Slide bổ sung điện tử từ trường lecture 5 bjts circuits

Lecture 05 BJTs Circuits Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt topics • Large-signal operation • BJT circuits at DC • BJT biasing schemes Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Large-signal Ỉ Bias (DC) + signal (AC) Bias + signal vo = Vcc − iC Rc = Vcc − Rc I S e vBE VT vBE = VBE + vi Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt DC load line : VBB = I B × RB + VBE Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt VCC = I C × RC + VCE Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt head room (small) Leg room (small) VCC = I C × RCA + VCE → QA VCC = I C × RCB + VCE → QB RCB > RCA Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt BJT operate as a switch Switch off: vI < 0.5V →i B = →i C = →v C = VCC Switch on: vC = 0.2V ≈ 0V Switch on Ỉ saturation mode Switch off Æ cut-off mode Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.3 BJT work in saturation mode VC = VCE ( sat ) = 0.2V 50 < β 10 − 0.2 I C ( sat ) = = 9.8mA 1k I C ( sat ) 9.8m I B (max) = = = 0.196mA 50 β < 150 I C ( sat ) 9.8m I B (min) = = = 0.0653mA 150 β max I B = I B (max) × overdrive RB = factor − 0.7 4.3 = = 2.2k IB 1.96 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.4 (DC analysis) Reverse bias β = 100 forward bias Assume BJT in active mode : VE = − 0.7V = 3.3V IE = VE 3.3 = = 1mA RE 3.3k 100 ×1mA = 0.99mA I C = αI E = 100 + I B = I E − I C= 0.01mA Active mode check VC = 10 − I C × 4.7 k = 5.3V Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.5 (DC analysis) Assume BJT in active mode : VE = − 0.7V = 5.3V β = 100 VE IE = = = 1.6mA RE 3.3k 100 ×1.6mA = 1.584mA I C = αI E = 100 + I B = I E − I C= 0.016mA VC = 10 − I C × 4.7 k = 2.48V JC : forward bias Not in active mode JE : forward bias 10 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.9 (DC analysis) β = 30 Assume BJT in active mode : VE = VEB + VB RB l arg e → I B ≈ − 0.7 = 4.3mA 1k I C ≈ I E = 4.3V → VC = 10k × 4.3m − 5V = 38V (impossible) VE ≈ 0.7V → I E = I C (max) = 0.5mA → VC = 0V 15 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.10 (DC analysis) Reverse bias β = 100 forward bias Thevenin’s equivalent circuit VBB = I B RBB + VBE + I E RE VBB = 15V VBB = I B RBB + VBE + ( βI B + I B ) RE RBB ⇒ I B = 0.0128mA 50k = 5V 100k + 50k = 100k // 50k = 33.3k Assume BJT in active mode : ⇒ I E = 101× I B = 1.29mA ⇒ I C = 1.28mA 16 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.11 (DC analysis) β = 100 β = 100 15V = ( I C1 + I B ) RC1 + VC1 ≈ I C1 RC1 + VC1 ⇒ VC1 ≈ 8.6V start VE = VC1 + 0.7V ≈ 9.3V 15 − 9.3 ≈ 2.85mA 2k = αI E ≈ 2.82mA IE2 = IC with I B = 0.028mA Find correct current by iteration VC = I C × 2.7 k ≈ 7.62V I B2 = IE2 ≈ 0.028mA 101 CuuDuongThanCong.com 17 Microelectronic Circuit by meiling CHEN https://fb.com/tailieudientucntt Exercise 5.30 (DC analysis) β = 100 β = 100 IC3 VC 18 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.12 (DC analysis) β = 100 β = 100 Q1 and Q2 cannot be conducting at same time If Q1 ON than Q2 OFF, and vice versa Assume Q1 on and Q2 off : 19 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt BJT’s biasing schemes self-bias Base fixed bias Collector-feedback bias Two power supply version bias Constant current bias 20 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Why we need good biasing scheme? 