Lecture 12: FET AC Properties

Transcription

1 Lecture 1: FET AC Properties Lecture 11, Hih Spee evices 016 1

2 Lecture 1: FET AC Properties Quasi-static operation iffusive an Ballistic FETs y-parameters Hybri p-moel Non-quasi Static effects Reain uie for chapter 6: , ,. Omit any specific MESFET relate topics. Non-quasi static effects briefly. Note that Liu only covers the iffusive FETs Lecture 11, Hih Spee evices 016

3 Time varyin voltaes i (t) The transistor is chare neutral: Q G (t)+q CH (t)=0 I +i (t) The channel chare is ivie into two parts: Source an rain chare: Q CH =Q S +Q + When V s chanes (reasonable slowly): I (conuction current) chanes The amount of channel chare chanes -> chares has to be supplie throuh i an i s! i supplies the ate chare! i (t)=q / nee to etermine Q (v,v,v s ) Lecture 11, Hih Spee evices 016 3

4 Quasi-Static Approximation Quasi-static approximation: the chare istribution in the channel reacts instantiounsly to any chanes in v S, v GS. tr, Q I ch L v sat, L v x, L v t Vali for time scales loner than Given by the time it takes for the electrons to propaate from source to rain f = 1 πτ tr 5 THz for a ballistic fet with L =60 nm. Ex let v GS chane from below V T to abiove V S >V S,sat n(x) t=t 3 v s (t) I t=t 3 t=t V T t=t 1 t=t t=t 1 x=l x Lecture 11, Hih Spee evices 016 4

5 Chares an currents Gate Current rain Current Source Current i = Q G(t) i (t) = I t + Q t i S = I t + Q S t i : total instantaneous rain current I : Quasi-static rain current (From previous lectures!) /: capacitive/charin current from chane in channel/ate chare storae. Flows only when the voltae chanes! The evice is chare neutral: Q G = Q S + Q Lecture 11, Hih Spee evices 016 5

6 Charin Currents V S i Q G V G V Charin currents: Q t + J x = 0 i s Q S Q i i G = Q G t Q S t = V Q S + V G Q S + V S Q S = i t V t V G t V s S Q t = V Q + V G Q + V S Q = i t V t V G t V S Q G t = V Q G + V G Q G + V S Q G = i t V t V G t V S Q S t + Q t + Q G t = 0 Quasi-static: Chare neutral evice KCL also hols i S = Q s t i = Q t efinition of transcapacitances C ii = Q i V i C jk = Q j V k This makes most capacitances positive Lecture 11, Hih Spee evices 016 6

7 Charin Currents - capacitances V S V G V Q t + J x = 0 i s i i = C SS C S C SG C S C C G C GS C G C GG v s v v Ex: Voltae is applie to the ate terminal only i = C GG v i = C G v i s = C SG v V GS i C GG If all voltaes are equal: i s = i = i = 0 If one voltae is applie: i s + i + i = 0 Rows/ columns sum to zero Lecture 11, Hih Spee evices i s C SG i V GS C G i

8 Charin Current / total Current i s i i = C SS C S C SG C S C C G C GS C G C GG v s v v i Common Source C G C i = C GG C G v GS t v S t We nee to calculate Q G (V GS,V S ) an Q (V GS,V S ) Lare Sinal, Quasi-static FET moel v GS i = I v GS, v S + C v GS, v S t C v S G v GS, v S t i G = C GG v GS, v S v GS t C v S G v GS, v S t Lecture 11, Hih Spee evices 016 8

9 iffusive FET Gate Chare: Q G Neative channel chares ivies onto rain an source chares Positive chare on ate terminal to make evice neutral Q S Q G Q More positive ate bias more positive chare on the ate Chare neutral evice: More neative chare in the channel V S,sat = (V GS V T ) u ch x, t = u ch 0, t 1 x L 1 α α = 1 v S V S,sat, V S < V S,sat 0 n x, t = WC G v GS t V T 1 x L 1 α Q Total Gate chare: G ( t) L L ' x ' 1 WCox uch x, t x uch (0, t) 1 1v s, v, v WLCox vs VT L Lecture 11, Hih Spee evices 016 9

10 Ballistic FET Gate Chare: Q G E FS Simplest possible assumption: n x = n 0 = C G (v GS V T ) C G = 1 C ox + C q + 1 C c 1 L G L G Q G = n x x 0 Q G = WL C G (v GS V T ) A real ballistic FET has a complicate ban profile Crue approximation : constant potential until rain reservior Lecture 11, Hih Spee evices

