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LECTURE 4: BJT AMPLIFIER FREQUENCY RESPONSE By: Syahrul Ashikin Azmi PPKSE Lecture’s content Objectives: – Discuss the effect of circuit capacitances on frequency response of an amplifier – Analyze low and high-frequency response of amplifiers Topic to be covered: – – – – – Basic concepts The Decibel Low-frequency amplifier response High-frequency amplifier response Total amplifier frequency response Basic concepts Assumption: Coupling capacitors and bypass capacitors act as short circuits to the signal voltages and open circuits to dc voltages. Capacitors do not change instantaneously from a short circuit to an open circuit as the frequency approaches zero. Assumption: transistors are ideal in that output signal respond instantaneously to input signals. There are internal capacitances in bipolar transistor. Frequency response of amplifier circuits due to circuit capacitors and transistor capacitances. The Decibel The decibel is a common unit of measurement of voltage gain and frequency response. It is a logarithmic measurement of the ratio of one power to another or one voltage to another. The formulas below are used for calculation of decibels for power gain and voltage gain. Ap(db) = 10 log Ap Av(db) = 20 log Av If Av is > 1, dB gain is +ve. If Av is < 1, dB gain is –ve and usually called attenuation. The Decibel -0 dB reference0 dB (actual voltage gain =1) used as a reference gain with which to compare other values of gains. Maximum gain occurs in range of frequencies between upper and lower critical frequencies; is called midrange gain and assigned as 0 dB value. Any value of gain below midrange is expressed as a –ve dB value. If midrange Av is 100, gain at a certain frequency below midrange is 50, this reduced Av is expressed as 20log(50/100)= -6 dB. This indicates 6 dB below 0 dB reference. Halving o/p voltage for a steady i/p voltage is always 6 dB reduction in gain. Doubling o/p voltage is always 6 dB increase in gain. The Decibel -0 dB referenceFigure shows a normalized gain vs frequency curve for several dB points. Normalized means midrange voltage gain is assigned a value of 1 or 0 dB. The Decibel -Critical frequency & power measurement in dBmThe critical frequency, also known as the cutoff frequency or corner frequency, is the frequency at which the output power drops by 3 dB, which represents one-half of its midrange value. Ap(dB) = 10 log(0.5) = -3dB An output voltage drop of 3 dB represents about a 70.7% drop from the midrange value. Av(dB) = 20 log(0.707) = -3dB Power is often measured in units of dBm. This is decibels with reference to 1 mW of power. This means that 0 dBm = 1 mW. Each 3dBm increase corresponds to doubling of power, 3 dBm decrease corresponds to a halving of the power. Basic concepts cont.. -Amplifier gain vs frequencyCc and CE short cct & Cπ and Cμ open cct. Gain almost constant. Midband Gain falls of due to the effects of C and C Gain falls of due to the effects of CC and CE -Amplifier gain vs frequencyLOW FREQUENCY RANGE gain decreases as the frequency decreases due to coupling and bypass capacitor effects. HIGH FREQUENCY RANGE gain decreases as the frequency increases due to stray capacitance and transistor capacitance effects. MIDBAND RANGE gain is almost constant – coupling and bypass capacitors act as short circuits and stray and transistor capacitances act as open circuits. Definitions Frequency response of an amplifier is the graph of its gain versus the frequency. Cutoff frequencies : the frequencies at which the voltage gain equals 0.707 of its maximum value. Midband : the band of frequencies between 10f1 and 0.1f2. The voltage gain is maximum. Bandwidth : the band between upper and lower cutoff frequencies Basic concepts cont.. -Low Frequency RangeAt low frequency range, the gain falloff due to coupling capacitors and bypass capacitors. As signal frequency , the XC - no longer behave as short circuits. Short-circuit time-constant method (SCTC) To determine