By Frederick J. Hill, Gerald R. Peterson
Tied to no specific set of computer-aided common sense layout instruments, it advocates the recent emphasis in VLSI layout. contains aid of structure synthesis from description in a check in move point language in addition to from layout trap. incorporates a distinct advent to Boolean algebra, Karnaugh maps and sequential circuits. during this variation dialogue of mix common sense has been prolonged; switching circuits up-to-date; a finished therapy of try out new release for VLSI integrated.
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Extra info for Computer Aided Logical Design with Emphasis on VLSI
Then its resistance must be 200/5 X 10~3 = 40 000 Ù. With the voltmeter across the anode resistor and the same supply voltage of 350 the current would be 7-78 mA and this divides between the anode resistor and the meter in inverse ratio to their resistance. Thus in this particular case the current through the meter would be 7-78/2 = 3-89 mA. Now 15 mA corresponds to 450 V on the meter scale so that 3-89 mA will correspond to 116-7 V, which will be the reading instead of the 150 that ought to be obtained.
The current passing the resistance-voltmeter combinations is 250 mA, the voltmeter resistance is 8-67 Ù and it reads 0-084 V. What is the 'true value' of resistance and to what order of accuracy is it measured? Answer. Let R = unknown resistance a n d / ^ = voltmeter resistance. Then resistance of combination is RyR Rv+R. C. RESISTANCE The voltage across the combination is given as 0-084 and the current passing the whole is 250 mA. Hence RVR _ 0-084 X 1000 250 Rv+R But Rv = 8-67 Ù. o 67 R Thus , , ' n £4.
This difficulty is overcome in the Sullivan bridge by arranging the S resistance of the unit on the decade principle, so that the first set of plugs give 0 to 9 Ù, the second 10 to 90, and so on. The arrangement is of the form shown in Fig. 5. È È LLä L_ QSSZ Z 100 ( 10 in in in m io too 1000 io ooo 100 l>g i m n? J Z Z3 m Ð5—ÀÃë ir\ HE ÓÆÆ i3 m ð; ES JZZZ2 5H 13 □ l SÜD HS ? 3 □ ESC ttel 3 m E5ZZZ2 5H ECZ3 a E5Z È C3Ü3C2Ü3 10 000 1000 _J! 1000 fed 100 an ne BE 1000 È È 3Ð 3á 10 000 Ô are terminals Ñ and Q are the ratio arms Fig.