ES 330 Electronics II Homework # 1 (Fall 2016 SOLUTIONS)

Transcription

1 SOLUTIONS ES 330 Electronics II Homework # 1 (Fall 2016 SOLUTIONS) Problem 1 (20 points) We know that a pn junction diode has an exponential I-V behavior when forward biased. The diode equation relating current I and voltage V is given by (see Section 4.2 of Sedra & Smith, pp ) I = I S (e (qv /kt) 1) where I S is the saturation current of the diode, V is the applied voltage across the pn junction, and kt/q (= V T ) is the thermal voltage (kt/q which we will take to equal 26 mv at so-called room temperature ). (a) Using the above expression for I, find the incremental (i.e., small-signal) resistance r d under forward bias. You should remember that resistance r d is the derivative (dv /di ). exp( / 1 exp( / I I V V I V V S T S T dv di IS d exp( V / VT IS exp( V / VT ; now rearranging gives V r d dv V 1 V di I exp( V / V I T T S T (b) What is the small-signal diode resistance r d at a forward current of I = 1 milliampere? r d = V T /I = 26 mv/1 ma = 26 ohms (c) Suppose the diode is forward biased at a current I 0. How much must one increase the forward voltage across this diode to increase the current ten-fold in magnitude (that is, I 1 = 10 I 0 where V 1 = V 0 + ΔV increase )? 1 0 T by substitution I 0 I 1 I 1 V0 VT ln, and V 1 VT ln ; V 1 V0 VT ln IS I S I 0 We are given the condition I 10I 10I V V V V V V V V 59.9 mv 60 mv T ln T ln10 26 mv mv I 0 1 0

2 Note: This is usually called the 60-millivolt rule. (d) From the expression for diode current, I = I S (e (qv/kt) 1), what does the minus one term physically represent in the equation for I? The factor of -1 accounts for minority carriers at the edge of the delpetion layer being swept across the junction because the direction of the built-in electric field within the depletion layer aids the transport of minority carriers in crossing the depletion layer. For a reverse biased junction, the -1 accounts for the reverse leadkage current of the junction. Problem 2 (18 points) Continuing with the diode in Problem 1, the saturation current I S = ampere. Plot the diode s forward current I (on the ordinate) versus the diode s forward voltage V (on the abscissa). For the plot use four different forward voltage values (for example, you could use V = 0.5 volt, 0.6 volt, 0.7 volt and 0.8 volt) on the semi-log graph provided below. Plot your four calculated data points on the graph and then fit a line to the data. (a) Plot your four data points on the grid provided below. Evaluating the four bias points gives iode voltage V iode current I 0.5 volt (exp(0.500/0.026)) = A 0.6 volt (exp(0.600/0.026)) = A 0.7 volt (exp(0.700/0.026)) = A 0.8 volt (exp(0.800/0.026)) = A Next, we plot these four points on the graph provided: See the graph on next page.

3 (b) What is the slope of the curve you plotted (remember this is a semi-log graph)? The slope is ΔI / ΔV is the conductance of the junction and its value varies with current. You can see this by looking at a linear current versus linear voltage plot. For example, Linear scales

4 You can see that the slope (i.e., a straight line tangent to the curve) depends upon what current you are on the curve. We often say the current increases by a factor-of-ten (i.e., one decade) for each 60 millivolt change in voltage., or whatever the voltage increment. For example, each increase of 10 millivolts forward bias to a diode results in an increase in current by a factor of 1.47 times. (c) oes the -millivolt rule calculated in Problem 1, part (c), also hold true in this plot? Indicate this on the plot. You found the 60-millivolt rule. The answer is yes and is so indicated on the plot. (d) How much does the current increase for a voltage increase of 100 mv? Recall Now we set V 1 V V 1V0 VT ln I millivolts, 100 mv I I 100 mv 26 mv 26 mv I 1 1 ln giving exp 46.8 times higher I0 I0 (e) What is the physical reason for the exponential I-V behavior? (Hint: Think about the derivation of the diode I -V equation covered in ES 230 Electronics I. Remember when you apply forward bias the potential barrier of the pn junction is lowered. Therefore, what is the reason this yields an exponential I -V characteristic?) The energy distribution of the majority carriers (and minority carriers also) on both sides of the pn junction have a Boltzmann (that is, exponential distribution) energy distribution. Therefore, as the potential barrier of the junction is reduced, the number of carriers having sufficient energy to pass over the junction barrier increases exponentially. That is why the diode current changes exponentially with linear changes in voltage. (f) We neglected any to include a parasitic series resistance in the diode (actually, all physically real fabricated diodes have some parasitic series resistance). How would you expect the inclusion of this series resistance to change the I -V curve on the plot above? (Note: If you want you can show this effect on the plot.) We would expect some of the applied voltage across the junction to fall across the resistive regions of the diode and therefore not contribute to the lowering the junction barrier potential. The dashed line in part (a) above shows the change in the I-V curve due to ohmic losses. Problem 3 (12 points) In this problem we are presented with an NPN BJT transistor with a saturation current I S with a value of I S = ampere and its common-emitter current gain varies within

5 a range of 50 to 200. (Note: This is not uncommon for discrete transistors because they want to sell all the devices they can across a wafer.) Suppose you bias the emitterbase junction at V BE = volt, exactly; fiind the values of i C, i B and i E for low = 50 and high = 200. (Note: Assume kt/q = V T = volt in your calculations) 15 i I exp( v / V 810 exp(0.700 / ampere i C S BE T C ampere 3.94 ma To fill out the table we use i i and i i i C B E C B Parameter = 50 = 200 i C 3.94 ma 3.94 ma i B 0.08 ma 0.02 ma i E 4.02 ma 3.96 ma Problem 4 (15 points) You are presented with an NMOS transistor with a threshold voltage of volt, a process transconductance parameter n C ox = 400 A/V 2 and a (W/L) ratio of (5 m/0.4 m). The C power supplies are plus/minus 1 volt, respectively, and drain and source resistors are included. esign problem: Using the parametric values for the NMOS device given above, determine the resistance values for R and R S which meet the criteria of a drain current I = 0.1 ma and a drain node voltage V = volt. (Please show your work.)

