1 Second Clock with 5" LED Displays Circuit Evaluation

Discussion in 'General Electronics Chat' started by CoachKalk, Oct 15, 2011.

  1. CoachKalk

    Thread Starter Member

    Sep 20, 2011
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    The second major (for me anyway) part of my project is using Bill's 1 second clock circuit to feed a counter/5" LED display. This is version 1 that I made from combining Bill's "second" portion and the 4553/4543 with 5" display portion.

    I have a few questions to start with and then I would appreciate some expert eyes looking over the setup to find my errors.

    [​IMG]

    I left off several features that I think Bill had for testing purposes. Of course I could be wrong, so I want to list them and get verification one way or the other.

    1. 4060 Pin 9 - Did not include the 32768Hz test point - should this and all unused pin go to ground? Good left alone?

    2. The 2222/Resistor/Red LED was not included coming from Pins 5 and 2 on the 4013. I just go straight to the 4553. Or is the Vcc associated with the Red LED required for the 4013 and the red LED verifies it is powered up? Seems like it should be obvious, but I am running myself in circles ...

    3. Someone had already answered my question about the bypass capacitors noted to the side on Bill's circuit - I just added the 1 to the 4060. Should the 4013 have 1 as well? Where would it be placed?

    4. I left off all details related to the Quad 4081 used by Bill - I think it was only used for minutes/hours.

    5. I tried to add in the ULN2803 suggested by elec mech. Please take a look at that area in particular. I am not sure whether I leave the 8th unused input/output as is or is it generally expected to ground ALL unused pins? Based on info from elec mech, the 5" LED's require approx. 17.6 VDC per seg. I used that figure for the LED's, 1.6 V (drop across 2803 - or should I use 1.1V?) and .6V across the 3906. Based on these numbers, I think the 20V wall wart is too close to the requirement - 19.8V. Please let me know if my figures are whacked. I know I mentioned I also found a 24V wall wart, but the 20V is packaged so much better so if at all possible it would be my first choice.

    [​IMG]

    Thanks!
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You will break the counter chip, as you have a powersupply of 9 Volts and a display voltage of 24 Volts.
    The absolute maximum voltage of the counter chip is 18 Volts (see datasheet).
    If you want to drive the 24 Volts displays, you will need a level translator.

    Bertus
     
  3. elec_mech

    Senior Member

    Nov 12, 2008
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    Yes and no. CD4060 is good as you've drawn it - no need to add or change anything. You only want to tie INPUT pins to ground, NOT output pins. Pins 7, 8, & 10 of CD4013 are all unused input pins and should be tied to ground.

    No need for the LED, it is only used as a visual indicator that the clock is working if you want to "see" it. You can add it in your breadboard/trial circuit as a debugging tool if you don't have an oscilloscope.

    Yes, all CMOS ICs should have a 0.1uF (ceramic or Mylar) capacitor between Vcc and ground, at least as close as you can get. So, include one each for the 4060, 4013, 4553, and 4543. No need for any on the ULN2803 - that is a transistor array in effect, not a logic IC.

    What you have for the 4060 and 4013 is fine. I'm using that same portion in a circuit I'm working on right now and it's working beautifully.

    No - the ULN2803 is not a logic IC, more of a transistor array. No need to connect anything to the unused pins on it. Won't hurt on the input side, but definitely don't want to connect anything to the output side lest you risk a short. I'd just leave any unused pins on the 2803 unconnected.

    Here's the fun part about designing a circuit and building one to work - you'll need to experiment a little to figure out what the actual voltage drops will be across the 3906s, 2803s and the 7-segments. Your calculated values look good. Depending on your power supply, you'll see different voltage drops and current consumption. I'd give the 20V supply a shot. Worst thing to happen is the display is very dim or doesn't come on at all - you won't hurt anything.

    I'm not sure what Bertus is referring to, but certainly take heed in case I'm missing something. If you're putting the 20V+ supply through a 9 or 12V regulator to feed Vcc to all of your logic IC's as shown, you should be okay. If you're using another 9V supply to power the logic ICs, just be sure the ground from the 9V is tied to the ground from the 20V+ supply.

    Lastly, is that an SCR at the top? What is it being used for, to latch the input on? Is this being used to save power?
     