11 Ballistic FET Gate Chare: Q G E FS I W = C G 8ħ v GS V T 3m πq C G v GS V T I = qn 0 v(0) W E ΔE kin qv S x L G L G L G x A real ballistic FET has a complicate ban profile Simple approximation : quaratic rop between top-of-the-barrier an the rain Carriers ain extra kinetic enery Since J(x) = constant : n(x) ecreases mv x Lecture 11, Hih Spee evices = E 0 + ΔE kin

12 Ballistic FET Gate Chare: Q G I W = qn 0 v(0) ΔE kin = qv S x L G v x = v 0 + qv S m x L G x=0 x=l G Q G = L G n x x 0 Total Gate Chare n x = n 0 v 0 v 0 + qv S m x L G Q G = WL GC G V GS V T v 0 m qv S sinh 1 qv S m v 0 v 0 = 8ħ 3m qπ C G v GS V T n 0 = C G v GS V T Lecture 11, Hih Spee evices 016 1

13 Total channel / ate chare in saturation Ballistic - flat t ox = 5nm t w = 7 nm m * =0.03 m 0 e r =e ox =14 L =50 nm V S =0.5V V S =0.6V Ballistic - quaratic Q G ecreases with increasin V S for the ballistic FET in saturation Lecture 11, Hih Spee evices

14 Q G for a ballistic FET in saturation V S =0.5V V S =0.6V Ballistic - quaratic Q G ecrease with increasin V S since: A) Less back injecte rain chare B) More back injecte rain chare C) Larer carrier velocity towars the rain ) Smaller carrier velocity towars the rain E)??? nano.participoll.com A B C E Lecture 11, Hih Spee evices

15 Transcapacitances: C G an C GG How much oes the total channel chare chane as V G an V is varie iffusive FET: C C Q V G G Q V WLC G ' ox WLC 3 ' ox Ballistic FET constant potential (Saturation) Ballistic FET quaratic potential (Saturation) C = WL G C G C = 0 C = Q G V G = δ xy : = 1 if x=y = -1 if x y C xy = δ xy Q x V y C = Q G V > Lecture 11, Hih Spee evices

16 Transcapacitances in saturation: C GG iffusive V S =0.5V V S =0.6V Ballistic - quaratic Lecture 11, Hih Spee evices

17 Transcapacitances C G C GG V C S =0.5V Ballistic - G quaratic V S =0.6V iffusive Lecture 11, Hih Spee evices

18 rain Chares Ballistic FET Neative channel chares ivies onto rain an source chares Q G + Q S + Q = 0 Q S Q G Q Q : When Q G chanes how much of that chare is supplie from the rain? Q S chare irectly supplie from the source Q chare irectly supplie from the rain Q S J Q J n x = J v + (0) + qv S m x L G n x J v 0 + qv S m x L G Lecture 11, Hih Spee evices

19 rain Chares iffusive FET Neative channel chares ivies onto rain an source chares Q G = Q S + Q Q S Q G Q Q : When Q G chanes how much of that chare is supplie from the rain? efinition of rain chare: Q L ' WCox ( t) xuch ox s T L Q x i t I Q t ' xx WLC v V V s =0.5V Q G V s =.0V Q Q x V =0.7V Q Q G V =0.3V V T =-0.5V In saturation: =0 Q /Q S =40/ Q S V GS (V) V S (V) Lecture 11, Hih Spee evices Q S

20 Transcapacitances iffusive FET C C Q V G G Q V WLC G ' ox WLC 3 ' ox C Q V G WLC ' ox C xy = δ xy Q x V y C Q V WLC ' ox δ xy : = 1 if x=y = -1 if x y Note that C C Lecture 11, Hih Spee evices 016 0

21 Transcapacitances Ballistic FET in saturation C GG Q G = WL GC G V GS V T v 0 m qv S sinh 1 qv S m v 0 Q S + Q G = 0 Q 0 C G C GG = Q G V G = WL G C G C,C G C G = Q G V = WL G C G C G = Q V G 0 C = Q V 0 Short channel effects can lea to neative C G! Lecture 11, Hih Spee evices 016 1

