555 in astable mode - Control pin role

Discussion in 'Homework Help' started by PsySc0rpi0n, Oct 18, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, my teacher is asking us to connect a DC power supply, varying from 1 V up to 5 V, and a 2.2 kΩ in series, between the Control pin and the GND, like:

    CTRL-----2.2kΩ----PS------GND


    In terms of circuit between CTRL and GND, I think I get something like this:
    2.3-circ.png

    By inspection and performing a nodal analysis, where i1 is the current flowing through R1, i2, the current flowing through R2 + R3 and i3, the current flowing through R4, and finally vx the voltage at the node where the 3 currents meet, I get:

    i2 = i1 + i3
    ⇔ (vx/(R2 + R3)) = (V2 - vx)/R1 + (V1 - vx)/R4

    using the calculator I get the correct result for vx, but when I tried to solve the equation prior to replacing the letters by their respective values, I can't get to the same result.

    My work is this:
    [​IMG]

    But I think is gone wrong somewhere!

    Can anyone help me find where?
     
  2. WBahn

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    Mar 31, 2012
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    Your problem is with 7th grade algebra and the concept of putting fractions over a common denominator.

    If you multiply both sides of your starting equation by (R1 + R4), what do you get on the right hand side?

    Whenever you reach a point where you suspect you have done something wrong, go back and check every step very carefully to ensure that you are not making some bonehead mistake (which we all make from time to time).
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, let me try again! (Everyone is allowed to make mistakes, no matter how big they are)!
     
  4. WBahn

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    Not only is every allowed to make mistakes, but it is all-but-guaranteed that everyone WILL make mistakes. That's why it's so important to develop sound error-detection skills and strategies. Remember that, in the real world, you will almost never have "the correct answer" available to compare your answer to.

    So let's consider a couple more error-detection approaches that fall into the category of asking if the answer makes sense.

    There are three obvious sanity-check situations for this circuit.

    If you set R1 = 0, then v1 = V2, but your final result doesn't reduce to that.

    If you set R4 = 0, then v1 = V1, but again your final result doesn't reduce to that.

    If you set R4 = ∞, then v1 = 2/3 V2, but again your final result doesn't reduce to that (in this case it drives v1 = 0V).

    As an additional note, it is a bit dangerous to have a supply named V1 and an unrelated node voltage named v1 -- that's just asking for trouble.
     
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    That's why I think that nobody should judge other's mistakes as being silly or not!

    Anyway,I think I got a solution.

    I took as example, V1 = 1V

    [​IMG]

    As for the V1 and v1, of course you're right. Just kept it that way because I only named that node as v1 after start addressing the problem.

    Results from LTSpice:
    equivalent-555-control-plot.png
     
  6. WBahn

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    Who judged your mistake as being silly -- I merely pointed out the kind of mistake you made to help you focus on what to look for in finding it.

    Let's now check your final equation against the three sanity check questions I posed.

    R1 = 0: v1 = V2 (check)
    R4 = 0: v1 = V1 (check)
    R4 = ∞: v1 = 2/3 V2 (after setting R1=R2=R3) (check)

    It's good that you solved the problem symbolically before plugging in values as the last step, but you still fail to recognize the important and value of tracking your units. Apparently you just don't care about being able to detect your mistakes.
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, anyway... We have our own opinions and they worth just what they worth for each one of us! I appreciate you tried to point me in the direction of my mistake, which I detected and give it another try. However I don't agree very much with the words used. I got the idea you ridiculed me because the mistake I made was a 7th grade mistake.

    About the sanity check, I just didn't followed the thought.
    Like, why should v1 = V2 if R1 = 0 Ω and the same for the other 2 sanity checks?

    About the units, do you mean in the equation where I replaced the general letters by actual values? I have not forgotten them in the final result: 1.92 V...
    Also I didn't understood why keep tracking of units says anything about me caring about detecting errors!
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Anyway, I have more questions if you will to help...

    I'm trying to understand the influence that the voltage at the Control pin in the output pulse Duty Cycle.

    I could already check with LTSpice that when we increase the voltage at this Control pin, the Duty Cycle also increases. I was expecting the other way around, but obviously I'm wrong.

    My thought was that as there is more current flowing through the resistors that are charging the capacitor, this would charge quicker, hence, the Duty Cycle would be lower...

    Can you drive my thought in the right path?
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    The only difference with a control voltage input is that it changes the two voltage limit points...the 1/3 and 2/3 of Vcc change to other values. Calculate the values using the two source resistor network and then use the same formulas as in the other thread, except use whatever fractions you come out with for the new voltage levels. Derive a general formula.

    If you like, you can prove that if the voltages are not symmetrical with respect to Vcc and ground (lower limit same voltage difference from ground as the upper limit is from Vcc) then the timing is not independent of the supply voltage, and conversely, if they are symmetrical then the timing is independent of the supply voltage if the control voltage is in ratiometric proportion to the supply voltage.
     
  10. WBahn

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    If R1 = 0 Ω, then the node v1 is connected directly to the V2 supply. Hence it doesn't matter what the value of V1 is or of any of the other resistors (with the fine point that R4 can't also be 0).

    Similarly, if R4 = 0 Ω, then the node v1 is connected directly to the V1 supply.

    If R4 = ∞, then it is effectively removed from the circuit and you are left with V2 and a voltage divider consisting of three equal resistors.

