Slide điện tử từ trường lecture 12 bjts differential pair

Lecture 12 BJT’s Differential Pair Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt topics • Ideal characteristics of differential amplifier – – – – Input differential resistance Input common-mode resistance Differential voltage gain CMRR • Non-ideal characteristics of differential amplifier – Input offset voltage – Input biasing and offset current • Differential Amplifier with active load Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Differential pair Figure 7.12 The basic BJT differential-pair configuration Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Common mode operation Q Q1 = Q2 ∴ vB1 = vB = vCM ⇒ iE1 = iE I = ⇒ vC1 = vC = VCC I − α RC vC1 − vC = Reject common mode input Figure 7.13 Different modes of operation of the BJT differential pair: (a) The differential pair with a common-mode input signal vCM Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt The differential pair with a “large” differential input signal (1)VB1 >> VB VB1 = +1V , VB = Q1 on → VE1 = 0.3V = VE → Q2 off VC1 = VCC − αIRC , VC = VCC VC1 − VC = −αIRC Figure 7.13 Different modes of operation of the BJT differential pair: (b) The differential pair with a “large” differential input signal Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt (2)VB1 VB VB1 = small , VB = I I I E1 = + ΔI , I E = − ΔI 2 I VC1 = VCC − α RC − αΔIRC I VC = VCC − α RC + αΔIRC VC1 − VC = vo = −2αΔIRC ⇒ vo = f (ΔI ) Figure 7.13 (Continued) (d) The differential pair with a small differential input signal vi Note that we have assumed the bias current source I to be ideal (i.e., it has an infinite output resistance) and thus I remains constant with the change in vCM Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Exercise 7.7 let α ≈ 1, vBE = 0.7V find vE , vC1 and vC − 0.7 I= = 4.3mA 1k + 0.7 − Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Large signal operation iE = IS e ( vB1 −vE ) / VT iE = IS e ( vB −vE ) / VT α α iE = e ( vB1 −vB ) / VT iE iE1 = iE1 + iE + e ( vB −vB1 ) / VT iE = iE1 + iE + e ( vB1 −vB ) / VT iE + iE = I iE = iE I + e −vid / VT I = + e vid / VT Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt iC1 ≈ I + e −vid / VT + e −vid / VT iC I = ⇒ ≈ vid / VT I + e vid / VT 1+ e I iE = iE ⇒ How to enhance linear region？ Figure 7.14 Transfer characteristics of the BJT differential pair of Fig 7.12 assuming α 10 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Differential mode vo vo RS vs RC αie RC vs RS g m vπ RE re re = Rid = (1 + β )(50 + 150) RE VT 25m = = 50 I E 0.5m 150 Rid = 2(1 + β )(50 + 150) = 40k v c1 − α (1 + β )ib 10k − αie Rc = = vs Rs ib + 0.2ie 5kib + 0.2k (1 + β )ib v c1 − α (1 + β )10k = vs 5k + 0.2k (1 + β ) Ad = Rid v c1 −v c v − α (1 + β )10k = 40 = c1 = vs vs 5k + 0.2k (1 + β ) 26 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Common mode vo RS vicm vo RS RC vicm αie αie ib 0.05k re g m vπ 0.15k RE ro = rO RC 10k RE Ricm REE Ricm ie 400k if VA VA = = 200k I IC 2 REE RC1 = Rc + 1% ≠ RC = Rc − 1% vc1 = − βib ( RC ) vc = − βib ( RC + ΔRC ) Ricm ≈ (1 + β )(400k // 200k ) ΔRC = RC × 0.02 Ricm ≈ (1 + β )(400k // 200k ) = 6.7 M Microelectronic Circuits by Meiling CHEN 27 vicm = 5kib + (1 + β )ib (400.2k ) Acm = vc1 − vc β (ΔRC ) = vicm 5k + (1 + β )(400.2k ) CuuDuongThanCong.com https://fb.com/tailieudientucntt 7-4.2 Input offset voltage Vos ≡ V o vi = Ad Solution : Add a -Vos − − Vos Ad Vo = Vos Ad + (−Vos ) Ad = + 28 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt 7-4.2 Input offset voltage Case : different RC Case : different Q let RC1 ≠ RC , Q1 = Q2 ΔRC RC1 = RC + ΔR RC = RC − C ΔR αI VC1 = VCC − ( )( RC + C ) 2 ΔR αI VC = VCC − ( )( RC − C ) 2 αI Vo = VC1 − VC = ΔRC αI αI ΔRC ΔRC Vo Vos ≡ = = IE Ad g m RC RC VT Vos = VT ΔRC RC 29 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Q1 ≠ Q ⇒ I S ≠ I S consider IC = ISe V BE VT ΔIS ΔIS IS2 = IS − Q V BE = V BE internal Consider Q and RC Vos = (VT I S1 = I S + ∴ I E1 = ΔRC ΔI ) + (VT S ) RC IS Solution : Add a -Vos ΔIS I (1 + ) 2IS ∴ IE2 ΔIS I = (1 − ) 2IS ⇒ VO I ΔIS =α RC IS V os = V T ( ΔIS ) IS