  4. #12

    Expert

    Nov 30, 2010
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    I noticed the scr. They have a minimum "on" current or they won't hold in the on condition. You didn't fill in all the +9 supply connections, so, at first I thought it was only supplying the 4060 chip which has an idle current of 10ua.

    You might want to double check that the scr has enough load to keep it latched on.
     
  5. CoachKalk

    Thread Starter Member

    Sep 20, 2011
    139
    2
    Thank you - I will have all of the datasheets handy when the components come in and I am ready to start - Mental Note Taken - Ground All Unused Inputs

    I have the 4060 and 4543 taken care of. To be honest, I just copied the 4553 and 4013 from other circuits. Now I realize how incomplete the circuit looks. The 4553 Datasheet has Pin 16 as the Vdd pin, but on all examples it doesn't even show where it is. Would the 4013 be similiar? The Vdd pin not drawn because everyone BUT complete NOOBS (me) should know to include it? I will def. add the necessary 0.1uF capacitors once I locate the stupid supply pins!

    I am thinking my incomplete drawing is causing confusion. First, about the SCR. For purely cosmetic reasons, the clock will be started and stopped with large mom. pushbuttons. After reading up on SCR's. I figured I could use one to make the pushbuttons work the clock. The start button trigger the SCR - which hopefully stays latched until the Stop pushbutton kills power. I also have a pushbutton (NO) on the 4553 reset pin. Hopefully what I am about to say doesn't impact the feasibility of the circuit/components, but while most of my circuit was taken from Bill's 1-second clock project, my setup will not run more than 120 seconds (rough guess) before being stopped and being reset. It is simply a way to time how long it takes my kids to get through the lights and back.

    As far as powering the entire circuit, I had been thinking about/planning to have one 9V source for all of the 9V requirements and a separate, higher supply (would be nice if the 20V works) specifically for the 5" displays. BUT, I must say I hadn't thought of using 1, higher supply then use a regulator to get down to the 9V requirements where needed. I would be interested in your opinions regarding the pros/cons of each option. Is it a matter of what someone has handy or is there significant trade offs?

    Make no mistake, I freaked out a bit when I read Bertus's comment. Hopefully he will stop back in so I can know for sure if trouble is around the corner.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    May I suggest a slight change in the schematic. I would add a toggle 555 circuit to turn of the clock instead of the power supply, if you are interested I will draw it up.
     
  7. CoachKalk

    Thread Starter Member

    Sep 20, 2011
    139
    2
    @Bill - I am total interested in your suggestion and help. To be honest, I am not sure what you have in mind, but would it still allow the kids to start and stop the clock from pushbuttons?

    I will just be patient and wait for what you have in mind ...

    Thanks
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    After I wrote the last post I realized you had a spare gate (a flip flop), so I used it instead. You can also wire that same flip flop where there is only one button, push once it starts, push again it stops. I'll draw that variation if you're interested.

    [​IMG]

    If this looks familiar it is easier to modify an existing schematic than to draw from scratch.
     
  9. CoachKalk

    Thread Starter Member

    Sep 20, 2011
    139
    2
    Bill,
    Thank you for your help. I like the idea of 2 different buttons (green - go, red - stop) so I used the setup you drew up - as is. I think I redrew it correctly, but please take a look to make sure I didn't miss anything.

    I would appreciate any insight you may have on the current design regarding the high voltage required for the 5" displays. If you scan up in the thread, you will see some discussion whether this setup would work or fry the 4543. Also, I am planning to use 1 supply for all the IC's and 1 separate supply to the 5" displays. Is there any benefit/harm if I use 1 24V supply and regulate down to 9V for the IC's?

    Thanks again for your help.

    [​IMG]
     
  10. elec_mech

    Senior Member

    Nov 12, 2008
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    I'll let Bill handle the switching circuit - I'm sure I could come up with something but he's already got some ideas and more experience than I.

    Regarding power, I don't see the harm in either powering the logic with a separate supply or using the same supply and using a regulator to drop the voltage.

