CHAPTER.4: Transistor at low frequencies

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1 CHAPTER.4: Transistor at low frequencies Introduction Amplification in the AC domain BJT transistor modeling The re Transistor Model The Hybrid equivalent Model Introduction There are three models commonly used in the small signal ac analysis of transistor networks: The re model The hybrid π model The hybrid equivalent model Amplification in the AC domain The transistor can be employed as an amplifying device, that is, the output ac power is greater than the input ac power. The factor that permits an ac power output greater than the input ac power is the applied DC power. The amplifier is initially biased for the required DC voltages and currents. Then the ac to be amplified is given as input to the amplifier. If the applied ac exceeds the limit set by dc level, clipping of the peak region will result in the output. Thus, proper (faithful) amplification design requires that the dc and ac components be sensitive to each other s requirements and limitations. The superposition theorem is applicable for the analysis and design of the dc and ac components of a BJT network, permitting the separation of the analysis of the dc and ac responses of the system. BJT Transistor modeling The key to transistor small-signal analysis is the use of the equivalent circuits (models). A MODEL IS A COMBINATION OF CIRCUIT ELEMENTS LIKE VOLTAGE OR CURRENT SOURCES, RESISTORS, CAPACITORS etc, that best approximates the behavior of a device under specific operating conditions. Once the model (ac equivalent circuit) is determined, the schematic symbol for the device can be replaced by the equivalent circuit and the basic methods of circuit analysis applied to determine the desired quantities of the network. Hybrid equivalent network employed initially. Drawback It is defined for a set of operating conditions that might not match the actual operating conditions. re model: desirable, but does not include feedback term Hybrid π model: model of choice. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 1

2 AC equivalent of a network AC equivalent of a network is obtained by: Setting all dc sources to zero and replacing them by a short circuit equivalent Replacing all capacitors by short circuit equivalent Removing all elements bypassed by the short circuit equivalents Redrawing the network in a more convenient and logical form. r e model In r e model, the transistor action has been replaced by a single diode between emitter and base terminals and a controlled current source between base and collector terminals. This is rather a simple equivalent circuit for a device VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 2

3 The Hybrid equivalent model For the hybrid equivalent model, the parameters are defined at an operating point. The quantities h ie, h re,h fe, and h oe are called hybrid parameters and are the components of a small signal equivalent circuit. The description of the hybrid equivalent model will begin with the general two port system. The set of equations in which the four variables can be related are: V i = h 11 I i + h 12 V o I o = h 21 I i + h 22 V o The four variables h 11, h 12, h 21 and h 22 are called hybrid parameters ( the mixture of variables in each equation results in a hybrid set of units of measurement for the h parameters. Set V o = 0, solving for h 11, h 11 = V i / I i Ohms This is the ratio of input voltage to the input current with the output terminals shorted. It is called Short circuit input impedance parameter. If I i is set equal to zero by opening the input leads, we get expression for h 12 : h 12 = V i / V o, This is called open circuit reverse voltage ratio. Again by setting V o to zero by shorting the output terminals, we get h 21 = I o / I i known as short circuit forward transfer current ratio. Again by setting I 1 = 0 by opening the input leads, h 22 = I o / V o. This is known as open circuit output admittance. This is represented as resistor ( 1/h 22 ) h 11 = h i = input resistance h 12 = h r = reverse transfer voltage ratio h 21 = h f = forward transfer current ratio h 22 = h o = Output conductance VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 3

4 Hybrid Input equivalent circuit Hybrid output equivalent circuit Complete hybrid equivalent circuit VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 4

5 Common Emitter Configuration - hybrid equivalent circuit Essentially, the transistor model is a three terminal two port system. The h parameters, however, will change with each configuration. To distinguish which parameter has been used or which is available, a second subscript has been added to the h parameter notation. For the common base configuration, the lowercase letter b is added, and for common emitter and common collector configurations, the letters e and c are used respectively. Common Base configuration - hybrid equivalent circuit Configuration I i I o V i V o Common emitter I b I c V be V ce Common base I e I c V eb V cb Common Collector I b I e V be V ec VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 5

