555 servo controller - please help!!

Discussion in 'The Projects Forum' started by allan991, Jan 5, 2008.

  1. allan991

    Thread Starter New Member

    Jan 5, 2008
    1
    0
    Hi guys,
    I have made a circuit using online sources and it doesnt work. What it should ! do, is move the servo from left to right then back to left in about 5 second pulsesThere are 3 parts. The first is a basic 555 timer which sends pulses to the second -
    which is a cd4017, when clocked changes the resistance to the 3rd circuit
    which is a 555 frequency circuit, which sends 1 - 2ms output to an rc servo.
    The cd4017 is supposed to go from zero resistance , then gradually up to the full variable 2.7k ohm then gradually back down, automartically reseting.
    When i try the pulsing 555 circuit, it works at about every 5 secs. Then i try the frequency circuit by itself and it works ( by changing the potentiometer 1 2.7k ) manually, i can get the servo to move from left to right. But when all is connected through the cd4017, it does nothing!
    Any help would be appreciated
    thanks very much
    allan
    (circuit attached)
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    When I bring up your diagram there are a couple of connections that appear to be missing or possibly cutoff.

    hgmjr
     
  3. hgmjr

    Moderator

    Jan 28, 2005
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    Now I see it.

    hgmjr
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    Can you provide a link to the source of the original circuit on the Internet? It might be helpful to refer to the information it contains.

    hgmjr
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    I'm sorry, but your Johnson counter just won't do that. It only counts in one direction. http://www.national.com/ds/CD/CD4017BC.pdf
     
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Like I said on the other website, the output resistance of an ordinary 4017 is too high. Use a 74HC4017 instead because its output resistance is much lower.

    Wait a minute. On a 4017 or 74HC4017, one output is high and 9 outputs are low. It won't do what you want.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,193
    1,757
    Hmm. Yep, only one high output at a time - the rest will be low.

    Right now, 9 out of the 10 resistors will be connected to ground, while one of them will be pulled near Vcc.

    (the following paragraph is no longer relevant)
    But if you have a Darlington array between the outputs of the 4017 and the resistors, only one of the Darlington pairs will be biased ON by the single high output, thus providing a ground for that output's resistor. There will still be a voltage offset due to the PN junction, but it will be far better than the current configuration. The remaining resistors will have a relatively high impedance path to ground. Now if you were hoping to supply a voltage near Vcc to the resistor - that's another matter.

    After perusing your schematic a bit more, I realize you actually want to supply Vcc via your selected resistor.

    You may be able to do this using your circuit with some VHC4066 ICs, which are high-perf CMOS quad bilateral switches. Tie the inputs of all of the switches to VCC, each output to a resistor, and the control lines go to the outputs of your 4017. When a VHC4066 gate is turned off, it's output is in a high impedance state; when on it's at a fairly low resistance depending upon the temp and the IC's Vcc.

    A better-suited quad switch for your application is Analog Devices' ADG713; Vcc to 5.5V and on-state resistance of 2.5 Ohms.
    http://www.analog.com/en/prod/0,,768_833_ADG713,00.html
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    As has been pointed out, the 4017 is going to drive one resistor high while holding the remaining resistors low. That means the output with zero ohms is going to be held low 90 percent of the time.

    sgtwookie's suggestion looks viable since it makes sure that one and only one resistor is enabled at a time.

    Another workaround would be to use three 74HC126 quad tristate buffers. The output lines of the 4017 would then be used to enable each of the HC126 buffers one at a time. The input of the HC126 would then be tied high so that when the buffer gets enabled it would go high. At any given time 9 of the 10 buffers would be tristated.

    hgmjr
     
  9. SgtWookie

    Expert

    Jul 17, 2007
    22,193
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    If you wanted a lower parts count than the idea I proposed with the quad analog switches, you could use a CMOS 4051 single 8-channel analog MUX/DEMUX with it's address lines driven using a binary 4-bit counter instead of the 4017, and 1 resistor tied to Vcc on each input. You would only have 8 steps instead of 10, but this would be workable.

    If you wanted more steps, instead of the 4051 you could use a 4067, which is a 16-to-1 line analog mux/demux. If you wanted exactly 10 steps, you could use a decade counter, or use an AND gate to reset the binary counter when it reached a count of 10 (synchronous reset) or 11 (for an asynchronous reset).
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    How about if you used 10 diodes, one in series with each of the resistors? The anode would be toward the 4017.

    hgmjr
     
  11. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    The diodes are a good idea, but the Johnson counter still needs to be replaced with an up/down counter.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,193
    1,757
    I think he was talking about using diodes between the output of the 4017 and the resistors. That would work to block the ground path.

    The remaining problem with the whole 4017 idea is the amount of current the 4017 can source. You'd have that problem with a counter as well.

    There doesn't seem to be a need for a down count here; just count one direction.

    Agree that the circuit would be simplified using a binary counter with binary weighted resistor values though.
     
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