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6.4 (a) Q(x) =WCox (VGS − V (x)− VTH) =WCox (VGS − VTH)−WCoxV (x) Increasing VDS L WCox (VGS − VTH) x Q (x ) The curve that intersects the axis at x = L (i.e., the curve for which the channel begins to pinch off) corresponds to VDS = VGS − VTH . (b) RLocal(x) ∝ 1 µQ(x) L x R L o c a l (x ) Increasing VDS Note that RLocal diverges at x = L when VDS = VGS − VTH . 6.15 VGS ID VTH Increasing VDS Initially, when VGS is small, the transistor is in cutoff and no current flows. Once VGS increases beyond VTH , the curves start following the square-law characteristic as the transistor enters saturation. However, once VGS increases past VDS + VTH (i.e., when VDS < VGS − VTH), the transistor goes into triode and the curves become linear. As we increase VDS , the transistor stays in saturation up to larger values of VGS , as expected. 6.17 ID = 1 2 µnCox W L (VGS − VTH) α , α < 2 gm , ∂ID ∂VGS = α 2 µnCox W L (VGS − VTH) α−1 = αID VGS − VTH 6.21 Since they’re being used as current sources, assume M1 and M2 are in saturation for this problem. To find the maximum allowable value of λ, we should evaluate λ when 0.99ID2 = ID1 and 1.01ID2 = ID1, i.e., at the limits of the allowable values for the currents. However, note that for any valid λ (remember, λ should be non-negative), we know that ID2 > ID1 (since VDS2 > VDS1), so the case where 1.01ID2 = ID1 (which implies ID2 < ID1) will produce an invalid value for λ (you can check this yourself). Thus, we need only consider the case when 0.99ID2 = ID1. 0.99ID2 = 0.99 1 2 µnCox W L (VB − VTH) 2 (1 + λVDS2) = ID1 = 1 2 µnCox W L (VB − VTH) 2 (1 + λVDS1) 0.99 (1 + λVDS2) = 1 + λVDS1 λ = 0.02 V−1 5.27 VDD − IDRD = VGS = VTH + √ 2ID µnCox W L 2ID µnCox W L = (VDD − VTH − IDRD) 2 ID = 1 2 µnCox W L [ (VDD − VTH) 2 − 2IDRD (VDD − VTH) + I 2 DR 2 D ] We can rearrange this to the standard quadratic form as follows:( 1 2 µnCox W L R2D ) I2D − ( µnCox W L RD (VDD − VTH) + 1 ) ID + 1 2 µnCox W L (VDD − VTH) 2 = 0 Applying the quadratic formula, we have: ID = ( µnCox W L RD (VDD − VTH) + 1 ) ± √( µnCox W L RD (VDD − VTH) + 1 )2 − 4 ( 1 2 µnCox W L RD (VDD − VTH) )2 2 ( 1 2 µnCox W L R2D ) = µnCox W L RD (VDD − VTH) + 1± √( µnCox W L RD (VDD − VTH) + 1 )2 − ( µnCox W L RD (VDD − VTH) )2 µnCox W L R2D = µnCox W L RD (VDD − VTH) + 1± √ 1 + 2µnCox W L RD (VDD − VTH) µnCox W L R2D Note that mathematically, there are two possible solutions for ID. However, since M1 is diode- connected, we know it will either be in saturation or cutoff. Thus, we must reject the value of ID that does not match these conditions (for example, a negative value of ID would not match cutoff or saturation, so it would be rejected in favor of a positive value). 6.33 (a) Assume M1 is operating in saturation. VGS = 1 V VDS = VDD − IDRD = VDD − 1 2 µnCox W L (VGS − VTH) 2 (1 + λVDS)RD VDS = 1.35 V > VGS − VTH , which verifies our assumption ID = 4.54 mA gm = µnCox W L (VGS − VTH) = 13.333 mS ro = 1 λID = 2.203 kΩ + vgs − gmvgs ro RD (b) Since M1 is diode-connected, we know it is operating in saturation. VGS = VDS = VDD − IDRD = VDD − 1 2 µnCox W L (VGS − VTH) 2 (1 + λVGS)RD VGS = VDS = 0.546 V ID = 251 µA gm = µnCox W L (VGS − VTH) = 3.251 mS ro = 1 λID = 39.881 kΩ + vgs − gmvgs ro RD (c) Since M1 is diode-connected, we know it is operating in saturation. ID = 1 mA gm = √ 2µnCox W L ID = 6.667 mS ro = 1 λID = 10 kΩ + vgs − gmvgs ro (d) Since M1 is diode-connected, we know it is operating in saturation. VGS = VDS VDD − VGS = ID(2 kΩ) = 1 2 µnCox W L (VGS − VTH) 2 (1 + λVGS) (2 kΩ) VGS = VDS = 0.623 V ID = 588 µA gm = µnCox W L (VGS − VTH) = 4.961 mS ro = 1 λID = 16.996 kΩ + vgs − gmvgs ro 2 kΩ (e) Since M1 is diode-connected, we know it is operating in saturation. ID = 0.5 mA gm = √ 2µnCox W L ID = 4.714 mS ro = 1 λID = 20 kΩ + vgs − gmvgs ro 6.38 (a) vin + vgs1 − gm1vgs1 ro1 − vgs2 + gm2vgs2 ro2 vout RD (b) vin + vgs1 − gm1vgs1 gm2vgs2 ro2 − vgs2 + RD vout ro1 (c) vin + vgs1 − gm1vgs1 ro1 vout RD + vgs2 − gm2vgs2 ro2 (d) vin + vgs1 − gm1vgs1 ro1 + vgs2 − vout gm2vgs2 ro2 (e) + vgs1 − gm1vgs1 vin gm2vgs2 ro2 − vgs2 + RD vout ro1 6.43 (a) Assume M1 is operating in triode (since |VGS | = 1.8 V is large). |VGS | = 1.8 V VDD − |VDS | = |ID| (500 Ω) = 1 2 µpCox W L [ 2 (|VGS | − |VTH |) |VDS | − |VDS | 2 ] (500 Ω) |VDS | = 0.418 V < |VGS | − |VTH | , which verifies our assumption |ID| = 2.764 mA (b) Since M1 is diode-connected, we know it is operating in saturation. |VGS | = |VDS | VDD − |VGS | = |ID|(1 kΩ) = 1 2 µpCox W L (|VGS | − |VTH |) 2 (1 kΩ) |VGS | = |VDS | = 0.952 V |ID| = 848 µA (c) Since M1 is diode-connected, we know it is operating in saturation. |VGS | = |VDS | |VGS | = VDD − |ID|(1 kΩ) = VDD − |ID|(1 kΩ) = 1 2 µpCox W L (|VGS | − |VTH |) 2 (1 kΩ) |VGS | = |VGS | = 0.952 V |ID| = 848 µA 6.44 (a) VX IX VDD − VTH VDD Saturation Cutoff M1 goes from saturation to cutoff when VX = VDD − VTH = 1.4 V. (b) VX IX 1 + VTH VDD Saturation Triode M1 goes from saturation to triode when VX = 1 + VTH = 1.4 V. (c) VX IX VDD − VTH VDD Saturation Cutoff M1 goes from saturation to cutoff when VX = VDD − VTH = 1.4 V. (d) VX IX VTH VDD Cutoff Saturation M1 goes from cutoff to saturation when VX = VTH = 0.4 V.
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