7k, R B = 250 and R E = 1.2k. = 90. Analyze the circuit. I B = ( V CC V BE ) / [ R B + ( R C + R E )] = A I C = ( I B ) = 1.07mA V CE = V CC I C ( R C + R E ) = 3.69V In the above circuit, Analyze the circuit if = 135 ( 50% increase). With the same procedure as followed in the previous problem, we get I B = 8.89 A I C = 1.2mA V CE = 2.92V 50% increase in resulted in 12.1% increase in I C and 20.9% decrease in V CEQ Problem 2: 17

18 Determine the DC level of I B and V C for the network shown: Solution: Open all the capacitors for DC analysis. Load line analysis R B = 91 k k = 201k I B = ( V CC V BE ) / [ R B + ( R C + R E )] = (18 0.7) / [ 201k + 75( )] = 35.5 A I C = I B = 2.66mA V CE = V CC (I C R C ) = 18 ( 2.66mA)(3.3k) = 9.22V The two extreme points of the load line I C,max and V CE, max are found in the same as a voltage divider circuit. I C,max = V CC / (R C + R E ) Saturation current V CE, max Cut off voltage Miscellaneous bias configurations 18

19 There are a number of BJT bias configurations that do not match the basic types of biasing that are discussed till now. Miscellaneous bias (1) Analyze the circuit in the next slide. Given = 120 Solution This circuit is same as DC bias with voltage feedback but with no emitter resistor. Thus the expression for I B is same except for R E term. I B = (V CC V BE ) / ( R B + R C ) = ( ) / [680k + (120)(4.7k)] = A I C = I B = 1.86mA V CE = V CC I C R C = 11.26V = V CE V B = V BE = 0.7V V BC = V B V C = 0.7V 11.26V = V Miscellaneous bias (2) 19

20 Equivalent circuit Input loop Output loop 20

21 Solution The above circuit is fixed bias circuit. Applying KVL to input loop: Miscellaneous bias (3) V EE = V BE + I B R B I B = ( V EE V BE ) / R B = 83 A I C = I B = 3.735mA V C = -I C R C = V V B = - I B R B = - 8.3V Determine V CE,Q and I E for the network. Given = 90 ( Note that the circuit given is common collector mode which can be identified by No resistance connected to the collector output taken at the emitter) 21

22 Input loop Writing KVL to input loop: V EE = I B R B + V BE + ( +1)I B R E I B = (V EE V BE ) / [R B + ( +1) R E ] = ( ) / [ 240K + (91)(2K)] = A I C = I B = 4.12mA 22

23 Output loop Applying KVL to the output loop: V EE = V CE + I E R E I E = ( +1) I B = 4.16mA, V EE = 20V V CE = V EE I E R E = 11.68V Miscellaneous bias (4) Find V CB and I B for the Common base configuration given: Given: = 60 Input loop 23

24 Applying KVL to input loop Output loop I E = ( V EE V BE ) / R E = 2.75mA I E = I C = 2.75mA I B = I C / = 45.8 A Applying KVL to output loop: Miscellaneous bias (5) V CC = I C R C + V CB V CB = V CC I C R C = 3.4V Determine VC and VB for the network given below. Given = 120 Note that this is voltage divider circuit with split supply. ( +V CC at the collector and V EE at the emitter) 24

25 Thevinin equivalent at the input R th = (8.2k)(2.2k) / [ 8.2k+2.2k] = 1.73k I = (V CC + V EE ) / [R 1 + R 2 ] = ( ) / ( 8.2K + 2.2K) = 3.85mA Equivalent circuit E th = IR 2 V EE = V 25

26 Applying KVL: V EE E th V BE ( +1)I B R E I B R th = 0 I B = ( V EE E th V BE ) / [( +1) R E + R th ] = A I C = I B = 4.25mA V C = V CC I C R C = 8.53V V B = - E th I B R th = V Design Operations: Designing a circuit requires Understanding of the characteristics of the device The basic equations for the network Understanding of Ohms law, KCL, KVL If the transistor and supplies are specified, the design process will simply determine the required resistors for a particular design. Once the theoretical values of the resistors are determined, the nearest standard commercial values are normally chosen. Operating point needs to be recalculated with the standard values of resistors chosen and generally the deviation expected would be less than or equal to 5%. Problem: Given I CQ = 2mA and V CEQ = 10V. Determine R 1 and R C for the network shown: Solution To find R 1 : 26

