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Cascode Amplifier Figure 1(A) Shows a Cascode Amplifier with Ideal Current Source Load

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Cascode Amplifier Figure 1(A) Shows a Cascode Amplifier with Ideal Current Source Load Cascode Amplifier Figure 1(a) shows a cascode amplifier with ideal current source load. Figure 1(b) shows the ideal current source is implemented by PMOS with constant gate to source voltage. (3) VDD |VTP0|+|∆V| VDD (7) ∆V VDD-(|VTP0|+|∆V|)=VG3 M3 (2) Vo Vo (6) M2 VG2 VTN0+VTN2+2∆V= VG2 ∆V M2 V +∆V TN2 (5) VTN0+∆V (1) M1 Vi VTN0+∆V= Vi M1 VTN0+∆V (4) VSS=0 (a) (b) Figure 1. Cascode amplifier with simple current load. 1. Low Frequency Small Signal Equivalent Circuit Figure 2(a) and 2(b) show its low frequency small signal equivalent circuit. Figure 2( c) shows its two-port representation and port variables assignment. 1 gm2vgs2 G1 D1 S2 D2 + + + gmb2vbs2 gds1 V S2 V Vi o g v m1 gs1 gds2 gds3 - S1 - - vgs2=-VS2; vbs2=-VS2 gm2VS2 G1 D1 S2 D2 + + + gmb2VS2 gds1 V S2 V Vi o g v m1 gs1 gds2 gds3 - - - S1 vgs2=-VS2; vbs2=-VS2 a I a b b I1 2 I1 I2 + + V a = V b a a 2 1 b V V1 Y Y i GL Z a b Z b o Zi o Figure 2. Cascode amplifier low frequency small signal equivalent circuit. 2 b b a YL = YL = g ds3 (or ZL = rds3 ); YS = YS = ∞(or ZS = 0) The current equation of network a is: a I1 = 0 a a a I 2 = g m1V1 + g ds1V2 The corresponding Y-parameter matrix is: a 0 0 Y = g m1 g ds1 The current equation of network b is: b b b I1 = (g m2 + g mb2 + g ds2 )V1 - g ds2 V2 b b b I 2 = −(g m2 + g mb2 + g ds2 )V1 + g ds2 V2 The corresponding Y-parameter matrix is: b g m2 + g mb2 + g ds2 − g ds2 Y = − (g m2 + g mb2 + g ds2 ) g ds2 The common gate stage gain is: b b − y 21 g m2 + g mb2 + g ds2 A V = b b = y 22 + YL g ds2 + g ds3 The input impedance of common gate stage (the load of common source stage) is: b b b a y 22 + YL g ds2 + g ds3 2 Zi = ZL = b b b = ≈ detY + y11YL (g m2 + g mb2 + g ds2 )g ds3 g m2 The gain of the common source stage is: a a − y 21 − g m1 − g m1 (g ds2 + g ds3 ) A V = a a = = y 22 + YL (g m2 + g mb2 + g ds2 )g ds3 g ds1 (g ds2 + g ds3 ) + (g m2 + g mb2 + g ds2 )g ds3 g ds1 + g ds2 + g ds3 − g (g + g ) - 2g = m1 ds2 ds3 ≈ m1 ≈ 2 g ds1g ds2 + g ds1g ds3 + g ds2g ds3 + (g m2 + g mb2 )g ds3 g m2 + g mb2 Assuming all gm are equal, and all gds are equal. In addition gm >>gmb, and gm>>gds. With a gain of 2 the miller capacitance Cgd1 of the common source stage is negligible. 3 The overall gain is: g + g + g − g (g + g ) b a m2 mb2 ds2 m1 ds2 ds3 A V = A V A V = g ds2 + g ds3 g ds1g ds2 + g ds1g ds3 + g ds2g ds3 + (g m2 + g mb2 )g ds3 - g (g + g + g ) − g = m1 m2 mb2 ds2 ≈ m g ds1g ds2 + g ds1g ds3 + (g m2 + g mb2 + g ds2 )g ds3 g ds The output impedance of the cascode amplifier is computed by obtaining the output impedance of the common source stage ( or the source impedance of the common gate stage) first. That is, a a a b y11 + YS 1 1 Zo = ZS = a a a = a = detY + y 22 YS y 22 g ds1 a The result is obtained by dividing the numerator and denominator by YS . The output impedance of the cascode is the output impedance of the common gate stage. b b b y11 + YS g m2 + g mb2 + g ds2 + g ds1 Zo = b b b = = (g m2 + g mb2 )rds2 rds1 + rds1 + rds2 ≈ g m2rds2 rds1 detY + y 22 YS g ds2g ds1 That is, the output impedance is equal to the output impedance, rds1 , of the first stage (common source) magnified by the gain, gm2rds2 , of the second stage (common gate ). The effective load is the parallel combination of output impedance of the cascode amplifier and the load. b Zo = Zo // ZL = ((g m2 + g mb2 )rds2 rds1 + rds1 + rds2 ) // rds3 ≈ (g m2rds2 rds1 )//rds3 ≈ rds3 The overall gain is approximately equal to : A V = −g m1Zo ≈ −g m1rds3 Although the output impedance has been magnified but the effective load impedance is determined by smaller impedance. That is, rds3. The overall gain of the cascode amplifier can be increased if we can increased rds3. This can be achieved by adding a cascode at the load. This is shown in Figure 3. Figure 4 shows its low frequency small signal equivalent circuit and its two-port representations and port variables assignment. There are three two-port networks. The Y-parameter matrices are derived as follows: The current equation of network a is: a I1 = 0 a a a I 2 = g m1V1 + g ds1V2 The corresponding Y-parameter matrix is: 4 a 0 0 Y = g m1 g ds1 The current equation of network b is: b b b I1 = (g m2 + g mb2 + g ds2 )V1 - g ds2 V2 b b b I 2 = −(g m2 + g mb2 + g ds2 )V1 + g ds2 V2 The corresponding Y-parameter matrix is: b g m2 + g mb2 + g ds2 − g ds2 Y = − (g m2 + g mb2 + g ds2 ) g ds2 The current equation of network c is: c c c I1 = (g m3 + g mb3 + g ds3 )V1 - g ds3V2 c c c I 2 = −(g m3 + g mb3 + g ds3 )V1 + g ds3V2 The corresponding Y-parameter matrix is: c g m3 + g mb3 + g ds3 − g ds3 Y = − (g m3 + g mb3 + g ds3 ) g ds3 The output impedance of network b has been obtained earlier it is, b Zo = (g m2 + g mb2 )rds2 rds1 + rds1 + rds2 ≈ g m2rds2 rds1 The output impedance of network c is computed as follows: c c c y11 + YS g m3 + g mb3 +g ds3 +g ds4 Zo = c c c = = (g m3 + g mb3 )rds3rds4 + rds3 + rds4 ≈ g m3rds3rds4 detY + y 22 YS g ds3g ds4 The effective load impedance is the parallel compbination of the output impedance of network b and c. g r 2 Z = Zb //Zc ≈ m ds o o o 2 Assuming all gm are equal and all gds are equal. 5 (3) VDD (|VTP0|+|∆V|) (9) VDD- (|VTP0|+|∆V|)= VG4 M4 (|VTP3|+|∆V|) (6) VDD (8) VDD-(|V |+|V |+2| V|)=V V TP0 TP3 ∆ G3 DD M3 (2) Vo Vo (7) VG2 VSS VTN0+VTN2+2∆V= VG2 VSS M2 M2 VTN2+∆V (5) (1) Vi V +∆V= Vi TN0 M1 M1 V +∆V TN0 (4) VSS=0 (a) (b) Figure 3. Cascode amplifier with cascode current load. 6 gm2VS2 G1 D1 S2 D2 + + + gmb2VS2 gds1 V S2 V Vi o gm1vgs1 gds2 - - - S1 vgs2=-VS2; vbs2=-VS2 a Y Yb gm3VS3 D4 S3 D3 + gmb3V3 gds4 VS3 gds3 - S4 Yd Yc a a b b I1 I2 I1 I2 + + + V a = V b a 2 1 V1 Ya Yb Vi Vo - b Zo Zo gds4 Yc c Zo Figure 4. Cascode amplifier with cascode load low frequency small signal equivalent circuit. 