340_Solutions

# 340_Solutions - 2.1(a k = 8.617 105 eV/K 15 3/2 ni(T = 300...

• Homework Help
• 703
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 1 out of 703 pages.

Unformatted text preview: 2.1 (a) k = 8.617 × 10−5 eV/K 15 3/2 ni (T = 300 K) = 1.66 × 10 (300 K)  exp − 0.66 eV 2 (8.617 × 10−5 eV/K) (300 K)  cm−3  exp − 0.66 eV 2 (8.617 × 10−5 eV/K) (600 K)  cm−3 = 2.465 × 1013 cm−3 15 3/2 ni (T = 600 K) = 1.66 × 10 (600 K) = 4.124 × 1016 cm−3 Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentration 13 in Ge at T = 300 K is 2.465×10 1.08×1010 = 2282 times higher than the intrinsic carrier concentration in Si at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is 26.8 times higher than that in Si. 4.124×1016 1.54×1015 = (b) Since phosphorus is a Group V element, it is a donor, meaning ND = 5 × 1016 cm−3 . For an n-type material, we have: n = ND = 5 × 1016 cm−3 2 [ni (T = 300 K)] = 1.215 × 1010 cm−3 n [ni (T = 600 K)]2 = 3.401 × 1016 cm−3 p(T = 600 K) = n p(T = 300 K) = 2.3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only to the drift component. Itot = Idrif t = q(nµn + pµp )AE n = 1017 cm−3 p = n2i /n = (1.08 × 1010 )2 /1017 = 1.17 × 103 cm−3 µn = 1350 cm2 /V · s µp = 480 cm2 /V · s 1V E = V /d = 0.1 µm = 105 V/cm A = 0.05 µm × 0.05 µm = 2.5 × 10−11 cm2 Since nµn ≫ pµp , we can write Itot ≈ qnµn AE = 54.1 µA (b) All of the parameters are the same except ni , which means we must re-calculate p. ni (T = 400 K) = 3.657 × 1012 cm−3 p = n2i /n = 1.337 × 108 cm−3 Since nµn ≫ pµp still holds (note that n is 9 orders of magnitude larger than p), the hole concentration once again drops out of the equation and we have Itot ≈ qnµn AE = 54.1 µA 2.4 (a) From Problem 1, we can calculate ni for Ge. ni (T = 300 K) = 2.465 × 1013 cm−3 Itot = q(nµn + pµp )AE n = 1017 cm−3 p = n2i /n = 6.076 × 109 cm−3 µn = 3900 cm2 /V · s µp = 1900 cm2 /V · s 1V E = V /d = 0.1 µm = 105 V/cm A = 0.05 µm × 0.05 µm = 2.5 × 10−11 cm2 Since nµn ≫ pµp , we can write Itot ≈ qnµn AE = 156 µA (b) All of the parameters are the same except ni , which means we must re-calculate p. ni (T = 400 K) = 9.230 × 1014 cm−3 p = n2i /n = 8.520 × 1012 cm−3 Since nµn ≫ pµp still holds (note that n is 5 orders of magnitude larger than p), the hole concentration once again drops out of the equation and we have Itot ≈ qnµn AE = 156 µA 2.5 Since there’s no electric field, the current is due entirely to diffusion. If we define the current as positive when