1 EE 560 MOS INVETES: STTIC CHCTEISTICS Kenneth. aker, University of Pennsylvania
IDE INVETE VOTGE TNSFE CHTEISTIC (VTC) B = 0 1 B 1 0 ogic 1 output ogic 0 output / Kenneth. aker, University of Pennsylvania
Vin CTU INVETE VOTGE TNSFE CHTEISTIC (VTC) OD + - V C out -> max output voltage when output is 1 V O -> min output voltage when output is 0 3 V I -> max input voltage which can be interprete as 0 V IH -> min input voltage which can be interprete as 1 V O = Kenneth. aker, University of Pennsylvania V I V th V IH
NOISE IMMUNITY ND NOISE MGINS 4 -> max output voltage when output is 1 V O -> min output voltage when output is 0 V I -> max input voltage which can be interprete as 0 V IH -> min input voltage which can be interprete as 1 max allowable 0 voltage min allowable 1 voltage V O V I V IH V O interconnect interconnect NOISE NOISE Kenneth. aker, University of Pennsylvania
NOISE IMMUNITY ND NOISE MGINS -> max output voltage when output is 1 V O -> min output voltage when output is 0 V I -> max input voltage which can be interprete as 0 V IH -> min input voltage which can be interprete as 1 5 V IH V I NM H Transition egion NM H = - V IH NM = V I - V O NM V O JUSTIFICTION FO V I, V IH : ET: = f( ), V out = f( + V noise ) ' = f( ) + 0 V V noise + H.O.term in Perturbe Output = Nominal Output + Gain x Ext. Perturbation Kenneth. aker, University of Pennsylvania
FIVE CITIC VOTGES: V O,, V I, V IH, V th etermine: --> DC Output Voltage Behavior --> Noise Margins --> ith an ocation of Transition egion 6 POE DISSIPTION ND DIE E Power Dissipation -> HET T j = T a + Θ P P --> P DC, P ynamic T j -> junction Temp T a -> ambient Temp Φ -> Thermal esistance P -> Power Dissipate P DC = C SSUME: = 1 50% of Op Time, 0 50% of Op Time Kenneth. aker, University of Pennsylvania P DC = [ I (V ="0")+ I (V ="1") DC in DC in ] DIE E --> MIN x
ESISTIVE-OD INVETE 7 = V GS + - V = V BS = 0 => V T,n = V T0,n = V DS CUTOFF: = V GS < V T0,n, = 0 INE: = V GS > V T0,n, = V DS < - V t0,n [ ] = (V V )V V in T 0,n out out STUTION: = V GS > V T0,n = V DS > - V T0,n = ( V T 0,n ) here: = µ n C ox, λ = 0 Kenneth. aker, University of Pennsylvania
= V GS + - V = = V DS B V O V I VIH = - V T0,n ST C IN 8 CCUTION OF : = < V T0,n => nmos Cut-off V T0,n = 0 = = 0 => = Kenneth. aker, University of Pennsylvania
B = - V T0,n ST C IN 8a V O V I VIH V T0,n = - V T0,n ST B C IN V O V I VIH V T0,n
+ - V = CCUTION OF V O : C C 9 = V GS = = V O where = V O = V DS V O = (V V )V V DD T 0,n O O V O V T 0,n + 1 V + O V DD = 0 [ ] where = = 0 < V O < V T0,n V O = V T 0,n + 1 ± V DD V T 0,n + 1 Kenneth. aker, University of Pennsylvania
