595 Shift Register Issue

Discussion in 'The Projects Forum' started by mxabeles, May 25, 2009.

  1. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    Hi,
    I am using two serial in parallel out shift registers. I am using 2 555 timers to clock in the latch and clock. Data is supplied from my microcontroller.
    It was working fine on day one, but today (day 2) the shift registers ONLY work if I touch them (which is probably not a good idea).
    I have been sniffin around the circuit with a multimeter. I am using a 7805 voltage reg. to power both 555s and both 595s. ALL grounds are tied (ground from micro, ground from 7805, and ground from 12v DC [used for high power lights that 595s are controlling and also used as input to 7805]).
    I mention all this because when I look at the voltage from the 7805, the input only reads around 1 volt and the output is even lower (0.5 volts).
    Why would this be? I thought maybe using a second power source just for the regulator would solve the issue, but w/ a seperate 9 volt still the same issue (but everything WAS working w/ this configuration yesterday).
    I am guessing the logic levels are too low for the 595, especially because output of the timers is even lower then their input voltage.
    So, it DOES work when i touch them, but that is not really feasible. I didn't use a socket for them, so i can't pop in new chips.

    Any suggestions, similar experiences? I am I just adding current by touching? If so how can I get more juice into/out of the 7805. It tried replacing the 7805 twice and still same result.


    Best,
    thanks for the patience,
    Max
    if i need to post a schematic thats fine...
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Yes, a schematic would be very helpful.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    If you're using a 9v "transistor" battery, it's probably dead. TTL circuits draw quite a bit of current, and will kill a 9v battery within a day quite easily.

    But post your circuit so that we can see what you're dealing with.

    Use the "Go Advanced" button on the bottom, then the "Manage Attachments" button to upload your images. I prefer .png format images because they are compact and very sharp. .jpg format images are "lossy" and hard to read.
     
  4. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    OK i'll post tomorrow. Just eagle light for Mac os x :D
    Thanks for the interest!

    Best,
    Max
     
  5. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Check the chip enable (CE) pin and the high impedance pin on the 595, one needs to be tied to gnd and the other tied to 5v for the outputs to work. The datasheet is your friend. :)

    Also covered in the datasheet is that you only need 1 clock source, you can tie both the serial clock input and latch clock together, and that way you only need one serial data in and one serial clock in, BUT you have to make a 9th clock pulse to operate the final latch event. This is how I like to use them, needing 2 PIC pins is better than 3.
     
  6. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    "you can tie both the serial clock input and latch clock together"
    i thought the latch pin shoves all 8 states out in parallel. If that happens every time the shift clocks, wouldn't all states only be on or off?? If you can clear me up on this a little it would be great because as of now I am using one 555 for shift clock and another whole 555 for latch clock (I fully realize I can't get any kind of predictable results with this setup. I am more interested with getting random/chaos like results anyhow).

    OK, some words about the schematic. Things get really messy when it gets to the outputs. All the outputs of the 595s (12, im not using 4 of them) are connected to base of 12 individual TIP120npn Transistors. The collector of each tip120 goes to one leg of a high power led, and the other leg is tied to 12v +.
    Second, I can't get the supply pins (8 and 16 i believe) to show up in Eagle. But trust me, those are connected to 5v+ and ground coming from the 7805 regulator.
    I really hope this helps. I am totally self taught in the realm of circuit building, it being only a part of my artistic practice. Thus I need some help when it gets this complicated (for me at least)! :confused:
    As always, thanks, thanks and thanks again.


    Best,
    Max:cool:
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    In Eagle, power/ground pins on IC's aren't shown in the schematic view. Use symbols from the supply1 or supply2 libraries (I like the latter better). Vcc and GND will be necessary. Use the GND symbol instead of running wires everywhere.

    Something you mustn't forget is to have one of each voltage symbol connected to a wirepad or other type of offboard connection, unless the voltage is going to be generated on the board itself. Otherwise, such connections won't show up in the PCB layout editor.

    If you need to use two same-named signals together (such as +5v and Vcc) you can use a wire jumper to connect the two.

    You're running some wires at odd angles. That makes the schematic hard to read. Try to stick with vertical and horizontal wires for the majority of the layout. Use 45° wires only when necessary.