flowing in the positive x direction, we can write   dn dp Itot = Idif f = AJdif f = Aq Dn − Dp dx dx A = 1 µm × 1 µm = 10−8 cm2 Dn = 34 cm2 /s Dp = 12 cm2 /s 5 × 1016 cm−3 dn =− = −2.5 × 1020 cm−4 dx 2 × 10−4 cm dp 2 × 1016 cm−3 = = 1020 cm−4 dx 2 × 10−4 cm       Itot = 10−8 cm2 1.602 × 10−19 C 34 cm2 /s −2.5 × 1020 cm−4 − 12 cm2 /s 1020 cm−4 = −15.54 µA 2.8 Assume the diffusion lengths Ln and Lp are associated with the electrons and holes, respectively, in this material and that Ln , Lp ≪ 2 µm. We can express the electron and hole concentrations as functions of x as follows: n(x) = N e−x/Ln p(x) = P e(x−2)/Lp Z 2 an(x)dx # of electrons = 0 = Z 2 aN e−x/Ln dx 0  2  = −aN Ln e−x/Ln 0   −2/Ln −1 = −aN Ln e Z 2 # of holes = ap(x)dx 0 = Z 2 aP e(x−2)/Lp dx 0  2  = aP Lp e(x−2)/Lp 0  −2/Lp = aP Lp 1 − e Due to our assumption that Ln , Lp ≪ 2 µm, we can write e−2/Ln ≈ 0 e−2/Lp ≈ 0 # of electrons ≈ aN Ln # of holes ≈ aP Lp 2.10 (a) nn = ND = 5 × 1017 cm−3 pn = n2i /nn = 233 cm−3 pp = NA = 4 × 1016 cm−3 np = n2i /pp = 2916 cm−3 (b) We can express the formula for V0 in its full form, showing its temperature dependence: " # NA ND kT ln V0 (T ) = q (5.2 × 1015 )2 T 3 e−Eg /kT V0 (T = 250 K) = 906 mV V0 (T = 300 K) = 849 mV V0 (T = 350 K) = 789 mV Looking at the expression for V0 (T ), we can expand it as follows: V0 (T ) =   kT  ln(NA ) + ln(ND ) − 2 ln 5.2 × 1015 − 3 ln(T ) + Eg /kT q Let’s take the derivative of this expression to get a better idea of how V0 varies with temperature.   k dV0 (T ) ln(NA ) + ln(ND ) − 2 ln 5.2 × 1015 − 3 ln(T ) − 3 = dT q  15 From this expression, we can see that if ln(N ) + ln(N ) < 2 ln 5.2 × 10 + 3 ln(T ) + 3, or A D h i  15 2 3 equivalently, if ln(NA ND ) < ln 5.2 × 10 T − 3, then V0 will decrease with temperature, which we observe in this case. In order for this not to be true (i.e., in order for V0 to increase with temperature), we must have either very high doping concentrations or very low temperatures. 2.11 Since the p-type side of the junction is undoped, its electron and hole concentrations are equal to the intrinsic carrier concentration. nn = ND = 3 × 1016 cm−3 pp = ni = 1.08 × 1010 cm−3   N D ni V0 = VT ln n2i   ND = (26 mV) ln ni = 386 mV 2.12 (a) r qǫSi NA ND 1 2 NA + ND V0 Cj0 Cj = p 1 − VR /V0 Cj0 = NA = 2 × 1015 cm−3 ND = 3 × 1016 cm−3 VR = −1.6 V   NA ND = 701 mV V0 = VT ln n2i Cj0 = 14.9 nF/cm2 Cj = 8.22 nF/cm2 = 0.082 fF/cm2 (b) Let’s write an equation for Cj′ in terms of Cj