Problem of Input matching Figure 5.9.(a)Use of resistive termination for matching,(b)simplified circuit. [View full size image] ￥DD RD手 2 Rsll Rp n,out 4kTY9m 4kT(Rsll Rp)○ M1 @ (b) Equation 5.17 = 4kT(RslRp)(gmRD)2 +4kTygmR+4kTRD. NF=1+ Rs yRs Rs RP 8m(RslIRp)2 8m(RslIRp)2Rp For Rp Rs,the NF exceeds 3 dB-perhaps substantially
Problem of Input Matching For RP ≈ RS, the NF exceeds 3 dB—perhaps substantially
Example 5.5. A student decides to defy the above observation by choosing a large Rp and transforming its value down to Rs.The resulting circuit is shown in Fig.5.10(a),where C1 represents the input capacitance of M1.(The input resistance of M1 is neglected.)Can this topology achieve a noise figure less than 3 dB? Figure 5.10.(a)Use of matching circuit to transform the value of Rp,(b)general representation of (a),(c) inclusion of noise of Rp,(d)simplified circuit of(c),(e)simplified circuit of(d). [View full size image] Rs 1 Rs H(s) o Vout Rp Rp (a) (b) Passive Rout Reciprocal Network Rs Rs Rs H(s) H(s) 4kTRs Rp Rs Rs Rs 4kTR● (c) d e 2 Equation 5.23 t,75 Rp ARs Rp A2= 4Rs 4kTRp Rout Rout Rp 1=k灯 Rout V匠mlRs= 4KTRs Rp 4Rs Rout =Rp. =kTRp. V匠ou IRP=kTRp
5.3 LNA Topologies---Common- source with inductive load Figure 5.11.(a)Inductively-loaded CS stage,(b) input impedance in the presence of CF,(c)equivalent circuit. Z L CE Vout Rs M专 a (b) (e) lAyl 8mRD Lis +Rs 2Ip VRD ZT= LICIs2 RsCis +1' VGS-VTH ID 2VRD VX= +k-gmVx)Z☑r CFS VGS -VTH Zin(s)= Vx= LI(C1 CF)s2 Rs(C1 CF)s+1 Ix [LIC1s2 +(RsC1 +8mL1)s +1+8mRs]CFs
5.3 LNA Topologies---Commonsource with inductive load Figure 5.11. (a) Inductively-loaded CS stage, (b) input impedance in the presence of CF , (c) equivalent circuit
CS with inductive load T Voo oVo 宁C M (b) Re(Zin]= [1 -LI(C1 CF)2]-(RsC1 +8mLI)]+Rs(C CF)(gmRs-L1C102+1)02 D where D is a positive quantity.It is thus possible to select the values so as to obtain Re{zin}=509
CS with inductive load