Experiment Determining the beta where it is stable.(6) Analysis and design of dc-biased transistor configurations (9)

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1 Visit for more resources DC BIASING BJTs (Analysis & Design) Design Procedure..(3) Limits of operation....(3) BJT modes of operation...(4) The Beta(h FE ).....(5) Experiment Determining the beta where it is stable.(6) Analysis and design of dc-biased transistor configurations (9) Fixed-Bias configuration...(9) Effect of elements.. (10) Design Example 1 with electronics workbench.....(12) Design Example 2 with ewb. (15) Transistor Inverter, design example(3) with ewb....(18) Fixed-Bias circuit with R E....(20) Effect of R E..(20) Design Example 4 with electronics workbench.....(21) Observation of stability between fixed-biased configuration and fixed-biased with R E configuration..(28) Voltage Divider Bias..(32) Design Example 5 with electronics workbench.....(33) Summary of voltage divider bias design..(36) 1

2 Even if you are designing a transistor circuit as a switch or as an amplifier, transistor has to be biased in desired region. a.c purpose circuits are also designed according to DC conditions. So DC biasing is very important in both ac purpose(amplifier) circuits and DC purpose(switch) circuits. There are some biasing configurations and in this tutorial these configuration will introduced to you. Question: Why we dont use a typical bias configuration instead of many configurations? Answer: There are many biasing configurations, each one has advantageous and disadvantageous.infact the main problem is stabilization.some configurations are more stable when environment conditions are changed. 2

3 DESIGN PROCEDURE Biasing a BJT means establishing the desired values of V CE and I C so that the amplifier will have the proper gain, input impedance, undistorted output voltage swing, etc. These values of VCE and IC are known as the quiescent operating point or Q-point. The values of VCE and IC required are determined from inspection of the BJT's data sheet and load line analysis. What s the first step? Firstly limits of operations have to be known otherwise design procedure may be more confusing.generally limits of operations are obtained from the manufacturer s datasheet.these datasheets can be found in manufacturers site.some general purpose transistor datasheets can be obtained from Limits of Operation The figure below illustrates a typical output characteristic of a transistor.the limits are often taken from the datasheet of the transistor Again, the maximum power dissipation is obtained from the datasheet and it is 300mW in this example.note that, P Cmax = V CE. I C.We can not use our transistor out of the P Cmax limits.operating point can be choosen at the points A,B & C.These points are at the active region of transistor.when amplification is intended, transistor have to work in active region and when switching operations are intended, the cut-off and saturation region is chosen for operating point of transistor. Figure 1.1 After determining the limits ( P C max, V CE max, I C max ), at least we will have an idea and know about the maximum values which must not be exceeded. Comments on point A,B and C: A : B : C : At lower I B levels spacing between I B curves is rapidly changing:beta,h FE, is not stable.moreover negative swing is limited Best point for small signal amplification(best symmetric swing occurs) Positive swing is limited 3

4 BJT MODES OF OPERATION Mode BE JUNCTION CB JUNCTION Cut-Off Reverse biased Reverse biased Active Mode Forward biased Reverse biased Saturation Mode Forward biased Forward biased *Forward biased! V P > V N *Reverse biased! V N > V P *For NPN transistors: Collector(n), Base(p), Emitter(n) * For PNP transistors: Collector(p), Base(n), Emitter(p) Active Region Condition: For NPN transistors: 1-Base-Collector junction must be reversed biased ( n>p ), ( V C > V B ) 2-Base-Emitter junction must be forward biased ( p > n), ( V B > V E ) V C > V B > V E V BE = 0.7V For PNP transistors: 1-Base-Collector junction must be reversed biased ( n>p ), ( V B > V C ) 2-Emitter-Base junction must be forward biased ( p > n), ( V E > V B ) V E > V B > V C V EB = 0.7V Cut-off region condition: For NPN transistors: V E > V B V C > V B For PNP transistors: V B > V E V B > V C Saturation region condition: For NPN transistors: V B > V E V B > V C V CE = 0.2V, V BE =0.8V For PNP transistors: V E > V B V C > V B V EC = 0.2V, V EB =0.8V For NPN Transistors : V BC Inverse-Active Saturation Cuttoff Forward-Active V BE 4

5 The Beta (β or h FE ) Beta (h FE ) is the D.C current gain. It s the ratio of the collector current to base current! h FE = I C / I B ( I C = h FE. I B ) h FE is not a constant value it changes in different operating points.as I C increases or decreases ( As well as I B ) h FE will change, but in somewhere it will be constant or more stable.unless some special purpose circuits intented to design, we will bias our transistor in this beta stabilized region. Figure 1.2 P2N2222 MOTOROLA DC CURRENT GAIN Figure 1.3 BC546 DC CURRENT GAIN As seen in the figure1.3, the beta is fairly constant when I C is between 10-2 ma and 10mA.Now we have an idea about collector current in which interval it must be.determining the collector current (I C ) is one of the most important step. 5