the lower-cutoff frequency having n coupling and bypass capacitors: n 1 L i 1 RiS Ci RiS = resistance at the terminals of the ith capacitor Ci with all the other capacitors replaced by short circuits. Basic concepts cont.. -Low Frequency RangeCoupling capacitors C1 and C3 limit the passage of very low frequencies. Emitter bypass C2 capacitor will have high reactance to low frequencies as well, limiting the gain. Also the capacitance causes a phase shift of the signal. Low frequency amplifier response Assume CC and CE are short-cct, midrange Av is: R1 R2 r Vo ro RC RL Av g m R1 R2 r RS Vs Low frequency amplifier response cont.. At the low frequency ac equivalent circuit of a capacitor coupled amplifier, we can see there are three RC circuits that will limit low frequency response. The input at the base, the output at the collector, and the emitter. Input RC Circuit Output RC Circuit Bypass RC Circuit Low frequency amplifier response cont.. -Input RC Circuit- Rin V in Vbase As frequency decreases, 2 2 R X in C1 XC1 increases less base voltage due to more voltage drop across C1. Thus, voltage gain is reduced. A critical point occurs when o/p voltage is 70.7% of its midrange value. This condition occur when XC1=Rin. Rin Rin Vin 0.707Vin Vbase Vin 2R R2 R2 in Vbase in in 20 log( 0.707 ) 3dB 20 log Vin Low frequency amplifier response cont.. -Input RC CircuitThe frequency at which the gain is down by 3 dB is called the lower critical frequency (fcl). This frequency can be determined by the formula below. X C1 1 2f cl ( input )C1 Rin 1 1 f cl ( input ) @ 2RinC1 2( RS Rin )C1 Rin @ Ri R1 R2 r Resistance of input source taken into account fcl also known as lower cutoff frequency, lower corner frequency or lower break frequency. Low frequency amplifier response cont.. -Input RC CircuitThe decrease in voltage gain with frequency is called the roll-off. A ten times change in frequency is called a decade. For each ten times reduction in frequency below fc, there is a 20 dB reduction in voltage gain. The attenuation measured in dB at each decade is dB/decade. Low frequency amplifier response cont.. -Input RC CircuitThis typical plot of dB Av vs frequency is called Bode plot. From Bode plot, it is flat (0dB) down to critical frequency, at which point gain drop at -20dB/ decade. Above fc are the midrange frequencies. Sometimes roll-off is expressed in dB/octave, which is a doubling or halving of the frequency. Midrange Low frequency Low frequency amplifier response cont.. -Input RC CircuitInput RC circuit also causes an increasing in phase shift through amplifier. At midrange, phase shift is approximately zero since XC1=0Ω. At lower frequencies, higher values of XC1 cause a phase shift and o/p voltage of RC circuit leads i/p voltage. 1 X C 1 Phase shift in input RC circuit is: tan Rin At critical frequency, XC1=Rin, so: 1 Rin tan 1 ( 1 ) 450 tan Rin Low frequency amplifier response cont.. -Input RC CircuitAs phase shift approach 90ο, frequency approaches zero. The input RC circuit causes base voltage to lead input voltage below midrange by an amount equal to circuit phase angle, θ. Low frequency amplifier response cont.. -Output RC CircuitThe output RC circuit affects the response similarly to the input RC circuit. The formula below is used to determine the cutoff frequency of the output circuit. f cl ( output ) 1 2( RC ro RL )C3 Low frequency amplifier response cont.. -Output RC CircuitPhase angle in o/p RC circuit is: X C3 tan RC RL 1 θ≈0o for midrange frequencies and approaches 90o as frequency approaches zero (XC3 ∞). At critical frequency, phase shift =45o. Low frequency amplifier response cont.. -Bypass RC CircuitAt low frequencies, XC2 is in parallel with RE creates an impedance that reduces the voltage gain. Example 1 VCC = 12V Given : Q-point values : 0.121mA, 11.27V R1 93.7k = 100, VA = 100V RS Therefore, r = 21.49k, ro =826k 0.5 k vS RC C2 6 k C1 vO 0.1 F RL 2 F R2 6.3k 100 k RE 0.4 k C3 10 F Low-frequency ac equivalent circuit C3 vo RS C1 RC vs RB RE C2 RL