MOSFET Drain Current Modeling

6.012 - Microelectronic Devices and Circuits MOSFET Drain Current Modeling In the Gradual Channel Model for the MOSFET we write the drain current, iD, as the product of qN* (y) , the inversion layer sheet charge density at position y along the channel; sy(y), the net drift velocity of the inversion layer carriers there (electrons in the n-channel device we are modeling), and W, the channel width: iD = - qN* (y) sy(y) W with εox dvCS(y) qN* (y) = - t [vGB - VT(y)] and sy(y) = - µe Ey = µe dy ox Substituting these expressions yields: * [v iD = W µ Cox GB - VT(y)] dvCS(y) dy * , as ε /t and where we have identified the gate capacitance per unit area, Cox ox ox where the threshold voltage is given by VT(y) = VFB + |2φp| +vCB(y) + Defining the body factor, γ, as vBS, we can rewrite this as 1 * 2εSi qNA [|2φp| + vCB(y) Cox * , and writing v (y) as v (y) 2εSi qNA /Cox CB CS VT(y) = VFB + |2φp| +vCS(y) - vBS + γ |2φp| + vCS(y) - vBS and thus we can write iD as * [v iD = W µe Cox GS -VFB - |2φp|- vCS(y) - γ |2φp| + vCS(y) - vBS ] dvCS(y) dy To proceed we integrate both sides for y = 0 to y = L, recognizing that the right-hand integral is equivalent to integrating with respect to vCS(y) from 0 to vDS: 1 L vDS 0 0 * ⌠ ⌡ [vGS - VFB - |2φp|- vCS - γ |2φp| + vCS - vBS ] dvCS i D⌠ ⌡ dy = W µe Cox The left-hand integral is iDL, so we can write iD as vDS ⌡ [vGS -VFB - |2φp|- vCS - γ |2φp| + vCS - vBS ] dvCS iD = K ⌠ 0 * . where K is defined as (W/L) µe Cox It is not hard to do the integral on the right-hand side of this equation, and you may want to do it as an exercise (it is done in the text and the result is given in Equation 10.9). The resulting expression is awkward and, most importantly, the threshold voltage, VT, is hard to identify in the expression and the role it plays in the current-voltage relationship is hard to understand; the result is not very intuitive. It will not be obvious to you until you get much more experience with MOSFETs, but it is very desirable from a modeling standpoint to do something to simplify the result and to get an expression that gives us more useful insight. Many texts simply ignore the vCS factor under the radical and write vDS ⌡ [vGS -VFB - |2φp|- vCS - γ |2φp|- vBS ] dvCS iD ≈ K ⌠ 0 which we can simplify as vDS iD = K ⌠ ⌡ [vGS - VT' - vCS ] dvCS 0 with VT' defined to be VFB + |2φp| + γ |2φp| - vBS . Doing the integral we get iD = K [(vGS - VT')vDS - vDS2 2 ] A more satisfying approach is to not ignor the vCS factor, but rather to try to linearize the dependance on it. The troublesome term is |2φp| + vCS(y) - vBS which can be written as 2 |2φp| + vCS - vBS vCS 1 + |2φ | - v p BS = |2φp| - vBS ≈ |2φp| - vBS [1 + 2(|2φp| - vBS) ] ≈ |2φp| - vBS + vCS vCS 2 |2φp| - vBS With this approximation, we next define 1/2 |2φp| - vBS be α, and write iD as to be δ and (1 + γδ) to vDS ⌡ [vGS -VFB - |2φp|- α vCS - γ |2φp|- vBS ] dvCS iD ≈ K ⌠ 0 Using our earlier definition for VT', this becomes vDS iD ≈ K ⌠ ⌡ [vGS - VT' - α vCS ] dvCS 0 and doing the integral yields iD = K [(vGS - VT')vDS - α vDS2 ] 2 In saturation, which now occurs for vDS > (vGS - VT')/α, we have K iD = 2α (vGS - VT') 2 These results are the same as those we obtained after ignoring vCS under the radical, except that we now have a factor of α appearing. To the extent that α is very near one, our earlier approximation is correct, inspite of it being rather adhoc. Collecting all the factors in α, we find it is α = 1+ 2εSi qNA * |2φ | - v 2 Cox p BS 3 Typically this is near 1, and it can be approximated as such. On the other hand, it is easy to leave α in the expression for iD since is such a minor complication. To summarize, our expressions for the drain current, when we retain α are iD = 0 for (vGS - VT')/α < 0 < vDS K iD = 2α (vGS - VT') 2 iD = K [(vGS - VT')vDS - α for 0 < (vGS - VT')/α < vDS vDS2 ] for 0 < vDS < (vGS - VT')/α 2 with K, VT’, γ, and α defined as * K ≡ (W/L) µe Cox VT' ≡ VFB + |2φp| + γ |2φp| - vBS γ ≡ α ≡ 1+ 2εSi qNA * Cox 2εSi qNA * |2φ | - v 2 Cox p BS 4 (Cutoff) (Saturation) (Linear region)