6 EXPERIMENT DETERMINING THE REGION WHERE THE BETA ( h FE ) IS STABLE Our first experiment will be finding the region where the beta is more stable.in other words we will search where the beta is constant. It s better for you to practice these experiments in your home or lab.you can learn many things when doing an experiment.moreover you may query yourself and try to make different configurations and see the results.this helps you to understand more clearly how things work. Even if you do not practice these exercises, read the theoritical informations and compare the measured results with the calculated ones. In this experiment the value of h FE ( β dc ) is examined and found where it is most stable.two transistor is used to show you different results.(bc546 & 2N2222) I B Figure 2.3 I C V CC = 9V R C = 2.2KΩ ( Measured as 2.17KΩ its real value ) R B is changed from 470Ω to 20.5MΩ for obtaining different base currents I B and the response of the I C to the base current I B Q1:BC546 After using BC546 use 2N222 instead of BC546 Q1:2N2222 Q - A - What s the saturation current? From equation (4) I C sat = V CC / R C I C = 9V / 2.17KΩ! I C sat = 4.14mA 2.2KΩ is measured as 2.17KΩ in this experiment. Q - A - What s the I B when R B is 470Ω From equaiton ( 2) I B = ( V CC 0.7V ) / R B I B = ( 9V 0.7V ) / 470Ω! I B = 17.65mA Check the results with measured values in the next page Q - What s the I C when I B is 5.55uA ua!microamper A - In order to answer this question we have to know the Beta (β) and if the the transistor works in active region. Lets give the answer of this question after the experiment results. 6

7 Experiment Results (R C = 2.2KΩ and I Csat 4.14mA from equation (4) ) BC546 Q2N222 R B I B I C β ( h FE ) # R B I B I C β ( h FE ) 470Ω 17.66mA 4.13mA Ω 17.61mA 4.13mA K 8.3mA 4.16mA K 8.3mA 4.16mA K 1.76mA 4.16mA K 1.76mA 4.16mA K 0.836mA 4.16mA K 0.836mA 4.16mA K 0.258mA 4.16mA K 0.258mA 4.16mA K 0.122mA 4.15mA K 0.122mA 4.13mA K 83.7uA 4.15mA K 83.7uA 4.12mA K 41.95uA 4.14mA K 41.95uA 4.11mA K 28.03uA 4.14mA K 28.03uA 4.09mA K 21.02uA 4.13mA K 21.02uA 3.70mA K 17.9uA 4.13mA K 18uA 3.14mA K 14.5uA 4.11mA K 14.5uA 2.53mA K 12.35uA 4.09mA K 12.35uA 2.24mA K 10.81uA 4.08mA K 10.81uA 1.90mA K 9.64uA 4.04mA K 9.64uA 1.77mA 184 1M 8.11uA 3.51mA M 8.11uA 1.43mA M 6.78uA 2.97mA M 6.78uA 1.25mA M 5.55uA 2.46mA M 5.55uA 986uA 177 2M 4.21uA 1.83mA M 4.21uA 746uA M 3.35uA 1.5mA M 3.35uA 589uA M 2.72uA 1.22mA M 2.72uA 488uA M 2.37uA 1.07mA M 2.37uA 414uA M 2.03uA 912uA M 2.03uA 368uA M 0.923uA 391uA M 0.923uA 147uA M 0.413uA 193uA M 0.413uA 78.3uA 189 Table 1 When V C > V B > V E transistor is active region.in the measurements BC546 entered active mode at row # 15, before # 15 it was in saturation region.the 2N2222 has entered active mode at row # 10. BC546 The β is fairly stable betwwen row # 15 and row # 24, and so it s β is determined as approximately 440 2N2222 The β is fairly stable betwwen row # 10 and row # 24, and so it s β is determined as approximately 180 Note that the saturation current is measured as approximately 4.16mA in both BC546 and 2N2222.It does not differ from transistor to transistor, saturation current is obtained from the equation (4) I C sat = V CC / R C I C sat = 9V / 2.17KΩ! I C = 4.14mA The calculated value is nearly same as measured value that is 4.16mA As you see the answer of questions are very near to results that measured in the experiment. 7

8 Now turning back to our last question ; Q - What s the I C when I B is 5.55uA (row #18) ua!microamper A - From equation (3) I C = β. I B BC546 I C = β. I B, I C = uA! I C =2.44mA ( Nearly same as measured value, look row # 18 ) 2N222 I C = β. I B, I C = uA! I C =999uA ( Nearly same as measured value, look row # 18 ) Q - What s the I C when I B is 1.76mA (row #3) A - From equation (3) I C = β. I B BC546 I C = β. I B, I C = mA! I C =774mA ( ERROR! ) 2N222 I C = β. I B, I C = mA! I C =316.8mA ( ERROR! ) The maximum current calculated and measured as 4.16mA, any value higher than I Csat =4.16mA can not exist! Equation (3) I C = β. I B is only applicable when transistor in active mode and where β is at its stable region Use this way, if equation (3) I C = β. I B satisfies or not : If I C (I C = β. I B ) is calculated higher than the saturation current I C sat ( I C sat = V CC / R C ) then the equation I C = β. I B is not applicable.infact it is applicable but the β is lower than its stable value. Q Isn t it a problem, beta is not stable always? A Yes, it is a big problem.but there are some beta-independent configurations which will introduce to you later. So we give the name of the problem, Beta Problem!This problem will be removed in more improved configurations. 8

9 ANALYSIS & DESIGN OF DC-BIASED TRANSISTOR CONFIGURATIONS In this section some transistor configurations( DC-Biased), thier designs & analysis, their advantages and disadvantages will be introduced to you. 1-FIXED-BIAS CIRCUIT Fixed-Bias is the simplest DC-biased transistor congfiguration. Figure 2.1 Loop - 1 Input Loop I B I C Loop - 2 Output Loop Figure 2.2 Fixed-Bias Configuration Kirchoff s Voltage Loops at input & output To get the neccessary equations we will use Kirchoff s voltage loops ( KVL ) at input & output From Loop-1 -V CC + I B.R B + V BE = 0 (1) V BE is 0.7 V I B = ( V CC 0.7V ) / R B (2) We have one more formula that is: I C = β. I B (3) (Active Region Equation) The equaiton (3) does not always satisfies I C = β. I B. It differs in saturation region. At Saturation: At saturation Vce 0.2V therefore KVL yields at output: I C = (V CC 0.2V) / R C (4) 9