Circuit for finding R1S RinCE RS R1S RC RL Replacing C2 and C3 by short circuits RB R1 S RS RB RinCE RS RB r 500 5900 2740 2371 RB R1 R2 5900 1 1 211 rad / s R1 S C1 2.371 k 2.00F Circuit for finding R2S RoutCE Replacing C1 and C3 by short circuits R2S RC RS RL RB R2 S RL RC RoutCE RL RC ro 100 k 6k 105k 106 k 1 1 94.63 rad / s R2 S C 2 106 k 0.100F Circuit for finding R3S RTH Replacing C1 and C2 by short circuits RC||RL RS RB RE RoutCC R3 S RE RoutCC RE r RTH 2740 460.94 400 29.37 1 101 RTH RS RB 460.94 1 1 3405 rad / s R3 S C 3 29.37 10F R3S Estimation of L 3 L i 1 1 211 94.63 3405 3710.63 rad / s RiS C i L fL 590.57 Hz 2 Total low-frequency response of amplifier Each RC circuit has a critical frequency determined by R and C values. If one of RC circuits has a critical frequency higher than the other two it is called dominant RC circuit. The dominant circuit determines the frequency at which the overall gain begins to drop at -20dB/decade. The other circuits each cause an additional 20dB/decade roll-off below their respective critical frequencies. Total low-frequency response of amplifier Refer to figure below, input RC circuit is dominant and bypass RC circuit has the lowest fc. The overall response is shown in blue line. Example 2 VCC = 10V Given : = 100, VA = 70 V R1 62 k Therefore, RS r = 1.62 k, ro = 43.75 k, gm = 61.54 mS 2.2 k C1 C2 vO 0.1 F RL 600 0.1 F vS R2 22 k Determine the total lowfrequency response of the amplifier. RC 10 k RE 1.0 k C3 10 F Example 2 Low frequency due to C1 and C2 C3 Low frequency due to C1 R1S RS RB r 600 16.24k 1.62k 2.07 k RB R1 R2 16.24 k f C1 1 1 768.86 Hz 769 Hz 2R1S C1 2 2.07 k 0.1F Low frequency due to C2 R2S RL RC ro 10 k 2.2k 43.75k 12.09 k f C2 1 1 131.64 Hz 132 Hz 2R2 S C2 2 12.09 k 0.1F Example 2 Low frequency due to C3 Low frequency due to C3 r RTH 1.62 k 0.58 k R3S RE 1k 21.32 1 101 RTH RS RB 0.58 k f C3 1 1 746.5 Hz 747 Hz 2R3S C3 2 21.32 10F -High Frequency Range Internal capacitance of BJTs and FETs comes into play at high frequencies limiting the gain. Remember reactance is low at high frequencies. The gain falls off at high frequency end due to the internal capacitances of the transistor. Transistors exhibit charge-storage phenomena that limit the speed and frequency of their operation. Small capacitances exist between the base and collector and between the base and emitter. These effect the frequency characteristics of the circuit. Cμ = Cbc ------ 2 pF ~ 50 pF Cπ = Cbe ------ 0.1 pF ~ 5 pF High frequency BJT amplifier response High-frequency response is limited by internal capacitances of the transistors. These act like shunts around the transistor. Miller’s theorem allows us to view the internal capacitances as external capacitors for better understanding of the effect they have on the frequency response. Miller’s Theorem This theorem simplifies the analysis of feedback amplifiers. The theorem states that if an impedance is connected between the input side and the output side of a voltage amplifier, this impedance can be replaced by two equivalent impedances, i.e. one connected across the input and the other connected across the output terminals. Miller theorem General case of Miller input and output capacitances. Miller equivalent circuit I1 V1 V2 Z V2 A V1 I1 V1 (1 A) Z Z I1 I2 I2 V2 V1 Z V2 A V1 -A V1 Z 1 A V1 V2 I2 I2 Input 1 V2 1 A Z V2 Z 1 1 A Output Miller equivalent circuit I2 I1 V1 Z 1 A V1 I1 Z 1 A ZM1 Z 1 A Input -A V 1 ZM1 ZM2 V 2 V2 I2 Z 1 1 A Z 1 1 A Z 1 1 A ZM 2 V2 Output Miller Capacitance Effect C I1 ZM1 I2 Z 1 A Z 1 1 A X CM 2 XC 1 1 A 1 CM 2 1 CM 2 -A V1 X CM 1 ZM 2 V2 XC 1 A 1 CM 1 1 C (1 A ) CM 1 C (1 A) Input -A V1 CM1 CM2 V2 C (1 C (1 Output 1 ) A 1 ) A High-frequency hybrid- model C B C + r V C gmV ro - C = Cbe E C = Cbc High-frequency hybrid- model with Miller effect B C r C CMi gmV ro CMo E CMi C 1 A Cbc 1 A 1 1 CMo C 1 Cbc 1 A A Cout CMo Cin C CMi A : midband gain Example 3 Given : VCC = 10V = 125, Cbe = 20 pF, Cbc = 2.4 pF, VA = 70V, VBE(on) = 0.7V RC Determine : R1 i-Upper