Few important advantages of FET over conventional Transistors 1. 2. Unipolar device i.
Download ReportTranscript Few important advantages of FET over conventional Transistors 1. 2. Unipolar device i.
Few important advantages of FET over conventional Transistors 1. 2. Unipolar device i. e. operation depends on only one type of charge carriers (h or e) Voltage controlled Device (gate voltage controls drain current) 3. Very high input impedance (109-1012 ) 4. Source and drain are interchangeable in most Low-frequency applications 5. Low Voltage Low Current Operation is possible (Low-power consumption) Less Noisy as Compared to BJT No minority carrier storage (Turn off is faster) Very small in size, occupies very small space in ICs Low voltage low current operation is possible in MOSFETS Zero temperature drift is possible 6. 7. 8. 9. 10. [email protected] October, 16 FET JFET MOSFET (IGFET) Enhancement MOSFET n-Channel EMOSFET p-Channel EMOSFET n-Channel JFET p-Channel JFET Depletion MOSFET n-Channel DMOSFET [email protected] p-Channel DMOSFET October, 16 The Junction Field Effect Transistor (JFET) Figure: n-Channel JFET. [email protected] October, 16 SYMBOLS Drain Drain Drain Gate Gate Gate Source n-channel JFET Source n-channel JFET Offset-gate symbol Source p-channel JFET [email protected] October, 16 Biasing the JFET Figure: n-Channel JFET and Biasing Circuit. [email protected] October, 16 Operation of JFET at Various Gate Bias Potentials Figure: The nonconductive depletion region becomes broader with increased reverse bias. (Note: The two gate regions of each FET are connected to each other.) [email protected] October, 16 Operation of a JFET Drain - N Gate + P P N + + - Source [email protected] October, 16 Output or Drain (VD-ID) Characteristics of n-JFET Figure: Circuit for drain characteristics of the n-channel JFET and its Drain characteristics. Non-saturation (Ohmic) Region: The drain current is given by V I DS Saturation (or Pinchoff) Region: I DS I DSS V2 P V V GS P 2I DSS V2 P V V DS P GS V2 V V DS V GS P DS 2 V 2 V V DS P GS V GS and I I 1 DS DSS V P 2 Where, IDSS is the short circuit drain current, VP is the pinch off voltage [email protected] October, 16 Simple Operation and Break down of n-Channel JFET Figure: n-Channel FET for vGS = 0. [email protected] October, 16 N-Channel JFET Characteristics and Breakdown Break Down Region Figure: If vDG exceeds the breakdown voltage VB, drain current increases rapidly. [email protected] October, 16 VD-ID Characteristics of EMOS FET Locus of pts where VDS VGS VP Saturation or Pinch off Reg. Figure: Typical drain characteristics of an n-channel JFET. [email protected] October, 16 Transfer (Mutual) Characteristics of n-Channel JFET V GS I I 1 DS DSS V P 2 IDSS VGS (off)=VP Figure: Transfer (or Mutual) Characteristics of n-Channel JFET [email protected] October, 16 [email protected] October, 16 Fixed bias circuit Self bias circuit Potential Divider bias circuit [email protected] October, 16 For Fixed Bias Circuit Applying KVL to gate circuit we get VGG I G RG VGS VGS Fixed, I G 0 2 V 1 GS I I DS DSS V P and 2 VGS I DS I DSS 1 VP and VDS VDD I DS RD Where, Vp=VGS-off & IDSS is Short ckt. IDS For Self Bias Circuit VGS I DS RS 0 I DS VGS RS October, 16 VGG + IGRG + VGS = 0 Since IG = 0 VGS = - VGG Since GS is reverse biased, IG = 0 and VRG = 0 RG is present to limit current in case VGG is connected with wrong polarity This would forward bias the gate-source junction causing high currents, which would destroy the transistor [email protected] October, 16 16 VGS ID = IDSS 1 VP 2 2 VGS = - VGG V I DS I DSS 1 GS VP and VDS VDD I DS RD [email protected] October, 16 17 or Fixed Bias Ckt. [email protected] October, 16 DC Equivalent Circuit [email protected] October, 16 19 IG = 0 VGS I DS RS 0 I DS VGS RS VRG = 0 [email protected] October, 16 20 JFET Self (or Source) Bias Circuit 2 V and I I 1 GS DS DSS V P V GS I 1 DSS V P V 2 GS R S V V GS GS I 1 2 V DSS V P P 2 VGS R 0 S This quadratic equation can be solved for VGS & IDS [email protected] October, 16 [email protected] October, 16 22 Input R2 