Frequency Response of BJT & JFET

Alexandria High Institute for Engineering & Technology Frequency Response of BJT & JFET Course ECE241- Chapter (2) Sameh Selim )1)

Logarithms To clarify the relationship between the variables of a logarithmic function, consider the mathematical equations: b is the base, x the power. If b = 10 and x = 2, a = b x = (10) 2 = 100 but x = log b a = log 10 100 = 2 For the electrical/electronics industry, the base in the logarithmic equation is limited to 10 and the number e = 2.71828 Common logarithms: logarithms taken to the base 10. Natural logarithms: logarithms taken to the base e. The two are related by In calculators, the common logarithm is denoted by "log" key and the natural logarithm by the "ln" key. Example 1: Using the calculator, determine the logarithm of the following numbers to the indicated base: a) Log 10 10 6 b) log e e 3 c) log 10 10-2 d) log e e -1 )2)

Example 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 N.B. the logarithm of a number does not increase in the same linear fashion as the number. The following table shows how the logarithm of a number increases only as the exponent of the number. If the antilogarithm of a number is desired, the 10 x or e x calculator functions are employed. Example 3: Using a calculator, determine the antilogarithm of the following expressions: a) 1.6 = log 10 a. b) 0.04 = log e a. Some properties of common logarithms: )3)

Example 4: Using calculator, determine the logarithm of the following numbers: a) log 10 (0.5) b) ( ) c) ( ) The use of log scales can significantly expand the range of variation of a particular variable on the graph. The semilog scale graph paper appears as in Figure below: )4)

DECIBELS The term "bel" (B) (derived from Alexander Graham Bell) was defined by the following equation to relate power levels P 1 and P 2 Practically, so the decibel (db) was defined such that: For a specified terminal (output) power (P 2 ) there must be a reference power level (P 1 ). The reference level is generally accepted to be 1mW, with an associated resistance of 600 (the characteristic impedance of audio transmission lines). There exists a second equation for decibels. For the system shown, V i equal to some value V1,. If V i is changed to some other level, V 2, then. To determine the resulting difference in decibels between the power levels, One of the advantages of the logarithmic relationship is the manner in which it can be applied to cascaded stages. )5)

For example, the magnitude of the overall voltage gain of cascaded system is given by Applying the proper logarithmic relationship results in In words, the equation states that the decibel gain of cascaded system is simply the sum of the decibel gain of each stage, i.e. Example 5: Find the magnitude gain corresponding to a decibel gain of 100. Check Example 6: The I/p power to a device is 10,000 W at a voltage of 1000 V. The O/P power is 500 W, while the o/p impedance is 20. a) Find the power gain in decibels. ( -13.01 db ) b) Find the voltage gain in decibels. ( - 20 db ) c) Explain why parts (a) and (b) agree or disagree.( Ri = 100 Ro) Example 7: An amplifier rated at 40-W O/P is connected to a 10- speaker. a) Calculate the I/P power required for full power O/P if the power gain is 25 db. ( 126.5 mw) b) c) Calculate the I/P voltage for rated O/P if the amplifier voltage gain is 40 db. ( 200 mv ) )6)

General Frequency considerations The magnitudes of the gain response of an RC coupled amplifier system is provided in the following Figure. Note that the horizontal scale is a logarithmic scale to permit a plot extending from the low- to the high-frequency regions. There is a band of frequencies in which the magnitude of the gain is either equal or relatively close to the midband value. To fix the frequency boundaries of relatively high gain, 0.707 A vmid was chosen to be the gain at the cutoff levels. The corresponding frequencies f 1 and f 2 are generally called the corner, cutoff, break, or half-power frequencies. The multiplier 0.707 was chosen because at this level the O/P power is half the midband power O/P, that is, at midfrequencies, and at the half-power frequencies, and P OHPF = 0.5 P omid (16) The bandwidth (or passband) of the system is determined by f 1 and f 2, that is, Bandwidth (BW) = f 2 f 1 (17) )7)

For applications of communications nature (audio, video), a decibel plot of the voltage gain vs frequency is more useful than that appearing in the above Figure: the curve is generally normalized as in the following Figure. In this Figure, the gain at each frequency is divided by the midband value. Obviously the midband value is then 1 as indicated. At halfpower frequencies, the resulting level is 0.707 =. A decibel plot can now be obtained by applying At midband frequencies, 20 log 10 1 = 0, and at the cutoff frequencies, 20 log 10 = -3 db. Both values are clearly indicated in the resulting decibel plot, as shown in the following Figure, Most amplifiers introduce a 180 o phase shift between I/P and O/P signals. In fact, this is the case only in the midband region. At low frequencies, there is a phase shift such that V o lags V i by an increased angle. At high frequencies, the phase shift will drop below 180 o. the following Figure is a standard phase plot for an RC coupled amplifier. )8)

