Analog expansion - -- = 2 2 1 ) ( DS DS TH GS ox n D V V V...

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Unformatted text preview: -- = 2 2 1 ) ( DS DS TH GS ox n D V V V V L W C I Note that L is the effective channel length. I D is constant along the channel: Figure: Drain current versus drain-source voltage in the triode region. Calculating I D / V DS , the reader can show that the peak of each parabola occurs at V DS = V GS V TH and the peak current is 2 max , ) ( 2 1 TH GS ox n D V V L W C I- = We call V GS V TH the overdrive voltage and W/L the aspect ratio. If V DS V GS V TH , we say the device operates in the triode region. If , V DS << 2(V GS V TH ), we have DS TH GS ox n D V V V L W C I ) (- The drain current is a linear function of V DS . The linear relationship implies that the path from the source to the drain can be represented by a linear resistor equal to ) ( 1 TH GS ox n on V V L W C R- = For V DS > V GS V TH . Figure: Saturation of drain current. Figure: Pinch-off behavior. x = 0 to x = L', where L is the point at which Q d drops to zero, and that on the right from V(x) = 0 to V(x) = V GS V TH . As a result: 2 ) ( 2 1 TH GS ox n D V V L W C I- = indicating that I D is relatively independent of V DS if L remains close to L. For PMOS devices, previous equations are respectively written as --- = 2 2 1 ) ( DS DS TH GS ox p D V V V V L W C I and 2 ) ( 2 1 TH GS ox p D V V L W C I- - = . , const VDS GS D m V I g = ) ( TH GS ox n V V L W C- = In a sense, g m represents the sensitivity of the device: for a high g m , a small change in V GS results in a large change in I D . Interestingly, g m in the saturation region is equal to the reverse of R on in deep triode region. The reader can prove that g m can also be expressed as D ox n m I L W C g 2 = TH GS D V V I- = 2 Second-Order Effects Body Effect Figure: NMOS device with negative bulk voltage. Figure: Variation of depletion region charge with bulk voltage.Figure: Variation of depletion region charge with bulk voltage....
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Analog expansion - -- = 2 2 1 ) ( DS DS TH GS ox n D V V V...

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