Georgia Institute of Technology. ECE 3400: Analog Electronics Summer 2016

Transcription

1 Georgia Institute of Technology SCHOOL OF ELECTRICAL AN COMPUTER ENGINEERING ECE 3400: Analog Electronics Summer 2016 Homework #1 Solutions 3.21 i I S exp v v 1 or ln 1 i n T n T I S For i I S, v ln i 1 or ln I v ln I S n T I S n T which is the equation of a straight line with slope 1/n T and xaxis intercept at ln (I S ). The values of n and I S can be found from any two points on the line in the figure: e. g. i = 10 4 A for v = 0.60 and i = 10 9 A for v = Then there are two equations in two unknowns: ln ln I S or ln I S n n ln ln I S or lni S n n Solving for n and I S yields n = 1.39 and I S = 3.17 x A = 3.17 pa T 1.38x I 1.60x10 19 I S exp n arying n and I S by trialanderror with a spreadsheet: n I S E15 I Measured I Calculated Error Squared E E E E E E E E E E E E E E E E E E10

2 E E E E E E E E E06 Total Squared Error E dv dt v G 3 T T m K 3.43 Since N A >> N, the depletion layer is all on the lightlydoped side of the junction. Also, R >> j, so j can be neglected. E max qn A x p S N A E 2 max S 2q R qn A w d S 3x10 5 = qn A S S q R N A 8.854x x x / cm L = C RFC C 10 H 10 H C C (b) C C jo 1 R j (a) C 39 pf pf f o 2 LC H26.0 pf 39 pf pf f o 2 LC MHz H10.6 pf MHz 3.58

3 (a) I 0 I 0.500mA I 0 5 Forward biased 0.5 I ma 10 4 (b) I 0 I 2.00mA I 0 6 In reverse breakdown 4 I ma 3k (c) I 0 I 1.00mA I 0 3 Reverse biased 3 I 0 i 2 ma 1 ma (c) Qpoint (a) Qpoint v (b) Qpoint 1 ma 2 ma 3.69 Using Thévenin equivalent circuits yields and then combining the sources I 1 k I 1.2 k 2.2 k (a) Ideal diode model: The source appears to be forward biasing the diode, so we will assume it is "on". Substituting the ideal diode model for the forward region yields I k ma. This current is greater than zero, which is consistent with the diode being "on". Thus the Qpt is (0, ma). I 2.2 k 0.6 on I 2.2 k Ideal iode: C:

4 (b) C model: The source appears to be forward biasing the diode so we will assume it is "on". Substituting the C model with on = 0.6 yields I 2.2k 136 A. This current is negative which is not consistent with the assumption that the diode is "on". Thus the diode must be off. The resulting Qpt is: (0 ma, ). I=0 2.2 k (c) The second estimate is more realistic. is not sufficient to forward bias the diode into significant conduction. For example, let us assume that I S = A, and assume that the full appears across the diode. Then i A exp pa, a very small current (d) Using the interative method from Prob with a starting guess of 1 fa yields a Qpoint of (0.163 na, 0.300) (a) 1 and 2 forward biased I k 400A I : 200 A, 0 2 : 400 A, 0 (b) 1 forward biased, 2 reverse biased I k 400A 1 : 400 A, : 0 A, 9 (c) 1 and 2 forward biased I k 600A I I : 200 A, 0 2 : 600 A, 0 k I 200A 2 k 200A

5 (d) 1 reverse biased, 2 forward biased 6 9 I 2 30k 500A I :0A, : 500 A, 0 (b) (a) 1 on, 2 on : I 2 = 400A I A 157A 15k 15k 1 : 157 A, : 400 A, 0.65 (b) 1 on, 2 off : I 2 0 I 1 = k 357A : 357 A, : 0 A, 9.65 (c) 1 on, 2 on : I 2 = 513A I 1 513A k 15k 70.0A 1 :70.0 A, : 513 A, 0.65 (d ) 1 off, 2 on : I 1 0 I 2 = 478A x10 3 I :0 A, k 2 : 478 A, (a) TH R TH I Z P =15I Z = 1.25 W (b) I Z ma P =15I Z =3.50 W ma *Problem S 2 1 C 0 AC 0 SIN( ) IOE IOE C U

6 C U RL 3 0 3K.MOEL IOE IS=1E15 RS=0.OPTIONS RELTOL=1E6.TRAN 0.1MS 100MS.PRINT TRAN (2,1) (3) I(S).PROBE (3) (2,1) I(S).EN 4.0k 3.0k v O 2.0k 1.0k 0 v S 1.0k Time 2.0k 0s 20ms 40ms 60ms 80ms 100ms Simulation Results: C = 2981, r = 63 The voltage doubler circuit is effectively two halfwave rectifiers connected in series. Each capacitor is discharged by I = 3000/3000 = 1 A for 1/60 second. The ripple voltage on each capacitor is With two capacitors in series, the output ripple should be 66.6, which is close to the simulation result i 0 5 I r 1k 5 ma I 5 F 1k ma 1k 3.6 ma 7ns S ln 1 4.4mA 5.59 ns 3.6mA k *Problem iode Switching elay PWL( N 5 10N N 3 20N 3) R K IOE.TRAN.01NS 20NS

7 .MOEL IOE TT=7NS IS=1E15.PROBE (1) (2) I(1).OPTIONS RELTOL=1E6.OP.EN 10 5 v 1 v 0 5 Time 10 0s 5ns 10ns 15ns 20ns Simulation results give S = 4.4 ns.