6 Normally h r is a relatively small quantity, its removal is approximated by h r 0 and h r V o = 0, resulting in a short circuit equivalent. The resistance determined by 1/h o is often large enough to be ignored in comparison to a parallel load, permitting its replacement by an open circuit equivalent. h-parameter Model v/s. r e Model h ie = βr e h fe = β ac VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 6

7 Common Base: r e v/s. h-parameter Model Common-Base configurations - h-parameters h ib = r e h fb = - α = -1 Problem Given I E = 3.2mA, h fe = 150, h oe = 25µS and h ob = 0.5 µs. Determine The common emitter hybrid equivalent The common base r e model Solution: We know that, h ie = β re and r e = 26mV/I E = 26mV/3.2mA = 8.125Ω β re = (150)(8.125) = kΩ r o = 1 /h oe = 1/25µS = 40kΩ VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 7

8 r e = 8.125Ω r o = 1/ h ob = 1/0.5µS = 2M Ω α 1 Small signal ac analysis includes determining the expressions for the following parameters in terms of Z i, Z o and A V in terms of β r e r o and R B, R C Also, finding the phase relation between input and output The values of β, r o are found in datasheet The value of r e must be determined in dc condition as r e = 26mV / I E Common Emitter - Fixed bias configuration Removing DC effects of V CC and Capacitors VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 8

9 r e model Small signal analysis fixed bias From the above re model, Z i = [R B βr e ] ohms If R B > 10 βr e, then, [R B βr e ] βr e Then, Z i βr e Z o is the output impedance when V i =0. When V i =0, i b =0, resulting in open circuit equivalence for the current source. Z o = [R C r o ] ohms A V V o = - βi b ( R C r o ) From the r e model, I b = V i / β r e thus, V o = - β (V i / β r e ) ( R C r o ) A V = V o / V i = - ( R C r o ) / r e VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 9

10 If r o >10R C, A V = - ( R C / r e ) The negative sign in the gain expression indicates that there exists 180o phase shift between the input and output. Common Emitter - Voltage-Divider Configuration The r e model is very similar to the fixed bias circuit except for R B is R 1 R 2 in the case of voltage divider bias. Expression for A V remains the same. Z i = R 1 R 2 β r e Z o = R C From the r e model, I b = V i / β r e thus, V o = - β (V i / β r e ) ( R C r o ) A V = V o / V i = - ( R C r o ) / r e VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 10

11 o If r o >10R C, A V = - ( R C / r e ) Common Emitter - Unbypassed Emitter-Bias Configuration Applying KVL to the input side: V i = I b βr e + I e R E V i = I b βr e +(β +1) I b R E Input impedance looking into the network to the right of RB is Z b = V i / I b = βr e + (β +1)R E Since β>>1, (β +1) = β VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 11

12 Thus, Z b = V i / I b = β (r e +R E ) Since R E is often much greater than r e, Z b = βr E, Z i = R B Z b Z o is determined by setting V i to zero, I b = 0 and β I b can be replaced by open circuit equivalent. The result is, Z o = R C A V : We know that, V o = - I o R C = - βi b R C = - β(v i /Z b )R C A V = V o / V i = - β(r C /Z b ) Substituting, Z b = β(r e + R E ) A V = V o / V i = - β[r C /(r e + R E )] R E >>r e, A V = V o / V i = - β[r C /R E ] Phase relation: The negative sign in the gain equation reveals a 180 o phase shift between input and output. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 12

13 Emitter follower r e model Z i = R B Z b Z b = βr e + (β +1)R E Z b = β(r e + R E ) Since R E is often much greater than r e, Z b = βr E To find Zo, it is required to find output equivalent circuit of the emitter follower at its input terminal. This can be done by writing the equation for the current Ib. I b = V i / Z b I e = (β +1)I b = (β +1) (V i / Z b ) We know that, Z b = βr e + (β +1)R E substituting this in the equation for Ie we get, VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 13

