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Hex inverter question

Discussion in 'Electronic Basics' started by Dave, Feb 8, 2008.

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  1. Dave

    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
     
  2. Eeyore

    Eeyore Guest

    Why are you applying a negative voltage here ?

    That's what Vss normally is.

    Graham
     
  3. Steven Swift

    Steven Swift Guest

    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.
     
  4. Dave

    Dave Guest

    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...
    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
     
  5. Dave

    Dave Guest

    Hey Steven,

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

    Thanks,

    Dave
     
  6. Dave

    Dave Guest

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

    Dave
     
  7. John Fields

    John Fields Guest

    ---
    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.
     
  8. Dave

    Dave Guest

    Thank you, John. Your patience with my ignorance is appreciated.

    Dave
     
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