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FEEDBACK AMPLIFIERS Xs + Xi Xo Xf βf * What is negative feedback? * * * * Adding the feedback signal to the input so as to partially cancel the input signal to the amplifier. What is feedback? Taking a portion of the signal arriving at the load and feeding it back to the input. Doesn’t this reduce the gain? Yes, this is the price we pay for using feedback. Why use feedback? Provides a series of benefits, such as improved bandwidth, that outweigh the costs in lost gain and increased complexity in amplifier design. Q. Why do you think that this type of feedback is called NEGATIVE? 1 Feedback Amplifier Analysis Xs + Xi Xo Xf βf X f = β f Xo where β f is called the feedback factor X o = AX i where A is the amplifier ' s gain , e.g . voltage gain Xi = Xs − X f where X i is the net input signal to the basic amplifier , X s = the signal from the source The amplifier ' s gain with feedback is given by Af = Xo AX i = Xs Xi + X = f 1+ A X A = f Xi 1+ β f Xo Xi = A 1+ β f A <A 2 Advantages of Negative Feedback * Gain desensitivity - less variation in amplifier gain with changes in β (current gain) of transistors due to dc bias, temperature, fabrication process variations, etc. * Bandwidth extension - extends dominant high and low frequency poles to higher and lower frequencies, respectively. ωL ω Hf = 1 + β f A ω H ω Lf = 1+ β f A ( ) ( ) * Noise reduction - improves signal-to-noise ratio * Improves amplifier linearity - reduces distortion in signal due to gain variations due to transistors Ì Ì Disadvantages of Negative Feedback Loss of gain, may require an added gain stage to compensate. Added complexity in design 3 Basic Types of Feedback Amplifiers * There are four types of feedback amplifiers. Why? Ì Output sampled can be a current or a voltage Ì Quantity fed back to input can be a current or a voltage Ì Four possible combinations of the type of output sampling and input feedback * One particular type of amplifier, e.g. voltage amplifier, current amplifier, etc. is used for each one of the four types of feedback amplifiers. * Feedback factor βf is a different type of quantity, e.g. voltage ratio, resistance, current ratio or conductance, for each feedback configuration. * Before analyzing the feedback amplifier’s performance, need to start by recognizing the type or configuration. * Terminology used to name types of feedback amplifier, e.g. Series-shunt Ì First term refers to nature of feedback connection at the input. Ì Second term refers to nature of sampling connection at the output. 4 Basic Types of Feedback Amplifiers Series - Shunt Shunt - Series Series - Series Shunt - Shunt 5 Method of Feedback Amplifier Analysis * Recognize the feedback amplifier’s configuration, e.g. Series-shunt * Calculate the appropriate gain A for the amplifier, e.g. voltage gain. Ì This includes the loading effects of the feedback circuit (some combination of resistors) on the amplifier input and output. * Calculate the feedback factor βf * Calculate the factor βf A and make sure that it is: 1) positive and 2) dimensionless A * Calculate the feedback amplifier’s gain with feedback Af using A f = 1 + β f A * Calculate the final gain of interest if different from the gain calculated, e.g. Current gain if voltage gain originally determined. * Determine the dominant low and high frequency poles for the original amplifier, but taking into account the loading effects of the feedback network. * Determine the final dominant low and high frequency poles of the ωL amplifier with feedback using ( ) ω Hf = 1 + β f A ω H ω Lf = (1 + β f A) 6 Series-Shunt Feedback Amplifier - Ideal Case * Assumes feedback circuit does not load down the basic amplifier A, i.e. doesn’t change its characteristics Doesn’t change gain A Doesn’t change pole frequencies of basic amplifier A Doesn’t change Ri and Ro * For the feedback amplifier as a whole, feedback does change the midband voltage gain from A to Af Basic Amplifier Feedback Circuit Af = * A 1+ β f A Does change input resistance from Ri to Rif ( Rif = Ri 1 + β f A Equivalent Circuit for Feedback Amplifier * Does change output resistance from Ro to Rof Rof = * ) Ro 1+ β f A Does change low and high frequency 3dB frequencies ( ) ω Hf = 1 + β f A ω H ω Lf = ωL 1+ β f A ( ) 7 Series-Shunt Feedback Amplifier - Ideal Case Midband Gain V A V AVf = o = V i = Vs Vi + V f Input Resistance AV AV AV = = Vf β f Vo 1 + β f AV 1+ 1+ Vi Vi Vi + V f Vi + β f Vo V = Rif = s = = Ri 1 + β f AV Ii Ii ⎛Vi ⎞ ⎜ R ⎟ i⎠ ⎝ ( ) Output Resistance It Vt V − AV Vi