hello, I have a 5.5v circut and a 7404 inverter. The chip functions as expected except for one thing. I cannot get the pin to not float with a resistor. Any resistor value will not result in a ground (0). I have used everything from 1k, 10k,48k,100k,1M and more in between. When I jump it from ground to the pin with just wire everything is as expected. But when I add my on off switch it hijacks the power and burns batteries since I have created a short. Things get hot.
The real question. How do I add a default state to a pin? I have to pull down to ground.
Please take a look at this document ti suggested to look at
https://www.ti.com/lit/an/scba004d/scba004d.pdf
So what i am getting from that it is recommended to use an ic that does that for you. Like the 7405.
100 ohm restister did the trick. I don’t know what that will do to battery life though.
It’s helpful to provide the specific part number with which one is working; it helps to avoid incorrect assumptions.
Looking at the datasheet for a current-production '7404 inverter, one can find a diagram of the gate structure, which is informative. To draw Input A to a logic “low” value requires sinking roughly a mA and change worth of current, an idea which is supported by the product characteristics table.
Apply Ohm’s law here, and one can figure that an input pull-down resistance in excess of a couple hundred ohms is probably not going to do the trick. But at the same time, hard-wiring the input to ground shouldn’t cause any serious issue either. If doing so does cause issue, it would seem a likely indicator that the device has been damaged, most probably as a result of ESD. If operating at the higher end of the recommended supply voltage range, issues of supply transients may also be a potential cause of problems.
