exp Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentration in Ge at T = 300 K is 2.

Transcription

1 .1 (a) k = ev/k n i (T = 300 K) = (300 K) 3/ 66 ev exp ( ev/k) (300 K) = cm 3 n i (T = 600 K) = (600 K) 3/ 66 ev exp ( ev/k) (600 K) = cm 3 cm 3 cm 3 Compared to the values obtained in Example.1, we can see that the intrinsic carrier concentration in Ge at T = 300 K is = 8 times higher than the intrinsic carrier concentration in 10 Si at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is = times higher than that in Si. (b) Since phosphorus is a Group V element, it is a donor, meaning N D = cm 3. For an n-type material, we have: n = N D = cm 3 p(t = 300 K) = [n i(t = 300 K)] = cm 3 n p(t = 600 K) = [n i(t = 600 K)] = cm 3 n

2 . (a) Mobility of electrons in Si = 1350 cm /V-s Mobility of holes in Si = 480 cm /V-s cm V velocity of electrons = μne = V- s μm 4 = m / s cm V velocity of holes = μpe = V- s μm 3 = m / s (b) Given E = 1 V/μm hole cunrent negligible μ n = 1350 cm /V-s μ p = 480 cm /V-s J tot = 1 ma/μm = q[μ n ne + μ p pe] < qμ n ne J 1 ma / μ tot m n = = q E -19 μ ( C)(1350 cm / V - s)(1 V / μm) n 17-3 = cm

3 .3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only to the drift component. Since nµ n pµ p, we can write I tot = I drift = q(nµ n + pµ p )AE n = cm 3 p = n i /n = ( ) /10 17 = cm 3 µ n = 1350 cm /V s µ p = 480 cm /V s E = V/d = 1V 1 µm = 10 5 V/cm A =05 µm 05 µm = cm I tot qnµ n AE = 54.1 µa (b) All of the parameters are the same except n i, which means we must re-calculate p. n i (T = 400 K) = cm 3 p = n i /n = cm 3 Since nµ n pµ p 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 I tot qnµ n AE = 54.1 µa

4 .4 (a) From Problem 1, we can calculate n i for Ge. Since nµ n pµ p, we can write n i (T = 300 K) = cm 3 I tot = q(nµ n + pµ p )AE n = cm 3 p = n i /n = cm 3 µ n = 3900 cm /V s µ p = 1900 cm /V s E = V/d = 1V 1 µm = 10 5 V/cm A =05 µm 05 µm = cm I tot qnµ n AE = 156 µa (b) All of the parameters are the same except n i, which means we must re-calculate p. n i (T = 400 K) = cm 3 p = n i /n = cm 3 Since nµ n pµ p 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 I tot qnµ n AE = 156 µa

5 .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 I tot = I diff = AJ diff = Aq D n dx D dp p dx A =1µm 1 µm = 10 8 cm D n = 34 cm /s D p = 1 cm /s dn dx = cm cm = cm 4 dp dx = 1016 cm cm = 100 cm 4 I tot = 10 8 cm C 34 cm /s cm 4 1 cm /s 10 0 cm 4 = µa

6 .6 Given Area = a find total e ectrons stored. N nx ( ) =- x+ N L N total electrons stored L N = a n( x) dx = a x N dx L x L anl = an - + x = L 0 0 L x

7 .7 Given Area = a find total electrons stored. -x n( x) = N exp Ld total electrons stored -x = a n( x) dx = a N exp dx 0 Ld -x = an - Ld exp = anl d. Ld 0 For the linear profile, the result depends on the length, L. For the exponential profile, the result is constant (since L d is constant).

8 .8 Assume the diffusion lengths L n and L p are associated with the electrons and holes, respectively, in this material and that L n,l p µm. We can express the electron and hole concentrations as functions of x as follows: # of electrons = n(x) =Ne x/ln p(x) =Pe (x )/Lp # of holes = 0 an(x)dx = ane x/ln dx 0 = anl n e x/ln 0 = anl n e /Ln 1 = 0 0 ap(x)dx ap e (x )/Lp dx = ap L p e (x )/Lp = ap L p 1 e /Lp Due to our assumption that L n,l p µm, we can write e /Ln 0 e /Lp 0 # of electrons anl n # of holes ap L p 0

9 .9 Drift is analogous to water flow in a river. Water flows from top of mountain to bottom because of gravitational field; electron flows from one terminal to the other because of electric field. DRIFT electrons electric field drift/current WATER FLOW water gravitational field water flow