It = t Ro But Vs = 0 so Vi = −V f and V f = β f Vo = β f Vt so It = = ( ) ( Vt − AV − V f Vt + AV β f Vt = Ro Ro Vt 1 + AV β f ( ) Ro V Ro so Rof = t = It 1 + AV β f ( 8 ) ) Practical Feedback Networks * Vi Vo Vf * * * * How do we take these loading effects into account? * Feedback networks consist of a set of resistors Ì Simplest case (only case considered here) Ì In general, can include C’s and L’s (not considered here) Ì Transistors sometimes used (gives variable amount of feedback) (not considered here) Feedback network needed to create Vf feedback signal at input (desirable) Feedback network has parasitic (loading) effects including: Feedback network loads down amplifier input Ì Adds a finite series resistance Ì Part of input signal Vs lost across this series resistance (undesirable), so Vi reduced Feedback network loads down amplifier output Ì Adds a finite shunt resistance Ì Part of output current lost through this shunt resistance so not all output current delivered to load RL (undesirable) 9 Series-Shunt Feedback Amplifier - Practical Case * Feedback network consists of a set of resistors * These resistors have loading effects on the basic amplifier, i.e they change its characteristics, such as the gain * Can incorporate loading effects in a modified basic amplifier. Basic gain of amplifier AV becomes a new, modified gain AV’ (incorporates loading effects). * Can then use feedback analysis from the ideal case. = Vf Vo AVf = =βf AV ' 1 + β f AV ' Rif = Ri (1 + β f AV ') Rof = Ro (1 + AV ' β f ) 10 Series-Shunt Feedback Amplifier - Practical Case Summary of Feedback Network Analysis 11 Example - Series-Shunt Feedback Amplifier * * Two stage amplifier Each stage a CE amplifier * Transistor parameters Given: β1= β2 =50 * Coupled by capacitors, dc biased separately * DC analysis: I IC1 = 0.94 mA, gm1 = C1 = 36 mA/ V , VT DC analysis for each stage can be done separately since stages are isolated (dc wise) by coupling capacitors. rπ1 = β1 gm1 = 1.4K I IC2 = 1.85 mA, gm2 = C2 = 71 mA/ V , VT rπ 2 = β2 gm2 = 0.7K 12 Example - Series-Shunt Feedback Amplifier * Redraw circuit to show Ì Ì Ì Feedback circuit Type of output sampling (voltage in this case = Vo) Type of feedback signal to input (voltage in this case = Vf) + _ Vi + Vf _ + Vo _ 13 Example - Series-Shunt Feedback Amplifier Input Loading Effects R1 = h11 = R f 1 R f 2 = 0.1K 4.7 K Vo=0 = 0.098 K Output Loading Effects R2 = h22 = R f 1 + R f 2 I1=0 = 4.7 K + 0.1K = 4.8K Amplifier with Loading Effects R2 R1 14 Example - Series-Shunt Feedback Amplifier * * Construct ac equivalent circuit at midband frequencies including loading effects of feedback network. Analyze circuit to find midband gain (voltage gain for this series-shunt configuration) R1 R2 R2 R1 15 Example - Series-Shunt Feedback Amplifier Midband Gain Analysis A Vo = ⎛ Vo Vo = ⎜⎜ Vs ⎝ Vπ 2 ⎞⎛ Vπ 2 ⎟⎜ ⎟⎜ V ⎠ ⎝ o1 ( Vo = − g m 2 RC 2 R2 Vπ 2 ( V o1 = − g m 1 R C 1 R 12 Vπ 1 )= ⎞ ⎛ V o1 ⎟⎜ ⎟⎜ V ⎠⎝ π 1 ⎞⎛ Vπ 1 ⎟⎜ ⎟⎜ V ⎠ ⎝ i1 ⎞ ⎛ V i1 ⎟⎜ ⎟⎜ V ⎠⎝ s ⎞ ⎟ ⎟ ⎠ ( − 71 mA / V 4 . 9 K 4 . 8 K R 22 rπ 2 )= )= − 172 ( Vπ 2 = 1 since V o1 − 36 mA / V 10 K 47 K 33 K 0 . 7 K )= rx 2 = 0 − 22 . 8 Vπ 1 I π 1 rπ 1 rπ 1 1 .4 K = = = = 0 . 22 V i1 I π 1 rπ 1 + (I π 1 + g m 1V π 1 )R 1 rπ 1 + (1 + β )R 1 1 . 4 K + 51 (0 . 098 K ) V i1 I r + (I π 1 + g m 1V π 1 )R 1 R i1 = = π1 π1 = rπ 1 + (1 + β )R 1 = 1 . 4 K + 51 (0 . 098 K ) = 6 . 4 K Iπ 1 Iπ 1 6 . 4 K 150 K 47 K R i 1 R 11 R 21 V i1 = = Vs R s + R i 1 R 11 R 21 5 K + 6 . 4 K 150 K 47 K A Vo = Vo = Vs (− 172 )(1 )(− A Vo ( dB ) = 20 log 449 22 . 8 )(0 . 22 )(0 . 52 ) = = 5 .4 K 10 . 4 K = 0 . 52 + 449 = 53 dB 16 Midband Gain with Feedback * βf = * Determine the feedback factor βf Xf Xo = Vf Vo ' = Rf1 Rf1 + Rf 2 = 0.1K = 0.021 0.1K + 4.7 K Calculate gain with feedback Avf β f AVo = 449(0.021) = 9.4 AVfo = AVo 449 449 = = = 43.1 1 + β f AVo 1 + 449(0.021) 10.4 AVfo (dB ) = 20 log 43.1 = 32.7 dB * Note Ì βf Avo > 0 as necessary for negative feedback Ì βf Avo is large so there is significant feedback. For βf Avo ¡ 0, there is almost no feedback. Ì Can change βf and the amount of feedback by changing Rf1 and/or Rf2. Ì NOTE: Since βf Avo >> 0 β f AVo = 449(0.021) = 9.4 AVfo ≈ 1 βf = Rf1 + Rf 2 Rf1 =1+ Rf 2 Rf1 AVfo = 1 1 AVo A ≈ Vo = = = 47.6 1 + β f AVo β f AVo β f 0.021 AVfo (dB ) = 20 log 47.6 = 33.6dB 17 Input and Output Resistances with Feedback * Determine input Ri and output Ro resistances with loading effects of feedback network. Ri = RS + RB1 Ri1 Ro = R2 RC 2 = 5K + 38.5K 6.4 K = 10.5K * = 4 .9 K 4 .8 K = 2 .4 K Calculate input Rif and output