7 2. High Frequency Small Signal Equivalent Circuit VDD VG2 M3 C Cgd3 db3 Vo C M2 gd2 Cdb2 VG1 CL Cgs2 Csb2 Cgd1 Cdb1 Vi M1 Cgs1 Figure 5. Cascode amplifier parasitic capacitances. Figure 5 shows all the parasitic capacitances needed for high frequency modelling. Figure 6(a) shows the high frequency small signal equivalent circuit of cascode amplifier with simple current load. Figure 6(b) its two-port representation and port variables assignment. b YL = YL = g ds3 + sC3 ; YS = ∞(or ZS = 0) C1 = Cgd1; C 2 = Cdb1 + Cgs2 + Csb2 ; C3 = Cgd2 + Cdb2 + Cdb3 + Cgd3 + C L The input capacitance Cgs1 is assummed to be part of the input voltage source. Cgs1 is shorted out by the input voltage source, it does not affect the circuit operation, hence can be ignored or deleted.. C3 is assummed to be part of the load. The current of the first stage (network a) is given by: a a a I1 = sC1V1 − sC1V2 a a a I 2 = (g m1 - sC1 )V1 + [g ds1 + s(C1 + C2 )]V2 8 C gm2VS2 G1 1 D1 S2 D2 + + + gds1 gmb2VS2 V VS2 i Vo C C2 C gs1 g v 3 m1 gs1 gds2 gds3 - - S1 - vgs2=-VS2; vbs2=-VS2 C =C C =C +C +C 1 gd1 2 db1 gs2 sb2 C3=Cgd2+Cdb2+Cdb3+Cgd3+CL a a b b I1 I2 I1 I2 + a b + V2 = V1 V a a b V 1 Y Y i GL Z b b i Zo Figure 6. Cascode amplifier high frequency equivalent circuit. The corresponding Y-parameter matrix is: a sC1 − sC1 Y = g m1 - sC1 [g ds1 + s(C1 + C 2 )] The current equation of network b is: 9 b a a I1 = (g m2 + g mb2 + g ds2 )V1 - g ds2 V2 b a a I 2 = −(g m2 + g mb2 + g ds2 )V1 + g ds2 V2 The corresponding Y-parameter matrix is : b g m2 + g mb2 + g ds2 - g ds2 Y = − (g m2 + g mb2 + g ds2 ) g ds2 The voltage gain of network b is: b b − y 21 g m2 + g mb2 + g ds2 A V = b b = y 22 + YL g ds2 + g ds3 + sC3 The input impedance of network b or load of network a is: b b b a y 22 + YL g ds2 + g ds3 + sC3 Zi = ZL = b b b = detY + y11YL (g m2 + g mb2 + g ds2 )(g ds3 + sC3 ) The voltage gain of network a is: a a - y 21 - (g m1 - sC1 ) A V = a a = y 22 + YL (g m2 + g mb2 + g ds2 )(g ds3 + sC3 ) g ds1 + s(C1 + C2 ) + g ds2 + g ds3 + sC3 - (g - sC )(g + g + sC ) = m1 1 ds2 ds3 3 (g ds1 + s(C1 + C 2 ))(g ds2 + g ds3 + sC3 ) + (g m2 + g mb2 + g ds2 )(g ds3 + sC3 ) The overall gain of the cascode amplifier is: 10 b a A V = A V A V g m2 + g mb2 + g ds2 = g ds2 + g ds3 + sC3 - (g m1 - sC1 )(g ds2 + g ds3 + sC3 ) (g ds1 + s(C1 + C2 ))(g ds2 + g ds3 + sC3 ) + (g m2 + g mb2 + g ds2 )(g ds3 + sC3 ) - (g - sC )(g + g + g ) = m1 1 m2 mb2 ds2 (g ds1 + s(C1 + C 2 ))(g ds2 + g ds3 + sC3 ) + (g m2 + g mb2 + g ds2 )(g ds3 + sC3 ) - (g - sC )(g + g + g ) = m1 1 m2 mb2 ds2 c[1+ bs + as 2 ] where : C (C + C ) a = 3 1 2 g ds1g ds2 + g ds3 (g m2 + g mb2 + g ds1 + g ds2 ) C (g + g + g + g ) + C (g + g ) + C (g + g ) b = 3 ds1 ds2 m2 mb2 2 ds2 ds3 1 ds2 ds3 g ds1g ds2 + g ds3 (g m2 + g mb2 + g ds1 + g ds2 ) c = g ds1g ds2 + g ds3 (g m2 + g mb2 + g ds1 + g ds2 ) If the poles are far apart, it can be approximated as follows: 2 s s 1 1 1 2 1 1 2 D(s) = 1+ bs + as = 1− 1− = 1− + s + s ≈ 1− s + s p1 p 2 p1 p 2 p1p 2 p1 p1p 2 where : | p 2 |>>| p1 | That is, −1 -[g ds1g ds2 + g ds3 (g m2 + g mb2 + g ds1 + g ds2 )] − g ds3 p1 = = ≈ b C3 (g ds1 + g ds2 + g m2 + g mb2 ) + C 2 (g ds2 + g ds3 ) + C1 (g ds2 + g ds3 ) C3 − b −[C3 (g ds1 + g ds2 + g m2 + g mb2 ) + C2 (g ds2 + g ds3 ) + C1 (g ds2 + g ds3 )] − g m2 p 2 = = ≈ a C3 (C1 + C 2 ) C1 + C 2 Note that the poles are associated with the inverse product of the resistance and capacitance of a node to ground.
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