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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 μ Note that L is the effective channel length. I D is constant along the channel: Figure: Drain current versus drainsource 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 = μ 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 ) ( ≈ μ 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 = μ For V DS > V GS – V TH . Figure: Saturation of drain current. Figure: Pinchoff 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 ′ = μ 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 μ and 2 ) ( 2 1 TH GS ox p D V V L W C I ′ = μ . , const VDS GS D m V I g ∂ ∂ = ) ( TH GS ox n V V L W C = μ 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 μ 2 = TH GS D V V I = 2 SecondOrder 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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 Spring '09
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 Electromagnet, Transistor, Volt, VDS

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