Rof resistances for the complete feedback amplifier. ( Rif = Ri 1 + β f AVo ) = 10.5 K [1 + 449(0.021)] = 109.5 K Rof = Ro 1 + β f AVo = 2.4 K = 0.23K 10.4 18 Equivalent Circuit for Series-Shunt Feedback Amplifier * * Voltage gain amplifier Modified voltage gain, input and output resistances Ì Included loading effects of feedback network Ì Included feedback effects of feedback network Ì Include source resistance effects * Significant feedback, i.e. βf Avo is large and positive β f AVo = 449(0.021) = 9.4 AVfo = ⎛V ⎞ AVo = 43.1 AVfo = ⎜⎜ o ⎟⎟ = ⎝ VS ⎠ f 1 + β f AVo AVfo (dB) = 20 log 43.1 = 32.7dB ( ) Rif = Ri 1 + β f AVo = 109.5 K Ro Rof = = 0.23K 1 + β f AVo = AVo 1 + β f AVo Rf1 + Rf 2 Rf1 ≈ AVo 1 = β f AVo β f = 1+ Rf 2 Rf1 = 47.6 AVf (dB) ≈ 20 log 47.6 = 33.6dB 19 Series-Series Feedback Amplifier - Ideal Case Voltage fedback to input * Output current sampling * * * Feedback circuit does not load down the basic amplifier A, i.e. doesn’t change its characteristics Doesn’t change gain A Doesn’t change pole frequencies of basic amplifier A Doesn’t change Ri and Ro For this configuration, the appropriate gain is the TRANSCONDUCTANCE GAIN A = ACo = Io/Vi For the feedback amplifier as a whole, feedback changes midband transconductance gain from ACo to ACfo ACo ACfo = 1 + β f ACo Feedback changes input resistance from Ri to Rif ( Rif = Ri 1 + β f ACo ) * Feedback changes output resistance from Ro to Rof * Feedback changes low and high frequency 3dB frequencies ( Rof = Ro 1 + β f ACo ( ) ω Hf = 1 + β f ACo ω H ω Lf = ) ωL 1 + β f ACo ( ) 20 Series-Series Feedback Amplifier - Ideal Case Gain (Transconductance Gain) ACfo = Io A V A ACo ACo = Co i = Co = = Vf β f I o 1 + β f ACo Vs Vi + V f 1+ 1+ Vi Vi Input Resistance Rif = Vs Vi + V f Vi + β f I o = Ri 1 + β f ACo = = Ii Ii ⎛Vi ⎞ ⎜ R⎟ i⎠ ⎝ ( ) Output Resistance V R of = It But V s = 0 so V i = −V f and V f = β f I o = β f I t so V i = − β f I t V = (I t − ACo V i )R o = (I t − ACo (− β f I t ))R o + V - = I t (1 + β f ACo )R o so R of = V V = R o (1 + β f ACo ) It 21 Series-Series Feedback Amplifier - Practical Case * * Feedback network consists of a set of resistors These resistors have loading effects on the basic amplifier, i.e they change its characteristics, such as the gain Ì Feedback factor βf given by = * * Vf Io =βf Can incorporate loading effects in a modified basic amplifier. Gain ACo becomes a new, modified gain ACo’. Can then use analysis from ideal case ACfo = ACo ' 1 + β f ACo ' Rif = Ri '(1 + β f ACo ') ω Hf = (1 + β f ACo ')ω H ω Lf = Rof = Ro ' (1 + β f ACo ') ωL (1 + β f ACo ') 22 Series-Series Feedback Amplifier - Practical Case Ì Find βf from βf = Vf/Io’ 23 Series-Series Feedback Amplifier - Practical Case * * Modified basic amplifier (including loading effects of feedback network) Can now use feedback amplifier equations derived ACfo = ACo ' 1 + β f ACo ' Rif = Ri '(1 + β f ACo ') ω Hf = (1 + β f ACo ')ω H * ω Lf = Rof = Ro ' (1 + β f ACo ') ωL (1 + β f ACo ') Note Ì ACo’ is the modified transconductance gain including the loading effects of RS and RL. Ì Ri’ and Ro’ are modified input and output resistances including loading effects. 24 Example - Series-Series Feedback Amplifier * * * * * Three stage amplifier Each stage a CE amplifier Transistor parameters Given: β1= β2 = β3 =100, rx1=rx2=rx3=0 Coupled by capacitors, dc biased separately DC analysis (given): IC1 = 0.60 mA, gm1 = rπ1 = β1 gm1 = 4.3K IC2 = 1.0 mA, gm2 = rπ 2 = Note: Biasing resistors for each stage are not shown for simplicity in the analysis. β2 gm2 β3 gm3 IC 2 = 39 mA/ V , VT = 2.6K IC3 = 4.0 mA, gm3 = rπ 3 = IC1 = 23 mA/ V , VT I C3 = 156 mA/ V , VT = 0.64K 25 Example - Series-Series Feedback Amplifier * Redraw circuit to show: Ì Feedback circuit Ì Type of output sampling (current in this case = Io) Collector resistor constitutes the load so Io ¡ Ic Emitter current Ie=(β +1) Ib = {(β +1)/ β} Ic ¡Ic = Io Ì Type of feedback signal to input (voltage in this case = Vf) Ic3 ≈ Io Voltage fedback to input Io Output current sampling 26 Example - Series-Series Feedback Amplifier Io Input Loading Effects R1 Output Loading Effects I2=0 R1 = z11 = RE1 [ RF + RE 2 ] = 0.1K [0.64 K + 0.1K ] = 0.088 K R2 I1=0 R2 = z 22 = RE 2 [ RF + RE1 ] = 0.1K [0.64 K + 0.1K ] = 0.088 K 27 Example - Series-Series Feedback Amplifier Voltage fedback to input Io Output current sampling Redrawn basic amplifier with loading effects, but not feedback. R1 R2 28 Example - Series-Series Feedback Amplifier IC3 * * Io= IE3 ≈ IC3 Construct ac equivalent circuit at midband frequencies including loading effects of feedback network. Analyze circuit to find MIDBAND GAIN (transconductance gain ACo for this series-series configuration) ACo = Io Vs Io VS R1 R2 29 Example - Series-Series Feedback Amplifier Midband Gain Analysis I1 I2 I3 Io VS Vi1 Vi3 Ri1 Io Vπ 3 Ri3 ⎛ I ⎞⎛ V ⎞⎛ V ⎞⎛ V Io = ⎜⎜ o ⎟⎟ ⎜⎜ π 3 ⎟⎟ ⎜⎜ i 3 ⎟⎟ ⎜⎜ π 2 Vs ⎝ Vπ 3 ⎠⎝ V i3 ⎠⎝ Vπ 2 ⎠⎝ Vπ 1 g V = m 3 π 3 = g m 3 = 156 mA / V Vπ 3 A Co = Note convention on Io is into the output of the last stage of the amplifier. ⎞⎛ Vπ 1 ⎟⎟ ⎜⎜ ⎠ ⎝ V i1 ⎞ ⎛ V i1 ⎟⎟ ⎜⎜ ⎠⎝ V s ⎞ ⎟⎟ ⎠ 0 . 