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I purchased SN74LS26N quadruple 2-input NAND Gates chips for my circuit. Before I insert any chip into circuit I test it on separate breadboard.

So I did with this chip and I get NO output when my inputs are LOW...

Does it make any sense to you? I am doing something wrong?

schematic

simulate this circuit – Schematic created using CircuitLab

I am not the best with schematics so I hope this is clean for you.. I tried adding bypass electrolytic caps 0.1uF but that didn't help

Thanks!

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    \$\begingroup\$ You have a dead-short between the +5 V and GND. It's the second vertical line from the right. \$\endgroup\$
    – Transistor
    Commented Apr 22, 2020 at 14:57
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    \$\begingroup\$ How do you know that you have "NO output"? Do you mean that the output never goes high? Are you relying on your LEDs or have you measured the voltage? Is this a real circuit or a simulation only? \$\endgroup\$ Commented Apr 22, 2020 at 14:58
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    \$\begingroup\$ LS chips do not deliver much current when driving. Maybe the LED is slightly glowing and you need some spectacles? \$\endgroup\$
    – Andy aka
    Commented Apr 22, 2020 at 14:59
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    \$\begingroup\$ @awjlogan - 0are connected to GND @Transistor - this is normally connected to GND on breadboard... as I said I'm not the best with schematics so it may looks wrong there... sorry @ElliotAlderson - I measured the voltage... 0.1V-0.3V @Andyaka - No, they are not "glowing" slightly. on the same set up SN74LS00N chip is working perfect. Pinout is the same so I am not sure what's wrong... [Data sheet ](ti.com/lit/ds/symlink/sn74ls26.pdf) says it's High voltage Positive NAND-Gate. Does it make a difference? \$\endgroup\$
    – aberforth
    Commented Apr 22, 2020 at 15:08
  • \$\begingroup\$ Yes it certainly makes a difference. Read up on Open Collector Outputs, which your 74LS26 has. If you have 74LS00 which are working fine, why didn't you buy those instead? \$\endgroup\$
    – StarCat
    Commented Apr 22, 2020 at 18:01

2 Answers 2

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The LS26 contains open-collector NAND gates. You must add pullup resistors to the outputs in order to see a high logic output. These gates are designed to interface to circuits operating at higher voltages, so there is no internal circuitry that pulls the output to 5V.

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  • \$\begingroup\$ so it has no use to me for 5V circuit if I wanted to integrate LS26 with other 5V Logic Chips? \$\endgroup\$
    – aberforth
    Commented Apr 22, 2020 at 15:22
  • \$\begingroup\$ Yes, it can be used at 5V but you need to add pullup resistors to the NAND gate outputs. \$\endgroup\$ Commented Apr 22, 2020 at 15:29
  • \$\begingroup\$ by pullup you mean to connect resistor from 5v to output of NAND gate? any idea what value should I use? \$\endgroup\$
    – aberforth
    Commented Apr 22, 2020 at 15:31
  • \$\begingroup\$ It looks like the chip can sink 4mA and still meet the TTL VOL specification, so the resistor should be no less than 1.25 kilohm. Larger values reduce power consumption but may cause slower rise time. \$\endgroup\$ Commented Apr 22, 2020 at 15:34
  • \$\begingroup\$ With 10k Ohm resistor output measure 4.59V. but output is too high: 0.8-0.95V... It tried different resitors but had no luck dropping LOW logic voltage down... \$\endgroup\$
    – aberforth
    Commented Apr 22, 2020 at 19:17
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The chip has open collector type outputs. It can only sink current. It cannot output current so the LEDs don't light up. If you want to use this chip and blink LEDs with it, you must change the circuit so that the LEDs are from VCC via ra resistor to the chip output, so the LEDs are on when chip output is low.

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