CuuDuongThanCong.com 30 Microelectronic Circuits by Meiling CHEN https://fb.com/tailieudientucntt Input offset current I /2 L symmetric case β +1 β1 ≠ β ⇒ I B1 ≠ I B I B1 = I B = let I os = I B1 − I B Δβ Δβ , β2 = β − 2 I E1 I I Δβ (1 − ) ⇒ I B1 = = ≈ (1 + β1 ) β + + Δβ / 2 β + 2β let β1 = β + ⇒ I B2 = I os = IE2 I I Δβ (1 + ) = ≈ (1 + β ) β + − Δβ / 2 β + 2β I ( Δβ ) 2( β + 1) β I +I I Q I B ≡ B1 B = 2( β + 1) Δβ ⇒ I os = I B ( ) β 31 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt 7-5.5 Differential amplifier with active load Active load Small-signal 32 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Active load Q3 Q4 Passive load RC Ad = − g m (ro // ro ) Ad = − g m RC Acm ≈ − αRC Acm ≈ 2REE CMRR ≈ g m REE Rid = (1 + β )(2re + 2RE ) = 2rπ + 2(1 + β ) RE r0 Ricm ≈ (1 + β )(REE // ) Ro = RC // r0 ro β REE CMRR ≈ g m (ro // ro ) β REE ro Rid = 2rπ Ricm ≈ Ro = ro // ro Improving: Differential gain Common-mode gain and CMRR Defect: Vos 33 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Differential amplifier with active load equivalent-circuit Differential-mode vo Q1 v B1 = vid Q3 Q4 vB2 = − vid 0v Vid Vo + rπ Vπ − Rid = Q2 + gm1Vπ1 ro1 re3 // ro3Vπ rπ − Ro = ro // ro − Vid + gm4Vπ ro ro gm2Vπ2 rπ Vπ − vid I vid = Irπ ⇒ Rid = 2rπ CuuDuongThanCong.com 34 Microelectronic Circuits by Meiling CHEN https://fb.com/tailieudientucntt vid v )(re // ro // ro1 // rπ ) ≈ − g m1re3 ( id ) 2 v id Q vb = vb ⇒ g m vb = − g m g m1re ( ) v vo = (− g m vπ − g m vπ )(r04 // ro ) io = g m ( id ) − g m vb vid = − − − v [ g v g ( )](r04 // ro ) vid vid o m π4 m ∴ io = g m ( ) + g m g m1re 2 v vπ = − g m vπ (r01 // re // ro // rπ ) ≈ − g m id re Q g m1 = g m = g m = g m vb = − g m1 ( I /2 gm ≈ VT re = α / g m ≈ / g m ⇒ GM = vid )(r04 // ro )[ g m (r01 // re // ro // rπ ) + 1] v vo ≈ g m ( id )(r04 // ro )[ g m re + 1] vo ≈ g m ( )(r04 // ro )[ g m re3 + 1] Ad = vid vo = g m ( io = gm vid Ad = vo = g m (r04 // ro ) vid 35 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Since four transistors have the same parameters i= vx v = x Ro 2ro ix = 2i + vx vx vx = + ro ro ro ⇒ Ro ≡ vx = ro // ro ix Ad ≡ vo = GM Ro = g m (ro // ro ) vid Q ro = ro = ro Ad ≡ g m ro Rid = 2rπ 36 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Common-mode gain at CMRR Differential amplifier with active load equivalent-circuit Common-mode i1 V icm ie1 ro1 vb Vo + rπ // // ro3 r π Vπ gm3 − gm4Vπ re1 i1 ro i2 ro ie i2 37 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com icm re 2 R EE R EE V https://fb.com/tailieudientucntt i1 ≈ i2 ≈ vicm REE vb = −i1 ( // rπ // ro // rπ ) g m3 Q3 ic = g m vb vo = (− g m vb − i2 )ro Acm ≡ =− let vo r // rπ // ro // rπ ) − 1] = o4 [ gm4 ( vicm REE g m3 ro REE 1 + + rπ rπ ro 1 g m3 + + + rπ rπ ro + vπ − rπ g m vπ ro gm ro g m = g m , rπ = rπ , ro >> rπ , ro >> rπ Acm ≡ rπ vo r r r = − o4 ≈ − o4 = − − o4 REE g + REE β vicm β REE m3 rπ CMRR ≡ Ad Acm = g m (ro // ro )( β REE ro ) + vπ − rπ when ro = ro = ro CMRR = β g m REE CuuDuongThanCong.com 38 Microelectronic Circuits by Meiling CHEN https://fb.com/tailieudientucntt Input offset voltage (systematic problem) I + I B + I 4= I ⇒ I3 + let I3 β + I 4= I I3 = α I I3 IB I4 I4 = L β P ≡ β3 = β I3 + / β P αI / I4 = 1+ / βP αI αI / αI / β P αI Δi = − = ≈ 1+ / βP 1+ / βP βP αI / β P Δi 2VT Vos = − =− =− αI / 2VT βP GM 39 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Exercise 7-13 + VCC + VCC Q3 Q4 v1 v2 Q1 Ro = ro Q2 Ro = ro + VCC I I Q5 Q6 g m vπ ro + vπ − + rπ vπ − rπ g m vπ ro − VEE 40 Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt ... current • Differential Amplifier with active load Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Differential pair Figure 7 .12 The basic BJT differential- pair. .. the BJT differential pair: (a) The differential pair with a common-mode input signal vCM Microelectronic Circuits by Meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt The differential. .. − VC = −αIRC Figure 7.13 Different modes of operation of the BJT differential pair: (b) The differential pair with a “large” differential input signal Microelectronic Circuits by Meiling CHEN