    Use the Erc (error check) icon to find out where you're missing connections.

    Pin 10 of IC3 isn't connected properly. Neither is Pin 12 of IC1. You're missing a junction on the resistor to the left of IC5.

    You don't show resistors being used on ANY of the bases of the TIP120's. This will overload the outputs of your 595 shift registers. Use at least 470 Ohm resistors between the 595 outputs and the bases of the TIP120's.

    If you want to simplify things and your current requirements are low enough, use ULN2003's or ULN2803's. Those are driver IC's which contain (respectively) seven and eight Darlington drivers, each capable of sinking 350mA current continuously, with incorporated input resistors.
     
  8. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    hey wookie,
    i forgot to put them into the schematic but I do have 1K series resistors between the base of tip120s and the 595 outputs.
    Thanks for the info on using eagle. Each high powered led array requires 150mA to operate. To get back to my original problem (the 595s not working unless I touched them, and seeing only 1.5 volts on the INPUT pin of my 7805 eventhough i am feeding it from 12v 1 Amp adapter), i think the crux of the matter is my power adapter is not powerful enough (only 1 Amp). Being that I need to power 12 150mA lights, 2 555s, and 2 595s, might this be the issue? Do i run the risk of frying the 7805 if I use a 12v 2 amp adapter?

    Thanks,
    Max

    * in terms of pin 10 being incorrectly placed, it is connected to pin 10 of the other shift register. Shouldnt that be ok since they are all going to 5v+ anyhow?

    * pin 12, latch pin, for both registers are supposed to share the same latch clock. How is this wrong? Sorry, i can't under stand.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    OK then, that's good. Be sure to add them to your schematic.

    It sounds to me like your 595's have an intermittent connection. If you're using a breadboard, that's quite likely.

    Your 555 timer's output may be why the 595's aren't clocking reliably, particularly if they're the BJT type. The outputs may not be going high enough. Note that the BJT 555's have a Darlington pair to pull pin 5 high. This causes the output to only be able to rise to around Vcc-1.3v. Since you're using a 5v supply, that would be 5v-1.3v = 3.7v.

    The 12 LED arrays alone will demand 1.8A when they are all operating. You'll need at least a 2A regulated supply.

    A 7805 regulator is only rated for 1A. You'll need a more capable regulator. Use a big heat sink.

    The problem is that you tried to connect the wire to the middle of the pin. It needs to be connected to the very end of the pin. ERC will also complain that the wire is overlapping the pin.

    Similar problem as pin 10; you tried to connect the wire to the middle of the pin. That just won't work if you try to use the schematic for PCB layout.
     
  10. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Hi, if you tie the register clock and latch clock, then every / clock edge the contents of the serial shift register are moved out to the latch register.
    This is an "old timers" trick, it means you only need one serial data input and one clock input.

    It has 2 drawbacks;
    1. you need an extra clock pulse to get the 8 bits output, so you need 9 clock pulses in total.
    2. while you are shifting in the 8 bits, they appear on the output for a short period before you shift in the next bit. Generally this is too fast for anyone to see and doesn't cause problems for things like display and multiplexing use.

    It's a popular enough "trick" it is covered in the datasheet.

    Also, I gotta ask.... "Schema blow up doll"???? :eek: :D
     
  11. mxabeles

    Thread Starter Active Member

    Apr 25, 2009
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    0
    Also, I gotta ask.... "Schema blow up doll"???? haha, just a whim - flight of fancy if you will.
    Thanks both of you. I bought a heat sink for my 7805, I might slap it on see if all as good.
    As for intermittent connection on the 595 circuit, I don't think so cause I have been using that circuit in other larger circuits w/out a problem. I used a conical tip that was much to narrow to solder the 555 circuit as well the rca connectors for this circuit.
    So, in light of all this helpful info here is my to do list:
    Use a higher current adapter (1.8 - 2 A)
    attach heat sink to my volt. regulator
    (If this doesnt fix the issue I will search for cold solder joints... ermm!)
    finally, i will bread board up a new circuit and try the single clock/ nine pulse trick. As well, I order a high current shift register from micrel that looks nice, but will probably need paralleling to be able to source 8 lights.

    Peace!
     
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