14 I e = (β +1) (V i / Z b ) = (β +1) (V i / βr e + (β +1)R E ) I e = V i / [βr e / (β +1)] + R E Since (β +1) = β, I e = V i / [r e + R E ] Using the equation I e = V i / [r e + R E ], we can write the output equivalent circuit as, As per the equivalent circuit, Z o = R E r e Since R E is typically much greater than r e, Z o r e A V Voltage gain: Using voltage divider rule for the equivalent circuit, V o = V i R E / (R E + r e ) A V = V o / V i = [R E / (R E + r e )] Since (R E + r e ) R E, A V [R E / (R E ] 1 Phase relationship As seen in the gain equation, output and input are in phase. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 14

15 Common base configuration r e model Small signal analysis Input Impedance: Z i = R E r e Output Impedance: Z o = R C To find, Output voltage, V o = - I o R C V o = - (-I C )R C = αi e R C o I e = V i / r e, substituting this in the above equation, V o = α (V i / r e ) R C V o = α (V i / r e ) R C VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 15

16 Voltage Gain: A V : A V = V o / V i = α (R C / r e ) α 1; A V = (R C / r e ) Current gain A i = I o / I i I o = - α I e = - α I i I o / I i = - α -1 Phase relation: Output and input are in phase. h-parameter Model vs. re Model CB re vs. h-parameter Model Common-Base h-parameters h h ib = r e = α 1 fb VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 16

17 Small signal ac analysis includes determining the expressions for the following parameters in terms of Z i, Z o and A V in terms of β r e r o and R B, R C Also, finding the phase relation between input and output The values of β, r o are found in datasheet The value of re must be determined in dc condition as r e = 26mV / I E Common Emitter Fixed bias configuration Removing DC effects of V CC and Capacitors VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 17

18 r e model Small signal analysis fixed bias Input impedance Z i : From the above r e model, is, Z i = [R B βr e ] ohms If R B > 10 βr e, then, [RB βr e ] βr e Then, Z i βr e Ouput impedance Z oi : Z o is the output impedance when V i = 0. When V i = 0, i b = 0, resulting in open circuit equivalence for the current source. Z o = [R C r o ] ohms VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 18

19 Voltage Gain A v : V o = - βi b ( R C r o ) From the re model, I b = V i / β r e thus, V o = - β (V i / β r e ) ( R C r o ) A V = V o / V i = - ( R C r o ) / r e If r o >10R C, A V = - ( R C / r e ) Phase Shift: The negative sign in the gain expression indicates that there exists 180 o phase shift between the input and output. Problem: Common Emitter - Voltage-Divider Configuration VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 19

20 Equivalent Circuit: The re model is very similar to the fixed bias circuit except for R B is R 1 R 2 in the case of voltage divider bias. Expression for A V remains the same. : Z i = R 1 R 2 β r e Z o = R C Voltage Gain, A V : From the r e model, I b = V i / β r e V o = - I o ( R C r o ), I o = β I b thus, V o = - β (V i / β r e ) ( R C r o ) A V = V o / V i = - ( R C r o ) / r e If r o >10R C, A V = - ( R C / r e ) VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 20

21 Problem: Given: β = 210, r o = 50kΩ. Determine: r e, Z i, Z o, A V. For the network given: To perform DC analysis, we need to find out whether to choose exact analysis or approximate analysis. This is done by checking whether βr E > 10R 2, if so, approximate analysis can be chosen. Here, βr E = (210)(0.68k) = 142.8kΩ. 10R2 = (10)(10k) = 100k. Thus, βre > 10R2. Therefore using approximate analysis, V B = V cc R 2 / (R 1 +R 2 ) = (16)(10k) / (90k+10k) = 1.6V V E = V B 0.7 = = 0.9V I E = V E / R E = 1.324mA r e = 26mV / 1.324mA = 19.64Ω Effect of r o can be neglected if r o 10( R C ). In the given circuit, 10R C is 22k, r o is 50K. Thus effect of r o can be neglected. Z i = ( R 1 R 2 βr E ) = [90k 10k (210)(0.68k)] = 8.47kΩ Z o = R C = 2.2 kω VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 21

