CMOS and TTL help

Discussion in 'Homework Help' started by GTeclips, Feb 22, 2012.

  1. GTeclips

    Thread Starter Member

    Feb 18, 2012
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    I have recently acquired a few chips, some of which are CMOS, and others TTL. Whenever I wire a TTL chip to the breadboard, it seems to work just fine, but when I wire a CMOS chip in the same fashion, it doesn't work. I searched Google for what the difference between them was, but it only disused Deep logical stuff.

    In all, I probably should have stuck with one or the other. So, my question is, is there some special that I would have to wire a CMOS NAND chip from a TTL NAND chip?

    Thanks
     
  2. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    Not all chips have the same pin-out. Check the data sheet for the chips as to the correct pin-out.
     
  3. GTeclips

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    Feb 18, 2012
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    Well I don't mean literally wiring them exactly the same, but rather wiring them in the same logical fashion to the corresponding pins.
     
  4. Wendy

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    Mar 24, 2008
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    TTL will assume an open input is a high.

    CMOS will dither, as in go analog state many times, it is an illegal condition. You must tie an input to a 1 or a 0, even if the gate is unused. It will affect the other gates on the chip otherwise.

    Schematics?
     
  5. GTeclips

    Thread Starter Member

    Feb 18, 2012
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    I'm no good with schematic diagrams, but ill try and explain it to you in words.

    Say you have a TTL single NAND gate. I would connect Vdd to + and ground to -, then I connect 2 switches that connect from + to the 2 inputs on the NAND gate, lastly, I would connect the output of the NAND gate to an led connected to a resistor and then to -.

    First off, would this work properly, and second, could this same setup be used on a CMOS chip of similar design?
     
  6. MrChips

    Moderator

    Oct 2, 2009
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    TTL gates and CMOS gates have totally different characteristics. They are not interchangeable. If you have a problem getting any one to work (TTL or CMOS) just draw a circuit and we will help resolve the problem for you.

    A picture is worth a thousand words.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    No to both cases. Guess you didn't understand Bill's post. :confused:

    As Bill explained, TTL goes high with an open circuit input, so when a switch opens, the input will still appear as a logic 1 and nothing changes. To set a TTL input to logic 0 you have to pull current from the input to ground. This can be done by connecting the switches to ground. Then when the switch is open you have a logic 1 and when the switch is closed you have a logic 0. For best noise immunity add a 10kΩ resistor from each TTL input to +Vdd

    CMOS has a very high impedance input and a floating input does not generate a defined logic state so when the switch is set to the open condition (not connected to +Vdd) then the CMOS input is undefined. To remedy that you can add a resistor (say 10kΩ) at each input to ground. That way when the switch is opened, the resistor will pull the input to ground (logic 0).
     
  8. GTeclips

    Thread Starter Member

    Feb 18, 2012
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    Oh, ok. When bill explained it, it just went a little over my head. Now I understand that it is not desirable to have a floating input for neither a CMOS nor a TTL. Now at this point I would state that if I did put an input to ground, couldn't an output from some other chip interfere and cause a high, but I'm assuming that's what the 10k resistor is for. Now the next problem I see is, how can I make a switch that when is 0 will have an input connected to ground and when it is 1, it will connect the input to Vcc?

    Thanks!
     
  9. crutschow

    Expert

    Mar 14, 2008
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    An output won't interfere with another input unless the input is floating.

    The answer to your last question is easy. Ever heard of SPDT switches?
     
  10. GTeclips

    Thread Starter Member

    Feb 18, 2012
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    Thank you for the SPDT switch info, and I'm guessing there are manually and electrically triggered SPDT's as well.

    But as for my first question. Say you have a CMOS NAND chip and some other separate chip on a breadboard and a 9V power supply. Say for instance that the misc chip has an output that is a constant high that is connected to the ground bus. If you connected one of the inputs of the NAND chip to the ground bus to give it a logic 0, wouldn't the output from the misc chip flow through the ground buss and trigger the input on the NAND chip, giving it a logic 1 when the desired state is 0, or is that what the 10k ohm resistor is for?
     
  11. crutschow

    Expert

    Mar 14, 2008
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    Electrically triggered SPDTs are called relays.

    If you I understand you correctly you are talking about connecting a constant high to the ground? :confused: That is something you should never do. It will draw high short-circuit output current and could zap the chip.
     
  12. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    :eek:

    Stop right there! Any output from any miscellaneous chip that is directly connected to ground, will let the magic smoke out quickly - especially if it is a constant high. That would be a short circuit :mad:

    This is shown in the top of the picture below
    [​IMG]


    The second picture shows how to connect an input to a switch so that is is not floating. If point B is connected to ground, the resistor is called a pull-down resistor; if it is connected to a logic 1 (the appropriate voltage for the type of logic you are using), it is called a pull up resistor. Point A then is connected to the opposite polarity.

    Let's consider the pull-up configuration. If the switch is open, then voltage flows through the resistor and causes a logical 1 to appear on the input of the logic gate (the unlabelled chip). Hence, the input is NOT floating. If the switch closes (and point A is connected to ground), then the input to the logic gate is a logical 0.

    Does this help?
     
    Last edited: May 7, 2012
    GTeclips likes this.
  13. GTeclips

    Thread Starter Member

    Feb 18, 2012
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    Lol. Sorry. If you haven't already guessed, I'm very new to all things electrical, but I must say I'm enjoying it! :)

    And yes, I think that diagram does help me, if I understand it right. So what your saying about the bottom diagram is: (I'm assuming that that is the input to the unlabeled chip) A is connected to the to Vcc and B is connected to ground, and when when the switch is open, any small amount of current in the ground buss will go through the resistor and trigger the input to 0, and when the switch is closed, the full power will flow into the input and trigger it to 1 and the resistor will prevent a short circuit (considering it is directly connected to ground).

    Do I have it right?
     
  14. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    I think you got it!
     
  15. GTeclips

    Thread Starter Member

    Feb 18, 2012
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    Thank you very much for all the help from all of you! I appreciate it very much! I'm very glad I found a forum that is mature and fun and willing to help me, even when my questions are very simple! I think my electrical learning experience will be a fun one! :)
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    I love digital myself. When it works it tends to do exactly what you expected, unlike analog electronics.
     
  17. BSomer

    Member

    Dec 28, 2011
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    I like the drawing of the "magic smoke". Nice representation of what could happen. :)
     
  18. dobidust

    New Member

    May 6, 2012
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    Does TTL & CMOS interface directly in anyway ?
     
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