    Which merely means that you ignored any and all mistakes you might have made and tacked the units that you HOPED were how they worked out onto the final result. Poor engineering.

    Because, as I have explained to you on several occasions (and which have clearly made zero impression on you), units tracking is perhaps the single most effective error-detection tool available to the engineer. Most mistakes you make will mess up the units, allowing you to almost immediately identify that an error has occurred and to track it down. By forfeiting that ability, you indicate that having that ability is unimportant to you.
     
  11. WBahn

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    How would changing the voltage on the control pin affect the current flowing through the resistors that are charging the capacitor?

    What node in the charging/discharging circuit is affected by changing the control voltage? What purpose does that node serve in determining the duty cycle of the circuit?
     
  12. WBahn

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    What does it mean for one voltage, say Va, to be in ratiometric proportion to another voltage, say Vb?
     
  13. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I think I already sorted this out too but I still have a few questions.

    I know now that the internal Ra is kinda discarded when using a voltage at the Control pin.

    Also know that when not using the Control pin, we have 3 resistors, dividing Vcc into 3 equal parts. When using the Control pin, only 2 resistor are splitting (in half), the Control pin voltage (CV)!

    Also acknowledge that when using the Control pin, as Ra is kinda discarded, there is less current charging the capacitor, so it takes longer to charge to CV (Control Voltage applied to Control pin), hence the Duty Cycle also increases.

    Lastly, the discharge time remains roughly the same.

    Questions:
    Why Ra is overridden when using Control pin?
    Why the discharge time remains roughly the same?
    When using Control pin, does Vcc takes any role on charging the capacitor?

    Going to bed now! I'll keep going with this tomorrow!
     
  14. WBahn

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    Assuming that by Ra you are referring to the resistor you have labeled R1 in your diagram above, you are correct but only to the extent that R4 is small relative to R1. If R4 is large, the R1 is not "discarded" and very much plays a role. In fact, R1 plays a role as long as R4 is non-zero, but the smaller R4 is, the less of a role it plays. Do the math.

    Why and how are you concluding that "discarding" Ra results in less current charging the capacitor. How does the value of any voltage anywhere in the control voltage circuit have any impact on the current that is flowing in the capacitor (other than altering WHEN the circuit switches between charging and discharging modes)?

    What happens if you set the control voltage to 75% of Vcc? What happens if you set it to 25% of Vcc? What observation can you make regarding your claim as a result?

    Correct.

    Look at the circuit! If you apply a voltage directly to the control pin, then what role does Ra play in determining the switching threshold?

    Do the math! Whatever the voltage is at the control voltage (whether it is being overridden or not), the voltage at the trigger threshold is half of that. How long does it take a first-order circuit with a time constant tau to go from V1 to 0.5·V1?

    Yes. And to see how, do the math!
     
  15. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    It just means that one voltage is proportional to the other. Usually the proportional voltage is lower than the other which distinguishes it from amplification, but i wont stress that.
    So if Va=10 and Vb=2, then if Va where to change to 20 and as a result Vb changed to 4, and if Va were to change to 30 and as a result Vb changed to 6, then Vb would be said to be ratiometric to Va, or in ratiometric proportion to Va, or just proportional to Va. The term ratiometric is usually used when the supply voltage changes output in proportion, but i wont stress that too strongly either as i am not completely sure that should be the only usage.

    Here's a quote from a patent that uses the same phrase:
    "Preferably, the receptor target carried by the carrier particle is identifiable by the ratiometric proportion of the fluorophores attached thereto, incorporated therein or contained thereby."
     
  16. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    This might assist in the discussion

    [​IMG]
     
  17. WBahn

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    Ah, I see what you are saying. I was interpreting what you were saying in the narrow context of a fixed supply voltage and that the time would be independent of the supply voltage only if the control voltage was chosen so as to be in ratiometric proportion to it, which didn't make any sense to me. But I see now that you were talking about if you change the supply voltage, that the control voltage must change proportionately in order for the timing to be unaffected. I definitely agree -- in fact the timing is best derived using the ratio Vctrl/Vcc as the control parameter.
     
  18. WBahn

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    I've never like this diagram, as common as it is, because it doesn't give enough detail to understand the workings. You need to know the polarity of the comparator inputs and the details of the FF inputs for that. One that shows the needed detail is

    upload_2015-10-18_20-35-50.png

    About the only thing missing here is an indication that the Qbar output of the FF is current limited, which could have been illustrated by putting a base resistor on the discharge transistor.
     
  19. JoeJester

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    Well, your diagram is better. My point was everyone, especially Psy, should have worked from a common diagram.
     
  20. MrAl

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    Hi,

    Yeah, i never liked that diagram either. The one that Wbahn posted is MUCH better because we can then really understand the internal workings of the chip. Radio Shack used to have that less detailed picture on their little 'card' that the 555 they sold came in back in the 80's i think it was. I used to hate that diagram :)

    That was a long time ago, and today some of the data sheets now show the internal diagram using NOR gates as the flip flop which helps a little too to understand what kind of flip flop is being used (ie how the S and R inputs really affect the output), but if anyone looks up an RS flip flop using NOR gates they will find the right way the flip flop operates also. Flip flops are sometimes peculiar how they work given a set of pins labeled as common as R and S and J and K and the like.
     
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