Lecture 8: Ballistic FET I-V 1
Lecture 1: Ballistic FETs Jena: 61-70
Diffusive Field Effect Transistor Source Gate L g >> l Drain Source V GS Gate Drain I D Mean free path much shorter than channel length Diffusive transport drift and diffusion x x J qnn µ n =qt/m * ' x, t C U x 3 q' ch U CH ox x x U 0 1 1 What happens when an electron travels through the channel without any scattering events? CH U CH =V GS V T -V CS (x) 1 ch c qn L n
Diffusive Field Effect Transistor Source Gate Drain L>>l 0 L<<l 0 Mean free path much shorter than channel length Diffusive transport drift and diffusion Mean free path much longer than channel length Ballistic transport Fully ballistic transport will give the upper limit of a transistor s performance! How do we calculate the ballistic current flowing through a FET? 4
Top of the barrier limited transport t t E f,s E f,s E E(0) E f,d qv ds E E f,d E v k k k * m 1 E k d J x = 0 dx k E 1 k k We model the source/drain regions as electron reservoirs Electrons with positive k-values moves right Electrons with negative k-values moves left 5
Top of the barrier limited transport E f,s E(0) qv ds E f,d E v k k k E 0 * m Ek k E Injected from the source: E f,s -E(0) Injected from the drain: E f,d -E(0) E f,s More electrons moving towards the source net electron current! E f,d E(0) k E(0) is mainly set by changing the gate voltage! 6
Net Electron current minute excercise E f,s qv ds E(0) E f,d Sketch the net electron current as V ds is increased! For simplicity: Assume that E(0) are independent of V ds! 0 ~ qe f,d 7 V ds
D Ballistic FET number of electrons n s = m πħ E C f 1 E f E de η Fs = E fs E 0 kt n s = N D [F 0 η F1 + F 0 (η F )] η Fd = E fs E 0 qv DS kt N D = m kt πħ C q = q m /(πħ ) E f,s Above threshold: n s C G (V GS V T ) 1 = 1 + 1 + 1 C G C ox C q C C E(0) qv ds E f,d 8
D Ballistic FET number of electrons n s = N D [F 0 η Fs + F 0 (η Fd )] N D = m kt πħ k x + k x E f,s E(0) E f,d V DS = 0V n s C G (V GS V T ) 1 = 1 + 1 + 1 C G C ox C q C C V DS >> 0V n s C G (V GS V T ) 1 = 1 + + 1 C G C ox C q C C MOS-limit: C ox << C q : n s = n s + + n s stays constant for all V DS n s + increases with V DS. η Fs must increase, until V DS is large so that n s 0. E f,s E(0) qv ds E f,d 9
D Ballistic FET - Current I J = q ħ π dk xdk y k E k f 0 E k E fs J : Current in x direction. (+k x ) kx E k = ħ k x m k x + k y = ħ m k x = ħ m k cos θ η Fs = E fs E 0 k B T η Fd = E fs E 0 qv DS k B T J = qħ π m k=,θ= π k=0,θ= π kcos θ k 1 + exp E k E fs k B T dkdθ E = ħ k m k = m E ħ de = ħ k E m dk η = k B T 10
D Ballistic FET - current J = q m kt 3 π ħ 0 η 1 + e η η FS dη π F 1/ η fs J = q m kt 3 π π ħ F 1/ η fs = J 0 D F 1/ η fs η Fs = E fs E 0 k B T η Fd = E fs E 0 qv DS k B T J 4 0 J D 3 π η fs 3 = q m π ħ 4 3 E fs 3/ Degenerate: E fs >> J = q m kt 3 π π ħ F 1/ η fd = J 0 D F 1/ η fs v Drain to source current d 11
D Ballistic FET - Saturated Current We need to determine η Fs to calculate I D qn s C G (V GS V T ) In saturation: F 0 η Fd 0 and V GS >>V T qn s = qn D [F 0 η Fs + F 0 η Fd ] qn D η Fs = C q kt q η Fs η Fs = C G C q V GS V T V Th J J 4 0 J D 3 π η fs 3 I W = C G 8ħ V GS V T 3m qπ C G qn s < v x > V GS V T Since only +k x states are occupied. C q = C q should be used for the calculation of C G. 1
D Ballistic FET - I(V DS,V GS ) General (approximate) solution for V GS >V T Assuming MOS-limit: C ox << C q qn s = qn D F 0 (η) = ln(1 + e η ) [F 0 η Fs + F 0 η Fd v d ] = C G (V GS V T ) Valid if C G C G : C G <<C q C G C q V GS V T V Th = η G = ln e η G = ln 1 + e η Fs 1 + e η Fs v d η Fs = ln 1 + e v d + 4e v d e η G 1 1 + e v d ln 0 J = J D F1 η fs F1 η fs v d J 0 D 4/3 π η 1.5 Fs η Fs v 1.5 d Analytical but complicated expression for the ballistic current! 13
D Ballistic FET - Current II 0 < V GS < 0.4 V C q = 0.05 F/m C G = 0.01 F/m Similar shape as diffusive transport : but I (V GS 14
Back Scattering 1-T T T λ 0 L G + λ 0 Non-degenerate limit Small V DS L G λ 0 µ n kt L q 1 v th = kt v th m π I W = C G 8ħ V GS V T 3m qπ C G V GS V T No scattering I W λ 0 8ħ C L G + λ G V GS V T 0 3m qπ C G V GS V T Approximate effect of scattering! This relates the mobility (long channel property) to the quasi-ballistic current! 15
D Ballistic FET current Fully Ballistic t w =10 nm t ox =4 nm (er=14) m*=0.03 µ n =3000 cm /Vs v sat =10 7 cm/s V GS -V T =0.5V Experimental L g =40 nm In 0.7 Ga 0.3 As HEMT I DS ~ 1 ma/µm (IEDM 011) (Lower I DS - Source/Drain resistance) 16
Below V T : Sub threshold current J = q m kt 3 π π ħ F 1/ η fs = J 0 N D = m kt D F 1/ η fs πħ Below threshold: n s 0, so that ψ s moves 1:1 with the applied V GS! (If no short channel effects) I D = q N D Wv T F 1/ η Fs F 1/ η Fs v d qw N D v T F 1/ η Fs I D = qw N D v Te q kt V GS V T If V DS large v T = kt πm Inverse subthreshold slope: SS = d dv GS log 10 I D 1 =.3 kt q = 60mV/decade Diffusive Transport gives the same subthreshold slope 17