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Non-changing current with C-EMF

Discussion in 'General Electronics Discussion' started by XRZ, Jul 30, 2014.

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  1. XRZ

    XRZ

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    Jul 30, 2014
    I'll go study this some more! The reason why I'm confused is because how can we maintain current and have voltage change without changing the resistance or increasing power(ohms law) how does a constant current work...that is what I need to study @Arouse1973 @Merlin3189 @(*steve*)

    But quick question, how fast does voltage change? What range are we talking here? Nano,micro,pica-seconds? And if I used a constant current source how fast could it detect change in current and react by changing the voltage?
     
    Last edited: Sep 22, 2014
  2. Arouse1973

    Arouse1973 Adam

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    In theory faster than you could ever measure.
    Adam
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Voltage and current changes cannot propagate along a conductor faster than the speed of light in that conductor. However, at a given point there is no limitation to the instantaneous rate of change for either of these.

    Once you start looking at rates of change which are fast compared to the rate of propagation along the conductor we can no longer assume many of the basic heuristics such as current or voltage being constant along a conductor. At this point we need to consider the wavelength of the signal.

    Indirectly this also provides a limit on the function of any constant voltage or current source. Since there is some propagation delay in the wires, the circuit can only react to the state that the current or voltage was some short time ago, and this change will not propagate to the measuring point for some short time.

    If we assume that the speed of light in a cable is half the speed of light in a vacuum, and that the length of the conductor between the control element and the measuring element is 10cm, then any change at the measuring point has to be communicated to the control element and then that change must arrive at the measuring point. (this ignores any delays in the control circuit itself -- let's assume it is instantaneous). So in this case, the circuit could not react faster than (0.1 * 2 * 2) / 3E8 seconds = 1.3E-9 seconds (1.3ns).

    In practice, the exact response will not be observed after 1.3ns. One of the following will happen:

    1. The control element (over)reacts, then continues to react, and there is overshoot, followed by undershoot followed by overshoot, with the excursions increasing until the output is slamming from one extreme to another (this is oscillation)
    2. The control element slightly overreacts, resulting in overshoot, followed by a smaller undershoot, followed by a yet smaller overshoot, until eventually the excursions are too small to be seen. This is called ringing. It's not good, but it is better than oscillation.
    3. The control element slightly underreacts, then it again slightly underreacts, this continues as the output gets asymtotically closer to the desired output until it is indistinguishable from it.
    Typically, in a good design) you'll get a version of (2) where there is a little overshoot, but the excursions are quickly damped.
     
  4. XRZ

    XRZ

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    Jul 30, 2014
    Well the concept of a constant current is starting to make a bit of sense. A power supply that will increase its internal resistance significantly to the point where the path of least resistance would be the load.
    But what happens to the voltage? It increases? Because, what makes sense to me is that more voltage is applied and current stays the same P = IV, that means more power is applied to the load?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    There is no "path of least resistance". This is a concept which simply doesn't apply to electronics. (Ok, there may be a path of least resistance, but not all the current flows through that path).

    Let's consider both the theoretical and a practical constant current device.

    In a theoretical device the voltage does exactly what is required to maintain a certain current flowing through the device. The voltage may rise, it may fall, it may even change polarity.

    In a practical device, there are some limitations. For example, the output voltage will have some defined practical maximum limit. If you need a higher voltage than that, it's simply not going to happen and the current will fall. It is quite common that the voltage can fall no further than zero (very few constant current devices can have an output which changes polarity whilst still delivering a particular current). If you hit this limitation the current may rise without control until some other limit is reached.

    It depends on the load. If it is inductive, or if it is storing power in some other way, that power may be delivered back to the current source later. When this happens, it is no longer "real" power" since when averaged ocer a cycle there are positive and negative excursions which cancel out (or nearly cancel out). See here for more information. Whilst the power is not "real" in that it's not consumed, it is very real to the source of that power which may have unusual peak demands and even the need to absorb power in parts of the waveform.

    This applies directly to magnetic fields because if you have a varying magnetic field impinging on a conductor, there will be a change in one direction as the magnetic field grows, which will possibly peak, stop and return to zero as the field becomes static.. Then as the magnetic field is removed, the process repeats in the opposite direction. Over a cycle, the net effect may be zero. (The effect won't be zero if you've taken advantage of it and extracted power or injected power -- in either case you have transferred force from or through the magnetic field, typically from or to whatever is causing the change in the field. Practically this means getting power from the shaft of a generator or applying power to the shaft of a motor)
     
  6. XRZ

    XRZ

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    Jul 30, 2014
    The reason I brought up the factor of time is thinking of how fast the constant-current source would react to the counter-emf. Would it be faster than back-emf or how would that work out?
    Because maybe due to some factors... back-emf would be much quicker than the source in reacting, or maybe... they are the same. @Arouse1973 @(*steve*) .

    I need to understand constant current sources more. I think the power factor that I'm concern about is nothing important.
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Nothing can exceed the speed of light. Even back-EMF.
     
  8. Arouse1973

    Arouse1973 Adam

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    I can't add anything to this, I think that is well explained Steve.
    Adam
     
  9. BobK

    BobK

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    You
    You calculation is wrong.

    To maintain current of 1000A across the resistor will always have the same voltage across the resistor. The back EMF is not added to the voltage needed without back EMF. This is what is causing your massive power.

