Anyone Built This CD4033 Counter Circuit?

Discussion in 'The Projects Forum' started by JDR04, Jun 18, 2011.

  1. JDR04

    Thread Starter Active Member

    May 5, 2011
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    Newbie struggling to get this circuit to work.

    www.circuitstoday.com/7-segment-counter-circuit

    Has anybody out there built this circuit and made it work? I've got the 555 part to pulse but just cannot get the CD4033 to do its thing....... I've read sometimes people publish circuits with mistakes on purpose etc. It's a bit frustrating for us newbies. Any help or comments out there??
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    What 7-sergment display are you using? The one in the example is a common cathode. You may be using a common anode 7-segment display.

    hgmjr
     
  3. bertus

    Administrator

    Apr 5, 2008
    15,648
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    Hello,

    As far as I know the 4026 and 4033 are not capable of driving led displays directly.
    In the datasheet the following circuits are given to drive led displays:

    [​IMG]

    I have attached the complete datasheet too.

    Bertus
     
  4. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    As far as I know, I'm using a common cathode display (RS Part number 235-8783) I'd be interested to know how one can physically check for the difference. Anyone out there know?
     
  5. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks Bertus. I checked out the datasheet but as a newbie I dont really undestand it all. So, is the schematic I posted wrong indeed and I've wasted a lot of time and money?? From looking at the datasheet am I correct in saying that each segment from A to G needs a tranny connected to it which is switched on by the 4033 output on each pin. If that is so, then I'm really stumped again as I tried measuring for a voltage on each output pin and there was nothing at all!!!! Have I messed up somewhere??? Thanks for your help.
     
  6. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    This circuit is driving me mad. I've checked that it is wired according to the schematic and cannot see what is wrong? Is this really a dud one and I'm wasting my time?

    One thing puzzles me. I've measured a pulse of about 3Vdc to pin 1 on the CD4033 IC (Clock In). I've measured a supply voltage of 5Vdc to pin 16. What's puzzling me is that there is absolutely no output on the pins connected to segments a,b,c,d,e,f and g.

    The other question I would like to ask somebody out there is how I can check if the CD4033 IC is OK. I dont have a fancy scope yet only a Fluke 179 meter. Can this be used to perform a check on the IC in question?

    Should I rather move on and write it off to my inexperience? :confused:

    Thanks guys, JDR04
     
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  7. SgtWookie

    Expert

    Jul 17, 2007
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    A few things offhand:
    1) The original schematic leaves Reset, pin 15 of the 4033 unconnected. Connect it to ground, or it may be causing continual resets. Make certain that it is not shorted to pin 16, which is Vdd.

    2) The stated supply voltage is 5v, and then there is a 1N4001 diode which drops that down to 4.3v - which really is not enough. You would get better results if it were operating from 6v to 8v.

    3) A standard bjt (transistorized) 555 timer has a Darlington follower output, which causes the output to be about 1.3v lower than it's supply voltage. The schematic shows a 220k resistor from the 555 pin 3 to ground; however that really isn't doing anything but loading the 555 output. Change it to 470 Ohms, disconnect the ground end and connect it to the + voltage supply; this will make the 555 output go much higher, making a more reliable clock signal for the 4033.

    4) If the outputs of the 4033 are always near 0v, then something else is wrong. Are you certain that pin 16 is connected to your + supply voltage?

    5) CMOS parts are very susceptible to damage by static electricity. If you are in an arid climate, you may have accidentally "zapped" the counter. If you can get a shock by walking across the room and touching a doorknob, you have a static electricity problem. It takes about 3,000 volts to be able to feel a static shock, but only about 30 volts to destroy a CMOS IC.
     
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  8. Wendy

    Moderator

    Mar 24, 2008
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    I wouldn't write it off, wiring errors can happen to the best of us, and I do mean everyone.

    You've done me a favor, I was starting an article on making a digital clock, a project for Volume 6 on the eBook out there, the 4026 or 4033 seem ideal for this application.

    There is an article on ESD in the same book (I wrote that chapter) you might want to review. Truth, I don't always follow ESD protocols myself, even though I know better. You really can blow a chip without knowing it.

    Chapter 9: PRACTICAL ANALOG SEMICONDUCTOR CIRCUITS
    So here is my question, why are you using the 4033 over the 4026? Just because the article you are referring to? Even with experience I find I'm always learning new things, so it doesn't hurt to ask the question.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    There's another problem with the original schematic; although it wouldn't keep it from displaying anything, they used just a single current limiting resistor on the cathode of the LED display. This will cause the display intensity to vary all over the place depending upon how many segments are lit.

    Each line (a-g) needs its' own current limiting resistor. As Bertus mentioned, the current source ability of the 4033 is quite limited, particularly at the low voltage that circuit is operating with.
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    We see that a lot. This chip resembles the 4017, it has the same current output specs. It is rated for 6.8ma, same as the 4017.

    I just spotted yet something else with this schematic. A diode in the power supply will drop the voltage below the rated voltage of the 555. It will work, but who ever designed this wasn't using their head.

    I plan on using 12VDC to power the chip, your circuit could also use either 9VDC (battery) or 12VDC. The more voltage feed into power supply the more current you can get out. For 12VDC it would be a 1.8KΩ resistor per display segment for a constant LED output level. Thank god for Washington University, this isn't the first datasheet they've provided.

    http://www.ee.washington.edu/stores/DataSheets/cd4000/cd4033.pdf

    Several questions...

    Are you sure the 555 circuit is working? You won't be able to see the pulse even if it is there, because it will be so narrow. This works for the application, but it is a question worth asking.

    Are you using a protoboard? It is probably the best method to go for this project. If you check out my 555 series of projects you will see how I like to wire them.

    Bill's Index

    The 555 Projects

    My Cookbook

    LEDs, 555s, Flashers, and Light Chasers
     
  11. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The schematic has many things wrong.

    Datasheets of Texas Instruments Cmos B-series logic have detailed graphs of output current.
    With a 5V supply the typical output current into a dead short is 4mA and is a minimum of only 2mA.
    Into a 2V red LED with no current-limiting resistor the current is typically only 3.5mA or a minimum of only 1.5mA.
    This circuit has a supply of only 4.3V so the currents are less.
     
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  12. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Hi Bill, thanks for your informative reply and I will check out the info you have posted. I only went for the 4033 because that was specified in the schematic. I have however changed everything and am now using a 4026 circuit. It worked first time!! I'm using a 555 clock with 3 x 4026 chips all powered off 5Vdc. I eventually want to build it up to 6 displays with an end objective of using it to test a pesky switch reliability problem.

    One thing puzzles me though, will I have to use some sort of a one shot circuit when I finally use it to check the switch? Does anybody have a proven one shot circuit for me please? Thanks for your help folks.

    I'd be in interested in any suggestions of a more accurate clock circuit if you have any.
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    Heh, the 555 is a monostable, that is its first function. It is why it is called a timer.

    555 Monostable

    I have a clock circuit under development in the Feedback and Suggestions forum, it is going to wind up in the AAC book if I get it working. The crystal oscillator is untried as of yet, but it will be. I give it a very high probability of working.

    Digital clocks
     
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