Low Frequency Analysis-Bode Plot: In the low-frequency region of the single-stage BJT or FET amplifier, it is the R-C combination formed by the network capacitors C c, C E and C s and the network resistive parameters that determine the cutoff frequencies. The series RC combination, as shown, and the development of a procedure that will result in a plot of frequency response with minimum of time and effort. At high frequencies, and the short-circuit equivalent can be substituted for the capacitor. The result is that V o V i at high frequencies. At f = 0 Hz, and the open-circuit approximation can be applied, with the result that V o = 0 V. Between the two extremes, the ratio Av = V o /V i will vary as shown in )9)

the following Figure. As the frequency increases, the capacitive reactance decreases and more of the input voltage appears across the O/P terminals. The O/P and I/P voltages are related by the voltage-divider rule as With magnitude of V o determined by For the special case where X C = R, And The level of which is indicated in on Figure above. In other words, at the frequency of which X C = R, the O/P will be 70.7% of the I/P for the RC network shown before. The frequency at which this occurs is determined from and in terms of logs, )11)

While at Av = Vo/Vi = 1 or Vo = Vi (the maximum value) If the gain equation is written as For the magnitude when f = f 1 In logarithmic form, the gain in db is For frequencies For where f f1 or (f1/f) 2 1, the equation above can be approximated by And finally, )11)

Ignoring the condition f f1 for the moment, a plot of the last equation on a frequency log scale will yield a result of a very useful nature for future decibel plots. The piecewise linear plot of the asymptotes and associated breakpoints is called a Bode Plot of magnitude vs frequency. The piecewise linear plot of the asymptotes and associated breakpoints is called a Bode plot of the magnitude versus frequency. Notes: A change in frequency by a factor of 2, i.e. one octave, results in 6-dB change in ratio, as shown by the gain change from f 1 /2 to f 1. A change in frequency by a factor of 10, i.e. one decade, results in 20- db change in ratio, as shown by the gain change from f 1 /10 to f 1. )12)

Steps for Bode plot: 1) Find f1 from the circuit parameters. 2) Sketch 2 asymptotes: one along the 0-dB and the other drawn thro' f1 sloped at 6 db/octave or 20 db/decade. 3) Find the 3-dB point corresponding to f1 and sketch the curve. Example 8: For the network shown in Figure: 1. Determine the break frequency. 2. Sketch the asymptotes and locate the -3 db point. 3. Sketch the frequency response curve. The Frequency response of the gain A v (db) for the R-C circuit is shown in Figure. The gain at any frequency can be determined from the frequency plot in the following manner: )13)

Check: Av(dB) = -3 db, Av = 0.707, and Av(dB) = - 1 db, (at f = 2f 1 ), Av = 0.891 The phase angle is determined from f f 1, = 90 o f 1 = 100 f, = 89.4 o f = f 1, = 45 o f f 1, = 0 o f = 100 f 1, = 0.573 o The phase response for the R-C circuit is shown. Low-Frequency Response (BJT) Amplifier In the analysis of the voltage-divider BJT, it will simply be necessary to find the appropriate equivalent resistance for the RC combination. Capacitors C s, C c, C E will determine the low-frequency response. We will examine the impact of each independently. Effect of Cs: the general form of the R-C configuration is established by the network of the following Figure. The total resistance is R s + R i. )14)

The cutoff frequency: At mid or high frequencies: The reactance of the capacitor will be short circuit approximate. The voltage V i is related to V s by: At f Ls, the voltage V i will be 70.7% of the value determined by this eqn., assuming that C s is the only capacitive element controlling the low frequency response. The ac equivalent for the i/p section of BJT amplifier: The value of R i is determined by: R i = R 1 //R 2 // r e The voltage V i applied to the i/p of the active device can be calculated using the voltage-divider rule: Effect of C C : Since the coupling capacitor is normally connected between the O/P of the active device and the applied load, the R-C configuration that determines the low cutoff frequency due to C C appears in the following Figure. The total series resistance is now R O + R L and the cutoff frequency due to C C is determined by: ( ) )15)