10 Use this formula when transistor works in saturation region and keep the value of I B high enough to prevent your transistor work in active region, or in other words high value of I B keeps transistor in saturation region I Bmax I Csat / β (The Level of I B in the active region just before the saturation) I B > I Csat / βmin (For the saturation level we must ensure this equation) Three important equations for fixed-bias configuration: I B = ( V CC 0.7V ) / R B (2) I C = β. I B (at active region) (3) At Saturation V CE =0.2V ; I C sat = ( V CC 0.2V )/ R C (4) EFFECT OF ELEMENTS EFFECT OF R B and I B : From equation (2), I B is determined by R B : I B = ( V CC 0.7V ) / R B (2) As R B decreases I B increases, As I B increases Q point shift up-left as seen in the figure2.3. R B R B I B Q-point shifts up-left I B Q-point shift down-right Vcc Rc Q-point Q-point Q-point Q-point I B1 Best Q point for small signal amp.because it is at the middle of the graph.therefore best symmetric swing occurs V CC at this point.see ac analysis. Figure 2.3 Adjusting R B for active mode and saturation mode operation, 1 st Way Q-point I B5 I B4 I B3 I B2 Higher the value of I B, more possible the transistor to be in saturation mode. Higher value of I B is obtained if lower value of R B is used Vcc Ic.R C = V CE (6).(at next page) I C = β.i B I Bmax = 12V 0.8V(V CE ) / β.r C (Maximum I B current when transistor works in acitve mode.) Why 0.8V? V BE = 0.7V, V E = 0V! V B = 0.7V V C > V B >V E.0.8V > 0.7V > 0V, V CE =V C = 0.8V border between active region and saturation region) I B = ( V CC 0.7V ) / R B (2) From Equation 2: R Bmin = 12V 0.7V(V BE ) / I Bmax (Minimum R B, transistor to be work in active mode.) If R B > R Bmin or I B < I Bmax then transistor will work in active mode. If R B < R Bmin or I B > I Bmax then transistor will work in saturation mode 10

11 Adjusting R B for active mode and saturation mode operation, 2 nd Way Active mode criteria : V C > V B > V E V C = V CC - I C R C V B = V CC I B R B V E = I E R E V C > V B! V CC I C R C > V CC I B R B! βi B R C < I B R B! R B > β.r C ( For R B > β.r C transistor works in active mode) R Bmin = β.r C (Note that For R B < β.r C transistor work in saturation mode) V B > V E! V CC I B R B > I E R E! V CC > I E R E + I B R B! V CC >I B [β+1)r E + R B ] (Transistor works in active mode) (Note that For V CC < I B [β+1)r E + R B ] transistor work in saturation mode) EFFECT OF Rc: From loop-2 at page 5; I C = (V CC V CE ) / R C (6) This equation gives the DC load-line. Note that in active mode of operation I C is not determined by R C, I C is determined by equation (3) I C = β. I B. R C determines V CE =V C.See the example 1. For bigger Rc values, slope of the load line increases and Q-point shifts to left, this may limit the transistor operation in a smaller region. Changing Rc and keeping other variables constant shifts the Q point to the right or left on the same I B line.ic remains constant if R C < R Cmax Higher values of Rc may limit the transistor operation in a smaller region. Vcc Rc1 Vcc Rc2 Vcc Rc3 Q3 R C1 > R C2 > Rc 3 Q2 Q1 I B5 I B4 I B3 I B2 I B1 Adjusting R C for active mode and saturation mode operation, 1 st Way Max R C in active mode: In active region V C > V B > V E In fixed-biased configuration V E = 0V V BE = 0.7V therefore V B = 0.7V and V CE = V C I C = (V CC V CE ) / R C (6) R C = (V CC V CE ) / I C (7) In active mode for max R C, V CE must be minimum and greater than 0.7V : V CE = V C > 0.7 V ( V C > V B ) ( V C > 0.7V ) Assume V CE = 0.8V R C = (V CC 0.8V) / β.i B (8) Maximum value of R C in active region 11

12 Adjusting R C for active mode and saturation mode operation, 2 nd Way The equation R B > β.r C can be used for adjusting the value of R C R C < R B / β ( For R c < (R B / β) transistor works in active mode) (Note that For R c >( R B / β ) transistor work in saturation mode) R C max = R B / β Minimum R C : V CC / RC = I Cmax (9) R C = V CC / I Cmax (10) Minimum vaue of R C I Cmax can be found in the datasheet of the transistor.i Cmax is illustrated in figure1.1 EFFECT OF Vcc Lower values of Vcc limit the transistor operation in a smaller region. Note that, as Vcc decreases I B also decreases.therefore Q 2 and Q 3 points are in a lower I B level(not I B2 level, below I B2 level) This phenomenon is not illustared in the graph. EXAMPLE 1: Vcc1 Rc Vcc2 Rcc Vcc3 Rcc Q 3 V CC1 > V CC2 > V CC3 Q 2 Q 1 Vcc3 Vcc2 Vcc1 I B5 I B4 I B3 I B2 I B1 Vcc=12V, (β DC )h FE = 100 a) Determine I C, I B and V CE if R B =240K and R C =2.2K Determine the mode of operation b) Determine the maximum and minumum value of R C in active mode if R B =240K and I Cmax =100mA c) Repeat a) for R B =240K and R C = 5K comment on the result of a) d) Repeat a) for R B =240K and Rc =1K comment on the result of a) and c) Solution: a) Assuming transistor in active mode: V BE = 0.7V, V E = 0V! V B =0.7V I B = (12V 0.7V) / 240K! I B = 47uA I C = β.i B I C = 100x47uA! I C = 4.7mA I C = (V CC V CE ) / R C (6) From equation (6) : V CE = V CC I C.R C 12

