The said component is sn74lvc1g08 from TI, datasheet.

I was planning on connecting 3 1.5 volt AA batteries (in other words, 4.7 or 4.8 volts max in total) to one of the input pins (either A or B) and I'm quite sure although not 100% sure that a resistor is necessary, but I don't really know how to calculate the value of the resistor.

Please give a rather detailed procedure regarding the method of calculation. Thanks in advance.

I'm working on some safe-critical project and I'm ticking in "extremely meticulous" mode so please bear with me.

  • \$\begingroup\$ an AA battery at 0.5V is quite dead already, 3 batteries would probably never trigger a "low" input (or at least not reliably) \$\endgroup\$ – Wesley Lee Apr 12 '16 at 9:05
  • \$\begingroup\$ Could you give a bit more context - why are you connecting the batteries direct to an input pin? Are they the same batteries that power the circuit? \$\endgroup\$ – pjc50 Apr 12 '16 at 13:58
  • \$\begingroup\$ I failed to find the button to reply to each one of you so I'll make my response in one go: \$\endgroup\$ – user6107491 Apr 13 '16 at 1:16
  • \$\begingroup\$ I failed to find the button to reply to each one of you so I'll make my response in one go: I do not anticipate it to ever give a "low", I have other mechanisms to measure the battery level, this circuit in discussion is used to decide whether a battery was plucked out of its chamber, so there's the context. And no, it is not the batteries which power this particular circuit, it powers a "main circuit" other than this one, this one is used to monitor the 3 batteries and has its very own battery as an independent power source. \$\endgroup\$ – user6107491 Apr 13 '16 at 1:23

No resistor is needed, just make sure whatever voltages are presented to the input pins are either between 0 and the maximum voltage for a logic 0, or between the minimum voltage for a logic 1 and the supply voltage.

The maximum voltage for a logic 0 (3*VCC for a supply voltage between 4.5V and 5.5V), and the minimum voltage for a logic 1 (0.7*VCC for a supply voltage between 4.5V and 5.5V) will be in the datasheet.

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  • \$\begingroup\$ Thanks for your expert advice tcrosley, you have been a wonderful help. \$\endgroup\$ – user6107491 Apr 13 '16 at 1:14
  • \$\begingroup\$ Hello again tcrosley, per my question, can I ask a bit more? This is my understanding as to why no resistor is needed: You see, most logic devices are of the CMOS nature, ideally and theoretically, they do not consume power, that is, voltage signal in, and voltage signal out, but that's impossible since there are at least junction capacitance which will cause the current in related circuit to be above zero, let alone the fact that resistance of CMOS transistors is usually enormous. In other words, there is no need for an extra resistor since (continued in another comment) \$\endgroup\$ – user6107491 Apr 13 '16 at 1:43
  • \$\begingroup\$ since the resistance of CMOS devices is already very, very high which means there wouldn't be any large current so long as the input signals do not exceeds the threshold provided in the datasheet. Am I correct in assuming that? \$\endgroup\$ – user6107491 Apr 13 '16 at 1:47
  • \$\begingroup\$ @user6107491 You are correct, the input of a CMOS transistor looks like a 10\$^{12}\$Ω resistor shunted by a 5 pF capacitor. So the input current is less than 1 µA. Any additional resistor in front of the input would have no effect. \$\endgroup\$ – tcrosley Apr 13 '16 at 2:29

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