cutoff frequencies ii- Dominant upper cutoff frequency vS 22 k RS 2.2 k C1 C2 vO 10 F RL 600 10 F 2.2 k R2 4.7 k RE 470 C3 10 F Example 3 High-frequency hybrid- model with Miller effect for CE amplifier Ro Ri RS vs vo R1||R2 C CMi r ro gmV CMo RC||RL R1 R2 r r R R 56.36 midband gain A gm RS R1 R2 r o C L CMi Cbc 1 A 2.4 p 57.36 137.66 pF CMo Miller’s equivalent capacitor at the input 1 Miller’s equivalent Cbc 1 2.4 p 1.018 2.44 pF capacitor at the output A Example 3 Ri RS R1 R2 r 600 22k 4.7k 1.55k 389.47 Ro RC RL ro 2.2k 2.2k 47.62k 1.08k resistance at the input Thevenin’s equivalent resistance at the output Cin Cbe CMi 20 p 137.66 p 157.66 pF Cout CMo 2.44 pF Thevenin’s equivalent total input capacitance total output capacitance 1 1 f cu ( input ) 2.59MHz 2RiCin 2389.47157.66 p upper cutoff frequency introduced by input capacitance 1 1 upper cutoff frequency fcu ( output ) 60.39MHz introduced by output 2RoCout 21.08k 2.44 p capacitance Total high-frequency response of BJT amplifier The lowest of the two values of upper cutoff frequencies is the dominant frequency. Therefore, the upper cutoff frequency of the amplifier in previous example is: f H 2.59 MHz Refer to Bode plot in next slide, at fcu(input) voltage gain begins to roll-off at -20dB/decade. At fcu(output), gain dropping to -40dB/decade because each RC circuit provide a -20dB/decade roll-off. Total high-frequency response of BJT amplifier The Bode plot of the high frequency response shown shows the combined effects of each internal capacitance. Total Amplifier Frequency Response -3 break points at lower critical frequencies (fc1,fc2 and fc3) produced by 3 low-frequency RC circuits formed by Cc and CE -2 break points at upper frequencies fc4 and fc5 produced by 2 highfrequency RC circuits formed by transistor’s internal capacitance Total Amplifier Frequency Response -Bandwidth The range of frequencies lying between fcl(dom) and fcu(dom) bandwidth of amplifier. Only dominant cutoff frequency appear in response curve because they determine the bandwidth. The amplifier’s bandwidth expressed in hertz as: BW f cu ( dom ) f cl ( dom ) Example 4 Total Frequency Response of CE Amplifier VCC = 5V Given : = 120, Cbe = 2.2 pF, Cbc = 1 pF, VA = 100V, VBE(on) = 0.7V R1 33 k Determine : i-lower and upper cutoff frequencies RS 2 k ii- midband gain vS RC C2 4 k C1 2 F RL 1 F R2 22 k vO 5 k RE 4 k C3 10 F Example 4 Q-point values VBB VBE (on) IB 2.615A RB 1RE VBB R2 VCC 2V R1 R2 R1 R2 RB 13.2 k R1 R2 I CQ I B 0.314 mA Example 4 Transistor parameters value r VT I CQ 9.94 k VA ro 318.47 k I CQ gm I CQ VT 12.08 mS Example 4 Midband gain Amid g m R R S r B r RB r r o RC RL RB 9.94k 13.2k 5.67k RS r RB 2k 9.94k 13.2k 7.67k r o RC RL 318.47k 2.22k 2.18k 5.67k 2.18k 19.47 Amid 12.08m 7.67k Example 4 Lower cutoff frequency Due to C1 1 1 130.38 rad / s R1S C1 R1S RS RB r 7.67 k Due to C2 1 2 55.87 rad / s R2 S C2 R2S RL RC ro 8.95 k Due to C3 1 3 1060.9 rad / s R3S C3 R3S RE r R 3 SCTC method S RB 1 L 1 2 3 1247.15 rad / s i 1 Lower cutoff frequency L fL 198.49 Hz 2 94.26 Example 4 Upper cutoff frequency Miller Capacitance CMi Cbc 1 A 1 p 20.47 20.47 pF CMo 1 Cbc 1 1 p 1.051 1.05 pF A Cin Cbe CMi 22.67 pF Cout CMo 1.05 pF Input & output resistances Ri RS R1 R2 r 1.48 k Ro RC RL ro 2.18 k Example 4 upper cutoff frequency Input side Output side f Hi 1 1 4.74 MHz 2Ri Cin 2 1.48 k 22.67 p f Ho 1 1 69.53MHz 2RoCout 2 2.18k 1.05 p Upper cutoff frequency (the smallest value) f H 4.74MHz Summary The coupling and bypass capacitor of amplifier affect the low-frequency response. The internal capacitances affect high-frequency response. Each RC circuit causes voltage gain to drop at a rate of 20dB/decade. For low frequency RC circuit, the highest critical frequency is the dominant critical frequency. For high frequency RC circuit, the lowest critical frequency is the dominant critical frequency. The bandwidth of amplifier is the range of frequencies between dominant lower critical frequency and dominant upper critical frequency.