EE 321 Analog Electronics, Fall 2013 Homework #5 solution


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1 EE 321 Analog Electronics, Fall 2013 Homework #5 solution For the circuit shown in Fig. P3.26, both diodes are identical, conducting 10mA at 0.7V, and 100mA at 0.8V. Find the value of for which V = 80mV. V 1 is the voltage across diode 1, V 2 across diode 2, I 1 is current through diode 1, and I 2 through diode 2. hen we have (using exponential model I = I 1 +I 2 ( V2 I 2 = I s exp ( ( ( V2 V V2 I 1 = I s exp = I s exp exp V or alternatively, and thus and ( V I 2 = I 1 exp ( ] V I = I 1 +I 2 = I 1 [1+exp and I 1 = = V I 1 = V I I ( 1+exp V [ ( ] V 1+exp Now we only need to determine. Note that ( ID log = V I S 1
2 and log ( IDa I S log log ( IDa I Db ( IDb I S = V a V b = V a V b = V a V ( b I log Da I Db Now inserting V a = 0.8V, I a = 100mA, V b = 0.7V, and I b = 10mA, we get Inserting that in the expression for, we get [ 1+exp = = 0.1 log10 = 43.4mV ( ] = 58.5Ω A shunt regulator utilizing a zener diode with an incremental resistance of 5Ω is fed through an 82 Ω resistor. If the raw supply changes by 1.3V, what is the corresponding change in the regulated output voltage? he circuit looks like this, with L =. he output is across the Zener diode, so across the supply and r z. he relationship between the input and output voltages is v O = V Z0 +(V V Z0 +r z o get the regulation, the ratio of change in the output for a change in the output we take the derivative of the output with respect to the input. dv O dv = r z +r z r z 2
3 For some change in the input voltage, V = 1.3V, the output is r z v O = V +r z = =0.075 V Consider a halfwave rectifier circuit with a triangular wave input of 5 V peaktopeak amplitude and zero average and with = 1kΩ. Assume that the diode can be represented by the piecewiselinear model V D0 = 0.65V and r D = 20Ω. Find the aveage value of v o. he relationship between the input and the output is { (v I V D0 +r v o = D v I v D0 0 v I < v D0 If the period of the signal is, and the input voltage is v I = V sin ( t, then the diode is turned on between times t 1 and t 2, where ( t1 sin = V D0 0 t 1 V 4 and t 1 = sin 1 V D0 V = sin = t 2 = 2 t 1 = ( 1 = = π he average value of v O is found by integrating it over the period and dividing by the period, v o = 1 = 1 = 1 = 1 0 t2 v o dt v o dt t 1 t2 [ ( ] t V sin V D0 dt t 1 +r D [ t2 ( ] t V sin dt (t 2 t 1 V D0 +r D t 1 Change integration variable, x = t π , we get, dx = dt, x 1 = t 1 = , x 2 = t 2 = 3
4 Now inserting all the numbers v o = 1 [ V x2 ] sinxdx (t 2 t 1 V D0 +r D x 1 = 1 [ V x2 sinxdx ] +r D x 1 (x 2 x 1 V D0 = 1 [ x2 ] V sinxdx (x 2 x 1 V D0 +r D x 1 = 1 [ V [ cosx] x 2 ] x +r 1 (x 2 x 1 V D0 D v o = [5 (cos cos(π (π ] = V A fullwave bridge rectifier circuit witha1 kω load operates from a 120 V (rms 60 Hz household supply through a 10to1 stepdown transformer having a single secondary winding. It uses four diodes, each of which can be modeled to have a 0.7 V drop for any current. What is the peak value of the rectified voltage across the load? For what fraction of the cycle does each diode conduct? What is the average voltage across the load? What is the average current through the load? he peak value of the rectified voltage across the load is where V I = 120V 2 10 = 16.97V, so V O = V I 2v D V O = = 15.57V Each dioded conducts for a fraction, f, of time which is equal to the time the input voltage is greater than 2v D, or the fraction of the time that a sinusoid exceeds the value 2v D /V I. f = 1 ( ( π 2sin 1 2vD V I = 1 (π = =47.37% (the answer in the book is incorrect. he average voltage across the load is found by integrating the output voltage over a period and dividing by the period, 4
5 v o = 1 0 v o dt he relationship between the input and the output voltage is { v I 2V D v I > 2V D v o = 0 v I 2V D where v I = V I sin t. Because of the symmetry we can just integrate over the half period corresponding to the positive peak in v I, v o = 2 = 2 2 v o dt ( t V I sin 2V D dt t 1 0 t2 Now change variables, x = t, and thus dx = dt, x 1 = t 1, and x 2 = t 2, Now, x2 v o = 2 V I sin x 2V D dx x 1 = 1 [ VI [ cosx] x ] 2 x π 1 2(x 2 x 1 V D ( ( x 1 = sin 1 2VD = sin 1 = V I and x 2 = π x 1 = π = , and we can insert v o = 1 [ cos( (π ] = 9.44V π he average current is simply the average voltage divided by the load resistance, i o = 1 v o = = 9.44mA he op amp in the precision rectifier circuit of Fig P3.91 is ideal with output saturation levels of ±12 V. Assume that when conducting the diode exhibits a constant voltage drop of 0.7V. Find v, v a, and v A for: (a v I = +1V (b v I = +2V (c v I = 1V 5