13 V CE = 12V (4.7mA).(2.2 K)! V CE = 1.66V V E = 0V V CE = 1.66V! V C = 1.66V V C = 1.66V V B = 0.7V V E = 0V V C > V B > V E Transistor is in active mode, our assumption is true. Electronics Workbench Results for Example 1-a) b) From equation (8) R C = (V CC 0.8V) / β.i B ( Maximum value of R C in active region ) R C = (12V 0.8V) / (100x47uA)! R C max = 2.38K (Max value of R C in act. mode) or R C < R B / β R C max = R B / β R C max = 240K / 100 R C max = 2.4K IMPORTANT From equation (10) R C = V CC / I Cmax ( Minimum value of R C ) R C = (12V ) / (100mA)! R C = 120Ω (Min value of R C ) When R C smaller than 2.38K, transistor is in active mode and therefore I C is constant and it is determined by equation (3) I C = β. I B. 120Ω < R C < 2.38K When R C greater than 2.38K, transistor will not be in active mode therefore I C is determined by equation (6) I C = (V CC V CE ) / R C R C > 2.38K c) For R C = 5KΩ :We know transistor is not in active mode when R C > 2.38KΩ from b). But for practicing we will again assume transistor in active mode(you will see now, it is not in active mode) Assuming transistor in active mode : V BE = 0.7 V V B = 0.7V I B = 47uA I C =4.7mA V C =V CE = V CC - I C. R C V CE = 12V (4.7mA).(5KΩ)!V CE =V C = V V C < V B therefore our assumption is wrong!( transistor is not in active mode) Cut-off criteria: V E > V B V C > V B We know that V E = 0 and V B has a positive value V E < V B.Therefore transistor is not in cut-off mode.we have only one choice that is : Transistor is in saturation mode. Saturation Criteria: V B > V E V B > V C V CE = 0.2V, V BE =0.8V I B = (12V 0.8V) / 240K I C = (V CC V CE ) / R C (6) I C = (12V 0.2V ) / 5KΩ! I B = 46.6uA! I C = 2.36mA 13

14 Electronics Workbench Results for Example 1-c) I C is determined by R C only in saturation mode d) For R C = 1KΩ :We know transistor is in active mode when 120 <R C < 2.38KΩ from b). But for practicing we will again assume transistor in active mode(you will see now, it is in active mode) Assuming transistor in active mode : V BE = 0.7 V V B = 0.7V I B = 47uA I C =4.7mA V C =V CE = V CC - I C. R C V CE = 12V (4.7mA).(1KΩ)!V CE =V C = 7.3 V V C > V B > V E therefore our assumption is true( transistor is in active mode) As seen in part a) I C is same in both art a) and d) where R C < 2.38KΩ Remark:I C is not determined by R C in active region.it is determined by I C = β.i B R C determines V CE, I C is determined by R C only in saturation mode.if R C high enough transistor works in saturation mode. Electronics Workbench Results for Example 1-d) 14

15 EXAMPLE 2: Vcc=12V, (β DC )h FE = 300 Consider the same circuit in example 1, this time use β=300 in your solutions a) Determine I C, I B and V CE if R B =240K and R C =2.2K.Determine the mode of operation and max R C and min R C.( take I Cmax = 100mA) b) Adjust R B so that transistor with Beta=300 and R C =2.2K will work in active mode.determine I Bmax and R Bmin when R C is constant (2.2KΩ) Solution: a) Assuming transistor in active mode: V BE = 0.7V, V E = 0V I B = (12V 0.7V) / 240K! V B =0.7V! I B = 47uA I C = β.i B I C = uA! I C = mA V CE = V C = V CC I C.R C V C = 12V (14.100mA).(2.2K)! V C = V CE = V V C < V B Therefore, transistor is not in active mode.our asumption is wrong.we can determine whether transistor is in active mode or saturation mode by finding max R C as follows: max R C = (V CC 0.8V) / β.i B ( Equation 8, max R C in active region.note that 0.8 V comes from V C > V B, V B = 0.7 V, see notes) R C = (12V 0.8V) / (300x47uA)! R Cmax = 794Ω (Max value of R C in active mode) or R C max = R B / β R C max = 240K / 300 R C max = 800Ω min R C : R Cmin = V CC / I Cmax (Equation 10, min R C ) R C = (12V ) / (100mA)! R C = 120Ω (Min value of R C ) R C = 2.2K R C > R Cmax 2.2K > 794Ω (Transistor in sat. Mode) if R C > R Cmax then transistor works in saturation mode. Assuming transistor in saturation mode: Saturation Criteria: V B > V E V B > V C V CE = 0.2V, V BE =0.8V I C = (V CC V CE ) / R C (Equation 4, Saturation current, I Csat. I C β.i B ) I C = (12V 0.2V) / 2.2K! Ic sat = I C = 5.36mA 15

