I have a task to convert analogue signal to digital on multisim. But multisim have no Ic to convert 8 bit binary to BCD. Now I have to make logic circuits to convert 8 bit binary to bcd. I have used kmap technique to minimize first 4 bits than second four bits enter image description here

then I have designed logic circuits through minimized expressions. enter image description here I have made two sets of 4,4 bit converters on ADC output pins like this.

enter image description here I have made this circuit on multisim but this is not converting a complete 8 bit binary to 8 bit bcd and not showing correct value of analog to digital

how I could convert a complete 8 bit binary to 8 bit bcd through single combinational logic circuit ?

  • \$\begingroup\$ What is the range? 00 to FF = 000 to 255 : You need 3 BCD digits \$\endgroup\$
    – Grabul
    May 20, 2020 at 11:24
  • \$\begingroup\$ You need to create a truth table that has 8 inputs and 8 outputs. Then, use 8-variable Karnaugh maps to simplify. Or, just write the thing in an HDL and synthesize. \$\endgroup\$ May 20, 2020 at 11:40
  • \$\begingroup\$ What you are discovering hints at the reality that this is not a simple problem. Typically today it would be done in MCU software, as custom hardware implementations just aren't cost effective. I believe there exist panel meter ICs, but many cheap panel meters you buy now have an MCU in them... \$\endgroup\$ May 20, 2020 at 14:56
  • \$\begingroup\$ #Elliot Alderson 8-variable k map is too complex for human beings to make SOP \$\endgroup\$
    – Alex90
    May 20, 2020 at 16:07

2 Answers 2


First of all, you can't convert 8 bits binary to EXACTLY one BCD, you'll need 3 BCD converters: the maximum value with 8 bits is 255, supposing you have unsigned numbers, so 3 digits are needed.

You have two choices:

  1. Use asynchronous solution

    There's an algorithm "Double dabble" that does the conversion from binary to BCD. You can start from here: Double dabble

  2. Use synchronous solution

    Use an 8×12 bits RAM. The address is the 8 bit value, the content of the cell, and hence the output, will be your BCD value.

  • \$\begingroup\$ You mean a ROM not a RAM. Of course in a DIY implementation that's likely to be an EPROM or some comparable EEPROM device. In theory an old PROM could work, but finding one... \$\endgroup\$ May 20, 2020 at 14:58
  • \$\begingroup\$ I said RAM because he may want to change the content later, for another use of that IC. It will be better to use a ROM in the simulation program as you said \$\endgroup\$ May 20, 2020 at 15:49
  • 1
    \$\begingroup\$ The problem with a RAM is that you have to initialize the contents each time the system powers on. Most things that could do easily do that could simply solve the BCD problem themselves. \$\endgroup\$ May 20, 2020 at 16:01
  • \$\begingroup\$ Yes I forgot about that, thank you, but I suppose that an EEPROM would still be better \$\endgroup\$ May 20, 2020 at 16:27

One logic-based method is to detect inputs greater than 9, then add 6 and carry out the excess to the next digit. So, a full adder and some gates for each digit.


  • 10d = 1010b = 1 0000 bcd
  • so, 1010b + 0110b = 1 (carry out) + 0000b

The logic to detect greater than 9 is a couple of product terms. Use that to select adding 0 or 6 at the full adder, plus carry-in.


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