    In the presence of -2V counter EMF, the constant current source has to provide 2.0042V, which, at 1000A is 2004.2W, large, but not massive.

    Your example is not close to anything that would occur in the real world (outside a high power physics lab.) So what are you worrying about.

    Taking a more reasonable example, which you suggested earlier. 0.01Ω, 1V, 100A. this is 100W. Add -2V of counter EMF and the voltage rises to 3 or 300W. This is a more reasonable example.

    In fact, if you look at real-world electric motors we see this effect. When the motor is not moving the motor will draw many times the current that it would when the motor is running freely. For example, I am working on a robot car right now, and the motors require 200mA at 6V, but when stalled the draw 1A at 6V. So the power when running is .2 * 6 = 1.2W and the power when stalled is 6W.

    The 1A at 6V stall current implies that the resistance of the motor coils is 6 / 1 = 6 Ohms. So, when it is running, at 200mA, the voltage across the windings must be 6 * 0.2 or 1.2V, so there is 4.8V of counter EMF being produced by the motor. If I wanted to maintain the stall current when the motor was running I would have to supply 4.8V more or 10.8 volts to the motor to get 6V and 1A across the windings (this assumes the motor is still running at the same speed so the back EMF stays the same). So to maintain the constant current against the back EMF I am required to supply 10.8W to the motor instead of 1.2W when I do not try to keep the current constant.

    Bob
     
  10. XRZ

    XRZ

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    Jul 30, 2014
    @(*steve*) thank you for that detailed response even though some of it is more than what my plate can take! Still thanks, gives me more things to consider.

    @Arouse1973 I agree.

    @BobK if the resistance of the conductor itself is 4.2x10^-5, how can the CCS apply 2.0042V with only 2004.2W? What about P = V^2/R?
    What do you mean the term "across" the resistor?
     
  11. XRZ

    XRZ

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    Jul 30, 2014
    @BobK If the resistance is that low, and the voltage is quite large the power is also going to be quite large as well. But I guess that is the point you're making in my calculations is wrong.
     
  12. BobK

    BobK

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    Because you are not using the correct voltage across the resistor. The voltage is the voltage from the constant current source (2.0042V) plus the counter EMF of -2V for a net voltage across the resistor of 0.0042V.

    If the current through a resistor of R Ohms is I Amps, the voltage across it is I * R. It does not matter that the power supply is supplying 2.00042V, that is not the voltage across the resistor when counter EMF is present, the two voltages must be added.


    Bob
     
  13. XRZ

    XRZ

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    Jul 30, 2014
    Well that just changed everything, thank you so much BobK and everyone!!

    @BobK , to know that fact that I need only 2004.2W it mind boggling. Because I assumed I needed 2.0042^2/R, but in fact the only thing that will indeed increase is the voltage, and therefore, the power will increase by the factor of increase to the voltage. If the voltage needed is 10V then P= 10V x RequiredFixedCurrent, or P = 100V x RequiredFixedCurrent.

    Constant current source is the solution indeed!
     
  14. XRZ

    XRZ

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    Jul 30, 2014
    @Gryd3 , @Arouse1973, @(*steve*) , @BobK

    How can the constant current source detect the change in current, and change it's voltage at instantaneous speeds(microseconds)?
     
  15. Gryd3

    Gryd3

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    feedback, and filter components.
    Filter components like Capacitors and Inductors will greatly reduce the amount of change that actually occurs. Any change that is still present is picked up by a feedback portion of the circuit.
    Feedback is important, the constant current source will monitor the output and compare this to a reference. It will adjust as quickly as the internal circuitry allows when the output changes in comparison to the reference point.
     
  16. Arouse1973

    Arouse1973 Adam

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    What part of this are you having problems with. Explain your issues, dont worry I have to ask questions sometimes 100 times before I get it .
    Adam
     
  17. XRZ

    XRZ

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    Jul 30, 2014
    I pretty much got what I needed. My misconception of the additional power to increase voltage has been adjusted, how constant current works(kinda getting the idea), the rate of reaction which I believe is equal to counter-emf induction OR quicker? I mean, would it be fast enough to maintain current during the whole time?

    What I mean is, induced -V starts to "oppose" the current is the constant current really fast at reacting and increases the supply voltage to keep the current the same never decreasing during the whole time?

    Again, thank you guys for the help! It's amazing finding experts helping out.
     
    Last edited: Sep 26, 2014
  18. XRZ

    XRZ

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    Jul 30, 2014
    Well, that last post was interesting(the mistakes... we're to much!) Anyhow... I was wondering about the rate of reaction by the constant current source. How fast would it be? Equal to the rate of counter-emf's induction?

    What I mean by this is... if counter-emf is induced simultaneously the source-voltage from the constant current source will increase to oppose it and current stays the same throughout the whole process? Never decreasing?

    @Gryd3, @Arouse1973 , @(*steve*) , @BobK
     
  19. Arouse1973

    Arouse1973 Adam

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    "How fast would it be? Equal to the rate of counter-emf's induction?" In an ideal world yes.
    Adam
     
  20. XRZ

    XRZ

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    Jul 30, 2014
    What do you mean by the ideal world :p, is there a catch?! If I try to work on a model?
     
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