Philadelphia University Faculty of Engineering Communication and Electronics Engineering

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1 Module: Electronics II Module Number: 6503 Philadelphia University Faculty o Engineering Communication and Electronics Engineering Ampliier Circuits-II BJT and FET Frequency Response Characteristics: - Logarithms and Decibels: Logarithms taken to the base 0 are reerred to as common logarithms, while logarithms taken to the base e are reerred to as natural logarithms. In summary: Some relationships hold true or logarithms to any base The background surrounding the term decibel (db) has its origin in the established act that power and audio levels are related on a logarithmic basis. That is, an increase in power level, say 4 to 6 W, does not result in an audio level increase by a actor o 6/4 4. It will increase by a actor o as derived rom the power o 4 in the ollowing manner: (4) 6. The term bel was derived rom the surname o Alexander Graham Bell. For standardization, the bel (B) was deined by the ollowing equation to relate power levels P and P: Lecturer: Dr. Omar Daoud Part II

2 Module: Electronics II Module Number: 6503 It was ound, however, that the bel was too large a unit o measurement or practical purposes, so the decibel (db) was deined such that 0 decibels bel. Thereore, There exists a second equation or decibels that is applied requently. It can be best described through the system with Ri, as an input resistance. One o the advantages o the logarithmic relationship is the manner in which it can be applied to cascaded stages. In words, the equation states that the decibel gain o a cascaded system is simply the sum o the decibel gains o each stage. - General Frequency Considerations: The requency o the applied signal can have a pronounced eect on the response o a single-stage or multistage network. The analysis thus ar has been or the midrequency spectrum. At low requencies, we shall ind that the coupling and bypass capacitors can no longer be replaced by the short-circuit approximation because o the increase in reactance o these elements. The requency-dependent parameters o the small-signal equivalent circuits and the stray capacitive elements associated with the active device and the network will limit the high-requency response o the system. An increase in the number o stages o a cascaded system will also limit both the high- and low-requency responses. For any system, there is a band o requencies in which the magnitude o the gain is either equal or relatively close to the midband value. To ix the requency boundaries o relatively high gain, 0.707Avmid was chosen to be the gain at the cuto levels. The corresponding requencies and are generally called the corner, cuto, band, break, or hal-power requencies. The multiplier was chosen because at this level the output power is hal the midband power output, that is, at midrequencies. Lecturer: Dr. Omar Daoud Part II

3 Module: Electronics II Module Number: 6503 The bandwidth (or passband) o each system is determined by and, that is, For applications o a communications nature (audio, video), a decibel plot o the voltage gain versus requency is more useul. Beore obtaining the logarithmic plot, however, the curve is generally normalized as shown in Fig In this igure, the gain at each requency is divided by the midband value. Obviously, the midband value is then as indicated. At the hal-power requencies, the resulting level is /. Fig. 9.6 Normalized gain versus requency plot. Fig. 9.7 Decibel plot o the normalized gain versus requency plot o Fig Low Frequency Analysis: Lecturer: Dr. Omar Daoud Part II 3

4 Module: Electronics II Module Number: 6503 In the low-requency region o the single-stage ampliier, it is the R-C combinations ormed by the network capacitors C C, C E, and Cs and the network resistive parameters that determine the cuto requencies. The analysis, thereore, will begin with the series R-C combination o the given Fig. and the development o a procedure that will result in a plot o the requency response with a minimum o time and eort. ) At very high requencies, A short-circuit equivalent can be substituted or the capacitor. The result is that Vo Vi at high requencies. ) At 0 Hz, An open-circuit approximation can be applied, with the result that Vo 0 V. 3) Between the two extremes, the ratio Av Vo/Vi will vary as shown below. As the requency increases, the capacitive reactance decreases and more o the input voltage appears across the output terminals. The output and input voltages are related by the voltage-divider rule in the ollowing manner: Lecturer: Dr. Omar Daoud Part II 4

5 Module: Electronics II Module Number: 6503 Lecturer: Dr. Omar Daoud Part II 5 C i o X R RV V + The magnitude o Vo determined by C i o X R RV V + For the special case where XC R, db V V A V R R RV X R RV V i o v i i C i o The requency o which XC R (the output will be 70.7% o the input) is determined as: ( ) db j A CR j CR j R X j jx R R A CR R C X Vi Vo A v C C v C v + + leads phase by which magnitude o 0log / tan π ω π π For requencies where << or ( /) >>, the equation above can be approximated as db db A v 0log log 0 Ignoring the previous condition or a moment, a plot on a requency log scale will yield a result o a very useul nature or uture decibel plots.