64 K Vπ 3 I π 3 rπ 3 rπ 3 = = = = 0 . 067 V i3 I π 3 rπ 3 + R 2 (I π 3 + g m 3V π 3 ) rπ 3 + R 2 (1 + g m 3 rπ 3 ) 0 . 64 K + 0 . 088 K (101 ) − g m 2 V π 2 (R C 2 [rπ 3 + R 2 (1 + g m 3 rπ 3 )]) V i3 = = − 39 mA / V (5 K Vπ 2 Vπ 2 − g m 1V π 1 (R C 1 rπ 2 ) Vπ 2 = = − 23 mA / V (9 K 2 . 6 K Vπ 1 Vπ 1 )= Vπ 1 I π 1 rπ 1 4 .3 K = = V i1 I π 1 rπ 1 + R 1 (1 + g m 1 rπ 1 ) 4 . 3 K + 101 (0 . 088 K [0 . 64 K + 101 (0 . 088 K )]) = − 128 − 46 . 4 ) = 0 . 33 V i1 =1 VS A Co = Io = (156 mA / V Vs )(0 . 067 )(− 128 )(− 46 . 4 )(0 . 33 )(1 ) = + 2 . 05 x 10 4 mA / V = 20 . 5 A / V 30 Feedback Factor and Midband Gain with Feedback * Determine the feedback factor βf βf = Xf Xo = Vf ' Io ' = RE1 I f 1 Io ' RE 2 = RE 1 RE1 + RE 2 + RF Io ’ If1 VE2 0.1K = 0.1K = 0.012 K = 12Ω 0.1K + 0.1K + 0.64 K * Calculate gain with feedback ACfo β f ACo = 20.5 A / V (12Ω) = +246 ACfo = * ACo 20.5 A / V 20.5 A / V = = = 0.083 A / V = 83 mA / V 1 + β f ACo 1 + 20.5 A / V (12Ω) 247 VE 2 = I f 1 ( RF + RE1 ) = ( I o '− I f 1 ) RE 2 I f 1 ( RF + RE1 + RE 2 ) = I o ' RE 2 I f1 Io ' = RE 2 RF + RE1 + RE 2 Note Ì βf ACo > 0 as necessary for negative feedback and dimensionless Ì βf ACo is large so there is significant feedback. Ì βf has units of resistance (ohms); ACo has units of conductance (1/ohms) Ì Can change βf and the amount of feedback by changing RE1 , RF and/or RE2. Ì Gain is largely determined by ratio of feedback resistances 1 R + RE 2 + RF 0.1K + 0.1K + 0.64 K 1 ACfo ≈ = E1 = = 84 = 84 mA / V βf RE1RE 2 0.1K (0.1K ) K 31 Input and Output Resistances with Feedback I1 Io Vi1 I1(1+gm1r1) Ro Ri = Ri1 * Determine input Ri and output Ro resistances with loading effects of feedback network. Ri = Ri1 = * Vi1 = rπ 1 + (1 + g m1rπ 1 )R1 Iπ 1 Ro = RC 3 + ∞ = ∞ = 4.3K + 101(0.088 K ) = 13.2 K Calculate input Rif and output Rof resistances for the complete feedback amplifier. ( Rif = Ri 1 + β f ACo ) = 13.2 K [1 + 20.5 A / V (12Ω)] Rof = Ro (1 + β f ACo ) = ∞ = 13.2 K (247 ) = 3.26 MΩ 32 Voltage Gain for Transconductance Feedback Amplifier Io * Can calculate voltage gain after we calculate the transconductance gain! ⎛V ⎞ ⎛−I R ⎞ ⎛I ⎞ AVfo = ⎜⎜ o ⎟⎟ = ⎜⎜ o C 3 ⎟⎟ = − RC 3 ⎜⎜ o ⎟⎟ = − RC 3 ACfo = −600Ω(0.083 A / V ) = −49.8V / V ⎝ Vs ⎠ f ⎝ Vs ⎠ f ⎝ Vs ⎠ f AVfo (dB ) = 20 log(49.8) = 34dB * Note - can’t calculate the voltage gain as follows: Assume AVfo = Find AVo = AVo Correct voltage gain for the amplifier with feedback! 1 + β f AVo Vo − I o RC 3 = = − RC 3 ACo = −600Ω(20.5 A / V ) = −1.23 x10 4 V / V Vs Vs ( ) Calculate β f AVo = 12Ω − 1.23 x10 4 V / V = −1.48 x105 Ω (Note this has units (it should not!) and a negative sign ) Calculate voltage gain with feedback from AVfo = AVo 1 + β f AVo = − 1.23 x10 4 V / V 1 = 0.083 5 Ω 1 − 1.48 x10 Ω Magnitude is off by orders of magnitude and units are wrong! Wrong voltage gain! 33 Equivalent Circuit for Series-Series Feedback Amplifier * Transconductance gain amplifier A = Io/Vs * Feedback modified gain, input and output resistances Ì Included loading effects of feedback network Ì Included feedback effects of feedback network * Significant feedback, i.e. βf ACo is large and positive β f ACo = 20.5 A / V (12Ω) = 246 ACfo = = ⎛I ⎞ ACo ACfo = ⎜⎜ o ⎟⎟ = = 83 mA / V β V 1 + A f Co ⎝ S ⎠f AVfo = RC 3 ACfo = −49.8V / V ( ) ACo ACo 1 ≈ = 1 + β f ACo β f ACo β f 1 = 84 mA / V 0.012 K Rif = Ri 1 + β f ACo = 3.3 MΩ ( ) Rof = Ro 1 + β f ACo = ∞ 34 Shunt-Shunt Feedback Amplifier - Ideal Case * * * * Feedback circuit does not load down the basic amplifier A, i.e. doesn’t change its characteristics Doesn’t change gain A Doesn’t change pole frequencies of basic amplifier A Doesn’t change Ri and Ro For this configuration, the appropriate gain is the TRANSRESISTANCE GAIN A = ARo = Vo/Ii For the feedback amplifier as a whole, feedback changes midband transresistance gain from ARo to ARfo ARo ARfo = 1 + β f ARo Feedback changes input resistance from Ri to Rif Rif = * (1 + β f ARo ) Feedback changes output resistance from Ro to Rof Rof = * Ri Ro (1 + β f ARo ) Feedback