Some questions of 555.

Discussion in 'General Electronics Chat' started by ttttrigg3r, Oct 16, 2016.

  1. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    Hello, This thread is old, but I'd like to bring up some questions on how some of the components in this circuit works.
    This is new to me, and I'd like very much to know it thoroughly. The basic Monostable multivibrator with a 555 is attached as a picture. (I googled it.)
    A couple of questions:
    1) This circuit I've attached will consume power if left connect like this. Yes?
    2) In the 555 circuit you, crutshchow, attached, there is V+ and V2. What is V2? If I want to use just one power source that's V+, can I just replace V2 with V+?

    From what I've learned so far, the trigger Pin 2 needs a negative pulse to start charging of the timing capacitor. I see that you've fed the Output into a NPN and PNP combination. (Q1 and Q2)
    Can you please explain why you've done this and how Q1 and Q2 are functioning in this circuit?
    Thank you


    Mods note:
    Please don't hijack other member's thread, now you have your own.
    This thread was split from -
    555 monostable low power circuit.
    http://forum.allaboutcircuits.com/threads/555-monostable-low-power-circuit.101901/#post-1050619
     
  2. crutschow

    Expert

    Mar 14, 2008
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    1) Yes
    2) No.
    Notice the label inside the dotted-line box where V2 is located.
    The circuitry inside the dotted line including V2, which generates a pulse to simulate the momentary pushing of a button switch, is not part of the circuit you would build.
    The actual circuit would just have a momentary push-button switch that connects between V+ and the node labeled Vs.

    Explanation of operation:

    Q1 and Q2 are normally off so there is no power to the 555 circuit.

    When the PB is depressed it applies power to the 555 circuit through the node labeled Vs.

    As power is applied, the 555 is triggered because the TRIG input is momentarily held low by capacitor C2.

    The resulting high OUT pulse of the 555 applies voltage to the base of Q1 to turn it on, which then turns on Q2. This keeps the power applied to the 555 after the PB is released, until the 555 times out.

    After the 555 times out, the OUT voltage goes low, removing the voltage to Q1's base. This removes the circuit power and the circuit returns to the quiescent state with no power being drawn from the supply.
     
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  3. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    1) It's momentarily low because C2 is being charged up, right?

    2) Is R1 and C2 the components to control the time for Vout?
    3) What is purpose of C1, C3, and C4?

    Thank you. I shall build this circuit and try it out before building the one with 33 PBs.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    1) Yes. When the power is initially applied, C2 is not charged so its voltage is zero, which triggers the 555.
    C2 then charges up through R1 so the 555 doesn't retrigger at the end of the one-shot time-out.

    2) No. R1 and C2 are just for the trigger as described above.
    The R2 C1 time-constant determine the one-shot on-time.

    3) C3 and C4 are filter (decoupling) caps.
     
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  5. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    Ok. I've built the circuit with 3904 and 3906. I don't have a 2907 on hand. My output comes out to be this funky one. Nothing happens when I push the button. Am I doing something wrong.? 20161018_155942.jpg 20161018_160009.jpg
     
  6. crutschow

    Expert

    Mar 14, 2008
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    Obviously there's something wrong.
    There's likely a misconnection in your circuit but I don't have the patience to try to determine that from your picture.

    Carefully do a continuity check with an ohmmeter of all connections on the circuit and mark each one off on the schematic with a highlighter as you verify it.
     
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  7. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    Ok. I went over it very carefully by checking wire by wire, and the circuit is built correctly. Maybe my 555 is bad, I'm not sure. I'll try to build a simple model or buy a new one tomorrow.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    Post a picture of the schematic you used with all parameter values.
    Are you sure the transistors are correctly connected with the proper pinout?
     
  9. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    OK The circuit works. I didn't connect the ground to the OSCOPE, so that's why it looks funky. When I have the scope set up correctly, it works and the output is clean.
    However, after the one shot resets, it cannot be triggered again. I have to cycle the power supply to be able to trigger it again.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    After the one-shot period ends the power is removed from the 555 circuit so removing the power aupply should not make a difference.

    Monitor the voltage on C2 and voltage Vs during and after the one shot period and tell us what they do.
     
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  11. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    Here's Vout, Vs , and V(c2) from top to bottom. I notice that Vc2 and Vs both sits at above 2.2V after the reset. I think this is what's causing the problem. 20161020_093030.jpg
     
  12. crutschow

    Expert

    Mar 14, 2008
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    Add a 1kΩ resistor from Vs to common.
    That should pull everything to 0V at the end of the cycle.
     
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  13. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    Ok. I added a 1.1k resistor from Vs to Gnd. No change. I checked out the 3906 PNP transistor and looks good. My power supply is reading .002A after the cycle happens, so I know the circuit is consuming power after the push button and one shot cycle. Could this be some sort of leakage from the transistors? I probed the base of the NPN and it reads 0, so I know the NPN is not on. When I try to probe the base of the PNP, it turns the 555 circuit on and runs another cycle, so I think probing the base of the PNP triggers that transistor somehow.

    What is the name of the transistor configuration that is here?
     
  14. crutschow

    Expert

    Mar 14, 2008
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    You shouldn't have transistor leakage but try connecting about a ≅10kΩ resistor between the base and emitter of the PNP (Q2) and see if that makes a difference.

    Yes, when you probe the base of the PNP, the meter input impedance will conduct a small current from its base to ground, which is enough to turn it on and trigger the 555.

    The base voltage of the PNP should be measured between its base and emitter, not between its base and ground.
     
    Last edited: Oct 20, 2016
  15. ttttrigg3r

    Thread Starter New Member

    Oct 14, 2016
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    I removed the.01uF capacitor at Pin5 and it fixed the issue. The circuit now works like it's supposed to. Right now, Pin5 is floating in the circuit. I don't know why, but at the moment, leaving pin5 floating makes the circuit work. I will continue to test it
     
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