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i need some advice how to prepare digital electronic circuit.

I have 64 signal lines, every line have a NO tact switch with latch . I need desiplay how much "1 or ON" signal i have and display this value on two 7segment displays.

difficulty: i can't use microcontroller/microprocessor/fpga or equivalent device. only mux/demux/counter/decoder and other "simple digital stuff"

every tip will be helpful thanks!


EDIT:

My idea is: connect 65 resistors in series 64x100Ohm + 1x6.4kOhm(Like a shunt resitor). Every single resistor from 64 will be connected parrarel with button if i click the button then i will have a bigger voltage drop on 6.4kOhm resistor. The voltage drop on 6.4k resistor connect to flash adc and then move out signal from priority encoder to 2x7seg displays.

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    \$\begingroup\$ " i can't use microcontroller/microprocessor/fpga" Sounds like a school project. What have you tried thusfar? \$\endgroup\$ – Oldfart Mar 23 '20 at 16:30
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    \$\begingroup\$ I am not sure that your flash ADC counts as 'simple digital stuff' as it is 1/ Not completely digital, 2/ Not exactly simple. \$\endgroup\$ – Oldfart Mar 23 '20 at 17:18
  • \$\begingroup\$ What is a "tact switch with latch"? Why do you need a priority encoder...aren't you interested in how many switches are pressed? How do you intend to convert a 6-bit binary value to a 2-digit BCD value? \$\endgroup\$ – Elliot Alderson Mar 23 '20 at 17:56
  • \$\begingroup\$ :Transistor thanks for your attention. :Oldfart, Probably i used wrong words to specify my problem "simply digital stuff' i know its half analog half digital but i can use this. I am afraid about noise in this solution ~20mV when i use 5V cmos or ~60mV 15V cmos. I also thought about multiplexed all signals 64 to 1 with 9 (8 to 1) mux's, latch this signal from every line and count high signal. but i don't have sufficient skills to do second solution. Elliot Alderson: tact switch with latch or easier bistable switch \$\endgroup\$ – testiro Mar 23 '20 at 17:56
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    \$\begingroup\$ Also, you should sketch up your idea in a pseudo-schematic to make it easier for us to follow. You don't need to show all 64 switches and both 7-segment displays. Just a handful of switches (say 4) and 1 display would suffice. We're (well, a lot of us) aren't going to do your homework, but if you make a stab at things we'll help guide you and point out errosr or shortcoming with your approach. \$\endgroup\$ – SteveSh Mar 23 '20 at 18:16
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If you want to take many inputs that's not the proper way. As you have 64 input lines so you need a 64-bit ring counter and a clock source. As a 64-bit ring counter isn't available, you need to couple some 8-bit shift registers. Each output would be connected to a switch through a diode. And the other terminals of all switches need to be short-circuited and pulled down through one 10K resistor. The short-circuited bus is the output bus and the status of this bus is needed to be checked at the falling edge of each clock pulse. This is done through the AND product of the output bus and inverted clock signal. As you need to count the 1, so just connect the output of AND gate to a counter. The simulation result with diagrams is here.

In addition, the counter has to be refreshed after each 64 clock pulse elapsed. I can't help you with how to display the counter value in 7 segment display right now. But you will find enough resources about that on the internet.

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    \$\begingroup\$ I'll try to add the display tomorrow, as here it's 3.00 AM. \$\endgroup\$ – Sadat Rafi Mar 23 '20 at 21:12
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    \$\begingroup\$ And you can use a tristate buffer instead of 6 diodes. \$\endgroup\$ – Sadat Rafi Mar 23 '20 at 21:14
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@Sadat Rafi your concept sounds good :)

enter image description here

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    \$\begingroup\$ Is this supposed to be an answer or is it additional information which should be in your question? (You can only post answers in the "Your Answer" box. Posts move up and down with user votes and sorting preferences so any attempt at a conversation in the answers quickly turns into a mess. \$\endgroup\$ – Transistor Mar 29 '20 at 12:20
  • \$\begingroup\$ If you have 10 switches (or less), then there is Ring counter. \$\endgroup\$ – Sadat Rafi May 21 '20 at 15:32

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