10 .10 (a) n n = N D = cm 3 p n = n i /n n = 33 cm 3 p p = N A = cm 3 n p = n i /p p = 916 cm 3 (b) We can express the formula for V 0 in its full form, showing its temperature dependence: V 0 (T )= kt q ln N A N D ( ) T 3 e Eg/kT V 0 (T = 50 K) = 906 mv V 0 (T = 300 K) = 849 mv V 0 (T = 350 K) = 789 mv Looking at the expression for V 0 (T ), we can expand it as follows: V 0 (T )= kt q ln(na ) + ln(n D ) ln ln(t )+E g /kt Let s take the derivative of this expression to get a better idea of how V 0 varies with temperature. dv 0 (T ) dt = k q ln(na ) + ln(n D ) ln ln(t ) 3 From this expression, we can see that if ln(n A ) + ln(n D ) < ln ln(t ) + 3, or 5. equivalently, if ln(n A N D ) < ln T 3 3, then V 0 will decrease with temperature, which we observe in this case. In order for this not to be true (i.e., in order for V 0 to increase with temperature), we must have either very high doping concentrations or very low temperatures.

11 .11 Since the p-type side of the junction is undoped, its electron and hole concentrations are equal to the intrinsic carrier concentration. n n = N D = cm 3 p p = n i = cm 3 ND n i V 0 = V T ln n i ND = (6 mv) ln = 386 mv n i

12 .1 (a) C j0 = C j = qsi N A N D 1 N A + N D V 0 C j0 1 VR /V 0 N A = cm 3 N D = cm 3 V R = 1.6 V NA N D V 0 = V T ln = 701 mv n i C j0 = 14.9 nf/cm C j =8. nf/cm = 08 ff/cm (b) Let s write an equation for C j in terms of C j assuming that C j has an acceptor doping of N A. Cj =C j q Si NA N D 1 NA + N D V T ln(na N D/n i ) V =C j R q Si NA N D 1 NA + N D V T ln(na N D/n i ) V =4Cj R q Si N A N D =8C j (N A + N D)(V T ln(n A N D/n i ) V R) N A qsi N D 8C j (V T ln(n AN D /n i ) V R ) =8C j N D (V T ln(n AN D /n i ) V R ) NA 8Cj = N D(V T ln(na N D/n i ) V R) q Si N D 8Cj (V T ln(na N D/n i ) V R) We can solve this by iteration (you could use a numerical solver if you have one available). Starting with an initial guess of N A = 1015 cm 3, we plug this into the right hand side and solve to find a new value of N A = cm 3. Iterating twice more, the solution converges to N A = cm 3. Thus, we must increase the N A by a factor of N A /N A =

13 .13 c j V R C j V C j V 0 1 : =. 1 = /V 0. = V 0 = 0365 V /V Substitute V 0 into 1 : 05. C j0 = fF. / μm V N A N D = ( C ) N A + N D j0 V0 ε q si ff = ( V) cm.. m μ εsiq N Fix a value for A N D N A > y N A + N D N A = 10 cm = N D yn A N A - y ( cm )( 10 cm ) = ( )cm 17-3 < cm 17-3

14 .14 (a) In forward bias, I D = 1mA, V D = 750 mv V - V D T I I e = (1 ma) exp [-750 mv / 6 mv] s D -16 = A (b) Since I s & Area, doubling area implies doubling I s From (a), I D = 1 ma = I s e V D/V T ID 1 ma VD = VT ln = (6 mv) ln -16 I S A = 73 V

15 .15 (a) I tot VB D VB VT Itot = I + ID = IS1( e VB VT - 1) + IS ( e -1) VB VT = ( IS1 + IS)( e -1) Therefore, the parallel combination operates as an exponential device, with an equivalent saturation current of I S1 + I S. ID 1 ID VT ln = VT ln IS1 IS Also, I = I + I I = I -I tot D D tot ID 1 Itot - I VT ln = VT ln IS1 IS IS1 I = Itot IS1 + IS IS ID = Itot IS + I 1 S

16 .16 (a) The following figure shows the series diodes. V D + I D D 1 D Let V be the voltage drop across D 1 and V D be the voltage drop across D. Let I S1 = I S = I S, since the diodes are identical. V D = V + V D ID = V T ln + V T ln I S ID =V T ln I S I D = I S e VD/VT Thus, the diodes in series act like a single device with an exponential characteristic described by I D = I S e VD/VT. (b) Let V D be the amount of voltage required to get a current I D and VD the amount of voltage required to get a current 10I D. ID V D =V T ln I S VD =V 10ID T ln I S VD 10ID ID V D =V T ln ln =V T ln (10) = 10 mv I S ID I S I S

17 .17 + V - + V D - Find I B, V, V D in terms of V B, I, I S I B D 1 D + V B - IB IB By KVL,VB = V + VD = VT ln + VT ln IS1 IS IB VB = VT ln IS1 IS B IB = IS1 IS exp VB VT = IS1 IS exp VT D I S V = VT ln + I S1 I = VT ln I B V IB IS1 IS V B V = VT ln = VT ln exp I S1 IS1 V T V IS1 V = VT ln + I B IS1 IS VB = VT ln exp S IS V T S B

