Show the details of the derivation for Eq. (6.33) for the PMOS device.

Transcription

1 Problem 6.11 Rahul Mhatre Show the details of the derivation for Eq. (6.33) for the PMOS device. Since the device is a PMOS MOSFET, source and drain are p+ regions and the substrate is an nwell. Therefore, all the bulk is the most positive terminal and source is more positive than drain and gate. We follow the voltage convention of Figure 6.1 in the book. Also, the threshold for PMOS device (V THP ) is a negative quantity. Consider the Figure 6.11, where V SG > V THP, so that the surface under the oxide is inverted and V SD > 0, causing a drift current to flow from the source to the drain. V BS V SG V SD Gate Source Drain FOX n+ p+ p+ V(y) FOX Body y+dy y n-well p-substrate Figure 6.11 Calculation of large-signal behavior of the PMOS MOSFET in the triode region V(y) is the channel voltage with respect to the source of the PMOS (a negative quantity) at a distance y away from the source. The potential difference between the Gate electrode and the channel is V SG V(y). The charge/unit area in the inversion layer is given by Q ch = C ox. [V SG - V(y) ] Eq (6.1)

2 Charge Q b (holes) is present in the inversion layer from the application of the threshold voltage, V THP, necessary for conduction between the drain and source. This charge is given by, Q b = C ox. V THP Eq(6.) The total charge available in the channel, for conduction of a current between the source and the drain, is given by the difference between Equation (6.1) and Equation (6.), which is Q I (y)= C ox. [ V SG - V(y) - V THP ] Eq(6.3) where is the charge in the inverted channel. The differential resistance of the channel region with a length dy and a width W is given by, dr =( 1/ µ p Q I (y) ).dy/w Eq(6.4) where µ p is the average hole mobility through the channel with units cm /V.sec. The differential voltage drop across this differential resistance is given by dv(y) = I D. dr = I D. dy / (W µ p Q I (y) ) Eq(6.5) Substituting Equation (6.) and rearranging I D. dy = W µ p C ox. [ V SG - V(y) - V THP ]. dv(y) Eq(6.6) We define the transconductance of the PMOS MOSFET as, KP p = µ p C ox = µ p ε ox / t ox Eq(6.7) The current can be found by integrating the left side of Equation (6.6) from source to drain, that is from 0 to L and the right side from 0 to V SD. This current flows from source to drain. This is as shown below: I D. 0 L dy = W KP p 0 VSD [ V SG - V(y) - V THP ]. dv(y) I D = KP p (W/L) [ (V SG - V THP ) V SD (V SD / ) ] for V SG > V THP and V SD V SG - V THP Eq(6.8) Eq(6.9) This current flows from Source to drain.

3 Problem 6.1 John Spratt Using Eq. (6.35) estimate the small-signal channel resistance (the change in the drain current with changes in the drain-source voltage) of a MOSFET operating in the triode region (the resistance between the drain and source. Eq 6.35: I D =β*[(v gs -V thn )V ds -V ds /] r = V ds / ID = (V ds1 - V ds ) / [(β*[(v gs -V thn )*( V ds1 )-( V ds1 ) /]- β*[(v gs -V thn )*( V ds )-( V ds ) /]] = (V ds1 - V ds ) / [(β*[(v gs -V thn )*( V ds1 - V ds )-( V ds1 ) /+( V ds ) /]] =1 / [(β*[(v gs -V thn )-( V ds1 + V ds )/]] r =1 / [β*(v gs -V thn - V ds) ]

4 Problem 6.13 Steve Bard Question: Show, using Eqs and 6.37, that the parallel connection of MOSFETs shown in Fig behaves as a single MOSFET with a width equal to the sum of the individual MOSFET s widths. D Id G Id1 Id Id3 Id4 S Equivalent of Fig From Kirchoff s Current Law, we know that Id = Id1 + Id + Id3 + Id4. So if each MOSFET has the same KP, L, V GS, V DS and V THN, equations 6.33 and 6.37 become: V ( V ) DS GS VTHN VDS W1+ W + W 3 + W 4 Id = KPn for Eq L [( V V ) ] W1+ W + W 3 + W 4 Id = KPn GS THN for Eq L This shows that the total drain current, Id, is equal to a single MOSFET with a width equal to W1 + W + W3 + W4.

5 Problem 6.14 Shambhu Roy Show that the bottom MOSFET. Fig 6.1. in a series connection of two MOSFETs cannot operate in the saturation region. Neglect the body effect. Hint: Show that M1 is always in either cutoff ( V GS1 < VTHN ) or triode ( VDS1 ` < VGS1 VTHN ). + M W L1 Vgs + Vgs1 M1 W L1 Vds1 - For V GS1 < VTHN both MOSFETs are in the cutoff region. When V > V GS1 THN VDS1 = VGS1 VGS Also VDS1 VGS1 VTHN VGS1 VGS VGS1 VTHN VGS < VTHN Which cannot be true, therefore if top MOSFET has to operate then the bottom MOSFET cannot be in Saturation.

6 Problem 6.15 Harish Reddy Singidi Show that the series connection of MOSFETs shown in fig. 6.1 behaves as a single MOSFET with Twice the length of the individual MOSFETs. Again neglect the body effect. V D M V D V G W/L V G V 1 W/(L 1 +L ) W/L 1 M1 Assuming both MOSFETs are in triode region For M 1 I D1 = I D I D1 = I D = KP n (W/L 1 ) [(V G V THN )V 1 - V 1 /] (I D L 1 )/ (KP n W) = [(V G V THN )V 1 - V 1 /] As Both MOSFETs are in series i.e., I D1 = I D = I D For M I D = I D I D = I D = KP n (W/L ) [(V G V 1 - V THN )(V D -V 1 ) - (V D -V 1 ) /] (I D L )/ (KP n W) = [(V G V 1 - V THN )(V D -V 1 ) - (V D -V 1 ) /] [(I D L 1 )/ (KP n W)] + [(I D L )/ (KP n W)] = [(V G V THN )V 1 - V 1 /] + [(V G V 1 - V THN )(V D -V 1 ) - (V D -V 1 ) /] [(I D (L 1 +L )/ (KP n W)] = [(V G V THN ) V D - (V D ) /] This is the current from drain to source for a single MOSFET with length (L 1 +L ) If L1= L = L [(I D L)/ (KP n W)] = [(V G V THN ) V D - (V D ) /] I D = [(KP n W)/L] [(V G V THN ) V D - (V D ) /]

7 Problem 7) Matlab code: % constants from table 6. KPn = 10e-6; W = 3e-6 L = 1e-6 Vthn = 0.8; lambda = 0.01; Vds = [0:0.01:5]; for Vgs = 1:5 Vdssat = Vgs - Vthn; IdSat(Vgs,:) = KPn/*W/L*(Vgs-Vthn)^*(1+lambda*(Vds(1,:)-Vdssat)); IdTri(Vgs,:) = KPn*W/L*((Vgs-Vthn)*Vds(1,:)-(Vds(1,:).^)/); for i = 1:length(Vds) if Vds(i)<=(Vgs - Vthn) Id(Vgs,i)=IdTri(Vgs,i); else Id(Vgs,i)=IdSat(Vgs,i); end end plot (Vds(1,:), Id(Vgs,:),'b') hold on end title('characteristics of a long-channel NMOS device'); xlabel('vds (V)'); ylabel('id (A)'); grid; 3.5 x 10-3 Characteristics of a long-channel NMOS device 3 VGS=5.5 Id (A) 1.5 VGS=4 1 VGS=3 0.5 VGS= VGS= Vds (V)