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Hi I am using a 74ls38 quad 2 input nand buffer with open collector outputs and as well as a 74ls74 Dual Positive-Edge-Triggered D Flip-Flops with Preset, Clear and Complementary Outputs. On the nand buffer I'm not using 2 of the nand gates so total of 4 inputs not used and 2 out puts on the chip. On the flip flop chip I'm not using 1 of the flipflops in the chip so it leaves 2 outputs and 4 other pins not in use. Just curious as to how I should handle the unsued pins. I'm sure leaving them all floating is not a wise idea. For the nand I figured ground the inputs and a high resistance pullup resistor on the output would be fine but the flip flop I'm not sure how you would handle that and looking at the datasheet I didn't get anything out of that.

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  • \$\begingroup\$ Leave the outputs unconnected. Be it a flip/flop or gate, putting a resistor on the output is not an issue as long as you don’t exceed the current specs. For TTL, the output drive is assy metric - they can sink more current than they can source. \$\endgroup\$
    – Kartman
    Commented May 26 at 13:19
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    \$\begingroup\$ FWIW, TTL inputs naturally "float" high. If you want to drive an input low, you do it by pulling current from the input pin, but when you drive an input high, there is little or no current. The takeaway: If you have an input pin whose state does not matter—the circuit will work whether it is held high or held low—then hold it high. Your circuit will use less power that way. \$\endgroup\$ Commented May 26 at 15:19
  • \$\begingroup\$ Curggles, you have some answers below. You've said nothing since. Is there something you still need to work out? Or are one of the answers below good enough? Just curious. \$\endgroup\$ Commented Jun 11 at 9:51

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Ground inputs that are supposed to be low.

For inputs you want to be high, a pullup resistor such as 1kΩ can be used. It was recommended procedure because 74 TTL inputs that had multiple emitters could be damaged by voltage transients.

However we seldom see 74 TTL these days, outside of repairs and maybe dusty parts in some school lab.

LSTTL is actually DTL so you can tie inputs directly to Vcc or ground. You can tie CMOS parts such as 74HC, 74HCT etc. directly to either supply.

You must tie CMOS parts to a valid logic level. You can leave TTL/LSTTL etc. inputs that are supposed to be high open but it's bad practice and they could pick up noise.

You might want to use a resistor anyway because it allows field enhancements to be accomplished more easily and resistors are cheap. A 0Ω resistor is appropriate to tie an input to GND.

Another possibility is to tie unused inputs to an existing output that has an appropriate logic level (keeping the fan-out rules for the logic family in mind). If appropriate you can tie unused inputs together, for example to turn a 4-input NAND into a 3-input NAND.

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For multi-emitter TTL inputs (commonly found with 74, 74S, 74LS, and 74ALS, for example), cannot tolerate higher than \$5.5\: \text{V}\$ at their inputs. So the minimum resistor size that is used for unused inputs must be based upon the maximum peak voltage at startup that may occur. If your power supply might peak up to say \$7\:\text{V}\$ before settling back down to \$5\:\text{V}\$ then you must have a resistor of at least \$\frac{7\:\text{V}-5.5\:\text{V}}{1\:\text{mA}}=1.5\:\text{k}\Omega\$.

Meanwhile, the input must be at least \$2.4\:\text{V}\$ to be seen as HI, so this means the maximum resistance is, assuming a minimum \$V_{_\text{CC}}\$ of \$4.75\:\text{V}\$, \$\frac{4.75\:\text{V}-2.4\:\text{V}}{N\,\cdot\,I_{_\text{IH}}}=11.75\:\text{k}\Omega\$ if there are \$N=5\$ inputs being serviced by the resistor and if \$I_{_\text{IH}}\approx 40\:\mu\text{A}\$, for example.

The above is assuming you want the inputs tied HI.

Finally, all parts of an IC must be connected. Even if you have, for example, two FFs in an IC and are using only one of them, you must make sure that the other unused FF has all of its inputs tied somewhere.

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Standard practice is to tie TTL inputs to VCC through a a resistor of 1kohms or so - the value isn't too critical. A floating input is nominally a '1', but a floating input can pick up noise. That will work for the NAND gates as well as the spare flipflop inputs.

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