assuming that Cj′ has an acceptor doping of NA′ . Cj′ = 2Cj s qǫSi NA′ ND 1 = 2Cj 2 NA′ + ND VT ln(NA′ ND /n2i ) − VR 1 qǫSi NA′ ND = 4Cj2 ′ ′ 2 NA + ND VT ln(NA ND /n2i ) − VR NA′  qǫSi NA′ ND = 8Cj2 (NA′ + ND )(VT ln(NA′ ND /n2i ) − VR )  qǫSi ND − 8Cj2 (VT ln(NA′ ND /n2i ) − VR ) = 8Cj2 ND (VT ln(NA′ ND /n2i ) − VR ) NA′ = 8Cj2 ND (VT ln(NA′ ND /n2i ) − VR ) qǫSi ND − 8Cj2 (VT ln(NA′ ND /n2i ) − VR ) We can solve this by iteration (you could use a numerical solver if you have one available). Starting with an initial guess of NA′ = 2 × 1015 cm−3 , we plug this into the right hand side and solve to find a new value of NA′ = 9.9976 × 1015 cm−3 . Iterating twice more, the solution converges to NA′ = 1.025 × 1016 cm−3 . Thus, we must increase the NA by a factor of NA′ /NA = 5.125 ≈ 5 . 2.16 (a) The following figure shows the series diodes. ID + D1 VD D2 − Let VD1 be the voltage drop across D1 and VD2 be the voltage drop across D2 . Let IS1 = IS2 = IS , since the diodes are identical. VD = VD1 + VD2     ID ID + VT ln = VT ln IS IS   ID = 2VT ln IS ID = IS eVD /2VT Thus, the diodes in series act like a single device with an exponential characteristic described by ID = IS eVD /2VT . (b) Let VD be the amount of voltage required to get a current ID and VD′ the amount of voltage required to get a current 10ID .   ID VD = 2VT ln IS   10ID ′ VD = 2VT ln IS      ID 10ID ′ − ln VD − VD = 2VT ln IS IS = 2VT ln (10) = 120 mV 2.19 VX = IX R1 + VD1  IX  IX = IX R1 + VT ln IS   VT IX VX − ln = R1 R1 IS For each value of VX , we can solve this equation for IX by iteration. Doing so, we find IX (VX = 0.5 V) = 0.435 µA IX (VX = 0.8 V) = 82.3 µA IX (VX = 1 V) = 173 µA IX (VX = 1.2 V) = 267 µA Once we have IX , we can compute VD via the equation VD = VT ln(IX /IS ). Doing so, we find VD (VX = 0.5 V) = 499 mV VD (VX = 0.8 V) = 635 mV VD (VX = 1 V) = 655 mV VD (VX = 1.2 V) = 666 mV As expected, VD varies very little despite rather large changes in ID (in particular, as ID experiences an increase by a factor of over 3, VD changes by about 5 %). This is due to the exponential behavior of the diode. As a result, a diode can allow very large currents to flow once it turns on, up until it begins to overheat. 