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4049 hex inverting ic

Discussion in 'General Electronics Discussion' started by Kstarloomo, Jul 5, 2012.

  1. Kstarloomo

    Kstarloomo

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    Jun 28, 2012
    can any one pls tell me how 4049 is used, expecially the nc terminals.
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    The NC terminals are Not Connected, so they AREN'T used!
    The 4049 is a hex inverting buffer. Each of the six elements inside it has an input and an output, and it drives the output to the opposite logical state from the input ("inversion").
    CMOS devices have nominal thresholds of 1/3 VDD and 2/3 VDD. This means that if the input voltage is above 2/3 of the supply voltage (VDD), it will be considered HIGH, and the 4049 will drive the output LOW. If the input voltage is below 1/3 of VDD, it will be considered LOW and the 4049 will drive the output HIGH. In between those thresholds is a no-man's land and if the input voltage stays in that range, this can cause significant current consumption and even heating in the device. In other words, it's not recommended.
     
  3. Kstarloomo

    Kstarloomo

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    Jun 28, 2012
    tanx, but i connected it without it and it fails to work, i mean not consistent-some times it works and some times not. what can be happening?
     
  4. davenn

    davenn Moderator

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    we would only be guessing without seeing how you have used it
    no power to it ?
    connected wrong way around?
    soldering fault?
    etc etc

    how about a circuit diag of your project

    Dave
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Right, as davenn says, we need a schematic, and preferably a photo of your circuit.
    As well as a description of what you're trying to do.
    What are you powering the 4049 from? You know it needs power?
    Also, the inputs of the inverters that you're not using should be connected to a supply pin, to stop them from floating and getting in between the thresholds.
     
  6. Kstarloomo

    Kstarloomo

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    Jun 28, 2012
    Also, the inputs of the inverters that you're not using should be connected to a supply pin, to stop them from floating and getting in between the thresholds.[/QUOTE]
     
  7. davenn

    davenn Moderator

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    where's the circuit diagram of your project we asked for
    we are not going to be able to help you unless you help us :)

    Dave
     
  8. CocaCola

    CocaCola

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    Kstarloomo if your previous post was supposed to be a question about what KrisBlueNZ suggested, Google pull-up or pull-down resistors... Using pull up or pull down resistors will 'lock' a pin into a certain state instead of possibly floating in between states... When the real logic is applied it will overcome the pulled logic applied through the resistor... This also helps avoid that dead zone that Kris talked about, or at least makes it harder to get stuck there accidently...

    Example pull the pin 'low' to ground with a 10K resistor, now the pin has zero logic and is held at zero logic... When a logic one (positive voltage) is applied to the pin from whatever it is supposed to invert, that logic one will overcome the 10k pull to zero... And the same could be done in reverse...

    Until then, waiting on pics as the others to actually answer you original question...
     
  9. Kstarloomo

    Kstarloomo

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    Jun 28, 2012
    here is the curcuit diagram
     

    Attached Files:

  10. BobK

    BobK

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  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If it were a line driver yes, but normal CMOS gates have a low enough current drive capability that you won't quickly damage the LED (or in some cases damage it at all).

    However it's not god practice and there is a spread of drive capabilities so you can't guarantee how bright the LED will be. Also, some gates might surprise you with additional drive capability in sinking or sourcing.

    I also advise playing it safe and using a resistor.
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That's not a complete circuit diagram.
    It should have worked, though, because the 4049's outputs have an inherent resistance that will prevent damage to the LEDs. BTW none of those resistors in the star are needed.
    There are much better ways to drive LEDs though, including directly from the 555, which itself has a much higher drive capability than the 4049's inverters! (Assuming that's pin 3 that you're feeding into the resistor star). If you drive the LEDs from the 555 you WILL need current limiting resistors. It would be better to connect several LEDs in series and use a single current limiting resistor. Find out the forward voltage of your LEDs, and put enough of them in series so they add up to a few volts short of the supply voltage, and calculate the limiting resistor based on the remaining voltage (minus a volt or so because the 555 output doesn't swing all the way to the positive supply).
    So what is the problem? You've given us a (very incomplete) circuit diagram but you haven't said what problem you are having with it.
    You have to do a lot better than this if you want anyone to be able to help you.
     
  13. David Sparks

    David Sparks

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    Jul 12, 2012
    I just recently downloaded a datasheet on the CD4049, as I have been considering it as a 3-channel oscillator. First of all, all of the above statements for using this part are true. What got my attention in the datasheet was a statement that even though the part has a working voltage of 3 to 12 VDC, input's MUST be driven with virtually a rail-to-rail voltage, or tied to the gnd or Vcc rail via a 1K to 10K resistor. The datasheet is from Harris semiconductor and it states that this part has a hi-to-low logic converter built in. It can convert CMOS to TTL levels, but not TTL to CMOS. I would follow all of the suggestions mentioned so far, plus make sure your Vcc and gnd pins have a bypass capacitor directly at the pins. Unless it is driving another CMOS device, you must buffer the output with a transistor through a 10K resistor.
    By the way, the Harris datasheet is about 882KB in size, so it has a lot of useful info.
     
