How do you make a 0-7 up/down counter using d-type flipflops and combinational logic gates?
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Three flops - A, B and C - A is the least significant bit.\nNew values for flops (inputs) are A', B' and C'.\nUp/down input is U - set to 1 for up\n\nA' = not A\nB' = B xor ((U and A) or (not U and not A))\nC' = C xor ((U and A and B) or (not U and not A and not B))
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The truth table is easy:\n\nABC . . . OUT\n000 . . . . 0\n001 . . . . 0\n010 . . . . 0\n011 . . . . 0\n100 . . . . 1\n101 . . . . 0\n110 . . . . 0\n111 . . . . 0\n\nConstructing the circuitry from logic gates would be a bit more work.
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Look at each segment in the display and determine for which numbers it should be on (logic 1), and which it should be off (logic 0). \n\nMake a truth table for all 4 input lines for each segment.\n\nCreate an expression for each segment using you favourite technique, such as Karnaugh maps.\n\nChange the expression to only use nand gates.\n\nDo happy dance, cuz you're done.
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You can start off by declaring a new integer variable "low", initialize it to 0, and then after any input compare the new value to the value in low and if the new value is lower then set low equal to the new value.\nThe following sites should also be of use.
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"if all inputs are low then the output is low" is true for AND, OR, and XOR.\n\n"if all inputs are low then the output is high" is true for NAND, NOR, and XNOR.\n\nBy referring to the "remaining 6 LOGIC GATES", you must be including NOT. It allows only one input, but one could still say that "if all inputs are low then the output is high".
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After several attempts, it seems that it is rather impossible.  If you start in the bottom corner, you can get all but 3 squares rather easily.  If you're careful, you can have 1 remaining if you start diagonal to the corner.\n\nWhat are you trying to do this for?  Is it a website?  I'd love to try it again.
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Conductors and insulators are only building materials that provide the means for directing data from one point to another. The functionality of manipulating data comes from the special properties of semiconductors, which can be made to amplify a signal (i.e., a variation in voltage). \nWhen connected together, two very simple amplifiers (possibly as simple as one transistor each) can make a bistable circuit, one which always has one transistor conducting and the other cut off. These bistable circuits, called flipflops, can store one binary digit (bit) of information: one state, with transistor A on and B off, is considered "0" and the opposite state is "1". So a single flipflop can take the role of one bit of memory. A flipflop can be triggered to go from one state to another. When connected so one flipflop's change of state in one direction (say from 1 to 0) triggers a change (from 0 to 1 or from 1 to 0) in the next flipflop, you have the beginning of a binary counter. Similarly, flipflops in other connections can form arithmetic registers, in which two binary numbers can be compared, added, and otherise manipulated. Building on these simple forms, complex logical and arithmetic functions can be implemented.\nThe above only describes digital circuits. Semiconductor amplifiers are also used in analog and hybrid (part digital, part analog) microcircuits.
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If you can draw it like a circuit it should be a little easier to see.\nClick on this site and go through the 'prev page' and 'next pages' for samples of Boolean logic. \nhttp://computer.howstuffworks.com/boolean2.htm\nHope this helps.\nTry using the output as an input to keep track of state
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This is 8 bit binary system.\nand u know, there are only two possibilities for a switch (on/off or 0/1).\nin binary syatem, as i have to take 10101000, to convert into decimal.\nfrom right to left:\n0*2^0 + 0*2^1 + 0*2^2 + 1*2^3 + 0*2^4 + 1*2^5 + 0*2^6 + 1*2^7 = 168 (it is the decimal form).\n\nsimilarly as we take 10110001\n\nthen its decimal is :\nfrom right to left : and increase pawer of 2 from 0 step 1.\n1*2^0 + 0*2^1 + 0*2^2 + 0*2^3 + 1*2^4 + 1*2^5 + 0*2^6 + 1*2^7 =  177
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The setup time for a digital flip-flop is the amount of time prior to a change in clock signal that the input data signal must be stable (set to a specific voltage).  The hold time is the amount of time a data signal must remain stable following a change in clock signal.  These two times together determine the operational frequency of flip-flops or in other words how fast they can operate.
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http://computer.howstuffworks.com/boolean.htm check this site out.
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keep a table of binary 4 bit equivalents 0 1 2 ... F of hexadecimal degits.\nconvert each 4 bits of your binary no. to equivalent hex. no looking the table.\n\nyou got the answer
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A Rube Goldberg machine
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X = 24/Y  Just start plugging in Y and getting X, there are lots of possibilities, unless you restrict it to just whole numbers.
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1>INPUT 1\n2>STORE 1 in A\n3>STORE 1 in B\n4>ADD A to B\n5>STORE A\n6>DISPLAY A\n\nThat's total of 6 steps. If you have 2 different numbers than you will have one more step of INPUT 'X' so it will be total 7 steps in that case.
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you just add the inputs, and the result (the least significant digit) is the answer you need:\n\n0+0+0=0\n0+1+0=1+0+0=0+0+1=1\n1+1+0=0+1+1=1+0+1=10 ->0\n1+1+1=11 ->1\n\nIf you like to solve it by boolean 1+digit adding, then,\nlet you entries be a,b,c\nr1=a.not(b)+b.not(a)\nr=r1.not(c)+c.not(r1)\nwhere . means and
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"mode"\n"display digits"\n"float 12" (or however many you want - 12 is the highest)