    If you opt for the latter, I recommend looking at the datasheets for each IC and seeing what the highest operating voltage is (this should be a little below the maximum voltage input stated close to the top of the datasheet). From there, select a regulator that will allow you the highest voltage within reason. Why? Because the more you drop the voltage, the more power you waste as heat and greater the need will arise to add a heat sink to the regulator. And if you need a heat sink, you'll want to be sure there is some air flow available to help remove said heat which means you may need to drill some holes in your enclosure (assuming you put this circuit in an enclosure). In short, heat is bad and you can avoid or minimize this by making sure you drop the voltage as little as needed.

    So, using a 20-24VDC supply means you can use a standard regulator as low as 18VDC for a 20VDC supply I believe (will need to check the datasheet on the regulator). I know many CMOS ICs will accept up to 15VDC and 15VDC regulators are pretty easy to get. I know 18VDC is another common regulator, but you'll need to check to see what the highest normal (not maximum) operating voltage is allowed for each and every IC.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Oh, boy - I didn't notice the higher voltage for the displays before. If you tried building that, all three displays would be on at the same time, and you wouldn't be able to read them.

    The PNP transistors on the DS1-DS3 outputs of the 4553 won't get turned off, as DS1-DS3 are not open-drain outputs; they're active outputs.

    This needs some thought.... Oh, and how much current are you planning on running through each segment? The 2N3906 transistors have a practical limit of ~100mA; after that the Vce gets excessive. 2N4403 or 2N2907 would be a better choice.
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    Coach, you have a minor error with your schematic on another front. If you look on my schematic you will note that R3 and R4 connect the inputs of the flip flop to ground when the buttons are not pressed. CMOS must be connected to something, it does not assume a logic state (not true for every logic family, but true for CMOS). You have left your 4553 MR input open, which I believe is an illegal condition. I have not looked at the data sheets yet, but it is a fair bet.

    [​IMG]

    The above schematic shows what I am talking about.

    You also need to look at the datasheets of these two chips to see if there are other inputs you may have missed.
     
  13. elec_mech

    Senior Member

    Nov 12, 2008
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    SgtWookie, why wouldn't the PNPs get turned off exactly? Transistors are still an elusive subject for me. Two of the DS outputs should be high at all times. Is it the fact that the base voltage provided by the DS outputs (currently about 9V) is much lower than the voltage being controlled by the transistor (20-24V)?

    If yes, would it work if he increased the logic voltage to 15VDC or is the problem still there? Is there a way to use a transistor or pair of transistors to control a large voltage with a smaller one?

    Would a PNP MOSFET be better suited to this application? If yes, are there any common ones you would recommend?

    Regarding current consumption, I see your point. If CoachKalk keeps to the ratings of the displays, he's probably going to want 20-25mA per segment so it's nice and bright, drawing 140-175mA through each transistor which comes close to the max ratings of the 2N3906.

    CoachKalk, this is why I recommend waiting to order parts until the design is complete. If you already have them on order, RadioShack should carry larger transistors/MOSFETs if they are needed.
     
  14. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, that is it.
    If the DSx output is at 9v maximum, and the emitter of the PNP transistor is at 24v, that is a difference of 15v. In order to turn off that PNP transistor, the base has to be at 23.5v (Ve - 0.5v) or higher.

    The problem remains. Vbe would still be 24-15=-9v; Vbe needs to be -0.5 or less.
    Yes.

    Don't confuse your bjt terminology with MOSFET terminology.
    You don't call them PNP MOSFETs, as people won't know what you are talking about.

    You have P-ch and N-ch MOSFETs
    You have PNP and NPN bjt's (bipolar junction transistors)
    A P-ch MOSFET would have the same difficulty, Vgs would have to be below the MOSFET threshold in order to be considered turned off.

    Actually, I wouldn't push them more than 100mA. There's no reason to, as either of the other transistors (and many more) would be perfectly suited. The question now is how about the sink current of the 4543...

    Radio Shack carries 15-piece assortments of PNP and NPN transistors for a couple of bucks each pack.
    The PNP usually contains:
    5 - 2N2907
    5 - 2N4403
    5 - 2N3906
    The NPN usually contains:
    5 - 2N2222
    5 - 2N4401
    5 - 2N3904

    In MOSFETs, they only carry the IRF510, which is an N-ch standard MOSFET; Vgs needs to be 10v in order for it to be considered fully ON.
     