2.22 VX /2 = IX R1 = VD1 = VT ln(IX /IS ) VT IX = ln(IX /IS ) R1 IX = 367 µA (using iteration) VX = 2IX R1 = 1.47 V 3.1 (a) IX = ( VX R1 0 VX < 0 VX > 0 IX VX (V) Slope = 1/R1 3.2 IX = ( VX R1 0 VX < 0 VX > 0 Plotting IX (t), we have 0 0 −V0 /R1 −π/ω 0 t π/ω −V0 VX (t) (Dotted) IX (t) for VB = 1 V (Solid) V0 3.3 IX = ( 0 VX −VB R1 VX < VB VX > VB Plotting IX vs. VX for VB = −1 V and VB = 1 V, we get: IX VB = −1 V VB = 1 V Slope = 1/R1 −1 Slope = 1/R1 1 VX (V) 3.4 IX = ( 0 VX −VB R1 VX < VB VX > VB Let’s assume V0 > 1 V. Plotting IX (t) for VB = −1 V, we get (V0 − VB )/R1 0 0 VB −π/ω Plotting IX (t) for VB = 1 V, we get 0 t π/ω −V0 VX (t) (Dotted) IX (t) for VB = −1 V (Solid) V0 IX (t) for VB = 1 V (Solid) (V0 − VB )/R1 0 0 −π/ω 0 t π/ω −V0 VX (t) (Dotted) V0 VB 3.5 IX = ( VX −VB R1 ∞ VX < 0 VX > 0 Plotting IX vs. VX for VB = −1 V and VB = 1 V, we get: IX IX for VB = −1 V IX for VB = 1 V 1/R1 Slope = 1/R1 −1 VX (V) −1/R1 Slope = 1/R1 3.6 First, note that ID1 = 0 always, since D1 is reverse biased by VB (due to the assumption that VB > 0). We can write IX as IX = (VX − VB )/R1 Plotting this, we get: IX VB VX (V) Slope = 1/R1 3.7 IX = IR1 = ( VX −VB R1 VX −VB R1 kR2 VX < VB VX > VB VX − VB R1 Plotting IX and IR1 for VB = −1 V, we get: IX IX for VB = −1 V IR1 for VB = −1 V Slope = 1/R1 + 1/R2 −1 Slope = 1/R1 Plotting IX and IR1 for VB = 1 V, we get: VX (V) IX IX for VB = 1 V IR1 for VB = 1 V Slope = 1/R1 + 1/R2 1 VX (V) Slope = 1/R1 3.8 IX = ( 0 IR1 = ( VX R1 + VB R1 +R2 VX R1 VX −VB R2 VX < VX > VX < VX > VB R1 +R2 R1 VB R1 +R2 R1 VB R1 +R2 R1 VB R1 +R2 R1 Plotting IX and IR1 for VB = −1 V, we get: IX for VB = −1 V IR1 for VB = −1 V Slope = 1/R1 + 1/R2 −VB /R2 Slope = 1/R1 VB R R1 +R2 1 VB R1 +R2 Plotting IX and IR1 for VB = 1 V, we get: VX (V) IX for VB = 1 V IR1 for VB = 1 V Slope = 1/R1 + 1/R2 Slope = 1/R1 VB R1 +R2 VB R R1 +R2 1 VX (V) 3.9 (a) Vout (V) Vout = ( VB Vin Vin < VB Vin > VB 5 Slope = 1 4 3 2 1 0 −5 −4 −3 −2 −1 0 1 (b) Vout ( Vin − VB = 0 Vin < VB Vin > VB 2 3 4 5 Vin (V) Vout (V) 2 1 0 −5 −4 −3 −2 −1 0 1 2 3 4 −1 5 Vin (V) −2 −3 Slope = 1 −4 −5 −6 −7 (c) Vout (V) Vout = Vin − VB 3 Slope = 1 2 1 0 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 0 1 2 3 4 5 Vin (V) Vout = ( Vout (V) (d) 2 Vin VB Vin < VB Vin > VB 1 0 −5 −4 −3 −2 −1 0 1 −1 Slope = 1 −2 −3 −4 −5 (e) Vout ( 0 = Vin − VB Vin < VB Vin > VB 2 3 4 5 Vin (V) Vout (V) 3 Slope = 1 2 1 0 −5 −4 −3 −2 −1 0 1 2 3 4 5 Vin (V) 3.11 For each part, the dotted line indicates Vin (t), while the solid line indicates Vout (t). Assume V0 > VB . (a) Vout (t) (V) Vout = ( VB Vin Vin < VB Vin > VB V0 VB π/ω −π/ω t −V0 (b) Vout ( Vin − VB = 0 Vin < VB Vin > VB Vout (t) (V) V0 VB π/ω −π/ω t −V0 −V0 − VB (c) Vout (t) (V) Vout = Vin − VB V0 VB V0 − VB π/ω −π/ω t −V0 −V0 − VB (d) Vout (t) (V) Vout = ( Vin VB Vin < VB Vin > VB V0 VB π/ω −π/ω t −V0 (e) Vout ( 0 = Vin − VB Vin < VB Vin > VB Vout (t) (V) V0 VB V0 − VB π/ω −π/ω t −V0 3.12 For each part, the dotted line indicates Vin (t), while the solid line indicates Vout (t). Assume V0 > VB . (a) Vout (t) (V) Vout ( Vin − VB = 0 Vin < VB Vin > VB V0 VB π/ω −π/ω t −V0 −V0 − VB (b) Vout = ( Vin VB Vin < VB Vin > VB Vout (t) (V) V0 VB π/ω −π/ω t −V0 Vout ( 0 = Vin − VB Vout (t) (V) (c) V0 Vin < VB Vin > VB VB V0 − VB π/ω −π/ω t −V0 (d) Vout (t) (V) Vout = Vin − VB V0 VB V0 − VB π/ω −π/ω t −V0 −V0 − VB (e) Vout = ( VB Vin Vin < VB Vin > VB Vout (t) (V) V0 VB π/ω −π/ω t −V0 3.16 (a) IR1 = ( Iin VD,on R1 Iin < Iin > VD,on R1 VD,on R1 IR1 VD,on /R1 VD,on /R1 Iin Slope = 1 (b) IR1 = ( Iin VD,on +VB R1 Iin < Iin > VD,on +VB R1 VD,on +VB R1 IR1 (VD,on + VB ) /R1 (VD,on + VB ) /R1 Iin Slope = 1 (c) IR1 = ( Iin VD,on −VB R1 Iin < Iin > VD,on −VB R1 VD,on −VB R1 IR1 (VD,on − VB ) /R1 Iin (VD,on − VB ) /R1 Slope = 1 (d) IR1 = ( Iin VD,on R1 Iin < Iin > VD,on R1 VD,on R1 IR1 VD,on /R1 VD,on /R1 Iin Slope = 1 3.17 (a) Vout = ( Iin R1 VD,on Iin < Iin > VD,on R1 VD,on R1 VD,on /R1 VD,on 0 0 −I0 R1 0 t −π/ω π/ω (b) Vout = ( Iin R1 VD,on + VB Iin < Iin > VD,on +VB R1 VD,on +VB R1 −I0 Iin (t) (Dotted) Vout (t) (Solid) I0 I0 (VD,on+VB )/R1 0 0 Iin (t) (Dotted) Vout (t) (Solid) VD,on + VB −I0 R1 0 t −π/ω −I0 π/ω (c) Vout = ( Iin R1 + VB VD,on Iin < Iin > VD,on −VB R1 VD,on −VB R1 VD,on 0 0 −I0 R1 + VB (VD,on − VB ) /R1 −π/ω 0 t π/ω −I0 Iin (t) (Dotted) Vout (t) (Solid) I0 (d) Vout = ( Iin R1 + VB VD,on + VB Iin < Iin > VD,on R1 VD,on R1 I0 VD,on /R1 0 0 −I0 R1 + VB −π/ω 0 t π/ω −I0 Iin (t) (Dotted) Vout (t) (Solid) VD,on + VB 3.20 (a) Vout = ( Iin R1 VB − VD,on Iin > Iin < VB −VD,on R1 VB −VD,on R1 I0 I0 R1 VB − VD,on 0 0 0 t −π/ω π/ω (b) Vout = ( Iin R1 + VB −VD,on V +V B Iin > − D,on R1 VD,on +VB Iin < − R1 −I0 Iin (t) (Dotted) Vout (t) (Solid) (VB − VD,on ) /R1 I0 0 Iin (t) (Dotted) Vout (t) (Solid) I0 R1 + VB 0 −VD,on −(VD,on +VB )/R1 0 t −π/ω π/ω −I0 (c) Vout = ( Iin R1 + VB VB − VD,on V Iin > − D,on R1 V Iin < − D,on R1 I0 VB − VD,on 0 0 −VD,on /R1 −π/ω 0 t π/ω −I0 Iin (t) (Dotted) Vout (t) (Solid) I0 R1 + VB 3.23 (a) R2 R1 +R2 Vin VD,on Vout (V) Vout = ( Vin < Vin > R1 +R2 R2 VD,on R1 +R2 R2 VD,on VD,on R1 +R2 VD,on R2 Vin (V) Slope = R2 / (R1 + R2 ) (b) Vout = ( R2 R1 +R2 Vin Vin − VD,on Vin < Vin > R1 +R2 R1 VD,on R1 +R2 R1 VD,on Vout (V) Slope = 1 R2 V R1 D,on Slope = R2 / (R1 + R2 ) R1 +R2 VD,on R1 Vin (V) 3.24 (a) IR1 = ( Vin R1 +R2 Vin −VD,on R1 ID1 = ( 0 Vin −VD,on R1 Vin < Vin > − VD,on R2 R1 +R2 R2 VD,on R1 +R2 R2 VD,on Vin < Vin > R1 +R2 R2 VD,on R1 +R2 R2 VD,on IR1 ID1 Slope = 1/R1 Slope = 1/R1 VD,on /R2 R1 +R2 VD,on R2 Vin (V) Slope = 1/ (R1 + R2 ) (b) IR1 = ( Vin R1 +R2 VD,on R1 ID1 = ( 0 Vin < Vin > Vin −VD,on R2 − R1 +R2 R1 VD,on R1 +R2 R1 VD,on VD,on R1 Vin < Vin > R1 +R2 R1 VD,on R1 +R2 R1 VD,on VD,on /R1 IR1 ID1 Slope = 1/R2 R1 +R2 VD,on R1 Vin (V) Slope = 1/ (R1 + R2 ) 3.25 (a) 2 VB + R1R+R (Vin − VB ) 2 Vin − VD,on Vin < VB + Vin > VB + R1 +R2 R1 VD,on R1 +R2 R1 VD,on Vout (V) Vout = ( Slope = 1 VB + R2 V R1 D,on VB + Slope = R2 / (R1 + R2 ) R1 +R2 VD,on R1 Vin (V) (b) Vout = ( R2 R1 +R2 Vin Vin − VD,on − VB Vin < Vin > R1 +R2 R1 R1 +R2 R1 (VD,on + VB ) (VD,on + VB ) Vout (V) Slope = 1 R2 R1 (VD,on + VB ) VB + Slope = R2 / (R1 + R2 ) R1 +R2 R1 (VD,on + VB ) Vin (V) (c) R2 R1 +R2 (Vin − VB ) Vin > VB + Vin + VD,on − VB Vin < VB + Vout (V) Vout = ( R1 +R2 R1 VD,on R1 +R2 R1 VD,on Slope = R2 / (R1 + R2 ) R2 V R1 D,on VB + R1 +R2 VD,on R1 Vin (V) Slope = 1 (d) R2 R1 +R2 (Vin − VB ) Vin < VB + Vin − VD,on Vin > VB + R1 +R2 R1 R1 +R2 R1 Vout (V) Vout = ( (VD,on − VB ) (VD,on − VB ) Slope = 1 R2 V R1 D,on VB + Slope = R2 / (R1 + R2 ) R1 +R2 R1 (VD,on − VB ) Vin (V) 3.26 (a) IR1 = ( Vin −VB R1 +R2 VD,on R1 ID1 = ( 0 Vin < VB + Vin > VB + Vin −VD,on −VB R2 IR1 ID1 − R1 +R2 R1 VD,on R1 +R2 R1 VD,on Vin < VB + Vin > VB + VD,on R1 R1 +R2 R1 VD,on R1 +R2 R1 VD,on VD,on /R1 Slope = 1/R2 VB + R1 +R2 VD,on R1 Vin (V) Slope = 1/ (R1 + R2 ) (b) IR1 = ( Vin R1 +R2 VD,on +VB R1 ID1 = ( 0 Vin < Vin > Vin −VD,on −VB R2 − R1 +R2 R1 R1 +R2 R1 (VD,on + VB ) (VD,on + VB ) VD,on +VB R1 Vin < Vin > R1 +R2 R1 R1 +R2 R1 (VD,on + VB ) (VD,on + VB ) (VD,on + VB ) /R1 IR1 ID1 Slope = 1/R2 R1 +R2 R1 (VD,on + VB ) Vin (V) Slope = 1/ (R1 + R2 ) (c) IR1 = ( Vin −VB R1 +R2 V − D,on R1 ID1 = ( 0 V +V +VB − in D,on − R2 Vin > VB − Vin < VB − R1 +R2 R1 VD,on R1 +R2 R1 VD,on VD,on R1 Vin > VB − Vin < VB − R1 +R2 R1 VD,on R1 +R2 R1 VD,on IR1 ID1 Slope = −1/R2 Slope = 1/ (R1 + R2 ) VB + R1 +R2 VD,on R1 Vin (V) −VD,on /R1 (d) IR1 = ( Vin −VB R1 +R2 VD,on −VB R1 ID1 = ( 0 Vin −VD,on R2 Vin < VB + Vin > VB + − VD,on −VB R1 R1 +R2 R1 R1 +R2 R1 (VD,on − VB ) (VD,on − VB ) Vin < VB + Vin > VB + R1 +R2 R1 R1 +R2 R1 (VD,on − VB ) (VD,on − VB ) IR1 ID1 VB + Slope = 1/R2 R1 +R2 R1 (VD,on − VB ) (VD,on − VB ) /R1 Slope = 1/ (R1 + R2 ) Vin (V) 3.27 (a) 0 R2 R1 +R2 (Vin − VD,on ) Vin < VD,on Vin > VD,on Vout (V) Vout = ( Slope = R2 / (R1 + R2 ) VD,on Vin (V) (b) Vout −VD,on 2 = R1R+R Vin 2 Vin − VD,on 2 Vin < − R1R+R VD,on 2 R1 +R2 − R2 VD,on < Vin < 2 Vin > R1R+R VD,on 1 R1 +R2 R1 VD,on Vout (V) Slope = 1 R2 V R1 D,on Slope = R2 / (R1 + R2 ) 2 − R1R+R VD,on 2 R1 +R2 VD,on R1 Vin (V) −VD,on (c) R2 R1 +R2 (Vin + VD,on ) − VD,on Vin Vin < −VD,on Vin > −VD,on Vout (V) Vout = ( Slope = 1 −VD,on −VD,on Slope = R2 / (R1 + R2 ) Vin (V) (d) 0 R2 R1 +R2 (Vin − VD,on ) Vin < VD,on Vin > VD,on Vout (V) Vout = ( Slope = R2 / (R1 + R2 ) VD,on VD,on Vin (V) (e) Vout = ( R2 R1 +R2 0 (Vin + VD,on ) Vin < −VD,on Vin > −VD,on Vout (V) −VD,on Vin (V) Slope = R2 / (R1 + R2 ) 3.28 (a) IR1 = ( 0 ID1 = ( 0 Vin −VD,on R1 +R2 Vin −VD,on R1 +R2 Vin < VD,on Vin > VD,on Vin < VD,on Vin > VD,on IR1 ID1 Slope = 1/ (R1 + R2 ) VD,on Vin (V) (b) IR1 = ID1 V +V in D,on R1 Vin R1 +R2 VD,on R1 0 = 0 Vin −VD,on R2 2 Vin < − R1R+R VD,on 2 R1 +R2 − R2 VD,on < Vin < 2 Vin > R1R+R VD,on 1 − VD,on R1 R1 +R2 R1 VD,on 2 Vin < − R1R+R VD,on 2 R1 +R2 − R2 VD,on < Vin < 2 Vin > R1R+R VD,on 1 R1 +R2 R1 VD,on IR1 ID1 VD,on /R1 Slope = 1/ (R1 + R2 ) Slope = 1 2 − R1R+R VD,on 2 −VD,on /R2 R1 +R2 VD,on R1 Vin (V) Slope = 1/R1 (c) IR1 = ( 0 ID1 = ( 0 Vin < −VD,on 0 Vin > −VD,on Vin +VD,on R1 +R2 Vin < −VD,on Vin > −VD,on IR1 ID1 −VD,on Vin (V) Slope = 1/ (R1 + R2 ) (d) IR1 = ( 0 ID1 = ( 0 Vin −VD,on R1 +R2 Vin −VD,on R1 +R2 Vin < VD,on Vin > VD,on Vin < VD,on Vin > VD,on IR1 ID1 Slope = 1/ (R1 + R2 ) VD,on Vin (V) (e) IR1 = ( 0 ID1 = ( 0 Vin < −VD,on 0 Vin > −VD,on Vin +VD,on R1 +R2 Vin < −VD,on Vin > −VD,on IR1 ID1 −VD,on Vin (V) Slope = 1/ (R1 + R2 ) 3.29 (a) Vout (V) Vout Vin < VD,on Vin 2 2 = VD,on + R1R+R (V − V ) VD,on < Vin < VD,on + R1R+R (VD,on + VB ) in D,on 2 1 R1 +R2 Vin − VD,on − VB Vin > VD,on + R1 (VD,on + VB ) Slope = 1 VD,on + R2 R1 (VD,on + VB ) Slope = R2 / (R1 + R2 ) VD,on Slope = 1 VD,on VD,on + R1 +R2 R1 (VD,on + VB ) Vin (V) (b) Vout = ( Vin + VD,on − VB R2 R1 +R2 (Vin − VD,on ) Vin < VD,on + Vin > VD,on + R1 +R2 R1 R1 +R2 R1 (VB − 2VD,on ) (VB − 2VD,on ) Vout (V) Slope = R2 / (R1 + R2 ) R2 R1 (VB − 2VD,on ) VD,on + R1 +R2 R1 (VB − 2VD,on ) Vin (V) Slope = 1 (c) Vout (V) Vout = ( Vin VD,on + VB Vin < VD,on + VB Vin > VD,on + VB VD,on + VB VD,on + VB Vin (V) Slope = 1 (d) Vout (V) Vout = 0 