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After executing <Ex2_4.CIR>, I(D) is plotted with the x-axis variable changed from time to vD V 1 V 2 V 1; 2 giving the static diode characteristic of Fig. 2-6(b). (b) Edit <Ex2_4.CIR> to move the asterisk preceding the second .MODEL statement to the rst .MODEL statement, thereby preparing for the ideal diode analysis. Setting the emission coe cient parameter (n) to a small value ensures a negligibly small forward voltage drop. Execute <Ex2_5.CIR> and plot the result as in part (a) to give the static characteristic of Fig. 2-6(c). Inspection of the marked points on the curve shows that the diode is approaching the ideal case of negligible reverse current and negligible forward voltage drop.
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A graphical solution necessarily assumes that the diode is resistive and therefore instantaneously characterized by its static iD -versus-vD curve. The balance of the network under study must be linear so that a Thevenin equivalent exists for it (Fig. 2-7). Then the two simultaneous equations to be solved graphically for iD and vD are the diode characteristic iD f1 vD and the load line iD f2 vD 1 v v Th RTh D RTh 2:4 2:3
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Load line
0 0.75 1 2 3
LTh L D, V
Fig. 2-7
Fig. 2-8
Example 2.5. In the circuit of Fig. 2-3(a), vs 6 V and R1 RS RL 500 . Determine iD and vD graphically, using the diode characteristic in Fig. 2-8. The circuit may be reduced to that of Fig. 2-7, with vTh R1 500 v 6 3V R1 RS S 500 500 500 500 500 750  500 500 The
RTh R1 kRS RL
Then, with these values the load line (2.4) must be superimposed on the diode characteristic, as in Fig. 2-8. desired solution, iD 3 mA and vD 0:75 V, is given by the point of intersection of the two plots.
Example 2.6. If all sources in the original linear portion of a network vary with time, then vTh is also a time varying source. In reduced form [Fig. 2-9(a)], one such network has a Thevenin voltage that is a triangular wave with a 2-V peak. Find iD and vD for this network.
SEMICONDUCTOR DIODES
[CHAP. 2
RTh = 50 W +
_ (a) iD, mA iD, mA
Diode characteristic
10 t5 _2 t4 t3 t2 t1 _1
Dynamic load line for LTh = 2 V
1 2 3
LD, V
_1 t1 t2 t3
LTh, V
t (b)
Fig. 2-9
In this case there is no unique value of iD that satis es the simultaneous equations (2.3) and (2.4); rather, there exists a value of iD corresponding to each value that vTh takes on. An acceptable solution for iD may be found by considering a nite number of values of vTh . Since vTh is repetitive, iD will be repetitive (with the same period), so only one cycle need be considered. As in Fig. 2-9(b), we begin by laying out a scaled plot of vTh versus time, with the vTh axis parallel to the vD axis of the diode characteristic. We then select a point on the vTh plot, such as vTh 0:5 V at t t1 . Considering time to be stopped at t t1 , we construct a load line for this value on the diode characteristic plot; it intersects the vD axis at vTh 0:5 V, and the iD axis at vTh =RTh 0:5=50 10 mA. We determine the value of iD at which this load line intersects the characteristic, and plot the point (t1 ; iD ) on a time-versus-iD coordinate system constructed to the left of the diode characteristic curve. We then let time progress to some new value, t t2 , and repeat the entire process. And we continue until one cycle of vTh is completed. Since the load line is continually changing, it is referred to as a dynamic load line. The solution, a plot of iD , di ers drastically in form from the plot of vTh because of the nonlinearity of the diode.
CHAP. 2]
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