hapter 2 The Well E E 480 Introduction to Analog and Digital VLSI Paul M. Furth New Mexico State University p+ sub ~ 150 m thick, p-epi ~ 30 m thick All transistors go in p- epi layer Typical p- doping is 1x10 15 atoms/cm 3 Notice parasitic diode. Normally p-substrate attached to lowest voltage, i.e., V SS = 0V So, resistor voltages must be 0V or higher to maintain diode reverse-biased 1
Masks are normally 10x or more enlarged opaque and transparent regions or patterns Patterns are focused onto the wafer using lenses Masks cost $millions The first layer of SiO 2 (silicon dioxide, quartz glass, a great insulator) is grown directly on the Si wafer. O 2 is used, so the Si in the SiO 2 comes from the wafer n-well region is lightly doped (n-) Typical n- doping 1x10 16 atoms/cm 3 > p- doping Also called an n-tub 2
L R t W R square L W R square t Never want parasitic BJT to become forward biased. Generally, p-sub attached to lowest voltage, V SS and n-well attached to highest voltage, V DD Parameter - resistivity of a material t - thickness or depth R square has units of /, Ohms per square Resistance of 2 adjacent sides of a square ~ 0.6 R square (a) R AB = R square + 0.6 R square + R square = 2.6 R square (b) R AB = 3 R unit (connected by low-resistance metal) Silicon (Si) has 4 valence e- => semiconductor In Si crystal, the valence e- are shared. Due to thermal energy (T > 0 o K), some e- move into conduction band. e- in conduction band increases conductivity of Si. e- move more freely in crystal than holes. 3
Silicon rystal In pure (instrinsic) Si, the number of free electrons (n) = the number of holes (p) 9 3 ni 14.510 carriers/cm at room temperature 2 np n MassAction Law i The concentration of Si atoms in a crystal is only N 21 atoms/cm 3 Si 5010 In n-type Si, crystal is doped with phosphorous (P) with 5 valence e-. N D is the e- Donor concentration. Suppose n = N D = 1x10 18 phosphorous atoms/cm 3, then p = n i2 / N D = 210 holes/cm 3 In p-type Si, crystal is doped with Boron (B) with 3 valence e-. N A is the e- Acceptor concentration. E V bi T fn E q fp N N V ln kt V T q D A 2 ni I D I S e Vd nvt 1 kt, VT 26mV, 1 n 2 q Doping N D (e.g., phosphorous) and N A (e.g., Boron) generally never same. In example above, N D > N A j0 j0sw j m V d Vd 1 1 Vbi Vbi (apacitance/area) WL j0 j0sw (apacitance/length) (2W 2L) m 0.5, msw(sidewall) 0.33 msw 4
As V D increases (negatively), depletion region becomes thicker (d increases), so j decreases A k A d d 0 When V IN = 10V, I D = (10 0.7) / 1k = 9.3 ma When V IN = -10V, I D = 0, but not immediately. At first, there is negative current, which removes the charge stored in the junction of the diode when it is forward biased. Vin + Vin R 1 1k V D I D D 1 n-well forms transmission line 5
t d = 0.7 R t rise = t fall = 2.2 R For distributed R model (accurate): t d 0.35 R square square n 2 n is the number of squares R lump = R square n lump = square n t d 0.35 R lump lump For lumped model: t d 0.7 R lump lump in R LUMP out LUMP R lump = 5k x 50 = 250 k lump = 5f x 50 = 250 ff td 0.35 R lump lump = 21.9 ns 130 nm process capable of (d), but we will only use (a) in lab Advantage of (d) is bulk (or body) terminals can be at different voltages for every transistor 6
Draw n-well cross sections below What does the 10k resistor and a reverse-biased diode act like? R circuit Importance of 1V D source? Reverse-bias diode What happens if D increases to 5V? hanges j 7