6 (d v I = 2V (a When v I > v D, the opamp will attempt to output current to raise v to v I by raising its output voltage. herefore I expect the diode to be conducting. In that case we have, and thus and i = v = v D v o = 2i = 2 v I = 2v I Inserting values we get v A = v O +V D v = v I = 1V v o = 2v I = 2 1 = 2V v A = v o +V D = = 2.7V (b In this case the derivation is exactly the same as for case (a, so v = v I = 2V v o = 2v I = 4V v A = v o +V D = = 4.7V (c In this case, the opamp output will attempt to draw current by lowering its voltage. It cannot draw current so the opamp output will go to negative rail. here is no current anywhere else in the circuit so v = v o = 0. hus, 6
7 v = 0V v o = 0V v A = 12V (d In this case the situation is identical to case (c, with the same voltages he opamp in the circuit of Fig P3.92 is ideal with saturation levels of ±12V. he diodes exhibit a constant 0.7V drop when conducting. Find v, v A, and v o for: (a v I = +1V (b v I = +2V (c v I = 1V (d v I = 2V (a In this case the input voltage is above ground, and the opamp will attempt to adjust by drawing current in. It can draw current through D 1, and then D 2 will not be conducting. hus, v A = V D = 0.7V, v = 0V. For the ground we realize that no current flows through the loop containing ground, and thus v o = v = 0V. (b his case is the same as case (a. he opamp is simply drawing twice as much current through D 1. he voltages are the same as case (a. (c In this case the input is below ground and the opamp will attempt to compensate by supplying current, raising its output voltage. In this case diode D 2 is conducting, and the opamp current will rise until v = 0. At that point, v o = v I = 1V, and v A = v o +V D = 1.7V. (d his case is similar to case (c. he opamp will output current through D 2 to make v = 0, and then v o = v I = 2V, and v A = v o +V D = 2.7V (a Use the EbersMoll expressions in Eqs and 5.27 to show that the i C v CB relationship sketch in Fig. 59. can be described by 7
8 ( 1 i C = α F I E I S α F e v BC V (b Calculate and sketch i C v CB curves for a transistor for which I S = A, α F 1, and = 0.1. Sketch graphs for I E = 0.1mA, 0.5mA, and 1mA. For each, give the values of v BC, v BE, and v CE for which (a i C = 0.5α F I E and (b i C = 0. (a he EbersMoll equations are i E = I ( S e v BE V 1 I S (e v BC V 1 α F ( i C = I S e v BE V 1 I ( S e v BC V 1 Eliminate e v BE V from the second equation by substituting the first equation into it. i C is then a function of i E (which the book assumes fixed biased so it calls it I E and v BC. earrange the first equation: Insert in the second equation ( I S e v BC V 1 = α F I E +α F I S (e v BC V 1 ( i C =α F I E +α F I S e v BC V 1 =α F I E I S ( 1 α F (b his plot shows i C as a function of v CB. e v BC V I ( S e v BC V 1 8
9 he solid curve is for i C = 1mA, the dotted is for i C = 0.5mA, and the dashed is for i C = 0.1mA. (a he value of v CB for which i C = 0.5α F I E can be found from he values are tabulated here: ( 1 0.5α F I E = α F I E I S α F e v BC V v BC = V ln ( 0.5α FI E 1 I S α F I E v BC 0.1mA 0.57V 0.5mA 0.61V 1mA 0.62V (b he value for v CB for which i C = 0 can be found from he values are tabulated here ( 1 0 = α F I E I S α F e v BC V v BC = V ln α F I ( E 1 I S α F 9
10 I E v BC 0.1mA 0.58V 0.5mA 0.62V 1mA 0.64V For the circuits in Fig P5.20, assume that the transistors have very large β. Some measurements have been made on these circuits, with the results indicated in the figure. Find the values of the other labeled voltages and currents. he hint that β is very large means that we can assume that i C = i E. (a I 1 = V CC V E E = = 10mA (b We can see that the transistor must be on, so V 2 = V B +V BE = = 2V (c he transistor is on because there is collector current flowing. I 3 = I E = I C = V C V CC = 0+10 = 1mA C 10 Since β is very large there is no current flowing in the base, so V 4 = V BB = 1V. (d Since β is very large there is no current flowing in the base and thus V B = V C, and we can write and thus V CC V EE = I 5 ( C + E +V BE I 5 = V CC V EE V BE C + E = = 0.97mA 10
11 V 6 = V CC I C C = = 4.6V 11
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