Begginer question about Schmitt trigger inverter

Discussion in 'The Projects Forum' started by mkgort, Jul 4, 2011.

  1. mkgort

    Thread Starter New Member

    Jul 4, 2011
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    Greetings.

    I am brand spanking new to the world of electronics, and could not be more excited! I wanted to get my feet wet with a very simple project, but I have already hit a wall. I am trying a simple signal conversion with a mercury switch and a Schmitt trigger inverter (SN74LS14N). If I hook the switch to an led and resistor and close the switch, I get light. However, when I connect pin 1 of the IC to the switch and output pin 2 to a led, it’s a no go for led number 2. I have not yet mastered the art of reading a data sheet, but if I had to guess, I am missing some kind of voltage threshold. I have tried various resistors to join the switch to pin 1, but no luck.

    I apologize for my lack of proper terminology. I hope this question is understandable. I literally received my first soldering iron five days ago and have several beginners’ books on order. I just couldn’t wait to power something up :D

    Thank you for your time, and thank you all for this fantastic website.



    MK
     
    Last edited: Jul 4, 2011
  2. #12

    Expert

    Nov 30, 2010
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    Welcome to AAC!

    First, it is helpful to provide a datasheet. It makes answering so much quicker for us.
    Second, a drawing of your circuit is almost always necessary.
    Third, you did not say you gave the IC a 5 volt power supply to use for its own needs.
    That's why a drawing is helpful. It eliminates a lot of dumb questions coming back at you.

    I can give you the datasheet. You should give me a drawing of your circuit.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Welcome to AAC, MK.

    The 74xx series is quite old, and the 74L and LS versions have lower voltage output than the C/H/HC versions. The 74xx, 74Lxx, 74LSxx versions need to operate on 4.5v to 5.5v (nominally 5v) or they either won't function properly or will burn up.

    To make things easier, you might consider getting some 4000-series CMOS IC's, as they operate over a much wider range of voltage; from approximately 3v up to around 16v; 18v in some cases.
    The 4106 is a hex Schmitt-trigger inverter that has the same pinout as the 74LS14.
    A very useful IC is the 4093; it is a quad Schmitt-trigger NAND; you can build any other logic gate using just NAND gates.
    See this Wiki entry: http://en.wikipedia.org/wiki/NAND_logic
    Consider getting several of these to experiment with.
     
  4. mkgort

    Thread Starter New Member

    Jul 4, 2011
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    I thoroughly appreciate the responses. I had a feeling I would forget some important information in my first post. The circuit is using 5v and I do have the IC properly powered and grounded (at least I think, nothing has blown up). I should have mentioned that I do indeed have the data sheet; I’m just not sure how to read it yet.:rolleyes: But, I will make sure I include one in future posts.



    Does the included diagram make any sense? This is my first attempt at one. Is there a particular program that will make it a little easier, or are most people just using a basic graphics program and cutting and pasting?


    It looks like I will give the 4093 a try. Would this be correct? http://www.mouser.com/ProductDetail...GAEpiMZZMtOXy69nW9rM%2bHWvLW4M5woYbE7VT26fco=


    Again, you have my deepest gratitude.

    MK
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    In the "General Electronics Chat" forum, there is a "sticky" titled "Decoupling and bypass capacitors - why?" - you should read it. Every logic IC requires at a minimum a 0.1uF cap across its' power/ground pins.

    Datasheets are a vital key in understanding how a part functions, and what its' limitations are. Some are better than others. You might look at datasheets from several different manufacturers, such as Texas Instruments, Motorola/OnSemi, Phillips/NXP, National Semiconductor, and others - to see what feels more comfortable to you.

    Sure, but...

    To the right of the power switch, you have an LED in series with the signal that you are using for an input to the inverter. The voltage drop across the LED may cause the logic level to the inverters' input to not be valid when the switch is closed. Also, the LED on the output of the inverter does not have a current limiting resistor.

    Bill_Marsden simply uses MSPaint and templates of various circuit elements to copy/paste together things. The advantage of that is a very "light" footprint and nothing to install. The downside is that there is no circuit simulation capability.

    I've been using Linear Technology's LTSpice for a few years. It's good and free; a quick Google search for "LTSpice download" will find it for you.
    Yahoo! Groups has LTSpice Users' Group. There are plenty of SPICE models and help available for simply creating a Yahoo! ID and requesting membership in the group.

    Circuitmaker Student is another SPICE program I use occasionally; it hasn't been supported for over a decade now, the components are rather dated, and it's limited to 50 parts in a schematic - but it can still be useful. You'll have to Google to find it.

    Cadsoft's Eagle is available in a freeware version; it doesn't support circuit simulation, but it's good for drawing schematics and making PCB's up to 3"x4". It has a rather steep learning curve, but it's quite powerful.

    That is a surface mount device. SMD's are difficult to use for protyping on breadboards. You really want a DIP (dual inline pins) package:
    http://www.mouser.com/ProductDetail...GAEpiMZZMtMa9lbYwD6ZFYFPHiRKj8UcfQsN%2bbIrPY=
     
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  6. mkgort

    Thread Starter New Member

    Jul 4, 2011
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    Lessons being learned….

