NMOS Cross SectionPoly-SiViSiO2Poly SiGaten+Sourcen+Drainp+BodyField OxideField OxideViaVia ViaInverted ChannelPWELL1The MOS Field-Effect TransistorV++++++++SiO2VGs++++++++d----------------------D(n+) S(n+)InversionDepletionDepletion2The MOS Field-Effect Transistor:RegionLinear TGDSAT:)V-V(VRegion Saturation =⎤⎡=Dinm1ZVCLZggμ2DSATinm(sat)ZVCμLZg =()⎥⎦⎤⎢⎣⎡−−=2inD21CLZIDDTGVVVVμ2DSATinD(sat)VCμLZI=IDPinch-offVD (Sat.)VG = +3V+1VG = +23VG= +1VDBasic MOSFET operation•PMOS–ap-type channel is created in a n-type substrate.a ptype channel is created in a ntype substrate.•NMOS an ntype channel is created in a ptype–an n-type channel is created in a p-type substrate.4Basic MOSFET operationIDPinch-offVD (Sat.)VG= +3VG=+1VG = +2VG 1VDICIB5VCEThe MOS Field-Effect TransistorVVxVDdVxXL0xxx +dx6The MOS Field-Effect Transistor:RegionLinear TGDSAT:)V-V(VRegion Saturation =⎤⎡=Dinm1ZVCLZggμ2DSATinm(sat)ZVCμLZg =()⎥⎦⎤⎢⎣⎡−−=2inD21CLZIDDTGVVVVμ2DSATinD(sat)VCμLZI=IDPinch-offVD (Sat.)VG = +3V+1VG = +27VG= +1VDWhat is “Pinch off”• For a given VG, you have a maximum amount of current you can flow through the ygchannel (regardless of VD).•When VD=VG-VT the channel is flowingWhen VD VGVT the channel is flowing as much current as it can so we call it pinched off (even though current continuespinched off (even though current continues to flow).8Delay in a CMOS inverterDl i h hi d i k f hDelay is how much time does it take for the gate to invert the signal.RC!VTVTDesigner chooseswidths.9nanospecTo design a simple current mirror, we need to know VT, KN, λ,,IN IOUT ROUT d A ll ifi ti10IN, IOUT, ROUT, and Area are all specificationsTo design a simple current mirror, we need to know VT, KN, λ,,IN IOUT ROUT d A ll ifi ti11IN, IOUT, ROUT, and Area are all specificationsCurrent Mirrors operate in Saturation12VT=.6Vλ can be quickly found from an ID, VDS curveλ017λ=.01713We are ready to begin?Specification:IREF=100uAIOUT 100 AIOUT=100uARo=107OhmsArea=500μm2Maximum Output swingswingThe ro at the IOUT we want14is only around 106Ω!Which is more important Ro or IREF? New spec: ro >106ΩTo design W, we need to know VGS. How do we pick it in a reasonable manner?Let’s try some even numbers:VGS 1VGS=115Channel Length Modulation• There is a depletion region near the Drain body interface that creeps under the gate.• This makes the actual channel length smaller than the length of the gate. If this depletion width is i il i i h l h h i ill ffsimilar in size to the gate length, then it will affect the transistors IV curve.Th l Vth l th d l ti i d•The larger VDthe larger the depletion region and the smaller the channel length: This leads to increased current (looks like the early effect)16increased current (looks like the early effect)Channel Length Modulation• The depletion regions of the two junctions can meet if the channel length is small enough. At this point the current is not controlled by the gate voltage but by V2/L3not controlled by the gate voltage but by VD2/L3.• Sub threshold current:– Some electrons are flowing before a channel is completely gpyinduced, and the is diffusion of electron from source to drain that can make it impossible to turn the device off.–Doping the source and drain heavily near the contacts and lightly pg y g ynear the gate can alleviate much of this problem. This lowers the electric field thus less electrons can be injected below threshold.– (Why? This is a diode question.) 17Channel Length ModulationSGDn+n+DepletionL'LΔ+++DGSn+n+pppDSGn+n+nnp18Channel Length Modulation5VWhat would VOUT be if VIN=5V?ID1kohmVOUTVINVOUTVGVD1/λ()()DTGinDVVVLZCIλμ+−=1219()()DTGDLSubstrate Bias Effect• We add a contact to the body (Normally the y( ysource and body are ties together at ground)+D +G S())(221NMOSVqNVas⎥⎤⎢⎡−=Δεn+n+()())(2)(212PMOSVqNVNMOSVCVBdsTBiT⎥⎦⎤⎢⎣⎡−=Δ⎥⎦⎢⎣=Δεnnp())(CBiT⎥⎦⎢⎣20-BMOSFETS• Measuring–KN, VT,γ,λKN, VT, γ, λ– Effective channel length and width21Regular MOSFETSThe width and the lengthchange from 5x5μm up to 80 8080x80μm.22Finding VT for a MOSFET• Set VDS to less than the Vbi of the Drain/Body Junction (.2-.5V)fbl dl• Sweep VGS from below VT to around 4-5 volts– If you are unsure of what VT is you can start at –5Volts•Measure ID•Measure ID• Plot gm(ΔID/ΔVG) vs. VGS•Find the xvalue for the maximum gm•Find the xvalue for the maximum gm• Find the tangent the of ID-VGS at the value point•VT is where the line formed by the tangent23VT is where the line formed by the tangent intersects the x-axis.IDTfIDTangent of ID atthe xvalue that give gm-maxIDgmgm-max24VTVT NumericalIDmax=The ID at the point of maximum gmVGmax=The gate voltage at the point of maximum gmgmmax=The largest value of ΔID/ΔVG()maxmaxmaxmaxgmVGgmIDVT×−−=25KNKN• Since Mobility varies with the electric field fG bdd difrom Gate to body and source to drain KN=μnCox varies as well. This means KN iOldkhis not a constant. One could take the average of the gm plot or use:eUDStTGScoxSiVUVVUtoxKNnewKN⎟⎟⎠⎞⎜⎜⎝⎛−−⋅⋅=⋅)()(εεThis requires extensive testing and a regression.26One could also report just gm-max.Do a ID, BGS plot for different VSB voltages.Body Effect (γ)27Body Effect (γ)TSBTVVVVφγ22)(0+−=FSBFVφϕ22−+)22()(VVVVφφγ−++=28)22()(0FFTSBTSBVVVVφφγ−++=Channel Width Modulation (λ)29Sample Regular MOSFETThe Body, GateSource and Drainare already labeled.The top number is the asdrawn width of the gate in The bottom number is theas drawn lengthof the gate in microns.microns.g30Effective Channel LengthMeasureID VGS flfor severalGate Lengths.31Effective Channel LengthCalculate gm,ddand record gm maxfor each channellength.g32Effective Channel LengthPlot 1/gm-max vs. das drawn channel length.Extrapolate this lineuntil it cross the x-axis.Where this line crosses theNote since this is a simulation the lineis perfectly straight.x-axis is the ΔL of the process.In measured devices you would have to do a regression. (This is easy to do in excel.)33For this example ΔL is around .1um.Field Oxide MOSFETsThese MOSFETS usethe field oxide for the tidSithgate oxide. Since the Filed oxide is much thicker that the active oxide layer, the VT should be much higher.However since we haveHowever since we have a SOG layer over the field oxide, which has Q th VT i ht ba Qss, the VT might beNegative! The μn x Coxwill be much