Chapter 4 BJT & FET Spring 2012 4 th Semester Mechatronics SZABIST, Karachi 2 Course Support humera.rafique@szabist.edu.pk Office: 100 Campus (404) Official: ZABdesk Subsidiary: https://sites.google.com/site/zabistmechatronics/home/spring-2012/ecd ebooks: https://sites.google.com/site/zabistmechatronics/home/ebooks 1
Chapter Contents 3 BJT & JFET Introduction Logarithms and Decibels General Considerations Bode plot Low Analysis Low BJT Amplifier Low FET Amplifier High BJT Amplifier High FET Amplifier Multistage Effects* 4 Introduction 2
Introduction 5 : Phase and amplitude plots and equations of an amplifier Prerequisites: 1. Logarithms 2. Semi-log plots 3. Decibels 4. Normalization 6 Logarithms 3
Logarithms 7 Logarithms: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. Common logarithms:, Natural logarithms: Relationship of CL and NL: 2.3 Benefits: Plotting of a variable between wide limits Compression of large data Logarithms: Broccoli, which grows in a logarithmic spiral A low pressure area over Iceland shows an approximately logarithmic spiral pattern Logarithms 8 The whirlpool Galaxy A nautilus displaying a logarithmic spiral 4
Logarithms 9 Example 9-1: Using the calculator, determine the logarithm of the following numbers to the base indicated: a. log 10 10 6 b. log e e 3 c. log 10 10 2 d. log e e 1 Example 9-2: Using the calculator, determine the logarithm of the following numbers: a. log 10 64 b. log e 64 c. log 10 1600 d. log 10 8000 Logarithms 1 0 Example 9-3: Using calculator, determine the antilogarithm of the following expressions: a. 1.6 = log 10 a b. 0.04 = log e a Example 9-4: Using calculator, determine the logarithm of the following numbers: a. log 10 0.5 b. log 10 (4000/250) c. log 10 (0.6 x 30) 1 0 1 5
11 Semi log Plots 2 Semilog graph paper Semilog Plots 12 1 Linear 30% log 10 2=0.3010 log 10 9 = 0.9543 log 10 8 = 0.9031 48% Log 10 3 = 0.4771 log 10 4 = 0.6021 ( 60%) log 10 7 = 0.8451 log 10 6 = 0.7781 log 10 5 = 0.6999 6
Semilog Plots 13 Identifying the numerical values of the tic marks on a log scale Semilog Plots 14 10 10 d 1 d 2 10 10 Example 9-5: Determine the value of the point appearing on the logarithmic plot in Fig. 9-4 using the measures made by a ruler (linear). 7
15 Decibels Decibels 16 10 10 20 = = + +..... + 8
Decibels 17 Example 9-6: Find the magnitude gain corresponding to a voltage gain of 100 db. Example 9-7: The input power to a device is 10,000 W at a voltage of 1000 V. The output power is 500 W and the output impedance is 20 Ω. Example 9-8: An amplifier rated at 40 W output is connected to a 10 Ω speaker. Calculate: a) The input power required for full power output if the power gain is 25 db b) The input voltage for rated output if the amplifier voltage gain is 40 db 18 General Considerations 9
General Considerations: Freq. Considerations 19 The frequency response of an amplifier refers to the frequency range in which the amplifier will operate with negligible effects from capacitors and device internal capacitance. This range of frequencies can be called the mid-range. At frequencies above and below the midrange, capacitance and any inductance will affect the gain of the amplifier. At low frequencies the coupling and bypass capacitors lower the gain. At high frequencies stray capacitances associated with the active device lower the gain. Also, cascading amplifiers limits the gain at high and low frequencies. General Considerations: A Bode plot indicates the frequency response of an Amplifier: The horizontal scale indicates the frequency (in Hz) and the vertical scale indicates the gain (in db) The mid-range frequency range of an amplifier is called the bandwidth of the amplifier The bandwidth is defined by the lower and upper cutoff frequencies Cutoff any frequency at which the gain has dropped by db Freq. Considerations 20 10
Freq. Considerations 21 General Considerations: 22 Normalization 11
Normalization 23 Normalization Process: In communication, a decibel plot vs frequency is normally provided rather than gain vs frequency A process in which the vertical parameter is divided by a specific level or quantity sensitive to a combination or variables of the system The band frequencies define a level where the gain or quantity of interest will be 70.7% or its maximum value Normalization 24 Normalization Process: Normalized gain versus frequency plot Decibels plots of the normalized gain versus frequency plot 12
Normalization 25 Example 9-9: Given the frequency response: a) Find the cutoff frequency f 1 and f 2 using the measurements provided b) Find the bandwidth of the response c) Sketch the normalized response Normalization 26 Example 9-9: 13
