Multiple Input Pull-down

F

FyberOptic

Guest
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!
 
On 2008-09-22, FyberOptic <fyberoptic@gmail.com> wrote:

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.


--

Bye.
Jasen
 
On Mon, 22 Sep 2008 00:34:33 -0700 (PDT), FyberOptic
<fyberoptic@gmail.com> wrote:

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.
---
View in Courier.

One of the gates in a 7400 2-input NAND looks like this, with the inputs
typically being the same for the entire 74XX TTL family.


Vcc>---------+--------+------+
| | |
| | [130]
[4k] [1.6K] |
| | C
| +----B Q3
| | E
B C |
A>----+--E E C----B Q2 [DIODE]
| | Q1 E |K
B>----|----+ | +----->Y
|K | | |
[DIODE] | | C
| |K +----B Q4
| [DIODE] | E
| | [1K] |
| | | |
GND>--+----+----------+------+

Note that when A and B float, the 4k resistor will allow current into
the base of Q2 through the collector-base diode of Q1.

That current turns Q2 on, which in turn turns Q3 off and Q4 on, forcing
the output, Y, low.

In order to turn Q2 off, either A or B must be pulled close enough to
ground to turn Q1 on, thereby grounding the base of Q2 through Q1,
turning it off.

However, using high-value resistors to pull the inputs down to ground
won't let the emitters of Q1 get close enough to ground to turn Q2 off.
---

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?
---
Yes; see the circuit description above.
---
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!
---
An easy solution would be to reverse the polarity of the I/O output and
to use an inverter between the I/O and the three inputs you've got
connected together.

JF
 
On 22 Sep 2008 08:43:59 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2008-09-22, FyberOptic <fyberoptic@gmail.com> wrote:

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

More like 330 ohms for classic TTL, 1/3 of that for three in parallel.
10K is OK for CMOS.

A single HCMOS buffer, ideally a Schmitt, in front of his TTL would be
a good fix.

John
 

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