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I am trying to build a half adder using a 74LS86 XOR gate. Initially I connected both input pins that I am using (9 & 10) on the 74LS86 to the ground rail using 10K ohm resistors. I then connected 5V to pin 9 and expected the LED fed from pin 8 to light. No luck. I then pulled pins 9 and 10 high by connecting the 10K ohm resistors to 5V. When I pulled pin 10 low by connecting it to ground, the LED lit. I substituted 330 ohm resistors for the 10K resistors and reconnected them to ground. Now the circuit works properly when I pull either pin 9 or 10 high. Is 330 ohms an appropriate choice? Are pins 9 and 10 sourcing current when they are not active (pulled high)? I read that low power schottky ICs can only source 8mA. I am calculating that with 5V and 330 ohms that the current draw could be 15mA. I am wondering where I should look on the datasheet to properly size pulldown resistor for TTL ICs.

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  • \$\begingroup\$ Welcome to EE.SE! Please add a schematic to your question, so it can be easier to understand and answer it. \$\endgroup\$ – anrieff Mar 19 '18 at 8:01
  • \$\begingroup\$ If you must use TTL there are rules about noise margin and fan-out capacity of 10 inputs which 15 mA exceeds, but works ok for an LED \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 19 '18 at 8:03
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Inputs on bipolar TTL (plain 74xx, 74LSxx, 74ALSxx, 74 anything without a "C") all source current, and must be pulled below 0.8 volts to be seen as a low.

Common practice for switch inputs was to put the switch between the input pin and ground, with a 5K or so resistor to +5V to ensure the input would really be seen as a High when the switch was open.

CMOS parts (anything with a "C" in the middle letters) have a very high input impedance, so can be used with a 5 - 10K pull-up resistor with a switch to ground, or 5 - 10K pull-down resistor, with a switch to +5V.

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See below clip of this datasheet:

enter image description here

The first number Vil is the maximum input voltage that is guaranteed to be interpreted as low. It is 800mV. However we almost always want a lower voltage than that to give some noise immunity. The next number Vol shows the guaranteed output voltage when sinking 4mA, which is 400mV, yielding 400mV of noise immunity. It would be good to at least match this, so input voltage between 0V and 400mV.

The final number Iil is the maximum input current when the input is at 400mV, -600uA.

So the maximum resistor value on a single input (on this particular chip) for 400mV of noise immunity will be (from Ohm's law) 400mV/0.6mA = 667\$\Omega\$.

If you just want to use the X-OR gate as an inverter, it's better to just tie the input to GND, of course, as that saves a part and gives you 800mV of noise immunity rather than 400.


Note that in some cases multiple inputs to a gate will source the same current as a single input. For example, the 74LS30 has the circuit shown below:

enter image description here

There is only a single resistor for all the inputs, so the input current is the same for one input low or 8 inputs low.

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I will expand on Peter Bennett's answer: Don't use use pulldowns at all.

Instead, use 1k pullups to +5, and ground the input for a low. In other words,

schematic

simulate this circuit – Schematic created using CircuitLab

Tieing inputs directly to ground will not cause excessive current to flow in (LS)TTL, although the same is not true for tieing to +5. So use pullups. For CMOS, you can tie inputs directly to ground and Vcc.

Also note the connection of the LED. As you have realized, TTL works much better as a current sink rather than a source, so go with it. If the polarity inversion (a low output turns the LED on) bothers you, buffer the LED with an inverter such as an LS04, or use another stage of the '86 to do the job.

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