Binary adder with 7-seg display

Discussion in 'The Projects Forum' started by pillyg, Sep 19, 2011.

  1. pillyg

    Thread Starter Member

    Sep 19, 2011
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    Ok sounds good. DO I need 12volts? I was planning to use 5. I will probably need to debounce 3 switches. 2 inputs and reset.

    I am going to use a transformer I found at radioshack for my power supply. It ouputs 6 and 12 volts. I will use a 7805 voltage regulator to drop that to 5. Should I get a pos. voltage reg or a neg. regulator. What is the difference?
     
  2. crutschow

    Expert

    Mar 14, 2008
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    5V is fine.

    If you use a transformer you also need two rectifiers to make a full-wave rectifier and a couple filter capacitors as well as a 7805 positive regulator.

    It would be a lot easier just to buy a wall-wort that outputs 5VDC (make sure it's has a regulated output).
     
  3. pillyg

    Thread Starter Member

    Sep 19, 2011
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    I was talking about elec's debouncer schematic. He used 12v. I guess I will use a 5v wall-wort. Should I get a voltage regulator? If so positive or negative?
     
  4. elec_mech

    Senior Member

    Nov 12, 2008
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    William,

    Again, I advise you hold off buying or specifying your parts until you have a schematic finished.

    I *think* you'll be able to power the whole circuit with 12VDC using a wallwart, without the need for a regulator which will save you parts, assembly time, and probably some money. This will change the values of your current-limiting resistors for the display, but, again, that is why a schematic is so important. If you use nothing but CMOS ICs (CD4XXX), you should be able to use anything from 5-15VDC, but you'll need to double-check the datasheet for each IC.

    When you do buy parts for a power supply (if you end up going that route), you'll want a positive regulator. Negative regulators are often used for circuits requiring both a positive and negative voltage, such as many operational amplifiers. For most circuits, you'll only need a positive voltage.
     
  5. pillyg

    Thread Starter Member

    Sep 19, 2011
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    I Will go through and check all the parts to make sure they will work with 12v DC. The 40110 will work, but the OR gate won't. Not sure about the displays though, couldn't find it. I started the schematic
     
  6. pillyg

    Thread Starter Member

    Sep 19, 2011
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    I finished the schematic for the power and debounce. I would like to use a voltage regulator because I cant find any wall-worts on futurlec (maybe im in the wrong spot). Capacitors only cost like $.05 so if I have the right thing, it should only add like $.40
     
  7. elec_mech

    Senior Member

    Nov 12, 2008
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    Looking good so far.

    Remember to tie any unused inputs to ground - pins 1 & 2 of CD4093 should be tied to ground if you're not going to use them.

    If you use a voltage regulator you still need to supply DC voltage to it. What are you planning to use to supply power to the regulator?

    If you have any old wallwarts from electronics you, your family, or your friends aren't using anymore, you should be able to use one of them for your power supply (with or without the regulator). These can come from cordless phones, cellphone chargers, etc. I have a sizeable collection I've gathered over the years. Most cordless phone power supplies are 9VDC and about 200mA which is probably just enough for your project. Cellphone chargers are usually 5VDC (no need for regulator!) and sometimes as high as 2A output. Ask your friends or family if they have any old electronics they no longer use and offer to "recycle" it for them - if they don't have any now, ask them to save any for you in the future instead of tossing them in the trash. You can also hack the electronics themselves for parts. Great way to build your electronics supply for free - assuming you want a supply for future projects.

    If you can't get power supply this way, Jameco, All Electronics, and Marlin P Jones offer wallwarts at decent prices:

    https://www.jameco.com/webapp/wcs/s...me=Power Supplies & Wall Adapters&category=45

    http://www.allelectronics.com/make-a-store/category/480/Power-Supplies/1.html

    http://www.mpja.com/products.asp?dept=37

    Unregulated will be cheaper, but you may need to play with LED resistor values since the voltage changes depending on the current draw. Regulated will keep things easier, but they cost more. If it doesn't say so in the description, assume the supply is unregulated.

    I'll try to get to explaining the auto counter later today or tomorrow, busy week!
     
  8. crutschow

    Expert

    Mar 14, 2008
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    If you use a CD4xxx family OR gate it will work at 15V.
     
  9. pillyg

    Thread Starter Member

    Sep 19, 2011
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    I have a wall-wort, but it isn't 5v. The regulator will make any wort I add to it 5v.

    I will start working on the input displays now.

    William
     
  10. elec_mech

    Senior Member

    Nov 12, 2008
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    What is the output of the wallwart? Please include voltage, current, and verify it is DC output (some wallwarts output AC).
     
  11. pillyg

    Thread Starter Member

    Sep 19, 2011
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    The one I will probably use is 9v 400ma DC but does it matter if I am using a regulator? My friend has one that is 5v DC (dont know amps) that I might use also.

