Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 Recitation 9: MOSFET VI CharacteristicsBefore the class first do an exercise on MOS capacitor.Under T.E., suppose we are under depletion, positive charges at M-O interface, negativecharges (Na−) at O-S interface & depletion region xdo.How does the C-V measurement curve look like? 1=Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 111=+CtotCoxCdoxsCox= ,Cd=toxxd(VGB)Useful relations:VFB= −(φgate− φbody)1VT(n+/p)=VFB− 2φp+2sqNa(−2φp)CoxCmin 1oxCox1+2C2(VT−VFB)qsNaWhere is Cmin? When VGBchanges, Coxdoes not change. Cdchanges due to xd(VGB).xd=0atVFB,xd= xd,maxat VT= Cmin⇒In tutorial, you can also find what the GV curves look like for p+− n MOS or p+− p MOS,or n+− n MOS.MOSFET Device• We only talked about 2 terminals in our MOS capacitor. Where are the other termi-nals? Source/Drain. In the MOS capacitor, S/D tie to bulk ground.→Figure 1: MOSFET: 4 terminal device2Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 • As we mentioned, VGB=⇒ VG− VB. In MOSFET, we usually have,VDS= VD− VS VGS= VG− VS You can do manipulation: VGD= VG− VD= VGS− VDS• If the substrate of MOSFET is p-type, what type of MOSFET device this is? n-MOSor p-MOS? It is n-MOS. MOSFET operates when it is in Inversion. So for n-MOS: Source/Drain are n+.Thuswehavetwop− n+junctions between source-substrate (bulk), n+(D) and p (B). When we apply biases, we try to keep VBS≤ 0,VBD≤ 0 otherwise the p− n+junction will conduct. • When we use n-MOS, we always try to use source as reference: VGS,VDSetc. To startwith, we let VBS=0 = VGB= VGS⇒From yesterday’s discussion or 6.002, what are the I-V characteristics (i.e. whenapplying VDS, what does IDSlook like) of a n-MOS?1. Remember we need to apply positive VGB(i.e. VGShere) in order to reachthreshold. Before threshold, no conduction.= VGS<VT,IDS= 0 always (cutoff)⇒2. VGS≥ VT, now we have inversion layer. If the VDS= 0, what is the inversionlayer charge density? |Qn| = Cox(VGS− VT) 3Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 When VDS> 0, how will this charge density change? Now from S to D, alongthe channel interface, potential is no longer 0.V (y)= 0(0 <y<L) at each location y∴,|Qn(y)| = Cox(VGS− V (y) − VT)Decrease from source(y =0)V (y) = 0 to minimum at D(y = L, V (L)=VDS).To calculate IDSremember current ∝ charge density, ∝ carrier velocity.IDS= W ·|Qn(y)|·vy(y)(vy(y) is velocity in the y direction at location y) (1)How to calculate vy(y)? v = μE. So need to know Ey(y). How to know Ey(y)?·Ey(y)=dV (y)(we have V (x, y) at each location:dVwill give Ex)dydxTherefore to plug everything in the equation (1)IDS= w · Cox(VGS− V (y) − VT) · μn·dVdy(y)4 Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 Integrating,yv(y)IDSdy = wμnCox(VGS− VT− V (y)) dV(y) (2)00IDS· y=(VGS− VT) V (y) −1V2(y) (3)wμnCox·2So we can solve the potential along each location y.V (y)=(VGS− VT) − (VGS− VT)22IDS·y−wμnCoxSince IDSshould be the same everywhere, when y = L, V (y)=VDS, plug in (3)IDS=wLμnCox(VGS− VT−V2DS) · VDSWhen VDSis small,wIDSLμnCox(VGS− VT) ·VDS→ linearGate voltage controlled resistorThen as VDSincreases, IDSbend over. When VDS= VGS− VT: IDSsaturatesw 1IDSAT= μnCox· (VGS− VT)2L2 (only depend on VGS)5Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 PMOSFET Case Will need VBS≥ 0,VBD≥ 0 always. (typically VBS= 0). Now in order to have inversion:VGB= VGS< 0In p-MOS, we useVSG= VS− VG> 0VSD= VS− VD> 06Recitation 9 MOSFET VI Characteristics 6.012 Spring 2009 When working with p-MOS, simply transform n −→ pVTn−→ − VTpIDn−→ − IDpVGS−→ VSGVDSTriode/linear: − IDp−→=VSDwLμpCoxVSG+ VTp−VSD2VSDVSD≤ VSG+ VTpSaturation: − IDp=w2LμpCox(VSG+ VTp)2VSD≥ VSG+ VTp7MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and CircuitsSpring 2009 For information about citing these materials or our Terms of Use, visit:
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