    I did have a capacitor attached to pin one of the IC, but did not think it mattered for testing purposes (and was not sure that was the place to put it). Also, I forgot to put the resistor image for the led attached to the output switch.

    I removed the forward led and will now attach directly to the switch.

    Perhaps these pics are a better explanation of my mess.

    I think I will spend the rest of the evening reading “Decoupling and bypass capacitors”, trying to figure out the role “hysteresis” in my inverter, and learning how to decrypt datasheets. I have a whole lot more to learn about ICs and the way voltages affect them and their functions.

    You have me started on the right track. I can't tell you how much I appreciate it.

    Humbly,
    MK
     
  7. #12

    Expert

    Nov 30, 2010
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    The local capacitor for the Schmidt chip (in the datasheet I provided) would be attached from pin 7 to pin 14. Its purpose is to supply a quick shot of current for the moment of switching because the wires have some inductance in them. Only a few inches of wire or circuit board trace is enough to interfere with the amazing speeds that IC's can do.

    I expect you'll read all about this in the "bypass and decoupling" document.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Here is a link to Phillips/NXP's datasheet for the HEF4093B:
    http://www.nxp.com/documents/data_sheet/HEF4093B.pdf

    Just so you know, manufacturers can have other prefixes than "CD" for the 4000-series CMOS (CD was the 1st, introduced by RCA many years ago)
    HEF = Phillips/NXP
    HCF = ST Microelectronics
    MC1 = OnSemi/Motorola
    There are other prefixes as well; this is just a sample.
     
  9. mkgort

    Thread Starter New Member

    Jul 4, 2011
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    Thank you, both of you. It looks like I've a bit more learning to do, and new parts to play with. In the interim, aside from the capacitor placement, was my bread board even close to working properly? I really thought turning a "yes" to a "no" to light a led would have been simpler! :)
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    When you want to drive LEDs from logic ICs, you're best off to use transistor drivers, as logic ICs generally don't have much current source or sink capability - particularly the 74Lx and 4000 series.

    74Lx are particularly anemic at sourcing output current; without looking at a datasheet it's something like 1.8mA at 3v, with Vcc=5v. 4000 series CMOS are pretty anemic, too.

    However, transistors are pretty cheap. You should also pick up some 2222 NPN's and 2907 PNP's. PN is the plastic TO-92 case prefix for those transistors. They are good for sinking or sourcing up to ~500mA current. They are generally considered to be complementary, although not a perfect match.

    You should pick up some 1/4 Watt carbon film resistors, too - a variety from around 10 Ohms to 100k Ohms for starters.

    Here is a decade table of resistor values:
    http://www.logwell.com/tech/components/resistor_values.html
    For starters, you could just get values in the E12 columns (yellow).
     
  11. mkgort

    Thread Starter New Member

    Jul 4, 2011
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    Ah! Light-bulb moment. I just finished a chapter on transistors (many actually, I had a hard time wrapping my head around it). That makes a lot more sense than trying to finesse this inverter. Time to get a bunch of parts!

    Thanks again,

    MK
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    There's lots to learn in electronics. ;)
    Even though your photos are pretty good, it's difficult to tell what all is connected to what.

    On the bottom of the 2nd photo, there are two rows of holes between the blue and red lines. The bottom row of holes are all connected together (near the red line) and the top row of holes (near the blue line) are connected together - but there is no connection between those two rows.

    On the bottom row, it looks like you have a larger black stranded wire jammed into it at column 24, next to the cap that is labeled .1 in red. Unless you have a jumper between the bottom row and the next row up, you should be aware that there will be no electrical connection between that capacitor lead and the bottom row.

    You should avoid jamming large wires/component leads into the breadboard, as you will permanently enlarge the hole; it won't be reliable in the future with the correct size wires and component leads.

    The upper lead of the .1uF cap and the right lead of the 1k resistor are connected to pin 1.
    Pin 7 of the 'LS14 is connected to the upper row via the green wire.
    Pin 2 of the 'LS14 is connected to the 220 Ohm resistor via the purple wire, and the 220 Ohm resistor is connected to one side of the green LED, but I can't tell if it's to the anode or the cathode. From the factory, those 5mm LEDs have a longer lead on the anode, and the cathode side usually has a flat spot on the raised rim.

    It's good that you used proper-sized jumpers so there aren't big loops of jumper wires everyplace.
     
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  13. mkgort

    Thread Starter New Member

    Jul 4, 2011
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    Duly noted about the larger wires. That did not occur to me. Thanks.

    Perhaps a better explanation of my thought process is in order. The idea is that when the circuit is interrupted by tilting the mercury switch (the black wires), the 'LS14 would be inverted and output an "on" (I was going to connect it to a keyboard controller so that when the connection was broken it would send a keystroke). I put a resistor between the switch and the 'LS14 because I had seen it done in a similar project and thought that is what would set the voltage low enough to trigger the 'LS14. :rolleyes: I double checked that the led is in the proper way. Anode (long wire) to the resistor and cathode to ground. I probably still have the cap in wrong place (more reading on that today). I went a little nuts at Mouser and Jameco last night and should soon have enough parts to try your other suggestions as well.

    Again, many thanks for your time and patience.
     
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