模拟集成电路的分析与设计：Chapter 5-Current Mirrors.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,*,Current Mirrors(,电流镜）,Basic current mirrors(1),As we seen,current sources are frequently used in analog circuit.,How can we supply an accurate I,SS,?If we have an accurate current source reference,we can copy it to the circuit.,Fig.5.5 Basic current mirror,Diode-connected to ensure M,1,always in saturation!,If not consider channel-length modulation:,Current Mirrors,Basic current mirrors(2),For example:,If consider channel-length modulation:,V,X,=V,GS0,=V,GST,while V,P,is dependent on the CM level and input signal of M,1,and M,2,and may not equal V,X.,In this case,I,T,not track I,REF,accurately,Current Mirrors,Cascode,current mirrors(1),Use,cascode,to get more accurate current:,Largely reduce the change so that V,Y,is more close to V,X,and hence,I,out,more closely track I,REF.,Refer to P65 Figure 3.23 and,eq,.(3.63),Current Mirrors,Cascode,current mirrors(2),More accurate but at cost of higher voltage headroom:,Minimum allowable voltage at node P is equal to:,.,One,V,gs,higher than basic current mirror,:,.,Current Mirrors,Cascode,current mirrors(3),When V,X,drops from high voltage to low voltage,M,3,enter triode region first,and then M,2,:,.,M,3,enter triode region,M,2,enter triode region,V,X,Large-Signal Analysis:,Current Mirrors,Cascode,current mirrors(4),Less one V,TH,So this can be used in low-voltage application,.,In order to solve the trade-off between accuracy and voltage headroom:,Minimum allowable V,P,becomes:,Current Mirrors,Active current mirrors(1),How about if we use a current mirror as load of a differential pair?,Current Mirrors,Active current mirrors(2),R,out,is equal to the output resistance of a CS with a degeneration resistance of 1/g,m1,.,Current Mirrors,Active current mirrors(3),Second approach to calculate|A,v,|:,Refer to,eq,.(3.110),page 80:,Then:,Current Mirrors,Active current mirrors(4),Second approach to calculate|A,v,|:,Refer to,eq,.(3.104),page 79:,Current Mirrors,Active current mirrors(5),Second approach to calculate|A,v,|:,Current Mirrors,Differential pair with,Active current mirrors(1),Problem of above differential pair:,Small-signal drain current of M,1,is“wasted”,Solve method:,Use differential pair with active current mirror to combine the small-signal current together.,Current Mirrors,Differential pair with,Active current mirrors(2),Large-signal analysis:,Current Mirrors,Differential pair with,Active current mirrors(3),Large-signal analysis:,If V,in1,-V,in2,=-,M,1,M,3,M,4,off,I,D4,=0,M,5,and M,2,are in deep triode region,V,out,=0;,When (,V,in1,-V,in2,),I,D1,=I,D3,=I,D4,V,out,;,When,V,in1,-V,in2,=0,Maximum current,I,D1,=I,D2,=I,D3,=I,D4,and hence maximum small-signal gain;,When,V,in1,-V,in2,0,I,D2,M,4,enter triode region,the gain decrease;,When V,in1,-V,in2,=,M,2,off and I,D2,=0,M,4,in deep triode region,V,out,=V,DD.,Current Mirrors,Differential pair with,Active current mirrors(4),Large-signal analysis result:,Always biased in the condition of V,in1,=V,in2,so that the gain of differential pair with active current mirrors keeps largest.,Current Mirrors,Differential pair with,Active current mirrors(5),Small-signal analysis:,Differential input and single-ended output,amplitude of,V,x,and,V,y,are not same,cannot use the concept of half circuit to analysis.,Current Mirrors,Differential pair with,Active current mirrors(6),There two approaches to calculate gain of the circuit:,One is to use the concept of G,m,and R,out,;,The second is to use,Thevenin,equivalent.,Please understand the first approach by your self(p151-152),Here I only introduce the second method.,Current Mirrors,Differential pair with,Active current mirrors(7),Use,Thevenin,equivalent to calculate gain of the circuit:,Current Mirrors,Differential pair with,Active current mirrors(8),Take point P as Virtual Ground,use half-circuit concept to calculate the,Thevenin,equivalent voltage:,While the equivalent impedance is:,Current Mirrors,Differential pair with,Active current mirrors(9),Then:,If the current flows through 1/g,m3,is mirrored into M,4,with unity gain:,!A little different from text book,Current Mirrors,Differential pair with,Active current mirrors(10),And:,Assuming 2r,o1,2,(1/g,m3,4,)|r,o3,4,and r,o3,4,1/g,m3,4,we obtain:,While the approximate result is same!,Current Mirrors,Differential pair with,Active current mirrors(11),Common-mode analysis:,Current Mirrors,Differential pair with,Active current mirrors(12),If symmetry,V,X,is always same as V,F,circuit can be simplified to be as Fig.5.30(a)and equivalent as Fig.5.30(b):,Here M,4,is thought to be also diode connected,Current Mirrors,Differential pair with,Active current mirrors(13),CM gain:,Common-mode rejection ratio:,Current Mirrors,Differential pair with,Active current mirrors(14),Consider the situation of g,m,mismatch:,When calculating,V,P,the small difference between V,F,and V,X,is neglected,M,1,and M,2,can be combined into one single transistor with g,m,=g,m1,+g,m2,Current Mirrors,Differential pair with,Active current mirrors(15),Current Mirrors,Differential pair with,Active current mirrors(16),If the effect of r,o1,and r,o2,is neglected;the difference between,I,D4,and I,D2,flows only through r,o4,so:,If,r,o3,1/g,m3,the equation can be rewritten:,Compare with symmetry case,The item(g,m1,-g,m2,)r,o3,is added in!,Questions for Chap.5(1),(1)Explain why use diode-connected for M,1,in Fig.5.5.,(2)In Fig.5.7,what will happen if consider channel-length modulation effect?,(3)Refer to Fig.5.9,what benefits do we have for using,cascode,structure in current source?And any drawbacks?,Fig.5.5 Basic current mirror,Questions for Chap.5(2),(4)What is the circuit structure in Fig.5.13 used for?,(5)What is the problem for the circuit in Fig.5.17?Any suggestions to solve it?,(6)What is the best bias condition for the circuit in Fig.5.21?,(7)What is CMRR?Write down the definition of it.,Fig.5.13,F.5.17,Fig.5.21,Questions for Chap.5(3),(8),Q5.1(p.158),(9),Q5.3(p.159),(10),Q5.5(p.159),(11),Q5.8(p.159),