模拟集成电路的分析与设计：Chapter 6-Frequency Response of Amplifiers.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,Frequency Response of Amplifiers,Bode diagram-first order(1),Frequency Response of Amplifiers,Bode diagram-first order(2),Frequency Response of Amplifiers,Bode diagram-first order(3),Frequency Response of Amplifiers,Bode diagram-first order(4),Frequency Response of Amplifiers,Bode diagram-first order(5),Frequency Response of Amplifiers,Bode diagram-first order(6),Frequency Response of Amplifiers,Bode diagram-first order(7),Frequency Response of Amplifiers,Bode diagram-first order(8),Frequency Response of Amplifiers,Bode diagram-feedback(1),Frequency Response of Amplifiers,Bode diagram-feedback(2),Frequency Response of Amplifiers,Miller Theorem (1),The current flowing through Z from X to Y is equal to the current flow through Z,1,so,That is,Similarity,Frequency Response of Amplifiers,Miller Theorem (2),Using Millers Theorem to calculate the input capacitance of Fig.6.2(a),The C,F,is multiplied by 1+A!explain please refer to P167,Frequency Response of Amplifiers,Miller Theorem (3),Millers Theorem is useful in cases where the impedance Z appears in parallel with the main signal.,Frequency Response of Amplifiers,Multiple poles in cascade amplifiers(1),Each node has one pole,j,=,j,-1,and,j,=R,j,C,j,While,C,j,and R,j,is the capacitance and resistance seen from the node to ground,for example in above circuit:,The input pole:,in,=1/(R,S,C,in,),The output pole:,out,=1/(R,2,C,p,),Frequency Response of Amplifiers,Multiple poles in cascade amplifiers(2),Another example:,C,in,=(1+A)C,F,R,in,=R,s,The input pole:,in,=1/R,S,(1+A)C,F,Frequency Response of Amplifiers,Common-Source Stage-Using Miller Effect (1),Low frequency gain:,Using Miller Theorem:,Therefore,the input pole is:,Back,Frequency Response of Amplifiers,Common-Source Stage-Using Miller Effect(2),If not consider the effect of R,s,and C,GS,on C,out,Using Miller Theorem again:,The output pole can be written as:,Back to Diff.Pair(1),Frequency Response of Amplifiers,Common-Source Stage-Considering effect of C,GS,on C,out,(1),If consider the effect of C,GS,and still ignore the effect of R,S,then Miller Theorem cant be used here.,In this case,the circuit is redrawn as:,The small-signal model for calculating Z,x,is as follow:,+,-,i,test,V,test,g,m1,V,GS,C,GD,C,GS,G,S,D,Frequency Response of Amplifiers,Common-Source Stage-Considering effect of C,GS,on C,out,(2),+,-,i,test,V,test,g,m1,V,gs,C,GD,C,GS,G,S,D,where:,Frequency Response of Amplifiers,Common-Source Stage-Considering effect of C,GS,on C,out,(3),The output node can be equivalent as:,R,eq,C,eq,C,DB,Output node,R,D,The output pole can be calculated as:,Frequency Response of Amplifiers,Common-Source Stage-Considering effect of C,GS,on C,out,(4),In this case,the output pole become as:,R,eq,C,eq,C,DB,Output node,R,D,r,o1,C,L,r,o1,and C,L,can be added to the equivalent circuit easily:,Frequency Response of Amplifiers,Common-Source Stage-Considering effect of C,GS,on C,out,(5),Once,in,and,out,are achieved,transfer function can be written as,:,Two poles,suppose,in,p1,:,Compare to the transfer function equation,p1,can be obtained by,Addition term compare to,in,calculated before,Frequency Response of Amplifiers,Common-Source Stage-Full calculation(4),p2,can also be obtained by,Very complex!Lets do some simplification.,Normally C,GS,is very large compare to other capacitance,suppose C,GS,(1+g,m,R,D,)C,GD,+R,D,(C,GD,+C,DB,)/R,S,and C,GS,C,DB,above equation can be simplified as,Frequency Response of Amplifiers,Common-Source Stage-Full calculation(5),Besides the two poles,a zero is found in the nominator of the transfer function:,The zero arises from direct coupling of the input to the output through C,GD,at high frequency,resulting in a slope in the frequency response that is less negative than 40 dB/dec and will introduce stability issues in feedback amplifiers.,-40 dB/dec,:,Less than,-40 dB/dec:,Back to Diff.Pair(1),Frequency Response of Amplifiers,Common-Source Stage-input impedance calculation(1),First-order approximation:,Miller effect,Back,Frequency Response of Amplifiers,Common-Source Stage-input impedance calculation(2),If consider the effect of the output node at high frequency:,C,GS,is connected in parallel with input node,so can be ignored in initial calculation,Frequency Response of Amplifiers,Common-Source Stage-input impedance calculation(3),C,GS,is added here!,Frequency Response of Amplifiers,Common-Source Stage-input impedance calculation(4),If the frequency is not high,|R,D,(C,GD,+C,DB,)s|1 and|R,D,C,DB,s|C,DB,and|R,D,C,DB,s|1+g,m,R,D,then:,Output node affect input impedance at high frequency,detail see p177,Low impedance path,Frequency Response of Amplifiers,Source Followers-poles(1),Sum the current at output node:,Beginning from V,in,add up all of the voltages:,Frequency Response of Amplifiers,Source Followers-poles(2),If,p1,|C,L,s|,and thus,The equivalent capacitance part equal to C,GS,g,mb,/(g,m,+g,mb,),same as Miller approximation C,GS,1-g,m,/(g,m,+g,mb,),For high frequency,g,mb,1/(C,GD,s),then C,GD,can be ignored.,The resistance 1/g,m,and capacitance C,L,C,SB,are parallel with the output node,not included in this calculation.,Frequency Response of Amplifiers,Source Followers-output impedance(2),From Figure 6.20(a):,At low-frequency:,At high-frequency:,If source follower is used as a buffer,the output resistance is normally low,that is 1/g,m,p1,then,Suppose C,E,is not too large and g,mP,r,oN,or g,mP,r,oP,1,then,Questions for Chap.6(1),(1)In Figure 6.1,write the equations of Z,1,and Z,2,.,(2)Write the equation of C,in,in Figure 6.2(a).,Questions for Chap.6(2),(3)Write the input pole of the circuit in Figure 6.8.,(4)Write the input and output pole of the circuit in Figure 6.10,ignoring the effect of C,Gs,and R,S,on C,out,.,(5)Write the transfer function at high-frequency V,out,(s)/Vin(s)of the circuit in Figure 6.23.,Questions for Chap.6(3),(6)Write the transfer function at high-frequency Vout(s)/Vin(s)of the circuit in Figure 6.25.,(7)Use Thevenin equivalent method to calculate the transfer function at high-frequency Vout(s)/Vin(s)of the circuit in Figure 6.30.,Assuming,p2,p1,s,uppose CE is not too large and g,mP,r,oN,or g,mP,r,oP,1,get,p1,p2,and,z,Questions for Chap.6(),Q6.3;,Q6.5;,Q6.8;,Q6.11;,Q6.16;,