Problem with 7490 and inputting square wave from zener

Discussion in 'General Electronics Chat' started by frankinaround, Jun 9, 2012.

  1. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    Hi. Im a newb, trying to build a digital clock. I found a website which gave me a diagram and taught me some stuff :

    http://electronics.howstuffworks.com/gadgets/clocks-watches/digital-clock4.htm

    But im having problems. Basically I made a 5 volt square wave with a zener diode (verified with my oscilloscope, which I just got yesterday). But I cant figure out how to transfer the signal to my 7490 counter.

    When I put my oscilloscope on both sides of the zener I see a 5 volt square wave. But every time I try to wire it up to the counter, when I put my oscilloscope probe between the counter output on pin 11 and the circuit ground im seeing a 28 volt sin wave like from my transformer without even going threw the zener...

    I have 1 transformer lead to a 1k resistor to reverse zener to the other transformer lead. This I verified to be 5 ohm sin wave.

    Then I have a divide by 10 decade counter wired up like it says here http://electronics.howstuffworks.com/gadgets/clocks-watches/digital-clock6.htm

    This decade counter has a plus 5 volt dc input from a different powersupply I made, and everything that says ground (from that last link) is connected to the dc output of that 5 volt input.

    How do I connect these 2 circuits properly to get the 5 volt square wave into the counter ?
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,783
    2,581
    This post had been automoderated (rendered invisible) due to the links. I have approved it.
     
  3. absf

    Senior Member

    Dec 29, 2010
    1,496
    375
    On both sides of the 7805, there are 2 capacitors with polarities. What are their values ?

    If you put your scope probes across the output of 7805, The +5V side, what is the waveform did you see? Was it a straight horizontal line?

    Allen
     
  4. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    Is your 5V power supply completely separate from the transformer with the zener?

    It sounds like you need to connect the bottom of the zener (the end that's connected to the transformer and not to the 1K resistor) to the ground of your 5V power supply. Then the pulse signal voltage at the top of your zener will be referenced to your 5V supply's ground level.
     
  5. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    thank you so much for replys.
    I think you're saying this: connect it from ac terminal to resister to the zener to the other ac terminal and ground of plus 5 volts.

    okay I will try tonight and post results.

    also I am not using 7805 I am using a different diy power supply
     
  6. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    I mean yes they are separate, the plus 5 and ac come from separate transformers.
     
  7. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    Ok so now the oscilliscope shows the 60hz signal from the chip input to ground. The chip is a 7490apc set up for divide by 10.

    But the oscilliscope on the output shows a flat line when positive probe to output and neg probe to ground, but with just positive probe attached no negative probe im seeing a sin wave at 60hz. Was expecting to see a 6hz wave. Any ideas? I tried looking up the datasheet, and i think i found it but for some reason the pdf i got wont open on my stupid nook...

    I have pin 12 connected to pin 1. Pin 2,3,6,7,10 go to ground. Input on pin 14, output on 11 and +5v on pin5
     
  8. BillB3857

    AAC Fanatic!

    Feb 28, 2009
    2,410
    349
    With the scope ground lead not connected to anything, you would see a 60HZ signal. Try just touching the scope probe tip with your finger and see what you get. Signal measurements almost always require a two wire connection of some kind.
     
  9. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    With positive to the output and the negative on ground its showing a flat line. When i touch the tip the frequency changes alot thats all
     
  10. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    Also im not a pro with an oscilliscope either. Its all x1 and i usually just press autoset and it works.
     
  11. BillB3857

    AAC Fanatic!

    Feb 28, 2009
    2,410
    349
    When you touch the tip of the probe, your body is acting as an antenna and feeding 60 Hz and any other noise that is around so you see a very ragged looking signal with the base sine wave being the 60 Hz. It should be 16.6 milliseconds per cycle. When you look at any point of your circuit with the scope ground wire not connected, you will see pretty much the same as just touching the probe.
     
  12. absf

    Senior Member

    Dec 29, 2010
    1,496
    375
    If everything is connected like the one in the attached file. You should be getting a 6 Hz square wave like mine...

    Allen
     
  13. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    Is vcc powering the chip? Also is pin 10 grounded here? Im going to try maybe dumping external and wiring exactly like this to see if it will work tonight. Thanks man
     
  14. absf

    Senior Member

    Dec 29, 2010
    1,496
    375
    Yes Vcc is powering the 7490 and pin 10 (Vss) should be grounded. In my simulator, the power supply pins are hidden.

    Allen
     
  15. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    Where is the 60hz from the zener going? One goes to the pin 14 input pidiagram n the other goes to where? In the diagram its going somewhere... but im not sure where. Also when i test the signal should my positive probe be on the output pin and negative probe to ground?
     
  16. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The extremely old 7490 IC needs a lot of logic-low input current that the 1k resistor cannot provide. The maximum input-low current is 1.6mA at 0.4V to 0.8V.
     
  17. absf

    Senior Member

    Dec 29, 2010
    1,496
    375
    The other one goes to the oscilloscope input for taking the wave form. You dont need to connectit if you are not measuring it, just go to pin 14 of 7490 is fine.

    Are you using 7490 or 74LS90? Or are there any 74HC90 around?

    Allen
     
  18. frankinaround

    Thread Starter New Member

    May 14, 2012
    15
    0
    Sry for late reply. It says 7490apc . Couldnt find datasheet. I trued measuring between pin 14 and 11 but no good.
     
  19. dataman19

    Member

    Dec 26, 2009
    136
    29
    So,
    Isn't the 1K resistor connected to one lead of the AC?
    Isn't this to just sample the 60Hz sine wave (as stated in the project text)?
    So the Zener is just clipping the AC Sinewave, not just providing a square wave.
    Would not the installation of a simple 1N914 or 1N4148 signal diode in-line with the 7490 clock input result in proper clocking? (eliminating the reverse voltage and providing a pulse more in-line with what the TTL clock signal required? Or would this just result in a 30Hz clocking signal?).
    ..
    The 7490 clocking isn't all that tedious, and it can be forgiving, but if the Zener isn't clipping enough, the AC will swing both ways anyway and possibly cause clocking errors. Since the TTL clocking is supposed to be a 5-Volt Clock signal 1/2 of the 10VAC would put it close (which again I would think that the 5.1V Zener is clipping it to keep it within the specs for TTL).
    ..
    So wouldn't it clock just fine with an AC clock signal anyway?
    ...
    Just my thoughts,
    ..
    Dave
    Phoenix, AZ
     
Loading...