0
\$\begingroup\$

I'm trying to get this circuit using a SN74HC191 chip to work but it's producing no output to the LEDs.

My current setup so far is:

enter image description here

What makes this more difficult is that the datasheet for the SN74HC191 has no truth/function table so I'm not sure which switches needs to be high/low. Please could anyone provide some help?

Thank you!

Edit: This is the datasheet.

I have now actually found another datasheet with a truth table. However, even with the same H/L settings for "Count up", it still does not produce any LED output.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ There are not pull-down resistors at the switches and no series resistors at the LEDs. Can the IC supply enough current for the LEDs? \$\endgroup\$
    – Curd
    Commented Jan 14, 2017 at 15:39
  • 1
    \$\begingroup\$ Have you applied +5v to Vcc pin? And connected GND pin to the supply as well? \$\endgroup\$
    – glen_geek
    Commented Jan 14, 2017 at 15:45

2 Answers 2

0
\$\begingroup\$

Your LEDs are backwards, and require ~330Ohm resistors for protection.

Also, your enable signal (~CTEN) is floating. Tie this to ground to enable the outputs. Tie all other logic inputs (D0-D3, D/~U, ~PL) that you're controlling with the DIP switch to ground with ~4.7k resistors so that they aren't floating, too.

\$\endgroup\$
0
\$\begingroup\$

enter image description here

This shows all the info you need for a truth table.

74HC' Ron driver resistance is approx 50 ohms at 5V to estimate your LED current, add suitable R for current limit. eg if Blue = 3V , then 2Vdrop/10mA=200 Ohm, so use 150 ohms. then verify voltage drop on R=150 = 1.5V. thus Voh =4.5V. (for example).

Include cap across Vss to Vdd for 5V and twisted pair for Clk //Gnd if using long wires.

  • Observe LED cathode towards gnd. (-ve is always the reflector cup side and may have marking (flat edge on body)

enter image description here

  • Never leave CMOS input floating. Pullup 1M to 10k.
  • Beware of ESD damage.
\$\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.