27 1. Find V B. And to find V B, find V E because, V B = V E + V BE 2. Thus, V E = I E R E and I E I C = 2mA = (2mA)(1.2k) = 2.4V 3. V B = = 3.1V 4. Also, V B = V CC R 2 /(R 1 + R 2 ) 3.1 = (18)(18k) / R1+18k Thus, R 1 = 86.52k To find R C : Voltage across R C = V CC ( V CE + I E R E ) = 18 [ 10 + (2mA)1.2k] = 5.6V R C = 5.6/2mA = 2.8K Nearest standard values are, R 1 = 82k k = 86.7 k where as calculated value is k. R C = 2.7k in series with 1k = 2.8k both would result in a very close value to the design level. Problem 2 The emitter bias circuit has the following specifications: I CQ = 1/2I sat, I sat = 8mA, V C = 18V, V CC = 18V and = 110. Determine R C, R E and R B. Solution: I CQ = 4mA V RC = (V CC V C ) = 10V R C = V RC / I CQ, = 10/4mA = 2.5k To find R E : I Csat = V CC / (R C + R E ) To find R B : Find I B where, I B = I C / = A Also, for an emitter bias circuit, I B = (V CC V BE ) / R B +( +1) R E Thus, R B = k Standard values: R C = 2.4 k, R E = 1 k, R B = 620 k 8mA = 28 / ( 2.5k + R E ) 27

28 Thus, R E = 1k Transistor switching networks: Through proper design transistors can be used as switches for computer and control applications. When the input voltage V B is high ( logic 1), the transistor is in saturation ( ON). And the output at its collector = V CE is almost 0V( Logic 0) Transistor as a switch When the base voltage V B is low( logic 0), i.e, 0V, the transistor is cutoff( Off) and I C is 0, drop across R C is 0 and therefore voltage at the collector is V CC.( logic 1) Thus transistor switch operates as an inverter. This circuit does not require any DC bias at the base of the transistor. Design When Vi ( V B ) is 5V, transistor is in saturation and I Csat Just before saturation, I B,max = I C,sat / DC Thus the base current must be greater than I B,max to make the transistor to work in saturation. Analysis When Vi = 5V, the resulting level of I B is I B = (Vi 0.7) / R B 28

29 = ( 5 0.7) / 68k = 63 A I Csat = V CC / R C = 5/0.82k = 6.1mA Verification ( I C,sat / ) = 48.8 A Thus I B > ( I C,sat / ) which is required for a transistor to be in saturation. A transistor can be replaced by a low resistance Rsat when in saturation ( switch on) Rsat = V CE sat / I Csat (V CE sat is very small and I Csat is I C,max is maximum current) A transistor can be replaced by a high resistance Rcutoff when in cutoff ( switch on) Problem Determine R B and R C for the inverter of figure: I C sat = V CC / R C 10mA = 10V/ R C R C = 1k I B just at saturation = I C sat / = 10mA / 250 = 40 A Choose I B > I C sat /, 60 A I B = (Vi 0.7) / R B 60 A = ( ) / R B 29

30 R B = 155k Choose R B = 150k, standard value, re calculate I B, we get I B = 62 A which is also > Thus, R C = 1k and R B = 155k I C sat / Switching Transistors Transistor ON time = delay time + Rise time Delay time is the time between the changing state of the input and the beginning of a response at the output. Rise time is the time from 10% to 90% of the final value. Transistor OFF time = Storage time + Fall time For an ON transistor, V BE should be around 0.7V For the transistor to be in active region, V CE is usually about 25% to 75% of V CC. If V CE = almost V CC, probable faults: the device is damaged connection in the collector emitter or base emitter circuit loop is open. One of the most common mistake in the lab is usage of wrong resistor value. Check various voltages with respect to ground. Calculate the current values using voltage readings rather than measuring current by breaking the circuit. Problem 1 Check the fault in the circuit given. Problem