Ignoring the effects of C S and C E the O/P voltage V o will be 70.7% of its midband value at f LC. For the network of the loaded BJT amplifier, the ac equivalent network for the O/P section with V i = 0 V appears in the following Figure. The resulting value of R o in the equation of f LC is then simply R o = R C //r o Effect of C E : To determine f LE, the network seen by C E must be determined as shown in the Figure below. Once the level of Re is established, the cutoff frequency due to CE can be determined using the following equation: For the BJT network, the ac equivalent as seen by C E appears in the following Figure. The value of R e is therefore determined by Where ( ) The effect of C E on the gain is best described by recalling that the gain for the configuration )16)

of the shown Figure is given by: The maximum gain is obviously available where R E is zero ohms. The higher f L will be the predominant factor in determining the lowfrequency response for the complete system. Example: For the network shown in Figure with the following parameters: Cs = 10 F, CE = 20 F, Cc = 1 F, Rs = 1 K, R1 = 40 K, R2 = 10 K, RE = 2 K, Rc = 4 K, RL = 2.2 K, = 100, ro =, Vcc = 20 V a) Determine r e b) Find A Vmid = V o /V i c) Calculate Z i. d) Find A Vmid = V o /V s. e) Determine f Ls, f Lc, and f LE. f) Determine the lower cutoff frequency. g) Sketch the asymptotes of the Bode plot defined by the cutoff frequencies. h) Sketch the low-frequency response for the amplifier using results of (f). a) Determine re for dc conditions: R E = (100)(2 K ) = 200 K >> 10 R 2 = 100 K, check V B = 4 V, I E = 1.65 ma, re 15.76 )17)

And r e = 100 (15.76 ) = 1576 = 1.576 K b) Midband gain, check -90 c) = 40K//10K//1.576K 1.32 K d) From the Figure shown: So that ( )( ) e) Effect of C s R i = R 1 //R 2 // r e, check Ri = 1.32 K Effect of C C ( ), check 6.86 Hz ( ), check 25.68 Hz Effect of C E = 1 K// 40 K // 10 K = 0.889 K ( ) 24.35 = ( )( )( ) 327 Hz f) f L = 327 Hz )18)

g), h) The low frequency plot for the network: Low-Frequency Response (FET) Amplifier The analysis is quite similar to that of the BJT amplifier. There are again capacitors C G, C C, and C S. The following Figure is used to establish the fundamental equations. Effect of C G : The ac equivalent network for the coupling capacitor between the source and the active device is shown in the following Figure. The cutoff frequency determined by CG will be: ( ) For the above network, )19)

Typically R G >> R sig and the lower cutoff frequency will be determined primarily by R G and C G. Effect of C C : For the coupling capacitor between the active device and the load the network of following Figure will result. The resulting cutoff frequency is For the given network, ( ) Effect of C S : For the source capacitor C S the resistance level of importance is defined by the following Figure. The cutoff frequency will be defined by For the above network, the resulting value of R eq ( ) ( ) Which for r d becomes Example: (a) Determine the lower cutoff frequency for the network of the above Fig, 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, g m = 2 ms I DSS = 8 ma, V P = -4 V, r d =, V DD = 20 V. (b) Sketch the frequency response using Bode plot. )21)

(a) Effect of C G : Using R sig =10 K, R i = R G = 1 M, check f LG 15.8 Hz Effect of C C : Using R O = R D //r d = 4.7 K, R L = 2.2 K, check f LC 46.13 Hz Effect of C C : Using R eq = R S //(1/g m ), check R eq = 333.33,and f Ls = 238.73 Hz (b) The midband gain is determined by: ( ), check A Vmid -3. Using the midband gain to normalize the response will result in the frequency plot of Figure. )21)

Miller Effect Capacitance: In high-frequency region, the capacitive elements of importance are the inter-electrode (between terminals) capacitances internal to the active device and the wiring capacitance between leads of the network. For inverting amplifiers, the I/P and O/P capacitance is increased by a capacitance level sensitive to the inter-electrode capacitance between the I/P and O/P terminals of the device and the gain of the amplifier. In Figure, this "feedback" capacitance is defined by C f. Miller I/P Capacitance: Applying KCL: I i = I 1 + I 2 Using Ohm's law: Substituting, we obtain Establishing the equivalent network shows that the I/P impedance includes R i with the addition of a feedback capacitor magnified by the )22)

gain of the amplifier. In general the Miller effect I/P capacitance is defined by C Mi = (1 - A v )C f This shows that: For any inverting amplifier, the I/P capacitance will be increased by a Miller effect capacitance sensitive to the gain of the amplifier and the inter-electrode (parasitic) capacitance between the I/P and O/P terminals of the active device. The reason for the constraint that the amplifier be of the inverting type is now more apparent when you examine the equation of C MI. A positive A v would result in a negative capacitance (for A v > 1). Miller O/P Capacitance: Applying KCL: I o = I 1 + I 2 Using Ohm's law: The resistance R o is usually sufficiently large to permit ignoring I 1,, substituting V i = V o /A v will result in Or Resulting in the Miller O/P capacitance: ( ) )23)