18 .18 + V - + V D - I B D 1 D + V B - IB I B I B VB = VT ln + VT ln = VT ln IS1 I S IS1 I S VB IB = IS1 IS exp V T Increase I I B by 10 times: = 10 I B, new B I B, new (10 I ) B VB, new = VT ln = VT ln IS1 I S IS1 IS I B = VT ln VT ln 100 IS1 I + S = V + V ln 100 V + 10 V V increases by 10 V B B T B

19 .19 V = I R 1 + V = I R 1 + V T ln I = V R 1 V T R 1 ln I I S I For each value of V, we can solve this equation for I by iteration. Doing so, we find I S I (V =5 V) = 435 µa I (V =8 V) = 8.3 µa I (V = 1 V) = 173 µa I (V =1. V) = 67 µa Once we have I, we can compute V D via the equation V D = V T ln(i /I S ). Doing so, we find V D (V =5 V) = 499 mv V D (V =8 V) = 635 mv V D (V = 1 V) = 655 mv V D (V =1. V) = 666 mv As expected, V D varies very little despite rather large changes in I D (in particular, as I D experiences an increase by a factor of over 3, V D 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.

20 .0 +-V I R 1 kω D 1 since I Area, I becomes: I S1 S1-15 V V T = 10 ( 10 A) ( e -1) I V = 8 V Suppose D is on. Assume V = 7 V 1 V - V 01V. V = 07V. I = = = 005mA. R kω V = 563 V 1 05 ma V = VT ln ( I I S 1) = ( 06 V) ln 563 V -15 = 0 10 A ( 8-563)V I = = 1 ma kω V 1 ma = ( 06 V) ln 585 V A V = 585 V I ( 8-585)V = = 11 ma kω V 11 ma = ( 06 V) ln 583 V A V = 583 V I ( 8-583)V = = 11 ma kω V = 583 V V 583 V I 11 ma

21 V = 1. V Suppose D is on. Use results from previous calculations as starting point. 1 V = 583 V I ( )V = = 31 ma kω V 31 ma = ( 06 V) ln 610 V A V = 610 V I ( )V = = 30 ma kω V 30 ma = ( 06 V) ln 609 V A V = 609 V I ( )V = = 30 ma kω V = 609 V V 609 V I 30 ma By increasing the cross-section area of D 1, intuitively this means D 1 can conduct same amount of current with less V, The results have shown that in this problem, V is less and I is more.

22 .1 +-I R 1 D 1 kω +V V = V,V = 850 mv 1 V - V I = = 58 ma R I IS = I V V e T ( - 1) [-V VT exp ] -18 = ( 58 ma) exp [-85 06] A

23 . V /=I R 1 = V = V T ln(i /I S ) I = V T R 1 ln(i /I S ) I = 367 µa (using iteration) V =I R 1 = 1.47 V

24 .3 +-V I R 1 Given: V = 1 V I = ma Find R and I. 1 V = V I = 5 ma S D 1 I By KVL, V = V - IR1 = VT ln IS ma 1 -( ma) R 1 = ( 06 V) ln IS 5 ma -( 5 ma) R 1 = ( 06 V) ln IS : 1 -( 3 ma) R 1 = ( 06 V) ln ( 06) V R1 = = 3. 5 kω 3 ma Substitute R 1 into 1: V - IR1 IS = I exp - VT 1 - ( m)(3. 5 k) -10 = ( ma) exp A 06 R 3.5 kω I 1 S A

25 .4 I R 1 1 kω D 1-16 Given I = 3 10 A find S V By KCL, I V I V I I T = + = ln + R1 R1 IS Since I, V T, R 1 and I S are known, this can be solved directly with special programs or graphing calculators. However, this can be also solved by iterations : Assume a V, calculate I, and re-iterate on V. Assume V = 7 V as starting point. I = 1 ma 07V. V = 7 V I = I - V R 1 = 1 ma - = 3 ma 1kΩ V = V ln (I I ) T S 3 ma = ( 06 V) ln 718 V A 718 V V = 718 V I = 1 ma - = 8 ma 1kΩ V 8 ma = ( 06 V) ln 717 V 3 10 A V V = 717 V I = 1 ma - = 8 ma 1kΩ V 717 V. V = 717 V

26 I = ma Assume V = 717 V from previous result. 717 V V = 717 V I = ma - = 1. 8 ma 1kΩ 1. 8 ma V = ( 06 V) ln 756 V A V V = 756 V I = ma - = 1. 4 ma 1kΩ 1. 4 ma V = ( 06 V) ln 755 V A 755 V V = 755 V I = ma - = 1. 4 ma 1kΩ V = 755 V V = 755 V

