01.MULTI STAGE AMPLIFIER (L 1).ppt
- 2. MULTISTAGE AMPLIFIER CASCADED CONNECTION OUTPUT OF ONE STAGE IS CONNECTED TO THE INPUT OF ANOTHER STAGE. Amplifier 1 Gain = AV1 Vin 1 Vo 1 Amplifier 2 Gain = AV2 Vo 2 Overall gain AV = VO2 / Vin 1 = Av1 AV2
- 4. Cascaded BJT Amplifier 297 V1o Cs1 Q1 C1 RE1 R2 Vi RC1 VCC R1 Gain AV1 = e L C o r R R V V 1 1 1 1 || Cs2 C2 Q2 RE2 R4 RC2 VCC R3 V2o Gain AV2 = e L C o o r R R V V 2 2 1 2 ||
- 5. Vi b R1 Ib2 ri2 R2 1Ib1 c Ic1 V1O RC1 R3 R4 e Ib1 ri1 c 2Ib2 V2O RC2 Cs1 Vi Q1 C1 V1O R2 R1 RC1 VCC RE1 Q2 C2 V2O R4 R3 RC2 VCC RE2 Cs2 e RL e1 e2 2 4 3 C1 v1 r r * β || R || R || R A e2 C2 v2 r R A e2 L C2 v2 r R || R A v2 v1 v A * A A gain, Overall
- 6. Vi b R1 Ib2 ri2 R2 1Ib1 c Ic1 V1O RC1 R3 R4 e Ib1 ri1 c 2Ib2 V2O RC2 e RL e1 1 2 1 i r * β * R * R Z Impedance, Input o c2 o r || R Z Impedance, Output
- 7. 20µf Vi Q1 10µf V1O 4.7K 15K 2.2K +20V 1K Q2 C2 V2O 6.8K 22K 1.5K 20V 1K 20µf Draw the (i) Input and (ii) output voltage waveshape for different values of Vi (a) 25µV (b) 1mV (c) X mV, where X is last two digits of your Roll No.
- 8. Cascaded FET Amplifier 504 Cs1 Q1 Ci Rs1 RG1 Vi RD1 VDD AV1 = -gm1RD1 Vi RG1 gmVgs D Ic1 V1O RD1 S G
- 10. Cascaded FET Amplifier 504 AV1 = -gm1RD1 Vi RG1 gmVgs D V1 O RD1 S G Vi RG2 gmVgs D V1O RD2 S G AV2 = -gm2RD2 Overall gain AV1AV2 = (-gm1RD1)(-gm2RD2) Input Impedance Zin = RG1 Output Impedance ZO = RD2
- 11. Cascaded FET Amplifier 504 Cs1 Q1 Ci Rs1 RG1 Vi RD1 VDD AV1 = -gm1RD1 AV2 = -gm2RD2 Overall gain AV1AV2 = (-gm1RD1)(-gm2RD2) Input Impedance Zin = RG1 Output Impedance ZO = RD2 Cs2 Q2 C2 Rs2 RG2 Vo1 RD2 VDD Vo2
- 15. CASCODE CONFIGURATION This is a CE – CB combination. This arrangement provides high input impedance but a low voltage gain. The low voltage gain reduces the Miller Input Capacitance therefore this combination works well in high frequency applications.
- 17. CASCODE CONFIGURATION Consider the cascode amplifier above a) Find the voltage gain Av = vo/vi b) Find the input resistance Ri c) Find the output resistance Ro
- 18. Darlington Connection 299 Overall = 1 2 Q1 Q2 Two transistors are connected in such a way that emitter current of one is fed to the base of another, so that the overall current gain is the product of their individual current gains. It is also known as superbeta transistor IE1 = (1+ 1 ) IB1 IB1 IE1 IE2 IE2 = (1+ 2 ) IB2 IE2 = (1+ 2 ) (1+ 1 ) IB1 IE2 2 1 IB1
- 19. 2N999 npn darlington transistor VBE = 1.8V at IC = 100mA βD = 4000 at IC = 10mA βD = 7000 at IC = 100mA Maximum βD = 70,000 at IC = 100mA
- 20. DC bias of Darlington Circuit RE RB VCC D IB IE IC E D B BE CC B R R V V I VBE B D B D E I I I 1 BE E B V V V E E E R I V
- 21. DC bias of Darlington Circuit 390R 3.3M 18V D IB IE IC A 2.56 390R * 8000 3.3M 1.6 18 IB VBE 20.48mA 2.56μ. * 8000 IE 6V . 9 6 . 1 8 B V 8V 390R * 20.48mA VE VBE = 1.6V βD =8000
- 22. AC analysis of Darlington Circuit Vi RE RB VCC D = 8000 VBE = 1.6V Ib IE IC Ci Ci VO Ib RE ri RB Vi VO DIb Ii
- 23. AC Input Impedance of Darlington Circuit Ib RE ri RB Vi VO DIb Ii i o i b r V V I E b D b o R I I V E D i b E D i b i R r I R r I V 1 E D i B i i i R r R I V Z || βD =8000 ri = 5K
- 24. AC Current Gain of Darlington Circuit b D b D b o I I I I i B E D i B b I R R r R I E D i b i R r I V i o i I I A Ib RE ri RB Vi VO DIb Ii Io i B E D B b I R R R I E D B B D i b b o i o i R R R I I I I I I A E D i B i i i R r R I V Z ||
- 25. AC Output Impedance of Darlington Circuit o o o I V Z RB ri RL VO DIb Ib VS RE ZO ri VO DIb RE IO b D i o E o o I r V R V I i o D i o E o o r V r V R V I i D i E o o o r r R I V Z 1 1 1
- 26. AC Voltage gain of Darlington Circuit i o V V V A E b D b o R I I V ri VO DIb Ib VS RE o i b i V r I V E D E i b i R R r I V b E D E o I R R V E D E E D E i i o R R R R r V V 1 E D E i E D E i o V R R r R R V V A
- 27. Summery of Darlington connection •Super beta transistor, D = 1 2. •High current gain, can amplify very small signal. •Increased input impedance. •Reduced output impedance. •Unity voltage gain.
- 28. CMOS
- 29. CMOS Circuit 594
- 30. CMOS Circuit 594
- 31. CMOS Circuit 594
- 32. CMOS Circuit 594
- 33. CMOS Circuit 594 Complementary Metal Oxide Semiconductor Field Effect Transistor VDD Vi Vo pMOS nMOS S S D D G G VGS = -5V VGS = 0V Vi = 0V Vo = 5V VDD = 5V VDD pMOS nMOS S S D D G G VGS = 0V VGS = 5V Vi = 5V Vo = 0V VDD = 5V CMOS is used as an inverter
- 34. CMOS Circuit 594
- 35. Current source RL = 1 IS = 10A IL = 10A RL = 2 IDSS= 8 mA VP = - 5 V RL =1K VDD =10V IL = 8 mA RL = 2K VDD = 24V VO= ? 2 1 P GS DSS D V V I I RL = 5K
- 36. BJT Current source 310 IC RE VB IE R2 -VEE R1 VCC RL EE B V R R R V 2 1 1 7 . 0 B E V V C E EE E E I R V V I ) (
- 37. BJT - Zener Current source 310 VCC RL C E BE Z E I R V V I IC RE VB IE VZ -VEE R1
- 38. Current Mirror Circuit 307 B C X I I I 2 VCC RX IC Q2 Q1 IX IC IE IE IB IB 2IB E E B I I I 1 E C I I E E X I I I 2 E E X I I I 2 V RL IL V1 RL1 IL1 L X I I