31 PNP transistors The analysis of PNP transistors follows the same pattern established for NPN transistors. The only difference between the resulting equations for a network in which an npn transistor has been replaced by a pnp transistor is the sign associated with particular quantities. PNP transistor in an emitter bias Applying KVL to Input loop: 31

32 V CC = I B R B +V BE +I E R E Thus, I B = (V CC V BE ) / [R B + ( +1) R E ] Applying KVL Output loop: V CE = - ( V CC I C R C ) Bias stabilization The stability of a system is a measure of the sensitivity of a network to variations in its parameters. In any amplifier employing a transistor the collector current I C is sensitive to each of the following parameters. increases with increase in temperature. Magnitude of V BE decreases about 2.5mV per degree Celsius increase in temperature. I CO doubles in value for every 10 degree Celsius increase in temperature. T (degree Celsius) Ico (na) V BE (V) x x Stability factors S (I CO ) = I C / I C0 S (V BE ) = I C / V BE S ( ) = I C / Networks that are quite stable and relatively insensitive to temperature variations have low stability factors. The higher the stability factor, the more sensitive is the network to variations in that parameter. 32

33 S( I CO ) Analyze S( I CO ) for emitter bias configuration fixed bias configuration Voltage divider configuration For the emitter bias configuration, S( I CO ) = ( + 1) [ 1 + R B / R E ] / [( + 1) + R B / R E ] If R B / R E >> ( + 1), then S( I CO ) = ( + 1) For R B / R E <<1, S( I CO ) 1 Thus, emitter bias configuration is quite stable when the ratio R B / R E is as small as possible. Emitter bias configuration is least stable when R B / R E approaches ( + 1). Fixed bias configuration S( I CO ) = ( + 1) [ 1 + R B / R E ] / [( + 1) + R B / R E ] = ( + 1) [R E + R B ] / [( + 1) R E + R B ] By plugging R E = 0, we get S( I CO ) = + 1 This indicates poor stability. Voltage divider configuration S( I CO ) = ( + 1) [ 1 + R B / R E ] / [( + 1) + R B / R E ] Here, replace R B with R th S( I CO ) = ( + 1) [ 1 + R th / R E ] / [( + 1) + R th / R E ] Thus, voltage divider bias configuration is quite stable when the ratio R th / R E is as small as possible. 33

34 Physical impact In a fixed bias circuit, I C increases due to increase in I C0. [I C = I B + ( +1) I C0 ] I B is fixed by V CC and R B. Thus level of I C would continue to rise with temperature a very unstable situation. In emitter bias circuit, as I C increases, I E increases, V E increases. Increase in V E reduces I B. I B = [V CC V BE V E ] / R B. A drop in I B reduces I C.Thus, this configuration is such that there is a reaction to an increase in I C that will tend to oppose the change in bias conditions. In the DC bias with voltage feedback, as I C increases, voltage across R C increases, thus reducing I B and causing I C to reduce. The most stable configuration is the voltage divider network. If the condition R E >>10R 2, the voltage V B will remain fairly constant for changing levels of I C. V BE = V B V E, as I C increases, V E increases, since V B is constant, V BE drops making I B to fall, which will try to offset the increases level of I C. S(V BE ) S(V BE ) = I C / V BE For an emitter bias circuit, S(V BE ) = - / [ R B + ( + 1)R E ] If R E =0 in the above equation, we get S(V BE ) for a fixed bias circuit as, S(V BE ) = - / R B. For an emitter bias, S(V BE ) = - / [ R B + ( + 1)R E ] can be rewritten as, S(V BE ) = - ( /R E )/ [R B /R E + ( + 1)] If ( + 1)>> R B /R E, then S(V BE ) = - ( /R E )/ ( + 1) = - 1/ R E The larger the R E, lower the S(V BE ) and more stable is the system. Total effect of all the three parameters on I C can be written as, I C = S(I CO ) I CO + S(V BE ) V BE + S( ) General conclusion: The ratio R B / R E or R th / R E should be as small as possible considering all aspects of design. 34

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