27 I = 4 ma Assume V = 755 V from previous result. 755 V V = 755 V I = 4 ma - = 3. 5 ma 1kΩ 3. 5 ma V = ( 06 V) ln 780 V A 780 V V = 780 V I = 4 ma - = 3. ma 1kΩ 3. ma V = ( 06 V) ln 780 V A V 780 V Note: As I increases, I increases, while (V /R 1 ) stays relatirely the same. Because of the exponential characteristic, the diode, once on, will absorb as much current as necessary to satisfy KCL.

28 .5 I R 1 1 kω D 1 Given I = 5 ma when I = 1. 3 ma, find I. S This means V = ( I -I ) R 1 I = I exp -V V S T [ ] [ ] -17 = (08mA)1k. Ω= 08V. = ( 5 ma) exp - 8 V 06 V A

29 .6 I R 1 1 kω Given I = I find I. I S -16 = 3 10 A I I V =,R1 = VT ln IS This can be solved directly with special programs or graphing calculators. Alternatively, one can solve this iteratirely by hand. Assume V = 8 V D V D 08V. V D = 08V. ( I ) = = = 08mA. R 1kΩ 1 VD ln I (0 06 V) ln 8 ma = V T = I - S A 744 V V D = 744 V I 744 V = = 744 ma 1kΩ V D 744 ma = ( 06 V) ln 74 V A V D = 74 V I 74 V = = 74 ma 1kΩ V D 74 ma = ( 06 V) ln 74 V A I = ( 74 ma) = ma

30 .7 I R 1 D 1 +-V Given I = 1 ma V = 1. V find R and I. 1 I = ma V = 1. 8 V S I = I -V R 1 (KCL) ID 1 I - V R1 By KVL, V = VT ln = VT ln IS IS (1 ma) - (1. V) R1 (1. V) = ( 06 V) ln IS ( ma) - (1. 8 V) R1 (1. 8 V) = ( 06 V) ln IS 1 ma V R1-1 : 6 V = ( 06 V) ln 1 ma - 1. V R1 1. exp[ 6 06] R 1 = 1k. Ω 1 ma exp ma -34 [ ] 18V. -18V. IS = ID exp [- V VT] = ma - exp 1. kω 06 V A

31 .8 D 1 I R 1 kω D Given I S D R1 = D 1-16 = 5 10 A with Find V for I = ma. V Current through the diodes = I = I - where V = voltage across R ID I VR1 VR1 = VT ln = VT ln - IS IS ISR1 R1 D R1 1 R1 This can be solved directly with special programs or graphing calculators or by hand iteratively. Assume a V R1, calculate I D, and re-iterate on new V R1 = ( V ) From experience, most diodes conduct at V D 7 V. Assume V R1 = 1.4 V. V R1 14V. V R1 = 1. 4 V ID = I - = ma - = 1. 3 ma R kω V R1 1 ID = VT ln IS 1. 3 ma = ( 06 V) ln V A

32 1. 49 V V R1 = V I D = ma - = 1. 6 ma kω 1. 6 ma V R1 = ( 06 V) ln V A V V R1 = V I D = ma - = 1. 6 ma kω V = V R1 voltage across R = V 1

33 .9 I D I R 1 D 1 1 ma I R1 D Given I S Find I = 5 ma, 1 R1-16 = 5 10 A for each diode R. By KCL,I = I - I = 5 ma D R1 I D 5 ma V = VD = V T ln = 06 ln I S A 718 V VR1 V ( 718 V) R 1 = = = =. 87 kω I I 5 ma R1 R1

34 .30 I R 1 +-V (a) Constant-voltage model : Consider, first, the extreme cases : when D 1 is off, we have the following: +I R1 V This implies V is linearly proportional to I -When D 1 is on, V is fixed (by KVL) by D 1 ( = V D, ON). This implies that any additional current from I cannot flow through R 1, which means D 1 will absorb all the currents to satisfy KVL. V forward-biased V D, ON R 1 I V D, ON R 1 D 1 reverse-biased

35 (b) exponential model: Assume I S negligible. I R 1 +-V D 1 When D 1 is off, most of I flows through R 1. When D 1 is on, V ( = V ) follows this relationship: ID 1 I V V = V = VT ln = VT ln - IS IS IR1 I = I exp ( V V ) + V R S T I exp ( V V ) S T when D is forward-biased ( V > V ) 1 1 T i.e. V V ln ( I I ) T S V f~i f (I )~ ln (I ) 0 I dominates I I R1 >> I (small window) D 1 reverse-biased