6 Module: Electronics II Module Number: 6503 At At At At 0 log() 0dB 0 log() 6dB 0 log(4) db 4 0 log(0) 0dB 0 - A change in requency by a actor o, equivalent to octave, results in a 6-dB change in the ratio as noted by the change in gain rom / to. - For a 0: change in requency, equivalent to decade, there is a 0- db change in the ratio as demonstrated between the requencies o Ex. For the given network: (a) Determine the break requency. (b) Sketch the asymptotes and locate the -3-dB point. (c) Sketch the requency response curve. Solution: (a) πrc π (b) See Figure below 3 6 ( 5 0 Ω)( 0. 0 F) 38.5Hz Lecturer: Dr. Omar Daoud Part II 6

7 Module: Electronics II Module Number: 6503 Fig..4 Frequency response or the RC circuit o the Ex. - Low Frequency Analysis-BJT Ampliiers: The analysis o this section will employ the loaded voltage-divider BJT bias coniguration, but the results can be applied to any BJT coniguration. It will simply be necessary to ind the appropriate equivalent resistance or the R-C combination ( or the capacitors Cs, CC, and CE, which will determine the low-requency response). ) The eect o C s : Since Cs is normally connected between the applied source and the active device, the total resistance is now Rs + Ri, and the cuto requency will be modiied to be as: Lecturer: Dr. Omar Daoud Part II 7

8 Module: Electronics II Module Number: 6503 Ls π ( R s + R i ) C s At mid or high requencies, the reactance o the capacitor will be suiciently small to permit a short-circuit approximation or the element. The voltage Vi will then be related to Vs by Ri V i mid Vs Ri + Rs The voltage Vi applied to the input o the active device can be calculated using the voltage-divider rule: Ri V i Vs R + R jx s ) The eect o C C : Since Cc the coupling capacitor is normally connected between the output o the active device and the applied load, the total resistance is now Ro + R L, and the cuto requency will be modiied to be as: i Cs LC π ( R o + R L ) C C Lecturer: Dr. Omar Daoud Part II 8

9 Module: Electronics II Module Number: ) The eect o C E : To determine LE, the network seen by CE must be determined as shown in the Fig. below. Once the level o Re is established, the cuto requency due to CE can be determined using the ollowing equation: LE π ( R e ) C E where R e could be calculated as: RS // R // R Re RE // + re β where R // R // R R S S Ex. (a) Determine the lower cuto requency or the voltage-divider BJT bias coniguration network using the ollowing parameters: Cs 0µF, CE 0µF, CC µf, Rs kω, R 40kΩ, R 0kΩ, RE kω, RC 4kΩ, RL. kω, β 00, ro, VCC 0V. (b) Sketch the requency response using a Bode plot. Solution: Lecturer: Dr. Omar Daoud Part II 9

10 Module: Electronics II Module Number: 6503 Lecturer: Dr. Omar Daoud Part II 0

11 Module: Electronics II Module Number: 6503 (b) - Low Frequency Analysis-FET Ampliiers: The analysis o the analysis o the FET ampliier in the lowrequency region will be quite similar to that o the BJT ampliier. There are again three capacitors o primary concern as appearing in the given network: CG, CC, and CS. ) The eect o C G : The cuto requency determined by CG will then be LG π ( R G + R sig ) C G Lecturer: Dr. Omar Daoud Part II

12 Module: Electronics II Module Number: 6503 ) The eect o C C : The cuto requency determined by CC will then be LC π where R o ( R + R ) L R D o // r d C C 3) The eect o C S : The cuto requency determined by CS will then be LS R ( R ) RS where Reg + d G which or r bcomes R eg π S d eg // g C m S RS ( + g mrd ) ( r + R // R ) L Ex. (a) Determine the lower cuto requency or the CS FET bias coniguration network using the ollowing parameters: CG 0.0µF, CC 0.5µF, CS µf, Rsig 0kΩ, RG MΩ, RD 4.7kΩ, RS kω, RL. kω, I DSS 8mA, Vp -4V, rd, VDD 0V. (b) Sketch the requency response using a Bode plot. Solution: Lecturer: Dr. Omar Daoud Part II

13 Module: Electronics II Module Number: 6503 Since LS is the largest o the three cuto requencies, it deines the low cuto requency or the network. Lecturer: Dr. Omar Daoud Part II 3

14 Module: Electronics II Module Number: Miller Eect capacitance: In the high-requency region, the capacitive elements o importance are the interelectrode (between terminals) capacitances internal to the active device and the wiring capacitance between leads o the network. The large capacitors o the network that controlled the low-requency response have all been replaced by their short-circuit equivalent due to their very low reactance levels. Miller input capacitance ( Av ) C CMi where C is the eedback capacitance. Miller output capacitance CMo C C A v >> A v where C is the eedback capacitance. - High Frequency Analysis-BJT Ampliiers: In the high-requency region, the RC network o concern has the coniguration appearing in given Fig. At increasing requencies, the reactance XC will decrease in magnitude, resulting in a shorting eect across the output and a decrease in gain. The derivation leading to the corner requency or this RC coniguration ollows along similar lines to that encountered or the low-requency region. The most signiicant dierence is in the general orm o Av appearing below: A v + j Lecturer: Dr. Omar Daoud Part II 4