1.Temperature change ỈCollector biasing current change 2.Device change Ỉ biasing current change iC T1 T2 T3 iC1 iC vBE iC = I S e VBE VT KT 1.38 ×10 − 23 ( o K ) VT = = q 1.6 × 10 −19 21 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt VBB − VBE IE = RE + 1R+Bβ Self-Bias Insensitive to T and β Constrains: VBB >> VBE RE >> Voltage-divider : RB 1+ β RR Q RB = R1 + R2 RE >> Suggestion: ( R1 + R2 ) × 0.1× I E = VCC CuuDuongThanCong.com The rule of thumb : (經驗法則) VBB = 13 VCC I C RC = 13 VCC ∴ R1 , R2 small → I B ↑ Trade-off RB 1+ β VCE (orVCB ) = 13 VCC 22 Microelectronic Circuit by meiling CHEN https://fb.com/tailieudientucntt Self-Bias (emitter feedback bias) VCC VCC − VBE IE = RB RE + 1+ β RC RB VE RE The rule of thumb : VBB = 13 VCC I C RC = 13 VCC VCE (orVCB ) = 13 VCC 23 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 5.13 design the following self bias circuit VBB − VBE IE = RB RE + 1+ β The rule of thumb : VB = 13 12 = 4V VE = − VBE = 3.3V given I E = 1mA VCC = 12V β = 100 ( R1 + R2 ) × 0.1× I E = VCC RE = VE 3.3 = = 3.3k I E 1m ⇒ ( R1 + R2 ) × 0.1× = 12 L (a ) RC = 12 = ≈ 4k αI E 0.99 ×1m R2 VCC L (b) VB = 4V ⇒ R1 + R2 R1 = 80k (a ), (b) ⇒ R2 = 40k 24 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Base fixed bias VCC RC RB Type IC = β (VBB − VBE ) RB Type Type IC = β (VCC − VBE ) RB IC = RB RB = RB1 // RB VBB Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com β (VBB − VBE ) RB = VCC RB1 + RB 25 https://fb.com/tailieudientucntt Collector-feedback bias (a) Constrains: RC >> VCC = I E RC + I BRB + VBE V −V I E = I B + I C = CC RBE RC + 1+Bβ RB 1+ β T ↑⇒ I C ↑⇒ I C RC ↑ ⇒ VCE ↓⇒ I B ↓⇒ I C ↓ Good biasing scheme 26 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Collector-feedback bias (b) VCC RB RC VCC = ( I B + I C ) RC + I BRB + VBE + ( I B + I C ) RE VCC − VBE I E = I B + IC = RC + RE + 1R+Bβ RE T ↑⇒ I C ↑⇒ I E ↑ ⇒ VCE ↓⇒ I B ↓⇒ I C ↓ Good biasing scheme 27 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Two-power supply version I B RB + VBE + I E RE = VEE VEE − VBE ⇒ IE = RE + 1R+Bβ Constrains: VBB >> VBE RB RE >> 1+ β 28 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Constant current bias by Current mirror I REF = I C1 + I B1 + I B Q Q1 ≡ Q2 ∴ I B1 = I B = I B I REF = I C1 + I B = ( β + 2) I B I = I C = I C1 = ( β + 2) I B I V − (−VEE ) − VBE I Re f = CC R VCC + VEE − VBE I = I Re f = R I REF = β ( β + 2) ≈β 29 Microelectronic Circuit by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt ... https://fb.com/tailieudientucntt Assume BJT in saturation mode : VE = − 0.7V = 5. 3V VC = VE + VCE ( sat ) = 5. 3 + 0.2 = 5. 5V IE = VE = = 1.6mA I E 3.3m 10 − 5. 5 IC = = 0.96mA I B = I E − I C= 0.64mA 11 Microelectronic... Example 5. 3 BJT work in saturation mode VC = VCE ( sat ) = 0.2V 50 < β 10 − 0.2 I C ( sat ) = = 9.8mA 1k I C ( sat ) 9.8m I B (max) = = = 0.196mA 50 β < 150 I C ( sat ) 9.8m I B (min) = = = 0.0 653 mA... equivalent circuit VBB = I B RBB + VBE + I E RE VBB = 15V VBB = I B RBB + VBE + ( βI B + I B ) RE RBB ⇒ I B = 0.0128mA 50 k = 5V 100k + 50 k = 100k // 50 k = 33.3k Assume BJT in active mode : ⇒ I E = 101×