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Small-Signal Techniques Small-signal analysis can be applied to the diode circuit of Fig 2-10 if the amplitude of the ac signal vTh is small enough so that the curvature of the diode characteristic over the range of operation (from b to a) may be neglected Then the diode voltage and current may each be written as the sum of a dc signal and an undistorted ac signal Furthermore, the ratio of the diode ac voltage vd to the diode ac current id will be constant and equal to vd 2Vdm vD ja vD jb vD dvD rd 2:5 id 2Idm iD ja iD jb iD Q diD Q where rd is known as the dynamic resistance of the diode It follows (from a linear circuit argument) that the ac signal components may be determined by analysis of the small-signal circuit of Fig.

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a[0] 44 a[1] 88 55 a[2] 55 88 a[3] 99 66 a[4] 66 99 33 a[5] 33 a[6] 22 a[7] 88 a[8] 77

2-12; if the frequency of the ac signal is large, a capacitor can be placed in parallel with rd to model the depletion or di usion capacitance as discussed in Section 23 The dc or quiescent signal components must generally be determined by graphical methods since, overall, the diode characteristic is nonlinear..

99 22

CHAP. 2]

Example 2.8. For the circuit of Fig. 2-10, determine iD . The Q-point current IDQ has been determined as 36 mA (see Example 2.7). The dynamic resistance of the diode at the Q point can be evaluated graphically: rd vD 0:37 0:33 2:5 iD 0:044 0:028

99 88

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Now the small-signal circuit of Fig. 2.12 can be analyzed to nd id : id vTh 0:1 sin !t 0:008 sin !t RTh rd 10 2:5 iD IDQ id 36 8 sin !t A

99 77

The total diode current is obtained by superposition and checks well with that found in Example 2.7: mA

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2 merge_sort calls merge_sort again, passing a list of the first two numbers in NUMS This will sort the front half of the list This is level 1 of recursion 3 Now merge_sort calls merge_sort again, passing only the first number in NUMS This is level 2 4 Now merge_sort simply returns; it s down to one element in the list, merge_sort returns to level 1 5 Now merge_sort calls merge_sort again, passing only the second of the first two numbers in NUMS This is level 2 6 Again, merge_sort simply returns; it s down to one element in the list, merge_sort returns to level 1 7 At level 1 of recursion, merge_sort now has result_A and result_B merge_sort calls merge to put those two numbers in order, and then it returns the sorted pair of numbers back to level 0 The first half of the list is sorted.

CHAP. 2]

88 33

Fig. 2-11

88 22

8 From level 0, merge_sort calls merge_sort again, passing a list of the last two numbers in NUMS This will sort the back half of NUMS It s back to level 1 of recursion 9 merge_sort calls merge_sort again, passing only the first of the last two numbers of NUMS This is level 2 of recursion again 10 Since the list contains only one number, merge_sort simply returns back to level 1 11 merge_sort calls merge_sort again, passing only the last of the numbers of NUMS This is level 2 of recursion again 12 Since the list contains only one number, merge_sort simply returns back to level 1 13 At level 1 of recursion, merge_sort now has result_A and result_B merge_sort calls merge to put the two lists in order, and then it returns the sorted set of two numbers back to level 0.

Fig. 2-12

88 77 77 88 88

Example 2.9. For the circuit of Fig. 2-10, determine iD if ! 108 rad/s and the di usion capacitance is known to be 5000 pF. From Example 2.8, rd 2:5 . The di usion capacitance Cd acts in parallel with rd to give the following equivalent impedance for the diode, as seen by the ac signal: 1 rd 2:5 Zd rd k jxd rd k j 1 j!Cd rd 1 j 108 5000 10 12 2:5 !Cd 1:56j 51:348 0:974 j1:218 In the frequency domain, the small-signal circuit (Fig. 2-12) yields I"d " 0:1j 908 0:1j 908 VTh 0:0091j 83:678 A RTh Zd 10 0:974 j1:218 11:041j 6:338 iD IDQ id 36 9:1 cos 108 t 83:678

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