16 Comment:Beta(β) affect the system if example 2 compared to example 1.When beta is 100 transistor works in active mode, when beta is 300 transistor works in saturation mode.to avoid saturation mode we have to use smaller R C (R C < 794Ω ), or smaller I B ( higher R B ).Note that smaller R C means higher V C and higher R B means lower V B.This settings adjust transistor s mode of operation.see active mode and saturation mode criteria) b) Vcc Ic.R C = V CE (6) From Equation 6: I C = β.i B I Bmax = 12V 0.8V(V CE ) / β.r C I B = ( V CC 0.7V ) / R B (2) From Equation 2: Electronics Workbench Results for Example 2-a) (Maximum I B current transistor to be work in acitve mode) R Bmin = 12V 0.7V(V BE ) / I Bmax I Bmax = 12V 0.8V / 300.(2.2K) R Bmin = 12V 0.7V / I Bmax R Bmin = 12V 0.7V / 16.97uA (Minimum R B, transistor to be work in active mode! I Bmax = 16.97uA! R Bmin = 665K or R Bmin = β.r C R Bmin = 300.(2.2K) R Bmin = 660K if we use R B > 665K transistor will work in active mode if we use R B < 665K transistor will work in saturation mode Use factor of safety and do not choose the values near the limits 16

17 For R B > R Bmin, transistor will work in active mode R B = 700K ( 700 K > 665K ), I B = (12V 0.7V) / 700K! I B = 16.14uA ( Note that I Bmax = 16.97uA ) I C = β.i B I C =(300).(16.14uA)! I C = 4.842mA ( Note that I Csat = 5.36mA ) Electronics Workbench Results for Example 2-b) for R B > R Bmin For R B < R Bmin, transistor will work in saturation mode R B = 600K ( 600 K < 665K ), I B = (12V 0.7V) / 600K! I B = 18.83uA ( Note that I Bmax = 16.97uA ) I C β.i B I C = (V CC V CE ) / R C (4) I C = (12V 0.2V) / 2.2K! Ic sat = I C = 5.36mA Electronics Workbench Results for Example 2-b) for R B < R Bmin 17

18 EXAMPLE 3: TRANSISTOR INVERTER V i 10V 0V 0V t V C 10V 10V 0V t a) Determine R B and R C for the transistor inverter of figure above if I Csat = 10mA, β=300 b) Determine maximum value of R C for transistor to be work in active mode if R B =155K. SOLUTION: a) At saturation I C sat = V CC V CE / R C (V CE =0.2V at saturation) I C sat = (Vcc 0.2V ) / R C 10mA = (10V 0.2V ) / R C At saturation : I B I C sat / β min Assume β = β min = 300 I B 10mA / 300! R C = 980Ω! I B 33.3uA Choosing I B = 60uA to ensure that saturation. I B = (V i 0.7V) / R B R B =(10V -0.7V) / 60uA R B = 155K Electronics Workbench Results for Example 3-a) 18

19 b) I B = (V i 0.7V ) / 155K!I B = 60uA R C max = (V CC 0.8V) / β.i B (8) Maximum value of R C in active region R C max = (10V 0.8V)/18mA!R C max = 511Ω For R C < R C max, transistor work in active mode For R C > R C max, transistor work in saturation mode another way for max R C : R C max = R B / β R C max = 155K / 300 R C max = 516Ω 19

20 FIXED-BIAS CIRCUIT with R E From input Loop: Vcc = I B.R B + V BE + I E.R E Vcc = I B.R B + V BE + (β+1)i B. R E At Active Region: Vcc VBE I B = RB + ( β +1) RE (at active region) (2.1) I C = β. I B (at active region) (2.2) From output Loop: Vcc = I C R C + V CE + I E R E At Saturation Region: (V CE =0.2V ) Additional Eqns: V C = V CC - I C R C V B = V CC I B R B V E = I E R E I C sat = ( V CC 0.2V )/ R C + R E (2.3) Vcc VBE Icsat. RE I B = (2.4) RB Note that R E is reflected back to the input by a factor ( β +1 ) in active region. Effect of elements are same with fixed bias circuit. Effect of R E R E improves stability. R E determines V E R E affect base current much more than R B. As R E increases I B will decrease As R E increases slope of dc load line will increase(same effect as R C in fixed-bias configuration)q point shifts down and left on the dc load line. R C and R E don t determines the I C in active mode.i C =β. I B At saturation mode I C sat = ( V CC 0.2V )/ R C + R E (2.3) Adjusting R B for active mode and saturation mode operation Active mode criteria : V C > V B > V E V C = V CC - I C R C V B = V CC I B R B V E = I E R E V C > V B! V CC I C R C > V CC I B R B! βi B R C < I B R B! R B > β.r C ( For R B > β.r C transistor works in active mode) R Bmin = β.r C (Note that For R B < β.r C transistor work in saturation mode) V B > V E! V CC I B R B > I E R E! V CC > I E R E + I B R B! V CC >I B [β+1)r E + R B ] (Transistor works in active mode) (Note that For V CC < I B [β+1)r E + R B ] transistor work in saturation mode) 20