15 Module: Electronics II Module Number: 6503 In the above Fig., the various parasitic capacitances (Cbe, Cbc, Cce) o the transistor have been included with the wiring capacitances (CWi, CWo) introduced during construction. In the high-requency equivalent model or the network, note the absence o the capacitors Cs, CC, and CE, which are all assumed to be in the shortcircuit state at these requencies. The capacitance Ci includes the input wiring capacitance CWi, the transition capacitance Cbe, and the Miller capacitance CMi. The capacitance Co includes the output wiring capacitance CWo, the parasitic capacitance Cce, and the output Miller capacitance CMo. In general, the capacitance Cbe is the largest o the parasitic capacitances, with Cce the smallest. In act, most speciication sheets simply provide the levels o Cbe and Cbc and do not include Cce unless it will aect the response o a particular type o transistor in a speciic area o application. Lecturer: Dr. Omar Daoud Part II 5

16 Module: Electronics II Module Number: 6503 ) For the input network, C i : For the input network, the -3-dB requency is deined by Hi πrthici where ) For the output network, C o : Ex. For the given network, with the ollowing parameters: Cs 0µF, CE 0µF, CC µf, Rs kω, R 40kΩ, R 0kΩ, RE kω, RC 4kΩ, RL. kω, β 00, ro, VCC 0V. with the addition o Cbe 36pF, Cbc 4pF, Cce pf, CWi 6pF, CWo 8pF (a) Determine Hi and Ho. (b) Sketch the total requency response or the low- and high-requency regions. Solution: (a) rom previous example Lecturer: Dr. Omar Daoud Part II 6

17 Module: Electronics II Module Number: 6503 (b) Lecturer: Dr. Omar Daoud Part II 7

18 Module: Electronics II Module Number: High Frequency Analysis-FET Ampliiers: The analysis o the high-requency response o the FET ampliier will proceed in a very similar manner to that encountered or the BJT ampliier. There are interelectrode and wiring capacitances that will determine the high-requency characteristics o the ampliier. The capacitors Cgs and Cgd typically vary rom to 0 pf, while the capacitance Cds is usually quite a bit smaller, ranging rom 0. to pf. At high requencies, Ci will approach a short-circuit equivalent and Vgs will drop in value and reduce the overall gain. At requencies where Co approaches its short circuit equivalent, the parallel output voltage Vo will drop in magnitude. Lecturer: Dr. Omar Daoud Part II 8

19 Module: Electronics II Module Number: 6503 Ex. (a) Determine the high cuto requencies or the CS FET bias coniguration network using the ollowing parameters: CG 0.0µF, CC 0.5µF, CS µf, Rsig 0kΩ, RG MΩ, RD 4.7kΩ, RS kω, RL. kω, I DSS 8mA, Vp -4V, rd, VDD 0V. Cgd pf, Cgs 4pF, Cds 0.5pF, CWi 5pF, CWo 6pF Solution: (a) From the previous example Lecturer: Dr. Omar Daoud Part II 9

20 Module: Electronics II Module Number: 6503 Lecturer: Dr. Omar Daoud Part II 0 - Multistage Frequency Eect: For Low Frequency region n For High Frequency region n

21 Module: Electronics II Module Number: Square wave Testing: Experimentally, the sense or the requency response can be determined by applying a square wave signal to the ampliier and noting the output response. The reason or choosing a square-wave signal or the testing process is best described by examining the Fourier series expansion o a square wave composed o a series o sinusoidal components o dierent magnitudes and requencies. The summation o the terms o the series will result in the original waveorm. In other words, even though a waveorm may not be sinusoidal, it can be reproduced by a series o sinusoidal terms o dierent requencies and magnitudes. Since the ninth harmonic has a magnitude greater than 0% o the undamental term, the undamental term through the ninth harmonic are the major contributors to the Fourier series expansion o the square-wave unction. Ex. For a speciic application (Audio ampliier with 0kHz Bandwidth), what is the maximum requency could be ampliied? 0kHz.kHz 9 Lecturer: Dr. Omar Daoud Part II

22 Module: Electronics II Module Number: 6503 I the response o an ampliier to an applied square wave is an undistorted replica o the input, the requency response (or BW) o the ampliier is obviously suicient or the applied requency. I the response is as shown in Fig..60a and b, the low requencies are not being ampliied properly and the low cuto requency has to be investigated. I the waveorm has the appearance o Fig..60c, the highrequency components are not receiving suicient ampliication and the high cuto requency (or BW) has to be reviewed. Fig..60 (a) Poor low-requency response; (b) very poor low-requency response; (c) poor high-requency response; (d) very poor high-requency response. The actual high cuto requency (or BW) can be determined rom the output waveorm by careully measuring the rise time deined between 0% and 90% o the peak value, as shown in the Fig. below. Substituting into the ollowing equation Lecturer: Dr. Omar Daoud Part II

23 Module: Electronics II Module Number: 6503 will provide the upper cuto requency, and since BW Hi - Lo Hi, the equation also provides an indication o the BW o the ampliier BW Hi t r The low cuto requency can be determined rom the output response by careully measuring the tilt and substituting into one o the ollowing equations: V V %tilt P% 00% V V V P (decimal orm) V The low cuto requency is determined rom P Lo s π then Ex. The application o a -mv, 5-kHz square wave to an ampliier resulted in the output waveorm o the given Fig. (a) Write the Fourier series expansion or the square wave through the ninth harmonic. (b) Determine the bandwidth o the ampliier. Solution: Lecturer: Dr. Omar Daoud Part II 3

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