22 A V = - R C / R E = If the same circuit is with emitter resistor bypassed, Then value of re remains same. Z i = ( R 1 R 2 βr e ) = 2.83 kω Z o = R C = 2.2 kω A V = - R C / r e = Common Emitter Un bypassed Emitter - Fixed Bias Configuration Equivalent Circuit: Applying KVL to the input side: V i = I b βr e + I e R E V i = I b βr e +(β +1) I b R E VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 22

23 Input impedance looking into the network to the right of R B is Since β>>1, Z b = V i / I b = βr e + (β +1)R E (β +1) = β Thus, Z b = V i / I b = β (r e +R E ) Since R E is often much greater than r e, Z b = βr E, Z i = R B Z b Z o is determined by setting V i to zero, I b = 0 and βi b can be replaced by open circuit equivalent. The result is, We know that, Z o = R C V o = - I o R C = - βi b R C = - β(v i /Z b )R C A V = V o / V i = - β(r C /Z b ) Substituting Z b = β(r e + R E ) A V = V o / V i = - β[r C /(r e + R E )] R E >>r e, A V = V o / V i = - β[r C /R E ] Phase relation: The negative sign in the gain equation reveals a 180 o phase shift between input and output. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 23

24 Problem: Given: β = 120, r o = 40kΩ. Determine: r e, Z i, Z o, A V. To find r e, it is required to perform DC analysis and find I E as r e = 26mV / I E To find I E, it is required to find I B. We know that, I B = (V CC V BE ) / [R B + (β+1)r E ] I B = (20 0.7) / [470k + (120+1)0.56k] = 35.89µA I E = (β+1)i B = 4.34mA r e = 26mV / I E = 5.99Ω Effect of r o can be neglected, if r o 10( R C + R E ) 10( R C + R E ) = 10( 2.2 kω k) = 27.6 kω and given that r o is 40 kω, thus effect of r o can be ignored. Z i = R B [β ( r e + R E )] = 470k [120 ( )] = 59.34Ω Z o = R C = 2.2 kω A V = - βr C / [β ( r e + R E )] = Analyzing the above circuit with Emitter resistor bypassed i.e., Common Emitter I B = (V CC V BE ) / [R B + (β+1)r E ] I B = (20 0.7) / [470k + (120+1)0.56k] = 35.89µA VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 24

25 I E = (β+1)i B = 4.34mA r e = 26mV / I E = 5.99Ω Z i = R B [βr e ] = Ω Z o = R C = 2.2 kω A V = - R C / r e = ( a significant increase) Emitter follower r e model Z i = R B Z b Z b = βr e + (β +1)R E Z b = β(r e + R E ) Since R E is often much greater than r e, Z b = βr E VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 25

26 To find Z o, it is required to find output equivalent circuit of the emitter follower at its input terminal. This can be done by writing the equation for the current I b. I b = V i / Z b I e = (β +1)I b We know that, = (β +1) (V i / Z b ) Z b = βr e + (β +1)R E substituting this in the equation for I e we get, I e = (β +1) (V i / Z b ) = (β +1) (V i / βr e + (β +1)R E ) dividing by (β +1), we get, I e = V i / [βr e / (β +1)] + R E Since (β +1) = β, I e = V i / [r e + R E ] Using the equation I e = V i / [r e + R E ], we can write the output equivalent circuit as, As per the equivalent circuit, Z o = R E r e Since R E is typically much greater than r e, Zo re VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 26