21 Adjusting R C for active mode and saturation mode operation Active mode criteria : V C > V B > V E V BE = 0.7V I C = (V CC V CE ) / R C + R E (This equation gives the DC load-line.) V C > V B : V CE > V BE V BE =0.7V In active mode for max R C, V CE must be minimum and greater than 0.7V : V CE > 0.7 V Assume V CE = 0.8V Vcc 0.8V RC max = RE (Equation for determining max value of R βi C ) B Note that aim of adjusting R C is changing the voltage V C The equation R B > β.r C can also be used for adjusting R C R C < ( R B / β ) ( For R c < (R B / β) transistor works in active mode) R C max = R B / β (Another equation for determining max value of R C ) (Note that For R c >( R B / β ) transistor work in saturation mode) The equation R B > β.r C can be used in either adjusting R B (keep R C constant) or adjusting R C ( keep R B constant) for determining the mode of operation. Minimum R C : V CC / RC = I Cmax R C = V CC / I Cmax (Minimum vaue of R C ) I Cmax can be found in the datasheet of the transistor.i Cmax is illustrated in figure1.1 at page 3 EXAMPLE 4: β = 300, R B = 470K, R C =2.2K, R E =1.2K, V CC =12V a) Determine mode of operation, I C and I B b) Adjust R B so that transistor works in active mode.what s the minumum value of R B if transistor wanted to be work in active mode? c) Adjust R C so that transistor works in active mode.what s the maximum value of R C if transistor wanted to be work in active mode? a) Assuming transistor works in active mode : Active Region Criteria: V C > V B > V E V BE = 0.7V I C = β.i B I B = R Vcc V B BE + ( β +1) R E (2.1) I B = [ 12V 0.7V ] / [ 470K + (301).(1.2K) ]! I B = 13.59uA I C = β.i B I C = 300.(13.59uA)! I C = 4.077mA 21

22 V C = V CC I C.R C V C =12V (4.077mA).(2.2K)! V C =3.03V V B = V CC I B.R B V B =12V (13.59uA).(470K)! V B =5.61V V B > V C therefore transistor does not work in active region, our assumption is wrong Assuming transistor works in saturation mode : Saturation Mode Criteria: V B > V E V B > V C V CE = 0.2V, V BE =0.8V I C sat = ( V CC 0.2V ) / R C + R E (2.3) I Csat = ( 12V 0.2V ) / (2.2K + 1.2K ) I Csat = 3.47mA Vcc VBE Icsat. RE I B = (2.4) RB I B = [12V 0.7V (3.47mA)(1.2K)] / 470K!I B =15.18uA Electronics Workbench Results for Example 4-a) b) Adjusting R B for active mode operation, 1 st Way: I Csat = 3.47mA I Bmax =I Csat / β I Bmax = 3.5mA / 300!I Bmax = 11.56uA Assumption: I B have to be smaller than I Bmax for transistor to be work in active mode.check at the end if this assumption is true. I B < I Bmax I B < 11.56uA I B = R Vcc V B BE + ( β +1) R E (2.1) 22

23 Finding R B min : 11.56uA = [ 12V 0.7V ] / [ R B + (301).(1.2K) ]! R B min = 616.3K Any value of R B greater than R B min will cause transistor to work in active mode(you must check if this assumption is true) Taking the value of I B = 11.00uA 11uA = [ 12V 0.7V ] / [ R B + (301).(1.2K) ]! R B = 666K R B > R B min 666K > 616.3K Checking the assumption: I B = 11uA for R B = 666K I C = β.i B I C = 300.(11uA)! I C =3.3mA V C = V CC I C.R C V C =12V (3.3mA).(2.2K)! V C =4.74V V B = V CC I B.R B V B =12V (11uA).(666K)! V B =4.67V V C > V B therefore our assumption is true.we can use R B =666K for active mode operation of transistor. Note that V C = 4.74V and V B = 4.67V are very close to each other.any small difference in the system may cause the transistor to work in saturation mode.to avoid from this situation keep R B high enough as compared to R Bmin.For example if we use R B = 800K, the difference between V C and V B will be more distinct then small changes will not change the mode of operation. Electronics Workbench Results for Example 4-b) for R B = 666K Now we will investigate the truth of our assumption if we choose R B more close to R Bmin. Taking the value of R B = 620K which is more close to R B min = 616.3K I B = [ 12V 0.7V ] / [ 620K + (301).(1.2K) ]! I B = 11.51uA I C = β. I B I C = (300)(11.51uA)! I C = 3.45mA V C = V CC I C.R C V C =12V (3.45mA).(2.2K)! V C =4.41V V B = V CC I B.R B V B =12V (11.51uA).(620K)! V B =4.863V V C < V B therefore our assumption is wrong! This way is not reliable when choosing values close to limits( For example R Bmin = 616K R B =620K, the values are very close) Use this way if you choose the value of R B distinct away from R Bmin ( For example R B =800K) 23

24 Adjusting R B for active mode operation, 2 nd Way: Active mode criteria : V C > V B > V E V C = V CC - I C R C V B = V CC I B R B V E = I E R E V C > V B! V CC I C R C > V CC I B R B! βi B R C < I B R B! R B > β.r C V B > V E! V CC I B R B > I E R E! V CC > I E R E + I B R B! V CC >I B [β+1)r E + R B ] R B min = (300)(2.2K) R B min = 660K ( Actual result for minimum value of R B ) R B > R B min for active mode of operation Choosing R B = 670 K I B = [ 12V 0.7V ] / [ 670K + (301).(1.2K) ]! I B = 10.95uA I C = β. I B I C = (300)(10.95uA)! I C = 3.28mA V C = V CC I C.R C V C =12V (3.28mA).(2.2K)! V C =4.784V V B = V CC I B.R B V B =12V (10.95uA).(670K)! V B =4.66V V C > V B V E I C. R E V E (3.28mA)(1.2K)! V E = V V B > V E V C > V B > V E : Transistor works in active mode It is an expected result because when R B = 666K transistor works in active mode as it is done in the previous page.if transistor works in active mode when R B = 666K, it will work in active mode for higher R B Electronics Workbench Results for Example 4-b) for R B = 670K 24

