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I'm building a simple 4-bit computer (a Nibbler) from 74 series ICs on a breadboard. Right now I'm working on 74LS173AN IC registers. I can't make it work. I have connected the input enable (pin 9,10), output control (1,2), and clear (15) to ground and some data inputs to vcc. The output pins (Q1-Q4) are connected to ground via LEDs. I'm simulating the clock signal by swithing a tact-switch on/off (with ground connected by 10k resistor and direct connection to VCC.

I tried different connections and searched web (many hopes with https://www.youtube.com/watch?v=9WE3Obdjtv0&t=14s, but it didn't help). Nothing works. I suspect there might be a problem with clock signal, but I would really appreciate the opinion from an experienced electronic specialist. My breadborard connections

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  • \$\begingroup\$ what value is your pull down resistor on the switch? (it looks like 10k) The clock a ttl input so it sources current. A resistor will prevent the clock being pulled low. \$\endgroup\$ – JIm Dearden Mar 3 '17 at 18:31
  • \$\begingroup\$ What happened to the series resistors to limit current through the LEDS? \$\endgroup\$ – JIm Dearden Mar 3 '17 at 18:37
  • \$\begingroup\$ I can not see by your image. Those leds... got Vcc or GND? and you need a resistor in there for those. \$\endgroup\$ – Trevor_G Mar 3 '17 at 18:48
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A number of tips about dealing with (LS)TTL:

1) Inputs should be either grounded or pulled high with a 1k pullup resistor.

2) Output LEDs should be connected to +5 with a 1k pullup resistor. And yes, this will produce a signal inversion in the sense that a LOW output will turn on the LED.

schematic

simulate this circuit – Schematic created using CircuitLab

3) You must decouple the power pin. Connect a 0.1 uF ceramic cap from pin 16 to pin 8, and don't use jumpers - plug the leads directly into the breadboard at the IC.

Make sure you test the LED polarity by unplugging the IC, applying power, and grounding the output contact. If the LED doesn't light up, it's bad or you've got it backwards. When you've checked out the LEDs, turn off power and plug in the IC.

Otherwise your circuit looks OK with one exception. You've missed the part in the Youtube video where it is pointed out that TTL inputs float high if no connection is made, so your circuit really ought to be turning on ALL the LEDs, not just the 2 that are connected to +5. With proper LED connections, leaving the inputs floating or pulled high should result in all LEDs being off.

Finally, using a switch to generate a clock to a flip-flop is fine for what you're doing now, but anything more complicated will require debouncing of the switch.

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  • \$\begingroup\$ Thanks. Reversal of LEDs has solved the problem. The other remarks should help in the future. I'm really grateful. \$\endgroup\$ – Karol Pogorzelski Mar 3 '17 at 21:35
  • \$\begingroup\$ Open TTL inputs do not float so they do not need pull-ups or to be grounded. Open TTL will act the same as a pulled high input. The value of the resistor should be selected based on how much current the output driver can sink and the desired LED luminous flux. \$\endgroup\$ – Misunderstood Mar 3 '17 at 22:33
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    \$\begingroup\$ While that may be true in a theoretical sense, practically speaking it is in an undefined logic state because an unconnected inputs tends to acts like an antenna for noise. According to Fairchild's FAST Application Note (1987), only a few hundred millivolts of noise could cause an unconnected input to go to the LOW state. Hence vendors of TTL, S and LS parts have always recommended connecting unused inputs to GND direcly, or to VCC either directly or through a resistor. I've never come across one that says it's OK to leave an input open. \$\endgroup\$ – SteveSh Dec 29 '19 at 18:07
  • \$\begingroup\$ @Misunderstood - SteveSh is correct, Another issue is "improvement". If you're going to work with TTL, eventually it will seem like a good idea to step up to CMOS, and at this point letting the inputs float will cause no end of heartache. I speak from experience. If it comes to that, even letting an unused CMOS gate float can cause the other gates on the chip to show bad behavior. Trust me on this. \$\endgroup\$ – WhatRoughBeast Jan 2 at 19:32
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I may be wrong here, but when you push the tact-switch, the metal contacts might bounce off each other more than once triggering more than a single clock pulse which will lead to an unexpected output.

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"The output pins (Q1-Q4) are connected to ground via LEDs."

There is your problem. By the spec sheet that device does not drive high.. IOH is -1.0 mA..( weird spec that is...)

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  • \$\begingroup\$ That's for the mil version, 54LS. 2.6 mA is the actual number, and that should be more than adequate for visible operation. \$\endgroup\$ – WhatRoughBeast Mar 3 '17 at 19:14

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