Multiple Input Pull-down

[FyberOptic]

I'm having a bit of a quandary. Since I'm still somewhat of a
beginner, I hope somebody more experienced can help me out!

I have three TTL logic chips, with one input on each one all connected
together to a single input signal. When that main signal wire is
connected to ground, the logic chip signals are all low as expected.
When I remove the ground connection though, they seem to all go high
on their own without being connected anywhere (aside from to each
other). This isn't my problem exactly, but I'm curious if someone can
explain it.

So I need a pull-down on that main input signal to keep the inputs
it's connected to forced low unless a high is explicitly introduced.
The main input signal itself will be connected to a pin coming out of
a chip elsewhere, which is switchable between being an input and
output. At startup, that pin is an input, so it's not outputting a
low like I want, hence the need for a pull-down.

Where the actual problem comes in is that when I add a "normal" pull-
down resistor value between the input signal and the ground, it
behaves as if it's not connected to ground at all. I've tried a few
different ones, from 100k, to 50k (didn't have a 47k), down to 4.7k.
I couldn't figure it out. Then finally I tried a 1k, and it seems to
properly pull it down low. Keep in mind that this isn't even
currently connected to the other chip I mentioned which will default
to an input at startup, this is just the input signal being connected
to ground or left floating. Is having to use such a low pull-down
resistor because I have three inputs connected together? Is there a
resistance inside the IC I'm not taking into account when connecting
it up like this? My assumption is that a "normal" 10k-47k pulldown
value assumes only a single input is being connected to. A possible
worry I have though is that the 1k resistor is kind of low and could
be wasting power (since it will be able to run from batteries), but I
dunno.

I'd appreciate any light anyone can shed on how these things work!

 

[Jasen Betts]

Where the actual problem comes in is that when I add a "normal" pull-
down resistor value between the input signal and the ground, it
behaves as if it's not connected to ground at all. I've tried a few
different ones, from 100k, to 50k (didn't have a 47k), down to 4.7k.
I couldn't figure it out. Then finally I tried a 1k, and it seems to
properly pull it down low.

10K for a normal TTL pull down, but with three inputs involved you
might need 1/3 of that or 3.3K

also the TTL thresholds are near grouind so you need to pull harder to
go down that you do to go up.

A possible worry I have though is that the 1k resistor is kind of low
and could be wasting power (since it will be able to run from
batteries), but I dunno.

if it's to be battery operated, CMOS might be a better choice than TTL.

 

[Jon Slaughter]

I'm having a bit of a quandary. Since I'm still somewhat of a
beginner, I hope somebody more experienced can help me out!

I have three TTL logic chips, with one input on each one all connected
together to a single input signal. When that main signal wire is
connected to ground, the logic chip signals are all low as expected.
When I remove the ground connection though, they seem to all go high
on their own without being connected anywhere (aside from to each
other). This isn't my problem exactly, but I'm curious if someone can
explain it.

It's called floating. The state is usually undefined but because of unknown
factors it could set it to either state. That or internally there is a
pullup.

(are you sure it's continuously high and not toggling back and forth very
rappidly?)

So I need a pull-down on that main input signal to keep the inputs
it's connected to forced low unless a high is explicitly introduced.
The main input signal itself will be connected to a pin coming out of
a chip elsewhere, which is switchable between being an input and
output. At startup, that pin is an input, so it's not outputting a
low like I want, hence the need for a pull-down.

Where the actual problem comes in is that when I add a "normal" pull-
down resistor value between the input signal and the ground, it
behaves as if it's not connected to ground at all. I've tried a few
different ones, from 100k, to 50k (didn't have a 47k), down to 4.7k.
I couldn't figure it out. Then finally I tried a 1k, and it seems to
properly pull it down low. Keep in mind that this isn't even
currently connected to the other chip I mentioned which will default
to an input at startup, this is just the input signal being connected
to ground or left floating. Is having to use such a low pull-down
resistor because I have three inputs connected together? Is there a
resistance inside the IC I'm not taking into account when connecting
it up like this? My assumption is that a "normal" 10k-47k pulldown
value assumes only a single input is being connected to. A possible
worry I have though is that the 1k resistor is kind of low and could
be wasting power (since it will be able to run from batteries), but I
dunno.

My bet is that there is an internal pullup. By adding your pulldown it works
in parallel resulting in a voltage divider.

I'd appreciate any light anyone can shed on how these things work!

What is the logic family? (e.g. what is the numbers on the chips)

Some chip families work better than others.

As Jasen mentioned, I'd end up using some form of low power cmos rather than
bipolar. You need to choose the right family if you are worried about power
loss. Some families are high speed which you might not need.

In any case the cmos families(or equivalents) are usually a direct
replacement so there is not that big a deal. (there are a few issues with
have to do with thresholds but I doubt you'll have those problems)

If I were you I'd probably look at a few alternatives to bipolar.

http://en.wikipedia.org/wiki/7400_series

somewhere there is also a chart that shows the power consumption
differences.