changes low and high frequency 3dB frequencies ω Hf = (1 + β f ARo )ω H ω Lf = ωL (1 + β f ARo ) 35 Shunt-Shunt Feedback Amplifier - Ideal Case Gain ARfo = Vo A I A ARo ARo = Ro i = Ro = = If β f Vo 1 + β f ARo I s Ii + I f 1+ 1+ Ii Ii Input Resistance Rif = = Is = 0 Vs Vs Vs = = I s Ii + I f I i + β f Vo Vs ⎛ V ⎞ I i ⎜⎜1 + β f o ⎟⎟ Ii ⎠ ⎝ = Ri (1 + β f ARo ) Output Resistance Io ’ _ + R of = Vo’ V o ' I o ' R o + A Ro I i I = = R o + A Ro i Io ' Io ' Io ' But I s = 0 so I i = − I f and I f = β f V o ' so I i = − β f V o ' − β f Vo ' Ii = = − β f R of Io ' Io ' ( R of = R o + A Ro − β f R of so R of = ) Ro (1 + β f ARo ) 36 Shunt-Shunt Feedback Amplifier - Practical Case * * * Feedback network consists of a set of resistors These resistors have loading effects on the basic amplifier, i.e they change its characteristics, such as the gain Can use y-parameter equivalent circuit for feedback network Ì Feedback factor βf given by = * * If Vo =βf Can incorporate loading effects in a modified basic amplifier. Gain ARo becomes a new, modified gain ARo’. Can then use analysis from ideal case ARfo = Rif = ARo ' 1 + β f ARo ' Ri Rof = (1 + β f ARo ') ω Hf = (1 + β f ARo ')ω H Ro 1 + β f ARo ' ( ω Lf = ) ωL 1 + β f ARo ' ( ) 37 Shunt-Shunt Feedback Amplifier - Practical Case 38 Example - Shunt-Shunt Feedback Amplifier * Single stage CE amplifier * Transistor parameters. Given: β =100, rx= 0 * No coupling or emitter bypass capacitors * DC analysis: VBE ,active = 0.7V 0.7V = 0.07 mA = 70µA I 47 K = I B + 0.07 mA 10K I C = βI B I 4.7 K = I C + ( I B + 0.07 mA) = (β + 1)I B + 0.07 mA I10 K = 12V = I 4.7 K 4.7 K + (I B + 0.07 mA)47 K + 0.7V 12V = ((β + 1)I B + 0.07 mA)4.7 K + (I B + 0.07 mA)47 K + 0.7V 12V − 0.33V − 3.3V = (101(4.7 K ) + 47 K )I B 8.37V = 0.016 mA = 16 µA I C = βI B = 100(0.016 mA) = 1.6 mA 522K 1.6 mA β I 100 = 63 mA / V rπ = = = 1.6 K gm = C = VT 0.0256 V g m 63 mA / V IB = 39 Example - Shunt-Shunt Feedback Amplifier * Redraw circuit to show Ì Ì Ì Feedback circuit Type of output sampling (voltage in this case = Vo) Type of feedback signal to input (current in this case = If) 40 Example - Shunt-Shunt Feedback Amplifier Input Loading Effects R1 = RF = 47 K Output Loading Effects R2 = RF = 47 K 41 Example - Shunt-Shunt Feedback Amplifier Original Feedback Amplifier Modified Amplifier with Loading Effects, but Without Feedback R2 R1 Note: We converted the signal source to a Norton equivalent current source since we need to calculate the gain ARfo = Vo ARo = I s 1 + β f ARo 42 Example - Shunt-Shunt Feedback Amplifier * * Construct ac equivalent circuit at midband frequencies including loading effects of feedback network. Analyze circuit to find midband gain (transresistance gain ARo for this shuntshunt configuration) ARo = Vo Is s 43 Example - Shunt-Shunt Feedback Amplifier Midband Gain Analysis A Ro = Vo = Vπ ⎛ V ⎞⎛ V Vo = ⎜⎜ o ⎟⎟ ⎜⎜ π Is ⎝ Vπ ⎠ ⎝ I s − g m Vπ R C R F ( Vπ ⎞ ⎟ ⎟ ⎠ ) = − g (R m C ) ( ) R F = − 63 mA / V 4 . 7 K 47 K = − 269 V / V ( ) Vπ = ⎛⎜ R S R F rπ ⎞⎟ = 10 K 47 K 1 . 6 K = 1 . 3 K ⎝ ⎠ IS A Ro = Vo = (− 269 )(1 . 3 K ) = − 350 K Is 44 Midband Gain with Feedback * * Determine the feedback factor βf X I ' If ' −1 = βf = f = f = X o Vo ' (− I f R f ) RF + _ V o’ −1 = = − 0.021 mA / V 47 K Calculate gain with feedback ARfo β f ARo = − 0 .021 mA / V ( − 350 K ) = + 7 .4 ARfo * − 350 K ARo = = = − 42 K 1 + β f ARo 1 + 7 .4 Note: The direction of If is always into the feedback network! Note Ì βf < 0 and has units of mA/V, ARo < 0 and has units of K> Ì βf ACo > 0 as necessary for negative feedback and dimensionless Ì βf ACo is large so there is significant feedback. Ì Can change βf and the amount of feedback by changing RF. Ì Gain is determined primarily by feedback resistance ARfo 1 1 ≈ = = − RF = − 47 K β f (−1 / RF ) 45 Input and Output Resistances with Feedback Ro Ri * Determine input Ri and output Ro resistances with loading effects of feedback network. Ri = RS RF rπ = 10 K 47 K 1.6 K = 1.3K * Ro = RC RF = 4.7 K 47 K = 4.3K Calculate input Rif and output Rof resistances for the complete feedback amplifier. Rif = = Ri (1 + β f ARo ) Rof = Ro (1 + β f ARo ) = 4.3K = 0.5 K 8.4 1.3K = 0.15K 8.4 46 Voltage Gain for Transresistance Feedback Amplifier * * Can calculate voltage gain after we calculate the transresistance gain! ARfo − 42 K ⎛ V ⎞ ⎛V ⎞ 1 ⎛V ⎞ = = − 4.2 V / V AVfo = ⎜⎜ o ⎟⎟ = ⎜⎜ o ⎟⎟ = ⎜⎜ o ⎟⎟ = V I R R I R 10 K s ⎝ s ⎠f s ⎝ s ⎠f ⎝ s s ⎠f Correct voltage gain AVfo (dB) = 20 log(4.2 ) = 12.5 dB Note - can’t calculate the voltage gain as follows: Assume Find AVfo = AVo = Calculate AVo 1 + β f AVo Vo V A − 350 K = o = Ro = = − 35 V / V Vs I s Rs Rs 10 K β f AVo = (− 0 .021 mA / V )(− 35 V / V ) = + 0 .74 mA / V (Note this has units; it should not! ) Calculate voltage gain with feedback from AVfo = AVo 1 + β f AVo = − 35 V / V 1 + 0 .74 mA / V Magnitude is off by nearly a factor of five and units are wrong! = − 20 V / V (?) Wrong voltage gain! 47 Equivalent Circuit for Shunt-Shunt Feedback Amplifier * * Rof Rif ARfoI i * ⎛V ⎞ ARo = − 42 K ARfo = ⎜⎜ o ⎟⎟ = ⎝ I S ⎠ f 1 + β f ARo Rif = Rof = Ri (1 + β f ARo ) Ro = (1 + β f ARo ) 1. 