22 minutes excercise C Q V I G WLC ' ox 3 1 =0 C Q V WLC 1 3 A B C E Lecture 11, Hih Spee evices 016 G ' ox V What is the physical reason that in saturation C =0 an C > 0 for an ieal, iffusive FET? A) The rain voltae oes not influence the total channel chare after pinch-off B) The ate voltae oes not influence the total channel chare after pinch-off C) Q S +Q +Q G =0 oes not hol in saturation. ) This is ue to approximations. C ij must always be C ji. E)??? nano.participoll.com 3 3 0

23 Lare Sinal Moel common source i i t Q v ( t), v ( t), v t I v ( t), v ( t) I C s s v s ( t) Q C s v v s, vs C v s, vs v v s v v s, vs Cv s, vs ( t), v ( t), v s s ( t) s Common source: vs v 0 v s v v s Gate Source v G C C v S C C v GS rain I (V GS, V S ) Lecture 11, Hih Spee evices 016 3

24 Small sinal C Moel Common Source 0 Gate rain i i 0 I V GS v s, V S v s v GS m V GS I I V V GS, VS vs vs GS V S I V S V GS Source m m m WCGn L WC v ' ox sat V V 1 GS T ' WCoxn L f ( V, V s s V ) GS V T Lon channel Velocity Saturation m = 3 C G W 8ħ 3m qπ C G (V GS V T ) > 0 Ballistic Lecture 11, Hih Spee evices 016 4

25 Quasi-static Common Source Small Sinal y-parameters i i t t C m v v s s C v s v C s v s C v s v i s v~ e ~ i e jt jt v i s v~ ~ i e s e jt jt ~ i ~ i y y,, y y,, v~ v~ s s y y y y,,,, jc m jc jc jc Lecture 11, Hih Spee evices 016 5

26 Small Sinal Hybri-p moel ~ i ~ i + - y y,, y y,, v~ v~ -y 1 i i 1 y 11 +y 1 (y 1-y 1)v s s v y y y y C -C =C s -C +C =C s,,,, jc m jc jc jc y +y 1 Gate C m v s rain C s C s jc m v s Source C m =C -C Lecture 11, Hih Spee evices 016 6

27 Small Sinal Hybri-p moel Gate C C s C s m v s jc m v s rain General FET Moel C m =C -C Source C Gate C s C s m jωc v s Ballistic FET in saturation Gate C s m + jωc v s iffusive FET in saturation Lecture 11, Hih Spee evices 016 7

28 Non-Quasi-Static For very short timescales: Chares has to be supplie from the source transient involves a charefront more complex math! tr, n(x) Q I ch u x L v sat, u x L v x L, ( V V n GS t x, t u x t n ch ch ch, iffusive Transistor: Continuity an rift equation T ) =0 t 1 t t 3 t 4 v s (t) I t = s V T t t 3 t 4 t=t 0 t 0 t 1 x=l x Lecture 11, Hih Spee evices 016 8

29 Small sinal non-quasi static u x U u x t x, t u x t n ch ch ch, ~ i ch CH j, I WC n I ' ox U CH ~ i ch Assume small sinal sinusoial perturbation: u i ch ch x, t UCH x u~ chxe ~ jt x, t I x i xe CH ch jt ~ i ch 3 3 U C U Iˆ ju 3/ C U Iˆ ju 3/ CH 1 CH 1/3 Î is the moifie Bessel function CH Express Î as a series expansion, an solve for u ch (x) from bounary conitions (very teious alebra, see pp ) CH 1/3 CH Lecture 11, Hih Spee evices 016 9

30 Non : Channel Resistance y y, N, N jc jc jc jc j j / j / 0 1 / j / 0 1 Essentially all corrections are zero/small if << 0... After plenty of alebra, one obtains the N y-parameters VGS V 0 n L (1 ) T C 1 0 s Source r ch Gate C s r ch important when m is small, i.e. Not fully linear reion / small V s Or at very hih frequencies! r Gate ch Re y m C s r ch s, N m m 1 + jω ω 0 τ 1 r rain m 1 j( / 0) Lecture 11, Hih Spee evices Source

31 Summary hybri p C Quasi-Static AC Gate rain Gate C rain v GS m V GS C s C s m ( 1 jcm) Source Source Gate C s r ch r + v 1 - rain m v 1 Non Quasi-Static AC Source Inorin C m Lecture 11, Hih Spee evices