Vin < VD,on 2 (Vin − VD,on ) VD,on < Vin < VD,on + R1R+R (VB + VD,on ) 2 R1 +R2 VD,on + VB Vin > VD,on + R2 (VB + VD,on ) R2 R1 +R2 VD,on + VB Slope = R2 / (R1 + R2 ) VD,on VD,on + R1 +R2 R2 (VB + VD,on ) Vin (V) 3.30 (a) IR1 = 0 ID1 = ( Vin −VD,on R1 +R2 VD,on +VB R1 Vin < VD,on 2 VD,on < Vin < VD,on + R1R+R (VD,on + VB ) 1 R1 +R2 Vin > VD,on + R1 (VD,on + VB ) 0 Vin −2VD,on −VB R2 − VD,on +VB R1 Vin < VD,on + Vin > VD,on + R1 +R2 R1 R1 +R2 R1 (VD,on + VB ) (VD,on + VB ) IR1 ID1 VD,on + VB Slope = 1/ (R1 + R2 ) Slope = 1/R2 VD,on VD,on + R1 +R2 R1 (VD,on + VB ) Vin (V) (b) If VB < 2VD,on : IR1 = ID1 = ( 0 Vin −VD,on R1 +R2 Vin < VD,on Vin > VD,on IR1 ID1 Slope = 1/ (R1 + R2 ) VD,on Vin (V) If VB > 2VD,on : IR1 = ID1 = ( VB −2VD,on R1 Vin −VD,on R1 +R2 Vin < VD,on + Vin > VD,on + R1 +R2 R1 R1 +R2 R1 (VB − 2VD,on ) (VB − 2VD,on ) IR1 ID1 Slope = 1/ (R1 + R2 ) VB −2VD,on R1 VD,on + R1 +R2 R1 (VB − 2VD,on ) Vin (V) (c) IR1 = ID1 ( 0 Vin −VD,on −VB R1 0 = 0 Vin −2VD,on −VB R2 Vin < VD,on + VB Vin > VD,on + VB Vin < VD,on + VB VD,on + VB < Vin < 2VD,on + VB Vin > 2VD,on + VB IR1 ID1 Slope = 1/R1 Slope = 1/R2 VD,on + VB 2VD,on + VB Vin (V) (d) IR1 = 0 Vin < VD,on 2 VD,on < Vin < VD,on + R1R+R (VB + VD,on ) 2 R1 +R2 Vin > VD,on + R2 (VB + VD,on ) ID1 = 0 Vin < VD,on 2 VD,on < Vin < VD,on + R1R+R (VB + VD,on ) 2 R1 +R2 Vin > VD,on + R2 (VB + VD,on ) Vin −VD,on R1 +R2 Vin −2V D,on −VB R1 Vin −VD,on R1 +R2 Vin −2V D,on −VB R1 IR1 ID1 Slope = 1/R2 VB +VD,on R2 Slope = 1/ (R1 + R2 ) VD,on VD,on + R1 +R2 R2 (VB + VD,on ) Vin (V) 3.31 (a) Vin − VD,on = 1.6 mA R1 VT = = 16.25 Ω ID1 R1 = ∆Vin = 98.40 mV rd + R1 ID1 = rd1 ∆Vout (b) Vin − 2VD,on = 0.8 mA R1 VT = rd2 = = 32.5 Ω ID1 R1 + rd2 = ∆Vin = 96.95 mV R1 + rd1 + rd2 ID1 = ID2 = rd1 ∆Vout (c) Vin − 2VD,on = 0.8 mA R1 VT = rd2 = = 32.5 Ω ID1 rd2 = ∆Vin = 3.05 mV rd1 + R1 + rd2 ID1 = ID2 = rd1 ∆Vout (d) Vin − VD,on VD,on − = 1.2 mA R1 R2 VT = = 21.67 Ω ID2 R2 k rd2 = ∆Vin = 2.10 mV R1 + R2 k rd2 ID2 = rd2 ∆Vout 3.32 (a) ∆Vout = ∆Iin R1 = 100 mV (b) ID1 = ID2 = Iin = 3 mA VT rd1 = rd2 = = 8.67 Ω ID1 ∆Vout = ∆Iin (R1 + rd2 ) = 100.867 mV (c) ID1 = ID2 = Iin = 3 mA VT rd1 = rd2 = = 8.67 Ω ID1 ∆Vout = ∆Iin rd2 = 0.867 mV (d) ID2 = Iin − rd2 = VD,on = 2.6 mA R2 VT = 10 Ω ID2 ∆Vout = ∆Iin (R2 k rd2 ) = 0....
View Full Document

• Fall '15
• Shiuh-huaWoodChiang
• Vin, VB Rocks, GM V platform, Rock and Republic

{[ snackBarMessage ]}

###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern