Lecture 3 Capacitance Calculation. References Detailed Load Capacitance Calculation (Hodges,Section 6.3) Detailed MOS Capacitance Model (West, Section.

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Presentation transcript:

Lecture 3 Capacitance Calculation

References Detailed Load Capacitance Calculation (Hodges,Section 6.3) Detailed MOS Capacitance Model (West, Section 2.3.2)

Identification of Various Capacitances

Data-Dependent Gate Capacitance Effective Gate Capacitance (i.e. capacitance into the gate) in a 0.35 um process

General Capacitance Model Typo: 2/3 General Capacitance Model: 1.Thin-Oxide Capacitance a. Voltage dependent (C gs, C gd, C gb ) b. Voltage independent: C ol 2.Junction Capacitance (C jsb, C jdb )

C gs, C gd and C gb Channel extends from source to drain Channel almost from source to drain C g in series with C j C(0V)=0.5C g Gate:-Q Under SiO 2 :Q

Intuition about Junction Capacitance The depletion region is narrower when the diode is forward biased, therefore the junction diode capacitance is higher. The depletion region is wider when the diode is reverse biased, therefore the junction diode capacitance is smaller.

Junction capacitance versus applied voltage

Junction Capacitance from Layout Data

Load Capacitance Calculation C load =C self +C wire +C fanout

Fanout Capacitance

Fanout Gate Capacitance C fanout : fanout capacitance due to the inputs of subsequent gates, C G. C fanout =C G1 +C G2 +C G3 …. Assumption: Each fanout is an inverter.

Input Capacitance Calculation C OL : overlap capacitance C GN, C GP : Thin Oxide Capacitance

Worst Case Analysis Assumption The thin-oxide capacitance is voltage dependent. The worst case analysis uses C ox WL to compute its worst case value.

Thin Oxide Capacitance:C g C G =WLC ox =WL(ε ox /t ox )=WC g Unit of C g : fF/μm [Worst Case Analysis]

CgCg toxLCg 110 nm51.61 fF/μm 7.5 nm0.35 μm1.65 fF/μm 2.2 nm0.1 μm1.61 fF/μm Cg is approximately 1.61 fF/μm for the last 25 years. Exception: the 0.18 μm process, which has a Cg of 1.0 fF/ μm. [Worst Case Analysis]

Thin Oxide Capacitance:C ol

Components of C ol C ol =C f +C ov C f :fringing capacitance C ov : overlap capacitance

Redefine C g For 0.13 μm, – C g (due to t ox alone): 1.6 fF/μm [Hodges, p.72] – C ol (due to C ov and C f ): 0.25 fF/ μm [Hodges, p.80] – Redefine Cg [Hodges, p.259] as C g =C ox L+2C ol C g =1.6 fF/μm fF/μm=2 fF/μm C g has been constant for over 20 years – Multipy Cg by W to obtain the total capacitance due to t ox, C ov and C f [Worst Case Analysis]

Gate Capacitance of an Inverter C G =C g (W n +W p ) C G =2fF/μm(W n +W p ) [Worst Case Analysis]

Fanout Gate capacitance of n Inverters C fanout =2fF/μm[(W n +W p ) 1 +(W n +W p ) 2 …(W n +W p ) n ] [Worst Case Analysis]

Self-Capacitance Calculation 1.Eliminate capacitors not connected to the output 2.Assume the transistors are either on (Saturation) or off (Cutoff). 3.C GD is negligible in either saturation or cutoff.

Miller Capacitance

Derivation of Miller Capacitance Z 1 =/Z(1-V y /V x ) Z 2 =/Z(1-V x /V y )

Calculation of Self-Capacitance C self =C DBn +C DBP +2C OL +2 C OL C DBn =C jn W n C DBp =C jp W p C OL =C ol W C self =C jn W n +C jp W p +2C ol ( W n +W p ) Assume C jn =C jp C self =C eff (W n +W p ) For 0.13: C eff =1 fF/μm [Hodges, p. 261]

Wire Capacitance Ignore wire capacitance if the length of a wire is less than a few microns. Include wires longer than a few microns – C wire =C int L wire – C int =0.2 fF/um For very long wires use distributed model