0
\$\begingroup\$

This circuit is based around the SN74HC283 4 bit full adder.

  • What purpose does the resistor network serve?
  • Will this circuit work without any resistors, i.e. using raw inputs and outputs?

Thanks.

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Having floating inputs MAY be OK for TTL inputs, but all these ICs are CMOS and must not float. Fresh out of college, I assisted a senior engineer debug about a 1,000-SSI data multiplexor for 10 or 15 sensors. I learned how CMOS worked ----- often erroneously ---- with no grounds or power to the ICs. Instead, the ESD diodes sufficed. Instead of 0/+12volts, the ICs often had +0.6 and +11.4 volts. Tooke a while to discover these; solution was to have the wire-wrap person use a magnifying glass on EACH of 1,000 ICs and ensure pins 7/14 or pins 8/16 were properly tied. A useful lesson for me. \$\endgroup\$ Mar 18, 2019 at 6:31

2 Answers 2

1
\$\begingroup\$

When the switches are closed, it is easy to see a direct connection from 5V to the input pins, which is read as a digital 1. If the switches were open, and those resistors weren't there, those nodes are left floating. This floating voltage could be read as a 1, it could be read as a 0. The adder would behave erratically. It would definitely read high signals correctly, but it wouldn't reliably read low signals. The resistors keep the voltage on the input pins from floating by pulling the signal to ground when the switches aren't closed.

If those switches were replaced with "Single Pole Double Throw" switches, it could work without those resistors. An example of how the single pole double throw switch would need hooked up is shown in the schematic below.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • \$\begingroup\$ right, that makes sense. Thanks for the swift reply. Sorry to be a pain, but the current limiting resistors (100ohm) are purely for the led's sake, right? The circuit will function without those? \$\endgroup\$
    – Steve
    Mar 18, 2019 at 10:05
  • \$\begingroup\$ @Steve: If you don't use the 100 Ohm resistors in series with the LEDs, you may destroy the LEDs. These resistors control the current drawn by the LEDs. \$\endgroup\$ Mar 18, 2019 at 15:51
0
\$\begingroup\$

The 10K pulldown resistors ensure that the inputs to the 'HC283 will be Low when the switches are open.

An unconnected CMOS input may float randomly between Low and High, as the input is very high impedance.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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