6 We are given that I 0.1 ma and te drain node voltage V 0.3 V 1 W 2 L 2 We start with I ncox VOV ; We want to determine VOV, 1 5 μm 0.1 ma ma/v V 0.4 μm V 0.2 volt OV GS t OV t Source node voltage V With a drain node voltage V S 2 2 OV V V V ;given V 0.5 volt yields V R R S volt 0.3 volt, 0.7 volt VS ( 1 V) ( 0.7 V) ( 1 V) 3,000 ohms I A 1 ( V ) 1 V (0.3 V) 7,000 ohms I A GS Problem 5 (15 points) For the NPN BJT in the schematic you can assume infinite current gain (i.e., or h FE ). The two current sources are ideal current sources. If V B is set to zero volts, find the values of voltages V C and V E at the collector and emitter nodes of the circuit. Clearly state any assumption you make in doing the problem. (Please show your work.)

7 We know that if is very large ( ie.., infinite), then alpha ( ) of the transistor is unity because by definition, = 1; thus I 1 To do this problem we assume V V C = volts V E = _ _ volt BE I volt [this is our assumption] VC By KCL at collector node: 4 ma IC 0, and 1000 VE by KCL at emitter node: 2 ma 2 ma 0, 1000 VC VE Thus, 4 ma 2 ma With the base node grounded (set to 0 volt), then V volt, and V C 2.7 volts. E E C Assumption used in work: We assumed that V BE = volt. The Challenge Problem on Homework #1 (forget Sedra & Smith for this problem): Problem 6 (20 points) A sinusoidal signal generator is used to (a) excite a lumped circuit consisting of capacitor C shunted by a fifty ohm resistor (upper figure), and (b) excite a distributed circuit consisting of a three foot length of fifty ohm coaxial cable terminated into a fifty ohm resistor (lower figure). The coaxial cable is RG 58A/U with a characteristic impedance of 50 ohms and 32 pf of capacitance per foot length as measured with a 1 MHz capacitance meter. So three feet of RG 58A/U will total 96 pf (= F).

8 In the lumped circuit C = 96 pf corresponding to the distributed circuit using three feet of coax cable. In other words, for both circuits the signal generator is loaded with a 96 pf of capacitor in parallel with 50 ohms. Now we want to compare the frequency response of the lumped and distributed circuits. (a) For the lumped component circuit, calculate the ratio of output voltage V OUT to input voltage V S (i.e., V OUT /V S ), in other words, calculate the voltage transfer function. What is the -3 db response frequency (that is, the frequency for which the response function falls 3 db below its low-frequency V OUT /V S magnitude)? Use C = 96 pf and both resistors are 50 ohms. In the lumped component circuit: (all 96 pf concentrated at one node) V 1 OUT 2 Hence, the -3 db frequency is VS RC 1 j given by f RC f Hz 66.3 MHz 12 RC 50 (9610 ) The -3 db frequency is 66.3 MHz. 1 R VS R I 1 jrc (b) Next, we now want to compare our result from part (a) with the distributed circuit with the coaxial transmission line. Ignoring any small resistive losses in the transmission line, we assume an ideal transmission line (of course, it does have 96 pf of total line V OUT R I jrc

9 capacitance because it is 3 feet long). If you were to measure the ratio V OUT /V S with increasing frequency, you would find that V OUT /V S is flat out to extremely high frequencies (that is, very far beyond the -3 db cutoff you calculated in part (a)). It turns out that there is a very important principle at work here. Why do you think the distributed case has a much higher cutoff frequency than the lumped case? The distributed circuit using the section of transmission line does not have a natural rolloff like a lumped RC network. The reason for this is the 96 pf inherent in the coaxial transmission line is distributed in time as it is charged because the voltage wave travel in time down the line. For this reason capacitance C is charged incrementally as the wave travels down the line. For a lumped capacitor it must be charged at a single node and must be charged completely during the time charge flows within the capacitor. Consider the model for the transmission line that partitions the line into small sections of series inductance and shunt capacitor. In the limit, for a very large number of sections, the capacitance is partitioned into an infinite number of infinitesimally small capacitance segments. This is schematically drawn as (we will ignore the losses from the series resistance and shunt conductance. The important point is this: As the signal wave travels down the transmission line only a small piece of the total line capacitance need be charged at each instant. Each infinitesimal section of the capacitance is charged sequentially in time so that the charging of the total line capacitance is distributed over time (i.e., the time it takes the signal to travel along the entire transmission line). Principle: Spreading out in time (temporal spreading) the charging of the capacitance allows extremely wide bandwidths to be achieved. Of course, this is impossible when the capacitance to be charged is concentrated at a single node in the circuit. That is at the heart of what we mean by lumped circuits.