25 c) Which I B will be used in the equation below?we can not use the I B which was was found I B =15.18uA in part a).because I B =15.18uA is the base current when transistor is in saturation mode.we need I B current when it is in active region.because, if the equaiton below is investigated β.i B is written for I C.We know that β.i B is true for only in active region. I B = R Vcc V B BE + ( β +1) R E (2.1) ( I B at active region) I B = [ 12V 0.7V ] / [ 470K + (301).(1.2K) ]! I B = 13.59uA R Vcc 0.8V = R C max E βi B (Any value of R C smaller than R Cmax will not affect I C and transistor will work in active region ) R Cmax = [(12V 0.8V) / 300.(13.59uA)] 1.2K R Cmax = 1.547K For R C smaller than 1.547K make transistor to work in active region or 2 nd way ( more easy) R C max = R B / β R C max = 470K / 300 R C max = 1.567K Choosing R C = 1K ( R C < R Cmax ) and investigating the system: Assuming transistor works in active mode : Active Region Criteria: V C > V B > V E V BE = 0.7V I C = β.i B Vcc VBE I B = RB + ( β +1) RE (2.1) I B = [ 12V 0.7V ] / [ 470K + (301).(1.2K) ]! I B = 13.59uA I C = β.i B I C = 300.(13.59uA)! I C = 4.077mA V C = V CC I C.R C V C =12V (4.077mA).(1K)! V C =7.92V V B = V CC I B.R B V B =12V (13.59uA).(470K)! V B =5.61V V E I C.R E V E =(4.077mA).(1.2K)! V E =4.89V V C > V B > V E therefore transistor works in active region 25

26 Electronics Workbench Results for Example 4-c) for R C < R C max R C=1K, R C max = 1.547K Choosing R C = 2K ( R C > R Cmax ) and investigating the system: We know transistor works in saturation mode because R C > R C max.but we will try to solve the problem assuming it is in active mode.it will be seen that our assumption will be wrong. Assuming transistor works in active mode : Active Region Criteria: V C > V B > V E V BE = 0.7V I C = β.i B Vcc VBE I B = RB + ( β +1) RE (2.1) I B = [ 12V 0.7V ] / [ 470K + (301).(2K) ]! I B = 10.54uA I C = β.i B I C = 300.(10.54uA)! I C = 3.162mA V C = V CC I C.R C V C =12V (3.162mA).(2K)! V C =5.676V V B = V CC I B.R B V B =12V (10.54uA).(470K)! V B =7.04V V C < V B therefore transistor does not work in active region.our assumption is wrong as it was foreseen Assuming transistor works in saturation mode: Saturation Mode Criteria: V B > V E V B > V C V CE = 0.2V, V BE =0.8V I C sat = ( V CC 0.2V ) / R C + R E (2.3) I Csat = ( 12V 0.2V ) / (2K + 1.2K ) I Csat = 3.68mA Vcc VBE Icsat. RE I B = (2.4) RB I B = [12V 0.7V (3.68mA)(1.2K)] / 470K!I B =14.64uA 26

27 V C = V CC I C.R C V C =12V (3.68mA).(2K)! V C =4.64V V B = V CC I B.R B V B =12V (14.64uA).(470K)! V B =5.12V Electronics Workbench Results for Example 4-c) for R C > R C max R C=2K, R C max = 1.547K 27

28 OBSERVATION OF STABILITY BETWEEN FIXED-BIASED CONFIGURATION AND FIXED-BIASED WITH R E CONFIGURATION Outside conditions such as change in temparature, beta and age of the device will affect the system.the two fixed-biased configuration with R E and without R E are compared when β=50 and β=100(100% change in beta) Fixed-Biased Configurations without R E : For β = 50 For β = 100 β I B (µa) I C (ma) V CE (V) β = 50! β = 100! 100% increase in I C 28

29 Fixed-Biased Configurations with R E = 2K, V E = 553.5mV : For β = 50 For β = 100 β I B (µa) I C (ma) V CE (V) β = 50! β = 100! 4% decrease in I B! 95% increase in I C (improved stability) 29

30 Fixed-Biased Configurations with R E = 2K, V E = : For β = 50 For β = 100 β I B (µa) I C (ma) V CE (V) β = 50! β = 100! 15% decrease in I B! 76% increase in I C (The higher V E the more stable system) As seen in the results, emitter resistor R E improves the stability.infact the stability is affected from V E.As V E increases the stability also increases.note that V E can not exist if R E does not exist.replacing a diode for R E for increasing V E will not make the system stable. We can generalize the subject and we can say that stability increases as V E increases and there must be a resistor R E for stability. We can adjust V E by changing the R E.(R B can also change the V E in active mode.) 30

31 Stability is determined directly by V E and R E and indirectly by R B.Observe the results. R E = 2K V E = 553.5mV! 95% increase in I C when 100% increase in beta (less stable system) R E = 2K V E = 1.790mV! 76% increase in I C when 100% increase in beta (more stable system) 31