3 K = 0.15 K 8.4 = Transresistance gain amplifier A = Vo/Is Feedback modified gain, input and output resistances Ì Included loading effects of feedback network Ì Included feedback effects of feedback network Significant feedback, i.e. βf ARo is large and positive β f ARo = −0.021mA / V (−350 K ) = +7.4 ARfo = AVfo = ARo 1 ≈ = − 47 K 1 + β f ARo β f ARfo RS = − 4.2 V / V 4 .3 K = 0.5 K 8.4 48 Shunt-Series Feedback Amplifier - Ideal Case * Current feedback Current sampling * * * Feedback circuit does not load down the basic amplifier A, i.e. doesn’t change its characteristics Doesn’t change gain A Doesn’t change pole frequencies of basic amplifier A Doesn’t change Ri and Ro For this configuration, the appropriate gain is the CURRENT GAIN A = AIo = Io/Ii For the feedback amplifier as a whole, feedback changes midband current gain from AIo to AIfo AIo AIfo = 1 + β f AIo Feedback changes input resistance from Ri to Rif Rif = * Ri (1 + β f AIo ) Feedback changes output resistance from Ro to Rof ( Rof = Ro 1 + β f AIo * ) Feedback changes low and high frequency 3dB frequencies ( ) ω Hf = 1 + β f AIo ω H ω Lf = ωL 1 + β f AIo ( ) 49 Shunt-Series Feedback Amplifier - Ideal Case Gain AIfo = Io A I = Io i = I s Ii + I f Input Resistance Rif = = AIo AIo AIo = = If β f I o 1 + β f AIo 1+ 1+ Ii Ii Vs Vs Vs = = I s Ii + I f Ii + β f Io Vs ⎛ I ⎞ I i ⎜⎜1 + β f o ⎟⎟ Ii ⎠ ⎝ = Ri (1 + β f AIo ) Output Resistance Is=0 Io ’ ⎛ V o ' R o (I o '− A Io I i ) I ⎞ = = R o ⎜⎜ 1 − A Io i ⎟⎟ Io ' Io ' Io ' ⎠ ⎝ But I s = 0 so I i = − I f R of = Vo’ and I f = β f I o ' so I i = − β f I o ' − β f Io ' Ii = = −β Io ' Io ' ( R of = R o 1 + β f A Io f ) 50 Shunt-Series Feedback Amplifier - Practical Case * * Feedback network consists of a set of resistors These resistors have loading effects on the basic amplifier, i.e they change its characteristics, such as the gain Ì Feedback factor βf given by If Io * * =βf Can incorporate loading effects in a modified basic amplifier. Gain AIo becomes a new, modified gain AIo’. Can then use analysis from ideal case AIfo = AIo ' 1 + β f AIo ' Rif = ω Hf = (1 + β f AIo ')ω H Ri (1 + β ω Lf = f AIo ') Rof = Ro (1 + β f AIo ') ωL (1 + β f AIo ') 51 Shunt-Series Feedback Amplifier - Practical Case 52 Example - Shunt-Series Feedback Amplifier * * * * * Two stage [CE+CE] amplifier Transistor parameters Given: β =100, rx= 0 Input and output coupling and emitter bypass capacitors, but direct coupling between stages Capacitor in feedback connection removes Rf from DC bias DC bias of two stages is coupled (bias of one affects the other) 53 DC Bias Analysis Given : β1 = β 2 = 100 VC1 ( I B1 = VB2 = VC1 = VBE2 + I E 2 RE2 = 0.7V + (β + 1)I B2 RE 2 RB2 15K 12V = 12V =1.6V RB1 + RB2 15K +100K ) IC1 = βI B1 =1008.7 µA = 0.87 mA (neglecting IB2 ) gm1 = IC1 0.87 mA = = 34 mA/V VT 0.0256 V rπ1 = β gm1 3.3V − 0.7V 2.6V = = 7.6 µA (<<IC1 =870 A) (β +1)RE2 101(3.4K ) I B2 = VTH1 −VBE1 1.6V − 0.7V = = 8.7 µA RTH1 + (β +1)RE1 13K + (101)0.87K ( ) VC1 = 12 V − IC1RC1 = 12 V − 0.87 mA10K = 3.3 V RTH1 = RB1 RB2 =100K 15K =13K VTH1 = rx1 = rx 2 = 0 VB2 = 100 = 2.9K 34mA/V ( ) IC 2 = βI B2 = 100 7.6 µA = 0.76 mA gm2 = rπ 2 = IC2 0.76 mA = = 30 mA/ V 0.0256V VT β gm2 = 100 = 3.3K 30 mA/ V 54 Example - Shunt-Series Feedback Amplifier * Redraw circuit to show Ì Feedback circuit Ì Type of output sampling (current in this case = Io) Ì Type of feedback signal to input (current in this case = If) Iout’ Io ’ Iout’ 55 Example - Shunt-Series Feedback Amplifier Iout’ Io ’ Input Loading Effects Io’= 0 Output Loading Effects R1 R1 = RF + R1 = 10 K + 3.4 K = 13.4 K R2 R2 = RF RE 2 = 10 K 3.4 K = 2.5 K 56 Example - Shunt-Series Feedback Amplifier Amplifier with Loading Effects but Without Feedback 57 Example - Shunt-Series Feedback Amplifier Midband Gain Analysis Ri2 Is IC’ Iout’ Vi2 R1 R2 I out ' ⎛ I out ' ⎞ ⎛ I c ' ⎞ ⎛ V π 2 ⎞ ⎛ V i 2 ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ = ⎜⎜ ' Is I V V ⎝ c ⎠⎝ π 2 ⎠⎝ i 2 ⎠⎝ Vπ 1 8K I out ' RC 2 = = = 0 . 