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Dave
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 Hex inverter question
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
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| 17 Mar 2008, 15:31 |
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Eeyore
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 Re: Hex inverter question
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
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| 17 Mar 2008, 15:31 |
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novatech
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 Re: Hex inverter question
"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.comNOVATECH INSTRUMENTS, INC. P.O. Box 55997 206.301.8986, fax 206.363.4367 Seattle, Washington 98155 USA
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| 17 Mar 2008, 15:31 |
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Dave
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 Re: Hex inverter question
"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 >
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?
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
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| 17 Mar 2008, 15:32 |
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Dave
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 Re: Hex inverter question
"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 > Hey Steven, And that pin would normally be Vcc, I am guessing? Thanks, Dave
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| 17 Mar 2008, 15:32 |
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Dave
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 Re: Hex inverter question
"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 > >
Nevermind. I found my CMOS Cookbook. It is ground. Sorry for the confusion and such.
Dave
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| 17 Mar 2008, 15:32 |
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John
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 Re: Hex inverter question
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.
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| 17 Mar 2008, 15:32 |
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Dave
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 Re: Hex inverter question
"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. >
Thank you, John. Your patience with my ignorance is appreciated.
Dave
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| 17 Mar 2008, 15:32 |
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