2007 03 12 Ling25 MOSFET Capacitances

8/3/2019 2007 03 12 Ling25 MOSFET Capacitances

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MOSFET Capacitances

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Device Capacitances

There is some capacitance between every

pair of MOSFET terminals.Only exception: We neglect the

capacitance between Source and Drain.


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Gate to Substrate capacitance

C1 is the oxide capacitance (between Gateand channel)

C1 = WLCOX

C2 is the depletion capacitance between

channel and Substrate

C2 = WL(qsiNsub/(4F))1/2 = Cd


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Gate-Drain and Gate-Source Overlap

Capacitance

C3

and C4

are due to overlap between the gate

poly-silicon and the Source and Drain regions.

No simple formulas for C3 and C4: It is incorrect to

write C3=C4=WLDCOX because of fringing electricfield lines.

We denote the overlap capacitance per unit widthas Cov


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Source-Substrate and Drain-Substrate

Junction Capacitances

Again, no simple formulas for C5 and C6:See right figure: Each capacitance should be

decomposed into two components:

Bottom-plate capacitance, denoted as Cj, which

depends on the junction area.

Side-wall capacitance Cjsw which depends on theperimeter of the junction.


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Source-Substrate and Drain-Substrate

Junction Capacitances

In MOSFET models Cj and Cjsw are usually given ascapacitance-per-unit-area, and capacitance-per-unit-

length, respectively.

Cj = Cj0 / [1+VR/B]m where VR is the junctions reversevoltage, B is the junction built-in potential, and m

typically ranges from 0.3 to 0.4


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Examples (both transistors have

the same Cj and Cjsw parameters)

Calculate CDB and CSB for each structure


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Left structure

CDB=CSB = WECj + 2(W+E)Cjsw


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Layout for Low Capacitance (see

folded structure on the right)

Two parallel transistors with onecommon Drain


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Layout for Low Capacitance (see

folded structure on the right)CDB=(W/2)ECj+2((W/2)+E)Cjsw

CSB=2[(W/2)ECj+2 ((W/2)+E)Cjsw]=

= WECj +2(W+2E)Cjsw

Compare to the left structure:

CDB=CSB = WECj + 2(W+E)Cjsw

CSB slightly larger, CDB a lot smaller. Both

transistors have same W/L


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CGS

,CGD

at Cutoff

CGD=CGS=CovW


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CGB

at Cutoff

C1 = WLCOX,C2= WL(qsiNsub/(4F))1/2 =CdCGB=C1C2/(C1+C2)= WLCOXCd/(WLCOX+Cd )

L is the effective length of the channel.


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CDB

and CSB

at all modes

Both depend on the reverse voltages Drain toSubstrate and Source to Substrate

Recall Cj = Cj0 / [1+VR/B]m


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CGS

,CGD

at Triode Mode

If VDS

is small enough, we can assume

VGSVGD Channel uniform Gate-Channel

capacitance WLCOX is distributed uniformly

CGD=CGS (WLCOX)/2+CovW


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CGS

,CGD

at Saturation Mode

CGD

CovW because there isnt much ofa channel near Drain.

(It can be proved):CGS=2WLeffCOX/3+WCov


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CGB at Triode and Saturation

Modes

CGB is negligible, because channel acts as ashield between Gate and Substrate. That is, if

VG varies, change of charge comes from Source

and Drain, and not from the Substrate!


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MOS Small Signal Model with

Capacitance


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C-V of NMOS

In Weak Inversion CGS is a series

combination of COX-capacitance and Cdep-

capacitance Low capacitance peak nearVGSVTH


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Gate Resistance

2007 03 12 Ling25 MOSFET Capacitances

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