= V GS + - V = CCUTION OF V I : = B 10 = ( V T 0,n ) Differentiate wrt to, i.e. B = 1 @ = V I => Kenneth. aker, University of Pennsylvania = V DS 1 = ( V T 0,n ) 1 ( 1) = (V I V T 0,n ) V I = V T 0,n + Fin : ( = V I ) ( = V I ) = 1 = (V V I T 0,n ) 1
= V GS + - V = CCUTION OF V IH : C C 11 = = V DS [ ] = (V V )V V in T 0,n out out Differentiate wrt to, i.e. 1-1 = k V IH n (V V ) in T 0,n V IH = V T 0,n + 1 = 1 @ = V IH => V O -1-1 V + out Kenneth. aker, University of Pennsylvania
CCUTION OF V IH : C V C DD Fin : = (V V ) V V IH T 0,n out out I where V IH = V T 0,n + 1 D = k n (V + V 1 T 0,n out V T 0,n ) 3 k V n out = 0 [ ] ( = V IH ) = 3 1 V IH = V T 0,n + 3 1 Kenneth. aker, University of Pennsylvania
CCUTION OF V th : = = V th => V DS = V GS > V GS - V T0,n B 13 = V th V th = ( V T 0,n ) V th = (V th V T 0,n ) V th V T 0,n 1 V + V th T 0,n = 0 V th = V T 0,n 1 ± V T0,n 1 + V T 0,n Kenneth. aker, University of Pennsylvania
SUMMY - ESISTIVE OD INVETE V th = V T 0,n 1 + V T0,n 1 + V T 0,n 1 V I = V T 0,n + 1 V out ( = V I ) = k V IH = V T 0,n + 1 n V 3 out ( = V IH ) = 3 V O = V T 0,n + 1 ± V DD V T 0,n + 1 14 = 5V = 5V V T0,n = 1 V = 8V -1 = V -1 = 4V -1 Kenneth. aker, University of Pennsylvania 5V
POE DISSIPTION - ESISTIVE OD INVETE 15 P DC = [ I (V ="0")+ I (V ="1") DC in DC in ] HEN = V O : DIVE nmos in CUT-OFF = = 0 => P( = 0) = 0 = 0 HEN = : C P DC (average) = V O Kenneth. aker, University of Pennsylvania
EXMPE 5.1 Consier the following inverter esign problem: 16 GIVEN: = 5 V, = 30 µ/v an V T0,n = 1.0 V Determine the / ratio of the river transistor an the value of the loa resistor to realize V O = 0. V. C For = V O => = = V O 5 0. = k ' n = (V V ) V V DD T 0,n O O 30x10 6 [ ] [ (5 1)0. (0.)] =.05x105 Ω NO UNIQUE /, Kenneth. aker, University of Pennsylvania
EXMPE 5.1 Cont. =.05x105 Ω (/) - TIO [kω] P DC (average) [µ] 17 1 3 4 5 6 05.0 10.5 68.4 51.3 41.0 34. 58.5 117.1 175.4 33.9 9.7 350.8 n + Polysilicon (ope) Metal 1 n + resistor 0-100 Ω/sq. Kenneth. aker, University of Pennsylvania GND
EXMPE 5. Consier a resistive-loa inverter with = 5 V, = 0 µ/v, V T0n = 0.8 V, = 00 kω, / = Calculate the critical voltages (V O,, V I, V IH ) on the VTC an etermine the noise margins. 18 = = 5 V = (/) = 40 µ/v => = 8 V -1 V O = V T 0,n + 1 ± V DD V T 0,n + 1 V O =0.147 V V I = V T 0,n + 1 V I =0.95 V Kenneth. aker, University of Pennsylvania
EXMPE 5. Cont. Consier a resistive-loa inverter with = 5 V, = 0 µ/v, V T0n = 0.8 V, = 00 kω, / = Calculate the critical voltages (V O,, V I, V IH ) on the VTC an etermine the noise margins. V IH = V T 0,n + 3 1 V IH =1.97 V 19 = = 5 V V O =0.147 V V I =0.95 V NM H = - V IH = 5V - 1.97V = 3.03V NM = V I - V O = 0.93V - 0.15V = 0.78V GOOD DESIGN => NM > /4 = 1.5 V Kenneth. aker, University of Pennsylvania