32 VOLTAGE DIVIDER BIAS This configuration is more stable then fixed-biased with R E configuration.i C and V CE will not affected by the change of beta if proper design is done. So we can call this configuration as Beta independent configuration. This configuration can be used for all BJT amplifiers (common emitter, common base, and common collector), although in the common-collector configuration we usually set R C = 0. Thevenin Equivalent : The only purpose of R E is to provide DC feedback to stabilize the Q-point against variations in the β (also known as h FE ) of the BJT. Since β can vary by as much as five to one from one BJT to another (even if the BJTs have the same 2N number), large unit-to-unit variations in Q-point could result unless this DC feedback is present. When the resistor values in the above figure are selected properly, β variations will have a negligibly small effect on the Q-point. VCC. To provide sufficient feedback, VE is usually chosen to be one-quarter to one-third of Because V BE =0.7 V (silicon devices), V E is set by the voltage divider formed by R1 and R2. However, reader will note from the above figure that R1 and R2 are not in series, and hence do not form a voltage divider. Only if the maximum value of I B is much less than I1(Current on R 1 ), so that I B can be neglected, will R1 and R2 act like a voltage divider. The circuit designer must establish this condition by the appropriate choice of values for R1 and R2. 32

33 The effect of elements are again same as explained in past sections (Fixed-biased and fixed-biased with R E configuration).this time R 1 and R 2 will be explained in the solution of example 1. The design procedure is explained in the example 1.And a summary of design procedure will be found at the end of this tutorial. EXAMPLE 5: V CC =12V β = 300 At I C =3 to 10mA maximum dc current gain occurs.choosing I C = 5mA a) Determine the value of the resistors for best symmetric swing at I C = 5mA b) Instead of R 2 (β min.r E ) / 10 (R 2 << β min.r E ) use R 2 (β min.r E ) / 60 ( R 2 <<< β min.r E ) SOLUTION: How we choose I CQ? I C is generally choosen by observing the datasheet of the transistor.generally choosen at where β is maximum and stable. Note that in the datasheets I C - h fe graph depends on the voltage level of V CE.So choosen V CEQ have to be suitable for desired I CQ a) 1-Determinig the Q point: For symmetric swing, suitable for calculated V CEQ ) Vcc - 0.2V VCEQ = (Check the datasheet if choosen I CQ is 2 V CEQ = (12V 0.2V) / 2!V CEQ = 5.9V I CQ = 5mA, V CEQ = 5.9V gives the Q point on the load line. 2-Determine V E for stability: For stable systems: ( V CC / 10 ) V E ( V CC / 3 ) ( As V E increases, stability also increases ) Choosing V E = V CC / 4 V E = 12V / 4! V E = 3V 3-Determine R E : R E = V E / I C ( if I C I E ) R E = 3V / 5mA! R E = 600Ω 33

34 4-Determine R C : R C =V RC / I C V RC = V CC V CEQ V E V RC = 12V 5.9V 3V! V RC = 3.1V R C = 3.1V / 5mA! R C = 620Ω 5-Determine V B : V B = V E + 0.7V V B = 3V + 0.7V! V B = 3.7V 6-Determine R 2 : R i = Equivalent resistance between base and ground looking into the base R i = ( β +1 )R E βr E If R i is much larger than the resistance R 2, the current I B will be much smaller than I 2 and I 2 will be approximately equal to I 1.If we accept the approximation that I B is essentially zero amperes compared to I 1 or I 2, then I 1 = I 2 and R 1 and R 2 can be considered series elements. R i >> R 2 β min R E 10R 2 R 2 (β min.r E ) / 10 Assuming β = β min = 300 R 2 = (300).600Ω / 10! R 2 = 18K 7-Determine R 1 : The voltage across R 2, which is actually the base voltage, can be determined using the voltage-divider rule. R 2Vcc VB = R 1 + R2 3.7V = (18K).(12V) / (R K) R 1 =40.3K 34

35 Electronics Workbench results for example 1-a) b) Same results are found until step 6. R E = 600Ω, R C = 620Ω, V E =3V, V B = 3.7V, I C = 5mA, V CEQ = 5.9V 6-Determine R 2 : R 2 (β min.r E ) / 60 more smaller) Assuming β = β min = 300 R 2 = (300).600Ω / 60 7-Determine R 1 : (this time denominator is higher(10!60) that is R 2 is! R 2 = 3K The voltage across R 2, which is actually the base voltage, can be determined using the voltage-divider rule. R 2Vcc VB = R 1 + R2 3.7V = (3K).(12V) / (R 1 + 3K)! R 1 =6.72K Electronics Workbench results for example 1-b) 35

36 As R 2 and R 1 gets smaller(i 1 and I 2 gets higher then I B has less effect on the system. ) the values for I CQ and V CEQ are more approximate to desired I CQ and V CEQ 1-Determinig the Q point: SUMMARY OF VOLTAGE-DIVIDER BIAS DESIGN Choose V CEQ as Vcc - 0.2V VCEQ = if symmetric swing desired. 2 Choose I CQ at the beta-stable region in the datasheet 2-Determine V E for stability: For stable systems: ( V CC / 10 ) V E ( V CC / 3 ) ( As V E increases, stability also increases ) 3-Determine R E : R E = V E / I C ( if I C I E ) 4-Determine R C : R C =V RC / I C V RC = V CC V CEQ V E 5-Determine V B : V B = V E + 0.7V 6-Determine R 2 : R 2 (β min.r E ) / 10 for more accurate design choose R 2 smaller like: R 2 (β min.r E ) / 30 7-Determine R 1 : R 2Vcc VB = R 1 + R2 36

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