27 A V Voltage gain: Using voltage divider rule for the equivalent circuit, V o = V i R E / (R E + r e ) A V = V o / V i = [R E / (R E + r e )] Since (R E + r e ) R E, A V [R E / (R E ] 1 Phase relationship As seen in the gain equation, output and input are in phase. Common base configuration VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 27

28 r e model Small signal analysis Z i = R E r e To find Z o = R C V o = - I o R C V o = - (-I C )R C = αi e R C Substituting this in the above equation, I e = V i / r e, V o = α (V i / r e ) R C V o = α (V i / r e ) R C A V = V o / V i = α (R C / r e ) α 1; A V = (R C / r e ) Current gain A i : A i = I o / I i I o = - α I e = - α I i I o / I i = - α -1 Phase relation: Output and input are in phase. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 28

29 Common Emitter - Collector Feedback Configuration r e Model Input Impedance: Z i Z i = V i / I i, I i = I b I, thus it is required to find expression for I in terms of known resistors. I = (V o V i )/ R F (1) V o = - I o R C I o = βi b + I Normally, I << βi b thus, I o = βi b, V o = - I o R C V o = - βi b R C, VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 29

30 Replacing Thus, I b by V i / βr e V o = - β (V i R C ) / βr e = - (V i R C ) / r e (2) Substituting (2) in (1): I = (V o V i )/ R F = (V o / R F ) - (V i / R F ) = - [(V i R C ) / R F r e ] - (V i / R F ) I = - V i /R F [ (R C / r e )+1] We know that, V i = I b βr e, I b = I i + I and, I = - V i /R F [ (R C / r e ) +1] Thus, V i = ( I i + I ) βr e = I i βr e + I βr e = I i βr e - (V i βr e )( 1/R F )[ (R C / r e )+1] Taking V i terms on left side: V i + (V i βr e )( 1/R F )[ (R C / r e )+1] = I i βr e V i [1 + (βr e )( 1/R F )[ (R C / r e ) +1] = I i βr e V i / I i = βr e / [1 + (βr e )( 1/R F )[ (R C / r e ) +1] But, [ (R C / r e )+1] R C / r e (because R C >> r e ) Thus, Z i = V i / I i = βr e / [1 + (βr e )( 1/R F )[ (R C / r e )] = βr e / [1 + (β)(r C /R F )] Thus, Z i = r e / [(1/β) + (R C /R F )] VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 30

31 To find Output Impedance Zo: Z o = R C R F ( Note that i b = 0, thus no effect of βr e on Z o ) Voltage Gain A V : V o = - I o R C = - βi b R C ( neglecting the value of I ) = - β(v i / βr e )R C A V = V o / V i = - (R C /r e ) Phase relation: - sign in A V indicates phase shift of 180 between input and output. Collector DC feedback configuration VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 31

32 r e model Z i = R F1 βr e Z o = R C R F2 r o, for r o 10R C, Z o = R C R F2 To find Voltage Gain A V : V o = - βi b (R F2 R C r o ), I b = V i / βr e for r o 10R C, V o = - β (V i / βr e )(R F2 R C r o ) V o / V i = - (R F2 R C r o ) / r e, A V = V o / V i = - (R F2 R C ) / r e Determining the current gain For each transistor configuration, the current gain can be determined directly from the voltage gain, the defined load, and the input impedance. We know that, current gain (A i ) = I o / I i I o = (V o / R L ) and I i = V i / Z i Thus, A i = - (V o /R L ) / (V i / Z i ) = - (V o Z i / V i R L ) Example: A i = - A V Z i / R L For a voltage divider network, we have found that, Z i = βr e Thus, A V = - R C / r e and R L = R C A i = - A V Z i / R L = - (- R C / r e )(βr e ) / R C A i = β VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 32

