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### lec12draincur

Course: CSAIL 6.012, Fall 2003
School: MIT
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Word Count: 829

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- 6.012 Electronic Devices and Circuits Fall 2003 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 = -...

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- 6.012 Electronic Devices and Circuits Fall 2003 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 dvCS(y) eox * [vGB - VT(y)] and sy (y) = - e Ey = e dy qN(y) = - t ox Substituting these expressions yields: * iD = W Cox [vGB - VT(y)] dvCS(y) dy * where we have identified the gate capacitance per unit area, Cox , as eox /tox and where the threshold voltage is given by 1 VT(y) = VFB + |2fp| +vCB(y) + * C 2eSi qNA [|2fp| + vCB(y) ox * Defining the body factor, g, as 2eSi qNA /Cox , and writing v CB(y) as v CS(y) vBS, we can rewrite this as VT(y) = VFB + |2fp| +vCS(y) - vBS + g |2fp| + vCS(y) - vBS and thus we can write iD as * iD = W e C ox [vGS -VFB - |2fp|- vCS(y) - g |2fp| + 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 * iD dy = W e C ox i [vGS - VFB - |2fp|- vCS - g |2fp| + vCS - vBS ] dvCS i 0 0 L vDS The left-hand integral is iDL, so we can write iD as vDS iD = K i [vGS -VFB - |2fp|- vCS - g |2fp| + vCS - vBS ] dvCS 0 * where K is defined as (W/L) e C ox . 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, V T, 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 result the and to get an expression that gives us more useful insight. Many texts simply ignor the vCS factor under the radical and write vDS iD K i [vGS -VFB - |2fp|- vCS - g |2fp|- vBS ] dvCS 0 which we can simplify as iD = K i [vGS - VT' - vCS ] dvCS 0 vDS with VT' defined to be VFB + |2f p| + g |2fp| - vBS . Doing the integral we get vDS2 iD = K [(vGS - VT')vDS - 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 |2fp| + vCS(y) - vBS 2 which can be written as |2fp| + vCS - vBS = |2fp| - vBS vCS 1 + |2f | - v p BS |2fp| - vBS [1 + 2(|2fp| - vBS) ] vCS 2 |2fp| - vBS to be d and (1 + gd) to vCS |2fp| - vBS + With this approximation, we next define 1/2 |2fp| - vBS be a, and write iD as vDS iD K i [vGS -VFB - |2fp|- a v CS - g |2fp|- vBS ] dvCS 0 Using our earlier definition for VT', this becomes iD and doing the integral yields iD = K [(vGS - VT')vDS - a vDS2 ] 2 K [vGS - VT' - a vCS ] dvCS i 0 vDS In saturation, which now occurs for vDS > (vGS - VT')/a, we have K iD = 2a (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 a appearing. To the extent that a is very near one, our earlier approximation is correct, inspite of it being rather adhoc. Collecting all the factors in a, we find it is a = 1+ 2eSi qNA * 2 C ox |2fp| - vBS 3 Typically this is near 1, and it can be approximated as such. On the other hand, it is easy to leave a in the expression for iD since is such a minor complication. To summarize, our expressions for the drain current, when we retain a are iD = 0 for (vGS - VT')/a < 0 < vDS for 0 < (vGS - VT')/a < vDS vDS2 ] for 0 < vDS < (vGS - VT')/a 2 (Cutoff) (Saturation) K 2 iD = 2a (vGS - VT') iD = K [(vGS - VT')vDS - a (Linear region) with K, VT', g, and a defined as * K (W/L) e C ox VT' VFB + |2fp| + g |2fp| - vBS g 2eSi qNA * C ox a 1+ 2eSi qNA * 2 C ox |2fp| - vBS 4
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