EE 321 Analog Electronics, Fall 2011 Homework #5 solution 3.37. Find the parameters of a piecewise-linear model of a diode for which v D 0.7 V at i D 1 ma and n 2. The model is to fit exactly at 1 ma and 10 ma. Calculate the error in millivolts in predicting v D using a piecewise-linear model at i D 0.5, 5 and, 14 ma We have i D I S e v D, so Next, I S i D e v D 1 10 3 e 0.7 2 25.2 10 3 0.93 na Then we have v D ln id I S 2 25.2 10 3 ln 10 10 3 0.82 V 0.93 10 9 and r D v D10 v D1 0.82 0.7 i D10 i D1 10 1 13 Ω v D0 v D1 r D i D1 0.7 13 1 10 3 0.69 V We can now calculate v D based on the piecewise linear model as or based on the exponential model v D v D0 + i D r D v D ln i D I S and those are tabulated here id v D exp v D lin error 0.5 0.67 0.70 0.03 5 0.78 0.76 0.02 14 0.83 0.87 0.04 3.54. In the circuit shown in Fig. P3.54, I is a DC current and v s is a sinusoidal signal. Capacitors C 1 and C 2 are very large; their function is to couple the signal to and from the diode but block the DC current from flowing into the signal source or the load not shown. Use the diode small-signal model to show that the signal component of the output voltage is v o v s + IR s If v s 10 mv, find v o for I 1 ma, 0.1 ma, and 1 µa. Let R s 1 kω and n 2. At what value of I does v o become one-half of v s? Note that this circuit functions 1
as a signal attenuator whith the attenuation factor controlled by the value of the DC current I. The signal portion is transferred from input to output according to a voltage division between R S and r D, where and thus r D v o v s R s + r D r D dv 1 1 D did I di D id I i d I I I v o v s R S + I IR s + Find v o for several values of I, and v s 10 mv, R s 1 kω, and n 2. It is tabulated below I ma v s mv 1 0.48 0.1 3.4 10 3 9.8 Value of I for which v o vs: 2 1 + IR s 2 I R s 2 25.2 10 3 1 10 3 5.0 10 5 A 50 µa 3.59 Consider the voltage-regulator crictuit show in Fig. P3.59. The value of R is selected to obtain an output voltage V o across the diode of 0.7 V. 2
a Use the diode small-signal model to show that the change in output voltage corresponding to a change of 1 V in V + is V o V + V + + 0.7 This quantity is known as the line regulation and is usually expressed in mv/v. b Generalize the expression above to the case of m diodes connected in seris and the value of R adjusted so that the voltage across each diode is 0.7 V and V o 0.7 m V. c Calculate the value of line regulation for the case V + 10 V nominally and i m 1, and ii m 3. Use n 2. a We can write i D V + v D R I S e v D and we are interested in finding. we can re-arrange and then compute V + vd R R + I Se v D v D + RI S e v D 1 + RI S e vd 3
Now, we are interested in evaluating this at a bias point v D V D, for which so we can insert that and get I D I S e V D 1 + RI D at that bias point we have RI D V + V D, so we can write and we are asked to compute 1 + V + V D dv 1 + 1 + V + V D + V + V D With the bias point V D 0.7 V this is identical to the expression we were asked to compute. b In this case we just write re-arrange and compute i D V + mv D R I S e v D V + mvd R R + I Se vd mv D + RI S e v D v O + RI S e v O m 1 + RI S m e vo m 1 + RI S m e vd Now note that at the bias point, v D V D V O m, we can substitute I D I S e v D Finally we can compute 1 + RI D m 1 + V + V O m 1 + V + mv D m 4
Again we use V D 0.7 V. dv 1 + 1 + V + mv D m m m + V + mv D c i ii 2 25.2 10 3 2 25.2 10 3 + 10 0.7 0.0054 3 2 25.2 10 3 3 2 25.2 10 3 + 10 3 0.7 0.019 3.61 Design a diode voltage regulator to supply 1.5 V to a 150 Ω load. Use two diodes specified to have a 0.7 V drop at a current of 10 ma and n 1. The diodes are to be connected to a +5 V supply through a resistor R. Specify the value of R. What is the diode current with the load connected? What is the increase resulting in the output voltage when the load is disconnected? What change results if the load resistance is reduced to 100 Ω? To 75 Ω? To 50 Ω? First we compute I S from the 10 ma point as I S i D 10 10 3 e v D 0.7 8.64 10 15 A e25.2 10 3 Next compute the amount of current which is required to produce a 0.75 V drop across one diode. i D I S e v D VT 0.75 8.64 10 15 e25.2 10 3 72.8 ma We have this current through the resistor, as well as the curent through the 150 Ω load resistor. The current through the load is i L v L 1.5 10 ma R L 150 and the size of the resistor can then be found from V R i D + i L + 2v D R V 2v D i D + i L 5 1.5 72.8 + 10 42.3 Ω If the load is disconnected let s assume that the additional small 10 ma goes through the diodes. Let s compute the diode resistance, r D, did r D 1 The change in voltage is then i D 72.8mA ID 5 I D 25.2 10 3 0.35 Ω 72.8 10 3
v O i D r D If the load is disconnected, i D 10 ma, and v D 10 10 3 0.35 3.5 mv If the load resistance is reduced to 100 Ω, i D 1.5 1.5 5 ma, and then v 150 100 D 5 10 3 0.35 1.8 mv. If the load resistance is reduced to 75 Ω, i D 1.5 1.5 10 ma, and then v 150 75 D 3.5 mv. If the load resistance is reduced to 50 Ω, i D 1.5 1.5 20 ma, and then v 150 50 D 7 mv. 6