1
\$\begingroup\$

enter image description hereI am hoping I am doing something dumb here. I know there are other posts about this but everything I have found leads me to believe my circuit should be working. Yes this is for homework. No, I am not looking for "here's the file so you can cheat. I feel either I my BCD converter is not working properly or I have my switches set up incorrectly.

  1. Construct a 4 bit 2’s complement adder circuit using discrete components (and, or, xor etc)

  2. Display the result on 2 seven segment displays, one for sign and one for number

  3. If the number if positive, use BCD to seven segment to display output

  4. If the number is negative, invert every signal, use another 4 bit adder and add a one, then use a BCD to seven segment to display output

  5. Describe operation of the circuit. Capture output of seven segment for max, min and in between values.

I have attached my schematic and truth table. I will also post my circuit in the comments. Any nudge would be greatly appreciated. enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ You have your design in a simulator...so use it! Look at internal signals in your design and figure out what isn't working. Start by looking at the connections to the switches and them work your way toward the displays. \$\endgroup\$ – Elliot Alderson Jul 8 '19 at 23:38
1
\$\begingroup\$

I have not analysed the circuit in any great detail, but unless I'm reading your circuit diagram wrong, all of your switches S1 - S8 are wired the wrong way round. In the position shown they all connect the 10V Vcc directly to ground, whilst the resistors R1 - R8 are open circuit leaving the inputs of the gates to which they are connected floating in an indeterminate state. Surely each of the the 1K resistors should be connected to the moving contact of their respective switches and Vcc and ground connected to the two fixed contacts.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ an addition to this answer: use pullup or pulldown resistors .... why? ... think about the state of the input when you transfer a switch \$\endgroup\$ – jsotola Jul 9 '19 at 0:10
  • \$\begingroup\$ @jsotola Agreed, I just thought "one thing at a time, let's at least give the circuit a chance of working!" \$\endgroup\$ – Peter Jennings Jul 9 '19 at 8:30
  • \$\begingroup\$ So, I have turned the switches around. That did not work. I have yet to add pull up or down resistors. I will try that later. I am starting to think the simulated adder does not work. I am going to do away with the adder and design the circuit with discrete only. I am curious if that is the issue because I am not getting a signal out of the adder. Thank every one for your help. I will update once I try. \$\endgroup\$ – Vincent Ledbetter Jul 9 '19 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.