STUTED ENHNCEMENT-OD INVETE 0 = V GS, + - = B = - V T0,n ST C IN = V DS, V BS, = 0 => V T, = V T0,n V BS, < 0 => V T, = V T0,n V O V I VIH V T0,n OD: V GS, = V DS, => ST conition is YS STISFIED = k ' n ( V GS, V T. ) = ' ( ) V T. Kenneth. aker, University of Pennsylvania
= V GS, + - = B C 1 = - V T0,n ST IN Driver -> Cutoff, oa -> Sat = k ' n ( V T. ) = 0 Driver -> Sat, oa -> Sat V I V T0,n VIH V O ' ( V T. ) = k ' n ( V T 0,n ) Driver -> inear, oa -> Sat ' ( V T. ) = k ' n [ V T 0,n ] Kenneth. aker, University of Pennsylvania = V DS, ( )
+ = V BS, = 0 => V T, = V T0,n V BS, = - => V T, = V T0,n - = V GS, CCUTION OF : = = 0 = k ' n = V DS, V T. 0 ( ) = 0 V GS, = - > V T, < - V T Kenneth. aker, University of Pennsylvania
= V GS, + - = B C 3 = - V T0,n ST V O V I VIH IN = V DS, CCUTION OF V O : C ' ( V O V T. (V O )) = k ' n = ( [ V T 0,n ]V O V O ) V T0,n C EQUIES:, V T, (V O ) Kenneth. aker, University of Pennsylvania
' ( V O V T. (V O )) (1) = k ' n ( [ V T 0,n ]V O V O ) () 4 SSUME V T, = V T0,n YES Kenneth. aker, University of Pennsylvania CCUTE V O EQ. (1) CCUTE V T, (V O ) EQ. () V T, > ε? V O NO SOVE FO V O THOUGH NUMEIC ITETIONS CONTINUE ITETION?
= V GS, + - = B C 5 = - V T0,n ST V O V I VIH IN = V DS, CCUTION OF V I : B ' ( ) = k ' n V T. Differentiating wrt : ' V T. ( ) Kenneth. aker, University of Pennsylvania V T. ' = V T0,n ( V T 0,n ) ( V T 0,n ) B
' ( ) V T. SOVING FO / : ' = ' 1+ V T. V T. V in ' = ( V T 0,n ) V V out T. ( ) ( V T 0,n ) 6 IF = V T. << 1 ' ( V T 0,n ) V DD V T. ' = ( ) = Since / =-const; thus, efine V I = V T0,n ' ' = I = k Kenneth. aker, University of Pennsylvania
= V GS, + - = 7 = - V T0,n B ST IN slope = k C ' Kenneth. aker, University of Pennsylvania = V DS, V O CCUTION OF V IH : C VIH V ' OH V I = V T0,n ( V T. ) = k ' n C ( [ V T 0,n ] ) Differentiating wrt : -1 ( V T. ) V I out D V in V ' IH = ( V T 0,n ) V -1-1 out V V out in
' ( V T. ) = k ' n ( [ V T 0,n ] ) (3) ' ( V T. ) V -1 out V (4) in V ' IH = ( V T 0,n ) V -1-1 out V V out in V SOVING FO V in IH V IH = V T 0,n V ( DD V T. )+ + V out ( = V IH ) (5) where V T. ( ) = V T 0,n + γ ( φ F + φ F ) (6) an where (sub V IH Eq (4) into Eq () an solving for ) ( = V IH ) = Kenneth. aker, University of Pennsylvania + 3 ( V T. ) (7) 8
9 SSUME V T, = V T0,n YES CCUTE EQ. (7) CCUTE V T, ( ) EQ. (6) V T, > ε? NO CCUTE V IH EQ. (5) CCUTE V IH THOUGH NUMEIC ITETIONS CONTINUE ITETION? Kenneth. aker, University of Pennsylvania
V GG INE ENHNCEMENT-OD INVETE 30 = - V T0,n + - = B ST C IN = = V DS, V GG > + V T, V BS, = 0 => V T, = V T0,n V BS, < 0 => V T, = V T0,n V O V I VIH = 0 V T0,n = => incease NM H vs. ST enh-loa inverter V O > 0; arge river-to-loa ratio k neee to reuce V O & increase NM, NM H. Kenneth. aker, University of Pennsylvania