VG = VDD R 1 + R 2 October, 16 Output -VG + VGS + IDRS = 0 VGS = VG - IDRS VDS = VDD - ID (RD + RS) [email protected] 23 The Potential (Voltage) Divider Bias V GS I 1 DSS V P 2 V G V GS R 0 S Solving this quadratic equation gives V GS and I DS [email protected] October, 16 A Simple CS Amplifier and Variation in IDS with Vgs [email protected] October, 16 FET Mid-frequency Analysis: VDD A common source (CS) amplifier is shown to the right. RD R1 io The mid-frequency circuit is drawn as follows: • the coupling capacitors (Ci and Co) and the bypass capacitor (CSS) are short circuits • short the DC supply voltage (superposition) • replace the FET with the hybrid-p model The resulting mid-frequency circuit is shown below. is ii + vs D ii Rs + vs RTh Ci vi S RSS io + gmvp rd RD RL vo _ _ s mid-frequency CE amplifier circuit Analysis of the CS mid-frequency circuit above yields: A vi = vo = -g m R 'L , where R 'L = rd R D R L vi A vs = Zi = vi = R Th , where R Th = R1 R 2 ii AI = Zo = vo io AP = = rd R D seen by R L Zi vo = A vi vs R s + Zi io = A vi ii vo R2 _ d vi = vp s + + _ g Co G RL + _ VDD Zi RL po = A vi A I pi [email protected] October, 16 CSS _ FET Mid-frequency Analysis: VDD A common source (CS) amplifier is shown to the right. RD R1 io D ii The mid-frequency circuit is drawn as follows: • the coupling capacitors (Ci and Co) and the bypass capacitor (CSS) are short circuits • short the DC supply voltage (superposition) • replace the FET with the hybrid-p model The resulting mid-frequency circuit is shown below. is ii + vs _ Ci S RL vs vi RSS _ io + gmvp rd RD RL vo _ _ s mid-frequency CE amplifier circuit Analysis of the CS mid-frequency circuit above yields: A vi = vo = -g m R 'L , where R 'L = rd R D R L vi A vs = Zi vo = A vi vs R s + Zi Zi = vi = R Th , where R Th = R1 R 2 ii AI = Z io = A vi i ii RL Zo = vo io AP = po = A vi A I pi = rd R D seen by R L vo R2 _ d vi = vp s + + g Co G + Rs + RTh VDD [email protected] October, 16 CSS _ Procedure: Analysis of an FET amplifier at mid-frequency: 1) Find the DC Q-point. This will insure that the FET is operating in the saturation region and these values are needed for the next step. 2) Find gm. If gm is not specified, calculate it using the DC values of VGS as follows: gm = 2I ID = DSS VGS - VP VGS VP2 gm = ID = K VGS - VT VGS (for JFET's and DM MOSFET's) (for EM MOSFET's) (Note: Uses DC value of VGS ) 3) Calculate the required values (typically Avi, Avs, AI, AP, Zi, and Zo. Use the formulas for the appropriate amplifier configuration (CS, CG, CD, etc). [email protected] October, 16 PE-Electrical Review Course - Class 4 (Transistors) 18 V Example 7: Find the mid-frequency values for Avi, Avs, AI, AP, Zi, and Zo for the amplifier shown below. Assume that Ci, Co, and CSS are large. Note that this is the same biasing circuit used in Ex. 2, so VGS = -0.178 V. The JFET has the following specifications: IDSS = 4 mA, VP = -1.46 V, rd = 50 k 18 V 500 800 k io D ii 10 k + vs Co G + + Ci S 8k vi vo 400 k _ 2k _ [email protected] October, 16 CSS _ VDD FET Amplifier Configurations and Relationships: VDD RD R1 io D ii CS Co G + Rs + + Ci S RL vs vo CSS RSS _ _ Common Source (CS) Amplifier ii D S + -g m R G + RD RSS RL _ R1 _ R 'L rd R D R L Zi R Th Zo rd R D C2 vo _ R2 VCC A vs Common Gate (CG) Amplifier VDD AI VDD R1 AP D ii Rs + vs gmR rd R D R L R SS io Ci vi A vi ' L 1 gm CD g m R 'L 1 g m R 'L R SS R L R Th Co + vs ' L R2 vi _ Rs CG rd R D R SS 1 gm Zi Zi Zi A vi A A vi vi R + Z R + Z R + Z i i i s s s Z A vi i RL A vi A I Z A vi i RL A vi A I Z A vi i RL A vi A I G + vi _ S Ci where R Th = R1 R 2 io Co R2 R SS _ + RL vo _ Note: The biasing circuit is the same for each amp. Common Drain (CD) Amplifier (also called “source follower”) [email protected] October, 16 Figure: Circuit symbol for an enhancement-mode n-channel MOSFET. [email protected] October, 16 Figure: n-Channel Enhancement MOSFET showing channel length L and channel width W. [email protected] October, 16 Figure: For vGS < Vto the pn junction between drain and body is reverse biased and i D=0. [email protected] October, 16 Figure: For vGS >Vto a channel of n-type material is induced in the region under the gate. As vGS increases, the channel becomes thicker. For small values of vDS ,iD is proportional to vDS. The device behaves as a resistor whose value depends on vGS. [email protected] October, 16 Figure: As vDS increases, the channel pinches down at the drain end and iD increases more slowly. Finally for vDS> vGS -Vto, iD becomes constant. [email protected] October, 16 Locus of points where [email protected] October, 16 Figure: Drain characteristics [email protected] October, 16 Figure: This circuit can be used to plot drain characteristics. [email protected] October, 16 Figure: Diodes protect the oxide layer from destruction by static electric charge. [email protected] October, 16 Figure: Simple NMOS amplifier circuit and Characteristics with load line. [email protected] October, 16 Figure: Drain characteristics and load line [email protected] October, 16 Figure vDS versus time for the circuit of Figure 5.13. [email protected] October, 16 Figure Fixed- plus self-bias circuit. [email protected] October, 16 Figure Graphical solution of Equations (5.17) and (5.18). [email protected] October, 16 Figure Fixed- plus self-biased circuit of Example 5.3. [email protected] October, 16 Figure The more nearly horizontal bias line results in less change in the Q-point. [email protected] October, 16 Figure Small-signal equivalent circuit for FETs. [email protected] October, 16 Figure FET small-signal equivalent circuit that accounts for the dependence of iD on vDS. [email protected] October, 16 Figure Determination of gm and rd. See Example 5.5. [email protected] October, 16 Figure Common-source amplifier. [email protected] October, 16 For drawing an a c equivalent circuit of Amp. •Assume all Capacitors C1, C2, Cs as short circuit elements for ac signal •Short circuit the d c supply •Replace the FET by its small signal model [email protected] October, 16 A C Equivalent Circuit Simplified A C Equivalent Circuit v Voltage gain, A o v v gs v i R g v R Out Zm gsr LR o put o imp., L o v d D r R d D r R d Input imp., Z in R R R G D A o g R , R R r v v m L L D d gs October, 16 1 2 Av gm(rd || RD) This is a CS amplifier configuration therefore the input is on the gate and the output is on the drain. Av gm(rd || RD) Av gmRD, r 10 R d Zi R1 || R2 Zo rd || RD Zo RD rd 10RD D October, 16 Figure vo(t) and vin(t) versus time for the common-source amplifier of Figure 5.28. [email protected] October, 16 An Amplifier Circuit using MOSFET(CS Amp.) Figure Common-source amplifier. [email protected] October, 16 A small signal equivalent circuit of CS Amp. Figure Small-signal equivalent circuit for the common-source amplifier. [email protected] October, 16 Figure vo(t) and vin(t) versus time for the common-source amplifier of Figure 5.28. [email protected] October, 16 Figure Gain magnitude versus frequency for the common-source amplifier of Figure 5.28. [email protected] October, 16 Figure Source follower. [email protected] October, 16 Figure Small-signal ac equivalent circuit for the source follower. [email protected] October, 16 Figure Equivalent circuit used to find the output resistance of the source follower. [email protected] October, 16 Figure Common-gate amplifier. [email protected] October, 16 Figure See Exercise 5.12. [email protected] October, 16 Figure Drain current versus drain-to-source voltage for zero gate-to-source voltage. [email protected] October, 16 Figure n-Channel depletion MOSFET. [email protected] October, 16 Figure Characteristic curves for an NMOS transistor. [email protected] October, 16 Figure Drain current versus vGS in the saturation region for n-channel devices. [email protected] October, 16 Figure p-Channel FET circuit symbols. These are the same as the circuit symbols for n-channel devices, except for the directions of the arrowheads. [email protected] October, 16 Figure Drain current versus vGS for several types of FETs. iD is referenced into the drain terminal for n-channel devices and out of the drain for p-channel devices. [email protected] October, 16