33 For a Common Base amplifier, Z i = r e, A V = R C / r e, R L = R C Effect of R L and R S : A i = - A V Z i / R L = - (R C / r e )(r e ) / R C = - 1 Voltage gain of an amplifier without considering load resistance (R L ) and source resistance (R S ) is A VNL. Voltage gain considering load resistance ( R L ) is A V < A VNL Voltage gain considering R L and R S is A VS, where A VS <A VNL < A V For a particular design, the larger the level of R L, the greater is the level of ac gain. Also, for a particular amplifier, the smaller the internal resistance of the signal source, the greater is the overall gain. Fixed bias with R S and R L : A V = - (R C R L ) / r e Z i = R B βr e Z o = R C r o To find the gain A VS, ( Z i and R S are in series and applying voltage divider rule) V i = V S Z i / ( Z i +R S ) V i / V S = Z i / ( Z i +R S ) A VS = V o / V S = (V o /V i ) (V i /V S ) A VS = A V [Z i / ( Z i +R S )] VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 33

34 Voltage divider with R S and R L Voltage gain: Input Impedance: Output Impedance: A V = - [R C R L ] / r e Z i = R 1 R 2 β r e Z o = R C R L r o Emitter follower with R S and R L r e model: VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 34

35 Voltage Gain: A V = (R E R L ) / [R E R L +r e ] Input Impedance: Z i = R B Z b Input Impedance seen at Base: Z b = β(r E R L ) Output Impedance Z o = r e Two port systems approach This is an alternative approach to the analysis of an amplifier. This is important where the designer works with packaged with packaged products rather than individual elements. An amplifier may be housed in a package along with the values of gain, input and output impedances. But those values are no load values and by using these values, it is required to find out the gain and various impedances under loaded conditions. This analysis assumes the output port of the amplifier to be seen as a voltage source. The value of this output voltage is obtained by Thevinising the output port of the amplifier. E th = A VNL V i Model of two port system Applying the load to the two port system Applying voltage divider in the above system: V o = A VNL V i R L / [ R L +R o ] VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 35

36 Including the effects of source resistance R S Applying voltage divider at the input side, we get: V i = V S R i /[R S +R i ] V o = A VNL V i V i = V S R i /[R S +R i ] V o = A VNL V S R i /[R S +R i ] V o / V S = A VS = A VNL R i /[R S +R i ] Two port system with R S and R L We know that, at the input side V i = V S R i /[R S +R i ] V i / V S = R i /[R S +R i ] At the output side, V o = A VNL V i R L / [ R L +R o ] V o / V i = A VNL R L / [ R L +R o ] Thus, considering both R S and R L : A V = V o / V s = [V o / V i ] [V i / V s ] VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 36

37 A V = (A VNL R L / [ R L +R o ]) (R i / [R S +R i ]) Example: Given an amplifier with the following details: R S = 0.2 kω, A VNL = - 480, Z i = 4 kω, Z o = 2 kω Determine: A V with R L =1.2kΩ A V and A i with R L = 5.6 kω, A VS with R L = 1.2 Solution: A V = A VNL R L / (R L + R o ) = (- 480)1.2k / (1.2k+2k) = With R L = 5.6k, A V = This shows that, larger the value of load resistor, the better is the gain. A VS = [R i /(R i +R S )] [ R L / (R L +R o )] A VNL = A i = - A V Z i /R L, here A V is the voltage gain when R L = 5.6k. A i = - A V Z i /R L = - ( )(4k/5.6k) = Hybrid π model This is more accurate model for high frequency effects. The capacitors that appear are stray parasitic capacitors between the various junctions of the device. These capacitances come into picture only at high frequencies. C bc or C u is usually few pico farads to few tens of pico farads. r bb includes the base contact, base bulk and base spreading resistances. VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 37

38 r be ( r π ), r bc, r ce are the resistances between the indicated terminals. r be ( r π ) is simply βr e introduced for the CE r e model. r bc is a large resistance that provides feedback between the output and the input. r π = βr e g m = 1/r e r o = 1/h oe h re = r π / (r π + r bc ) VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 38

39 VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS 39

KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D II )

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