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    If you're after multiple oscillators, you can get six out of a hex Schmitt trigger inverter like the 40106, 4584, or 74C14.
    No, not necessarily. The 4049 can drive an LED easily enough, and a lot of other circuit elements. It can't supply a great amount of current though. Whether a buffer is needed really depends on what you're driving.
     
  15. David Sparks

    David Sparks

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    Jul 12, 2012
    Kris, I should have clarified that this CD4049AE was one that I had "laying around", and was looking to see what tricks it could do for me. I have 555's, 556's, TL494's, etc, or I could just build a multivibrator, but I just wanted to see how much I could get out of this chip. You are correct in that in can drive the new hi-efficiency LED's in pull-down mode (I highly recommend the one's made by ITT of China, as they are VERY bright at 6ma), but not hi-powered LED's or a Mosfet for a SMPS. I should have been more clear as to why I was using such an ancient ic-lol.
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Just a comment on making oscillators from inverters.

    You typically see a 2 inverter circuit, however this is not recommended because 2 inverters must be made to oscillate -- do a thought experiment of what happens when the capacitor value falls to zero.

    A three inverter circuit (essentially any odd number) will always oscillate and the purpose of additional components is simply to slow it down.

    A single inverter does not make a good oscillator either because you risk it being biased into its linear region. A schmitt trigger is an exception to this because it cannot be biased into a linear region.

    Having said that, you will see an awful lot of 2 inverter oscillators, and you have to dig deeply into the application notes to find the advice warning against this practice.

    Let me do the digging for you! Here.
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Very good advice as usual (*steve*).
    The Schmitt trigger oscillator is a favourite of mine because it uses only 1/6th of an IC (or 1/4 of a quad Schmitt NAND gate, if you already have a need for one or more NAND gates), and there are often other uses for the other gates, such as buffering the oscillator output for increased drive. For example you can use a hex Schmitt inverter with one gate as the oscillator and the remaining unused inverters "paralleled" (not really paralleled, but it looks like it on the schematic) to reduce the output impedance. This is good for capacitor-diode charge pumps for generating a low-current negative rail or a low-current higher-than-VCC positive rail.
     
  18. gorgon

    gorgon

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    Jun 6, 2011
    The main problem with Schmitt trigger based oscillators, is that you can't be sure that the threshold levels are equal enough between batches from the same manufacturer and, certainly not between different manufacturer. This problem is of course only present if you plan to produce something over time, or you do the same design with the same component values with different chips/brands.

    This will be a problem if you need the frequency to be about the same from unit to unit. If you need real, accuracy you'll need to add some adjustment system to calibrate the oscillator.

    Regarding the cmos inverter like 4049, it comes in two versions. One with buffer, and one without an extra buffer, normally marked with an 'U' after the number. The 'U' part will be a better choice for an oscillator circuit, due to a better linearity.

    TOK ;)
     
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    gorgon, it's certainly true that you can't expect good frequency accuracy or repeatability from an oscillator based on a Schmitt trigger IC, but you can't expect particularly good accuracy or repeatability from any oscillator that relies on the value of a capacitor anyway. If you need predictable frequency accuracy you use a crystal or ceramic resonator, or software calibration.
    Unbuffered gates like the 4049U or 4069U are also preferred over B versions for use in Pierce crystal oscillators, for the same reason you gave.
     
  20. David Sparks

    David Sparks

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    Jul 12, 2012
    I agree with you 100% Steve. Two-transistor multivibrator's are also prone not oscillate if the capacitors are too small or they may take several cycles to stabilize. This can be fatal if you are using them to drive Mosfet's in a SMPS or ultra-sound generator. My intent is to wire the 2 inverter's as a simple Schmitt by adding a 1K resistor at the input connected through a 10K resistor to the output of the second inverter.
    Admittedly, this guarantees nothing, even though I only need them to work in the 15KHZ to 25KHZ range. I have LF353 hi-speed op-amps which, according to the mfg, are dependable and are as frequency stable as the capacitor you choose. Then my only issue is clamping or offsetting the -12 volt swings so they do not damage the pnp driver for the n-channel mosfet. At least I have an adjustable current limiter already built in, so the mosfet/transformer/piezo transducer do not burn up if the oscillator does not start. I start at 250ma and see how it goes.
    I highly recommend using adjustable current limiters when experimenting with oscillator's of dubious performance. Sharp rise and fall times are also important, or the transformer or transducer or "hi-power LED" could burn up. Been there-done that.
     
    Last edited: Jul 13, 2012
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