89 8 K + 1K Ic ' RC 2 + R L A Io = ⎞⎛ Vπ 1 ⎟⎟ ⎜⎜ ⎠⎝ I s ⎞ ⎟⎟ ⎠ Ic ' g V = m 2 π 2 = g m 2 = 30 mA / V Vπ 2 Vπ 2 Vπ 2 I π 2 rπ 2 r rπ 2 3 .3 K = = π2 = = 3 . 3 K + 101 (2 . 5 K Vi2 I π 2 rπ 2 + I π 2 (1 + g m 2 rπ 2 )R 2 Ri2 rπ 2 + (1 + g m 2 rπ 2 )R 2 ) = 0 . 013 Vi2 = − g m 1 R C 1 R i 2 = − g m 1 (R C 1 [rπ 2 + (1 + g m 2 rπ 2 )R 2 ]) = − 34 mA / V (10 K [3 . 3 K + 101 (2 . 5 K Vπ 1 ( ) ( )]) = − 327 ) Vπ = R S R 1 rπ 1 R B 1 = 10 K 13 . 4 K 2 . 9 K 13 K = 1 . 7 K IS A Io = I out ' = (0 . 89 )(30 mA / V Is )(0 . 013 )(− 327 )(1 . 7 K ) = − 193 58 Midband Gain with Feedback * Determine the feedback factor If βf = * Xf Xo = If ' Io ' = − RE 2 − 3.4 K = (RE 2 + R f ) 3.4 K + 10 K = − 0.25 + VE2 - Calculate gain with feedback AIfo β f AIo = − 193 ( − 0 .25 ) = 48 AIfo = AIo 1 + β f AIo = − 193 = − 3 .9 1 + 48 VE 2 = − I f ' R f = (I o '+ I f ')RE 2 − I f ' (R f + RE 2 ) = I o ' RE 2 AIfo (dB ) = 20 log 3 .9 = 11 .8 dB * If ' Io ' Note Ì Ì Ì Ì Ì = − RE 2 R f '+ RE 2 βf < 0 and AIo < 0 βf AIo > 0 as necessary for negative feedback and dimensionless βf AIo is large so there is significant feedback. Can change βf and the amount of feedback by changing RF. Gain is determined by feedback resistance AIfo ≈ 1 βf =− RE 2 + R f RE 2 = − 4.0 59 Input and Output Resistances with Feedback Ro Ri * Determine input Ri and output Ro resistances with loading effects of feedback network. Ri = RS R1 RB1 rπ 1 = 10 K 13.4 K 13K 2.9 K = 1.7 K * Ro = RC 2 RL + ∞ = ∞ Calculate input Rif and output Rof resistances for the complete feedback amplifier. Rif = = Ri (1 + β f AIo ) Rof = Ro (1 + β f AIo ) = ∞(49) = ∞ 1.7 K = 0.035 K 49 60 Voltage Gain for Current Gain Feedback Amplifier * Can calculate voltage gain − RL AIfo − 1K ⎛V ⎞ ⎛ − I 'R ⎞ − RL ⎛ I out ' ⎞ ⎜⎜ ⎟⎟ = (− 4.2) = + 0.42 V / V AVfo = ⎜⎜ o ⎟⎟ = ⎜⎜ out L ⎟⎟ = = V I R R I R 10 K s ⎝ s ⎠f s ⎝ s ⎠f ⎝ s s ⎠f AVfo (dB) = 20 log(0.42) = −7.5 dB * Note - can’t calculate the voltage gain as follows: Assume AVfo = Find AVo = AVo 1 + β f AVo Vo − I o RL − AIo RL − (−193)(1K ) = = = = +19.3 V / V Vs I s Rs Rs 10 K Calculate β f AVo = (− 0.25)(19.3 V / V ) = −4.8 Calculate voltage gain with feedback from AVfo = AVo 1 + β f AVo = − 19.3 V / V 1 − 4.8 = +5.8 V / V Magnitude is off by nearly a factor of ten! 61 Equivalent Circuit for Shunt-Series Feedback Amplifier * * * Current gain amplifier A = Io/Is Feedback modified gain, input and output resistances Ì Included loading effects of feedback network Ì Included feedback effects of feedback network Significant feedback, i.e. βf AIo is large and positive β f AIo = − 193 ( − 0 .25 ) = 48 Rif AIfoI i Rof AIfo = AIo 1 + β f AIo = − 193 = − 3 .9 1 + 48 AIfo (dB ) = 20 log 3 .9 = 11 .8 dB ARfo Rif = Ri (1 + β f AIo ) ⎛I = ⎜⎜ o ⎝ IS = 0.035 K ⎞ AIo ⎟⎟ = = − 3 .9 ⎠ f 1 + β f AIo AIfo ≈ 1 βf =− RE 2 + R f RE 2 = − 4 .0 Rof = Ro (1 + β f AIo ) = ∞ 62 Summary of Feedback Amplifier Analysis * Xs Xi Xo Xf βf * ωL = (1 + β f Ao ) * ω Hf = (1 + β f Ao )ω H * * ω Lf Identify the amplifier configuration by: Ì Output sampling Io = series configuration Vo = shunt configuration Ì Feedback to input Io = shunt configuration Vo = series configuration Calculate loading effects of feedback network Ì On input Ì On output Calculate appropriate midband gain A’ (modified by loading effects of feedback network) Calculate feedback factor βf. Calculate midband gain with feedback Af. A fo = Ao 1 + β f Ao 63 Summary of Feedback Amplifier Analysis 64 Feedback Amplifier Stability * A() Feedback analysis Ao Ì Midband gain with feedback A = fo Afo Ì New low and high 3dB frequencies ( Ì ω H ω Hf ωi ω Amplifier becomes unstable (oscillates) if at some frequency i we have * * ⎡ Im{β (ω i )A(ω i )}⎤ 0 and φ (ω i ) = tan −1 ⎢ ⎥ = −180 = −π radians ⎣ Re{β (ω i )A(ω i )}⎦ sin ce β (ω i )A(ω i ) = β (ω i )A(ω i ) e jφ (ω i ) = e − jπ = −1 so A f (ω i ) = A (ω i ) 1 + β (ω i )A(ω i ) = A (ω i ) 1−1 →∞ * ωL 1 + β f Ao ( ) Ri (1 + β f AIo ') ( Rof = Ro 1 + β f AIo ' ) Amplifier’s frequency characteristics A( s ) = AM FH ( s ) FL ( s ) Feedback amplifier’s gain Af (s) = β (ω i )A(ω i ) = 1 ω Lf = 1 + β f Ao Modified input and output resistances, e.g. Rif = ω Lf ω L ) ω Hf = 1 + β f Ao ω H Ao A (s) 1 + β f (s) A (s) → Af (ω) = A (ω) 1 + β f (ω) A (ω) Define Loop Gain as βfA β (ω )A(ω ) = β (ω )A(ω ) e jφ (ω ) Ì Ì Magnitude β (ω ) A(ω ) Phase ⎡ Im{β (ω )A(ω )}⎤ ⎥ ⎣ Re{β (ω )A(ω )}⎦ φ (ω ) = tan −1 ⎢ 65 Feedback Amplifier Instability βfA(dB) General form of Magnitude plot of βfA * Amplifier with one pole Ì Phase shift of - 900 is the maximum A(ω ) = ω1 φ (ω ) General form of Phase plot of