Hex inverter question

D

Dave

Guest
I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8 as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053 chips
I am working with? Or should pin 8 simply be ground?

Ignorantly yours,

Dave
 
Dave wrote:

I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8 as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053 chips
I am working with?
Click to expand...
Why are you applying a negative voltage here ?


Or should pin 8 simply be ground?
Click to expand...
That's what Vss normally is.

Graham
 
"Dave" <db5151@hotmail.com> writes:

I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8 as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053 chips
I am working with? Or should pin 8 simply be ground?

Ignorantly yours,

Dave
Click to expand...
This part is the same as all the other 4049 inverters. It is designed for
level shifting so it has a VCC and VDD pin. VSS is ground, or 0 volts.

Oops- the "U" means that pin 16 is open. The regular older 4049 required
both.
--
Steven D. Swift, novatech@eskimo.com, http://www.novatech-instr.com
NOVATECH INSTRUMENTS, INC. P.O. Box 55997
206.301.8986, fax 206.363.4367 Seattle, Washington 98155 USA
 
"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:47ACAE77.9A132DE6@hotmail.com...
Dave wrote:

I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8
as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053
chips
I am working with?

Why are you applying a negative voltage here ?


Or should pin 8 simply be ground?

That's what Vss normally is.

Graham
Click to expand...
Hey Graham,

Well, I was under the impression (maybe not correctly so, now that I look at
it again) that Vss *was* -5V. Here follows the answer to a post I made
earlier this year about the nature of Vss...

Dave wrote:

Not sure what to say. My "signal" (on the Ain/out and the A1/A0 etc.of
the
4053) is actually analog (thus my concerns about Ron), and the 74HC4053
(according to that datasheet) requires both a +5V (Vdd) and a -5V (Vee).
And (BTW) what is Vss?
Click to expand...
Vss is the negative logic supply rail. The control logic
operates between Vss and Vdd. The analog signals can swing
between Vee and Vcc, though Vee can be connected to Vss if
the analog signals stay between the logic rails. But
reducing the total difference between Vdd and Vee from 10
volts to 5 volts almost doubles the switch on resistance
(from something like 4o ohms to something like 70 ohms).

Sooo, I guess it could be ground, but it could also be -5V? (The basic
circuit did work when I hooked it up that way...) I guess that, if I am
correct, it might even operate better in some instances, judging from the
answer given above.

So, with the 74HC4053 I use -5V for Vss, and for the 4049 I use ground for
Vss? Somebody help me here (John?) I feel I ought to be consistant, at
least...

Many thanks,

Dave
 
"Steven Swift" <novatech@eskimo.com> wrote in message
news:foibo10rvc@enews1.newsguy.com...
"Dave" <db5151@hotmail.com> writes:

I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8
as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053
chips
I am working with? Or should pin 8 simply be ground?

Ignorantly yours,

Dave

This part is the same as all the other 4049 inverters. It is designed for
level shifting so it has a VCC and VDD pin. VSS is ground, or 0 volts.

Oops- the "U" means that pin 16 is open. The regular older 4049 required
both.
--
Steven D. Swift, novatech@eskimo.com, http://www.novatech-instr.com
NOVATECH INSTRUMENTS, INC. P.O. Box 55997
206.301.8986, fax 206.363.4367 Seattle, Washington 98155 USA
Click to expand...
Hey Steven,

And that pin would normally be Vcc, I am guessing?

Thanks,

Dave
 
"Dave" <db5151@hotmail.com> wrote in message
news:13qp8focnn90u2b@corp.supernews.com...
I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8
as Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053
chips I am working with? Or should pin 8 simply be ground?

Ignorantly yours,

Dave
Click to expand...
Nevermind. I found my CMOS Cookbook. It is ground. Sorry for the
confusion and such.

Dave
 
On Fri, 8 Feb 2008 12:39:49 -0600, "Dave" <db5151@hotmail.com>
wrote:

I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8 as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053 chips
I am working with? Or should pin 8 simply be ground?
Click to expand...
---
Motorola's MC14049 is equivalent to RCA's CD4049 (note the
similarity in the last four digits of their part numbers) and Vss
refers, in both cases, to logic ground.

The 4053's you've been working with are analog transmission gates
and, in order to be able to be able to pass a signal which goes more
negative than logic ground, must be connected to a supply which goes
more negative than logic ground and gives the AC signal something to
work against.

That supply is called 'Vee', and is usually connected to pin 7 of
the 4053 like this:

+------
+-------16|Vdd
|+ |
Vdd |
| |
+----+---8|Vss
|+ | |
Vee GND |
| |
+--------7|Vee
+-----

That way, when the logic switches turn the channels ON or OFF with
control signals which are either at Vdd or GND, analog signals with
amplitudes between Vdd and Vee will be allowed to either pass
through the switch or be blocked.
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:hvfpq3911n8d1laajnhn290fe61kqb8lum@4ax.com...
On Fri, 8 Feb 2008 12:39:49 -0600, "Dave" <db5151@hotmail.com
wrote:

I have a chip labeled MC14049U with the Motorola logo that I can't find a
datasheet for (specifically). The MC14049UB datasheets I find show pin 8
as
Vss. Does that mean it needs -5V applied to pin 8, like the 74HC4053
chips
I am working with? Or should pin 8 simply be ground?

---
Motorola's MC14049 is equivalent to RCA's CD4049 (note the
similarity in the last four digits of their part numbers) and Vss
refers, in both cases, to logic ground.

The 4053's you've been working with are analog transmission gates
and, in order to be able to be able to pass a signal which goes more
negative than logic ground, must be connected to a supply which goes
more negative than logic ground and gives the AC signal something to
work against.

That supply is called 'Vee', and is usually connected to pin 7 of
the 4053 like this:

+------
+-------16|Vdd
|+ |
Vdd |
| |
+----+---8|Vss
|+ | |
Vee GND |
| |
+--------7|Vee
+-----

That way, when the logic switches turn the channels ON or OFF with
control signals which are either at Vdd or GND, analog signals with
amplitudes between Vdd and Vee will be allowed to either pass
through the switch or be blocked.
Click to expand...
Thank you, John. Your patience with my ignorance is appreciated.

Dave
 
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