    Cant I use any wall-wort as long as it doesn't fry the regulator?
     
  12. elec_mech

    Senior Member

    Nov 12, 2008
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    No. You need to look at the datasheet of the regulator. For 78XX series regulators, they generally need an input of 2V or more higher than the output voltage. A quick look-up of the 7805 from one manufacturer states it needs at least 7.2VDC to provide 5VDC reliably. So, you could use the 9V adapter, but not the 5V.

    If the 5V adapter is regulated, then you can skip a 5V regulator. Check this by plugging in the 5V wallwart and measure the voltage output with a voltmeter. If the voltage is fairly close to 5V (4.9-5.1V), then it is regulated. If it is much higher, say 6V or more, then it is unregulated.

    If you stick with CMOS ICs, CD4XXX, you should be fine with the 9VDC wallwart without the use of a regulator. You'll still want a good electrolytic capacitor (say 47 to 100uF) at the power input terminals and 0.1uF ceramic capacitors across Vcc and GND at each of the ICs. The big circuit I provided earlier was powered using a 12VDC unregulated wallwart. There was a 3VDC regulator on it to power a remote circuit that only operated off of 3V, but all the circuit itself didn't need a regulator. If you're going to use TTL ICs, 74XXX, then you'll need a 5VDC regulator. You can make this circuit entirely with CMOS though and avoid the need for a regulator, your choice.
     
  13. elec_mech

    Senior Member

    Nov 12, 2008
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    As promised . . .

    This circuit will work for one of the two inputs. You can use it, twice, for both inputs. You can save some parts by tying the auto-counter outputs (555 circuits) to both sets of 40110s and adding a slider switch so the user can select between entering the first and second value using the same up and down buttons - if you choose. I can draw it out to better explain if needed.

    Look at it, study it, and let me know if you have any questions.
     
  14. pillyg

    Thread Starter Member

    Sep 19, 2011
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    Alright. I see how pretty much everything works (not why for some parts but oh well)

    Is the switch debounce output inputted into pin 4 on the 555?

    I will use my 9v wall wort and use a voltage regulator. What should the capacitors on each side be? What I meant by any wall wort was anything about the minimum.

    I need a capacitor for voltage and another one for ground for each IC?

    Also, what program do you use for schematics?


    William
     
  15. crutschow

    Expert

    Mar 14, 2008
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    You need one capacitor per IC connected directly from power to ground.
     
  16. elec_mech

    Senior Member

    Nov 12, 2008
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    Ditto. 0.1uF across the Vcc and ground connnections, or at least close as you get them, for each IC should do it.

    Yes. Reset is held low by the 100kΩ pull-down resistor when there is no high signal from the debounce circuit. This forces the 555 to be held in reset mode effectively keeping it off until a high signal is sent.

    Look at the datasheet for the regulator you plan to use. It should give you a range of recommended values, usually with a simple wiring diagram.

    I use Microsoft Visio to draw my schematics. You can also use Microsoft Paint (free) - Bill uses this for all of his schematics and he does way more than I. Using Visio or Paint will require you to draw your own parts such as resistors, ICs, etc. If you don't want to fuss with making your own parts, try out the free schematic capture (drawing program) offered by ExpressPCB: http://www.expresspcb.com/ExpressPCBHtm/Free_schematic_software.htm.
     
  17. pillyg

    Thread Starter Member

    Sep 19, 2011
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    So I would need 22 0.1uf capacitors. I would have 11 ICs with 2 for each. 7 BCD counters/drivers 1 OR gate 1 NAND gate and 2 555 timers.

    The voltage reglator data sheet shows I need a .22uf capacitor on the input side and a .1uf one on the output side also connected to ground.


    I think I know how to do everything now and I just need to make a arts list. I know how to get the power, increment a counter that will also increment a 2 digit display, how to debounce a circuit, and how to add the 2 incremented numbers and display them on a display.
     
  18. elec_mech

    Senior Member

    Nov 12, 2008
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    Nope - you only need one 0.1uF capacitor per logic IC. You place the capacitor BETWEEN Vcc and ground of each IC, not one on both. In a perfect world, you'd place it literally across the Vcc and ground pins, but this isn't practical. I usually try to physically put one lead of the capacitor at Vcc and connect the other lead to the closest ground connection I have in the circuit.

    Post your schematic once it is drawn and we can help you check it for errors and make suggestions for improvements before you order parts or build it. I know the first impulse is to buy a bunch of stuff and put it together, but posting a schematic will help save you time (and possible damage to components) in the long run.
     
  19. pillyg

    Thread Starter Member

    Sep 19, 2011
    52
    0
    Ok I think I understand. I put a capacitor between the voltage pin on the IC and the ground in my circuit. I do nothing with the ground pin on the chip. That means I only need 11 capacitors then.
     
  20. elec_mech

    Senior Member

    Nov 12, 2008
    1,501
    196
    You got it.
     
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