For the usual situation where A v >> 1, this equation reduces to C MO = C f High-Frequency Response (BJT) Amplifier: Network Parameters: In the high-frequency region, the RC network of concern has the configuration appearing in Figure, the general form of A v : ( ) This results in a magnitude plot that drops off at 6 db/ octave with increasing frequency. The high-frequency model for the network of the following Figure appears in the next Figure. (BJT network with capacitors that affect the high frequency response) )24)

In fact, most specification sheets provide levels of C be and C bc and do not include C ce. Determining the Thevenin equivalent circuit for the I/P and O/P networks of the above Figure will result in the configuration of the following Figures. For the I/P network, the -3 db frequency is defined by With And ( ) At very high frequencies, the effect of C i is to reduce the total impedance of the parallel combination of R 1, R 2, Ri and C i, which results in a reduced level of voltage across C i, a reduction in I b, and gain of the system. )25)

For the O/P network, And At very high frequencies, C o will decrease reducing the total impedance of the O/P parallel branches of the equivalent model. The net result is that V o will also decline toward zero as the reactance X C becomes smaller. The frequencies f Hi and f Ho will each define a -6 db/octave asymptote. )26)

Example For the network shown in the Figure with the following parameters: R s = 1 K, R 1 = 40 K, R 2 = 10 K, R E = 2 K, R C = 4 K, R L = 2.2 K, C S = 10 F, C C = 1 F, C E = 20 F, C be = 36 pf, C bc = 4 pf, C Ce = 1 pf, C Wi = 6 pf, C Wo = 8 pf, = 100, r o =, V CC = 20 V. a) Find the I/P resistance R i b) Find the voltage gain A vmid. c) Determine the Thevenin equivalent I/P resistance R thi. d) Determine the Thevenin equivalent O/P resistance R tho. e) Find the input capacitance C i. f) Find the output capacitor C o. g) Determine f Hi and f Ho. a) Get V B, check V B = 4 V, get r e = 15.76. r e = 1.576 K, R i = R 1 //R 2 // r e, check Ri = 1.32 K b), check -90 c) R thi = R s // R 1 // R 2 // R i, check 0.531 K d) R tho = R C // R L, check R tho = 1.419 K )27)

e) ( ), check C i = 406 pf f) = 8 pf + 1 pf + [1 (-1/90)]4 pf = 13.04 pf g), check f Hi = 738.24 KHz h), check f Ho = 8.6 MHz In general, the lowest of the upper-cutoff frequencies defines a maximum possible bandwidth for the system. )28)

High-Frequency Response FET Amplifier The shown network is an inverting amplifier, so Miller effect capacitance will appear in the high-frequency ac equivalent network. Thevenin's i/p circuit Thevenin's o/p circuit For the i/p circuit f Hi = 1/2 R thi C i R thi = R sig //R G C i = C wi + C gs + C Mi C Mi = (1 A v )C gd For the o/p circuit f Ho = 1/2 R tho C o R tho = R D //R L //r d C o = C wo + C ds + C Mo C Mo = (1 1/A v )C gd )29)

Example: Determine the high cutoff frequency for the network of the above Fig, 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, g m = 2 ms I DSS = 8 ma, V P = -4 V, r d =, V DD = 20 V. Cgd = 2 pf, Cgs = 4 pf, Cds = 0.5 pf, Cwi = 5 pf, Cwo = 6 pf. A v = -g m (R D //R L ), check A v = -3. R thi = R sig //R G = 9.9 K. C i = C wi + C gs + (1-A v )C gd, check = 17 pf. f Hi = 1/(2 R thi C i ) = 945.67 KHz R tho = R D //R L 1.5 K C o = C wo + C ds + C MO, check = 9.17 pf f Ho = 1/(2 )(1.5K )(9.17pF) = 11.57 MHz From which it is noticed that the i/p capacitance with its Miller capacitance will determine the upper cutoff frequency. )31)