Normalization 27 db Plot: A v /A vmid A v /A vmid db 1 0 0.707-3 0.5-6 0.35-9 0.25-12 Decibel plot of the normalized gain versus frequency plot 28 Low Analysis 14
20 3/29/2012 LF Analysis Bode Plot 29 Low RC Circuit Analysis: Low frequency response for the R-C circuit LF Analysis Bode Plot 30 Low RC Circuit Analysis: 0.707 1 2 1 1 20 1 1 15
LF Analysis Bode Plot 31 Low RC Circuit Analysis: f f 1 /f A v(db) f 1 1 0 ½ f 1 2-6 ¼ f 1 4-12 1/10 f 1 10-20 LF Analysis Bode Plot 32 Low RC Circuit Analysis: f f 1 /f A v(db) f 1 1 0 ½ f 1 2-6 ¼ f 1 4-12 1/10 f 1 10-20 16
LF Analysis Bode Plot 33 Low RC Circuit Analysis: The piecewise linear plot of the asymptotes and associated breakpoints is called a Bode plot of the magnitude versus frequency A change in frequency by a factor of 2, equivalent to 1 octave, results in a 6-dB change in the ratio as noted by the change in gain from f 1 /2 to f 1. For a 10:1 change in frequency, equivalent to one decade, there is a 20-dB change in the ratio as noted by the change in gain from f 1 /10 to f 1. 10 / LF Analysis Bode Plot 34 Low RC Circuit Analysis: Phase response for the RC circuit Example 9-9 17
LF Analysis Bode Plot 35 Example 9-10: For the network of fig. 9-20: (R = 5 kω, C = 0.1 µf) a) Determine the break frequency b) Sketch the asymptotes and locate the 3 db point c) Sketch the frequency response curve d) Find the gain at A v(db) = 6 db LF Analysis Bode Plot 36 Example 9-10: 18
Example 9-10: % bode plot of Example 9-10 f = 10:10^4; fo = 318.5; A = 20*log(1./(1+(fo./f).^2).^(1/2)); semilogx(f,a), xlabel('f (log scale)'), ylabel('av(db)') grid Av(dB) LF Analysis Bode Plot Computer Analysis 0-10 -20-30 -40 37-50 -60-70 10 1 10 2 10 3 10 4 f (log scale) 38 Low BJT amplifiers 19
LF BJT Amplifiers 39 BJT Amplifiers:. LF BJT Amplifiers 40 Effects of Cs on the LF response: 20
LF BJT Amplifiers 41 Effects of C C on the LF response: LF BJT Amplifiers 42 Effects of C E on the LF response: 21
LF BJT Amplifiers 43 Effects of Cs and C E on the LF response: The cutoff frequency due to C S can be calculated by where f Ls 1 = 2 π (R + R )C s i s R = R R βr i 1 2 e The cutoff frequency due to C C can be calculated with 1 flc = 2 π ( R + R )C o L c where R = R r o C o LF BJT Amplifiers 44 Example 9-11: a) Determine the lower cutoff frequency for the network of Fig. 9.23 using the following parameters: C S = 10 µf, C E = 20 µf, C C = 1 µf, R S = 1 kω, R 1 = 40 kω, R 2 = 10 kω, R E = 2 kω, R C = 4 kω, R L = 2.2 kω, β = 100, r o = Ω, V CC = 20 V a) Sketch the frequency response using a Bode plot b) Verify the result using a Simulator. 22
LF BJT Amplifiers 45 Example 9-11: 46 Low FET amplifiers 23
LF FET Amplifiers 47 FET Amplifiers: LF FET Amplifiers 48 FET Amplifiers: The cutoff frequency due to C G can be calculated with f LG 1 = 2 π (R + R )C sig i G where R i = R G The cutoff frequency due to C C can be calculated with f LC 1 = 2 π (R + R )C o L G where R = R r O D G 24
LF FET Amplifiers 49 FET Amplifiers: The cutoff frequency due to C S can be calculated with f LS where R 1 = 2 π R C eq = eq R S S 1 g r m d Ω R eq LF FET Amplifiers 50 Example 9-12: a) Determine the lower cutoff frequency for the network of Fig. 11.32 using the following parameters: C G = 0.01 F, C C = 0.5 F, C S = 2 F R sig = 10 k, R G = 1 M, R D = 4.7 k, R S = 1 k, R L = 2.2 k I DSS = 8mA, V P = 4 V r d = Ω, V DD = 20 V b) Sketch the frequency response using a Bode plot. 25
LF FET Amplifiers 51 Example 9-12: 52 High FET amplifiers 26
HF FET Amplifiers 53 FET Amplifiers: HF FET Amplifiers 54 FET Amplifiers: 27
HF FET Amplifiers 55 FET Amplifiers: Capacitances that affect the high-frequency response are Junction capacitances C gs, C gd, C ds Wiring capacitances C wi, C wo Coupling capacitors C G, C C Bypass capacitor C S HF FET Amplifiers 56 FET Amplifiers: f Hi 1 = 2πR C Thi i C = C + C + C i Wi gs Mi C Mi = (1 A v )Cgd R Thi R sig R G Figure 9-64 (a) & (b) = Ho f 1 = 2πR C Tho o C C o = CWo + Cds + CMo Mo 1 = 1 C Av R Tho = R D R L rd gd 28
HF FET Amplifiers 57 Example 9-14: 58 Square Wave Testing 29
Square Wave Testing 59 Square Wave Testing: Square Wave Testing 60 Square Wave Testing: 30
Square Wave Testing 61 Example 9-15: The application of a 1-mV, 5-kHz square wave to an amplifier resulted in the output waveform of Fig. 9-72. (a) Write the Fourier series expansion for the square wave through the ninth harmonic. (b) Determine the bandwidth of the amplifier (c) Calculate the low cutoff frequency. Home Task 62 Reading: 1. Summary 2. Equations 3. Computer analysis Problems: 1. Sec 8.2: (odd) 2. Sec 8.3: 17,18 3. Sec 8.4: 19,21 4. Sec 8.5:23,25 5. Sec 8.6: 27,29 6. Sec 8.7: 31 7. Sec 8.8: 33,35,37 8. Sec 8.10: 39,41 9. Sec 8.11: 43 10. Sec 8.12: 45 11. Sec 8.14: 47 12. Sec 8.15: 49 31
FET References 63 1. Bolestad 2. Paynter CH 1 32