βfA 00 - 450 - 900 - 1350 - 1800 1+ j ω ω HP1 Cannot get - 1800 phase shift. Ì No instability problem Amplifier with two poles Ì Phase shift of - 1800 the maximum Ì ω 0 dB Ao * A(ω ) = ω Ao ⎛ ω ⎞⎛ ω ⎜1 + j ⎟⎜1 + j ⎜ ω HP1 ⎟⎠⎜⎝ ω HP 2 ⎝ ⎞ ⎟ ⎟ ⎠ Can get - 1800 phase shift ! For stability analysis, we define two important frequencies Ì 1 is where magnitude of βfA goes to unity (0 dB) Ì 180 is where phase of βfA goes to -180o Ì Instability problem if 1 = 180 Ì ω180 * A f (ω1 = ω180 ) = A (ω1 ) 1 + β (ω1 ) A(ω1 ) = A (ω1 ) 1−1 →∞ 66 Gain and Phase Margins βfA(dB) * * ω180 0 dB φ (ω ) - 30 Gain margin = −1350 − (−1800 ) = +450 * ω1 00 ω180 - 450 Example Phase margin = φ (ω1 ) − φ (ω180 ) ω ω1 Gain and phase margins measure how far amplifier is from the instability condition Phase margin Phase margin = φ (ω1 ) − φ (ω180 ) Gain margin ω Gain margin = β A(ω1 ) − β A(ω180 ) Example Gain margin = β A(ω1 ) − β A(ω180 ) = 0 dB − (−30 dB) = +30 dB - 900 - 1350 - 1800 Phase margin Φ(ω1) Φ(ω180) * What are adequate margins? Ì Ì Phase margin = 500 (minimum) Gain margin = 10 dB (minimum) 67 Numerical Example - Gain and Phase Margins βfA(dB) 23.5 dB Ao = 150 ω1 Pole 1 Ao β f = 15 Pole 2 β f = 0.1 Ao β f (dB) = 23.5 dB 15 ω ⎞⎛ ω ⎞ ⎛ j + 1 ⎜1 + j ⎟ ⎜ ⎟ 8 x105 ⎠⎝ 3 x106 ⎠ ⎝ From plots of magnitude and phase Ao (ω ) β f = Gain margin ω1 = 6 x106 rad / s and ω180 = 3 x107 rad / s φ (ω1 ) = −1400 and φ (ω180 ) = −1800 Phase Shift (degrees) so phase margin is φ (ω1 ) − φ (ω180 ) = −1400 − (−1800 ) = +400 Ao β f (ω1 ) = 0 dB and Ao β f (ω180 ) = −28 dB ω180 Phase margin so gain margin is Ao β f (ω1 ) − Ao β f (ω180 ) = 0 − (−28 dB) = +28 dB Note that the gain and phase margins depend on the feedback factor βf and so the amount of feedback and feedback resistors. * Directly through the value of βf. * Indirectly since the gain Ao and pole frequencies are influenced by the feedback resistors, e.g. the loading effects analyzed previously. 68 Amplifier Design for Adequate Gain and Phase Margins A(dB) General form of Magnitude plot of A 1/βf(dB) A(ω180) 0 dB φ (ω ) 00 Gain margin General form of Phase plot of A * * * ω180 - 450 1 ω1 ω 180 ω1 βf ω ω - 900 - 1350 - 1800 Ì Ì * * Phase margin How much feedback to use? What βf ? A change of βf means redoing βf A() plots! Is there an alternate way to find the gain and phase margins? YES Ì Plot magnitude and phase of gain A() instead of βf A() vs frequency Ì Plot horizontal line at 1/ βf on magnitude plot (since βf is independent of frequency) Ì Why? At intersection of 1/ βf with A(), = 1 Φ(ω1) Φ(ω180) = A(ω ) so β f A(ω1 ) = 1 So this intersection point gives the value where ω = 1 ! Then we can find the gain and phase margins as before. If not adequate margins, pick another βf value, draw another 1/ βf line and repeat process. Note: βf is a measure of the amount of feedback. Larger βf means more feedback. Ì Recall for βf = 0, we have NO feedback βf = Ì Xf Xo so β f = 0 → X f = 0 Af = A →A 1+ β f A As βf increases, we have more feedback 69 Example - Gain and Phase Margins Alternate Method A(dB) Ao = 150 ω1 43.5 dB Ao (ω ) = 1/βf(dB) A(ω180) Gain margin β f = 0 .1 so 1 βf Ao ( dB ) = 43 .5 dB 150 ω ⎞⎛ ω ⎞ ⎛ 1+ j ⎟ ⎜1 + j 5 ⎟⎜ 8 x10 ⎠⎝ 3 x10 6 ⎠ ⎝ so 1 βf = 1 = 10 0 .1 ( dB ) = 20 log 10 = 20 dB From plots of magnitude and phase ω1 = 6 x10 6 rad / s and ω180 = 3 x10 7 rad / s φ (ω1 ) = −140 0 and φ (ω180 ) = −180 0 φ (ω ) ω180 Phase margin so phase margin is φ (ω1 ) − φ (ω180 ) = − 140 0 − ( − 180 0 ) = + 40 0 1 / β f (ω1 ) = 20 dB and Ao (ω180 ) = −8 dB so gain margin is 1 / β f (ω1 ) − Ao (ω180 ) = 20 − ( − 8 dB ) = +28 dB These are the same results as before for the gain and 70 phase margins . Summary of Feedback Amplifier Stability * Feedback amplifiers are potentially unstable Ì Can break into oscillation at a particular frequency ωi where A() Ao Afo ω Lf ω L β (ω i )A(ω i ) = 1 ω H ω Hf ⎡ Im{β (ω i )A(ω i )}⎤ 0 and φ (ω i ) = tan −1 ⎢ ⎥ = −180 = −π radians ⎣ Re{β (ω i )A(ω i )}⎦ sin ce ωi ω β (ω i )A(ω i ) = β (ω i )A(ω i ) e jφ (ω i ) = e − jπ = −1 so A f (ω i ) = Oscillation instability is a design problem * for any amplifier with two or more high frequency poles. This is the case for ALL amplifiers since each bipolar or MOSFET transistor has two capacitors and each capacitor gives rise to a high frequency pole! A (ω i ) 1 + β (ω i )A(ω i ) = A (ω i ) 1−1 →∞ Need to design feedback amplifier with adequate safety margin Ì Minimum gain margin of 10 dB Ì Minimum phase margin of 50o. Ì Adjust βf by varying size of Rf . Smaller Rf is, the larger is the amount of feedback. 71