  15. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    I already mentioned that a circuit is needed to drive the dispaly, as the DS output is not working with the connected transistor.
    Perhaps a circuit like this will help:

    [​IMG]

    Bertus
     
  16. elec_mech

    Senior Member

    Nov 12, 2008
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    Doh! :eek: I knew that (deep down), but wasn't thinking when I typed that up.

    Is it a general rule to use only half or so of a transistor's rated current, much like a resistor's power rating, or just so for the 3906?

    So, even if the transistors are switched out, the display won't work as the design stands now, correct?

    Bertus, I did not know what you were referring to earlier. So, you were saying the voltage of the ICs would have to match that of the displays in order to operate the transistors which in turn would have destroyed the ICs, correct?

    I'm not familar with Bertus's circuit. Could I ask someone to elaborate? How will this work if the base voltage has to be no lower than 0.5V of Ve?

    Aside from that, if the DS output sends a high signal (for off, since two of the DSs will always be high), then the NPN transistor conducts which pulls the base of the PNP low, thereby allowing it to conduct. Wouldn't this cause two of the digits to always be on and the digit that should be displayed forced off?

    If yes, could this work switching out the NPN transistor for a PNP?
     
  17. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    In the drawn circuit the signals are as followed.

    If the input at DSx is low, the NPN is not driven.
    The output current of the PNP is nihil,as the input is pulled high via the 10K.

    If the input of DSx is high, the NPN is driven and pulls the voltages low via the 2k2 at the collector.
    The output current of the PNP is flowing, as the 2k2 is putting current through the bas of the PNP.

    Bertus
     
  18. elec_mech

    Senior Member

    Nov 12, 2008
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    Bertus, thank you, that makes sense. I'm still not clear on why this works to control a higher voltage, but I'll learn with time.

    In any event, don't we want the PNP to conduct power to the display when DS output is low? DS are active low, correct?

    If yes, then two of the DS outputs are high at any given time while one is low (the one being displayed). This circuit, as I understand it, will cause the opposite effect - two displays will be turned on simulateously while the active one is turned off.

    Will this circuit work for this application if we change out the NPN transistor for a PNP one? This, in my limited experience with transistors, will allow the active low DS output to turn on the PNP going to the display and force the other two DS outputs, which are in high states, to be turned off.
     
  19. SgtWookie

    Expert

    Jul 17, 2007
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    Bertus, your idea won't work, as the display logic would be inverted; that is, the PNP transistors would source current at the same time. [eta] When display 2 was supposed to be on, displays 1 and 3 would be on instead. When display 3 was supposed to be on, displays 1 and 2 would be on instead; etc.

    I think the easiest fix is going to be to use some 12v & 5.1v Zener diodes in series with a current limiting resistor, along with a base return resistor for each PNP.

    See the attached for a schematic of what I'm talking about. ZD1, ZD2, R1 and R2 need to be used as shown. I chose those values because I know that Radio Shack carries them; but if one wanted to send away for them, some 16v Zeners would work.
     
    Last edited: Oct 18, 2011
  20. CoachKalk

    Thread Starter Member

    Sep 20, 2011
    139
    2
    First - All of you guys are AMAZING! My head is spinning and you guys are bouncing ideas off each other like its playing cards! Insane I say! I really appreciate all of the time all of you are spending getting me through this. As usual, I will update my circuit and throw it back up for you guys to take a look at.

    Who would be willing to explain the zener diode config for me? I had already read, and just reread the zener diode portion on this site, but SgtWookie has me all turned around (not difficult to do).

    If not reversed biased, the zener acts like a reg. diode (-.7V) across. When reversed biased, assuming the supply voltage exceeds the zener voltage, the zener only allows the zener voltage through. So, in SgtWookies circuit, the 5.1 zener comes before the 12 zener correct? And they are both reverse biased based on the 24V supply. So wouldn't the 5.1 reduce the voltage that the 12 sees down to 5.1V, therefore causing the 12V zener to act like a regular diode (-.7V)? Am I all horsed up or is that the desired effect?

    While I would appreciate some basic, NOOB level explaination, I would also understand a "Just shut up and build it!" anwser too!
     
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