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Supplement S6.1 - Noise Margins for the Saturated Load Inverter We will now calculate the values of the VIL, VOH, VIH and VOL for the inverter with a saturated load device. Remember that VIH and VIL are defined by the points in the transfer function at which the slope is equal to -1. In Fig. S6.1.1 the slope of the transfer function abruptly changes as MS begins to conduct at the point where vI = VTNS. This point defines VIL: VIL = VTNS = 1 V for VOH = VH = VDD - VTNL = 3.39 V. 5.0V NMOS Inverter with Saturated Load 4.0V O VOH = VH u t p u 3.0V t V o l t a 2.0V g e 1.0V VOL VL 0V 0V VIL 1.0V VIH 2.0V v 3.0V VH 4.0V 5.0V 6.0V I Figure S6.1.1 - PSPICE simulation of the voltage transfer function for the NMOS inverter with saturated load S6.1-1 Next let us find VIH. In order to find a relationship between vI and vO, we observe that the drain currents in the switching and load devices must be equal. At vI = VIH, the input will be at a relatively high voltage, and the output will be at a relatively low voltage. Thus, we can guess that MS will be in the linear region, and we already know that the circuit connection forces ML to operate in the saturation region. Equating drain currents in the switching and load transistors: iDS = iDL Ê v ˆ K 2 K S Á v I - VTNS - O ˜vO = L (VDD - vO - VTNL ) Ë 2¯ 2 with (S6.1.1) ÊW ˆ K S = K n' Á ˜ Ë L ¯S † and ÊW ˆ K L = K n' Á ˜ Ë L ¯L ∂v The point of interest is O = -1, but solving for the value of vO can be quite tedious. ∂v I † Since the derivatives are smooth, continuous and non-zero, we will assume that -1 ∂vO Ê ∂v I ˆ =Á ˜ ,†and will solve for vI in terms of vO: ∂v I Ë ∂vO ¯ v I = VTNS + † or † vO 1 K 2 + (VDD - vO - VTNL ) where K R = S 2 2K R vO KL 2 ˘ vO 1 È (VDD - VTNL ) Í v I = VTNS + + - 2(VDD - VTNL ) + vO ˙ 2 2K R ÍÎ vO ˙˚ and (S6.1.2) 2 ˘ ∂v I 1 1 È (VDD - VTNL ) Í˙ @ + + 1 ∂vO 2 2K R ÍÎ vO2 ˙˚ In this last expression, the dependence of VTNL on vO has been neglected for simplicity. (This approximation will be justified shortly.) † Setting the derivative equal to -1 at vI = VIH yields S6.1-2 2 ˘ 1 1 È (VDD - VTNL ) Í˙ -1 = + + 1 2 2K R ÍÎ vO2 ˙˚ and solving for VOL = vO for vI = VIH yields: VOL †V - VTNL = DD 1+ 3K R and V (V - VOL - VTNL ) + OL + DD 2 2K RVOL VIH = VTNS 2 (S6.1.3) For†the inverter design of Fig. 6.26 with VTNL = 1 V, we find VOL = VDD - VTNL = 1+ 3K R VIH = 1+ VDD - VTNL 1+ 3 (W /L) S (W /L) L = (5 -1)V = 0.66V 1+ 3( 3.53)( 3.39) 0.66V 1 1 2 + (5 - 0.66 -1) V 2 = 2.04V 2 2( 3.53)(3.39) 0.66V With these values we can check our assumption of the operating region of MS: VGS - VTNS = 2.04 - 1 = 1.04 V and VDS = 0.66 V . † Since VDS < VGS - VTN, the linear region assumption was correct. In Fig. S6.1.1, it can be seen that the calculated values of VIH and VIL agree well with SPICE simulation results. Thus, the approximation of neglecting the voltage dependence of the threshold voltage of the load device does not introduce significant error. However, if we so desire, we can use an iterative procedure to improve on the estimates above. The calculated value of VOL can be used to determine a better estimate for VTNL, and this new value of VTNL can then be used to improve the estimate of VIH. For VOL = 0.66V, VTNL = 1V + 0.5 V ( (0.66 + 0.6)V ) - 0.6V = 1.17 V and the new values of VOL and VIH from Eqn. (S6.1.3) are equal to † VOL = 0.63 V and S6.1-3 VIH = 1.99 V . This process can be continued, but it has already converged as indicated in Table 7.2 below. (Note that this procedure is another example of an algorithm that is very easy to implement using MATLAB or a spreadsheet.) Table S6.1.1 - Iterative Update of VOL and VIH Iteration # VOL VTNL V'OL VIH 0 --- 1.00 V 0.66 V 2.04 V 1 0.66 V 1.17 V 0.63 V 1.99 V 2 0.63 V 1.17 V 0.63 V 1.99 V NML = VIL - VOL = 1 - 0.63 = 0.37 V NMH = VOH - VIH = 3.39 – 1.99 = 1.40 V S6.1-4