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Capacitor circuit and current flow

Discussion in 'General Electronics Chat' started by nicksydney, Feb 20, 2012.

  1. colin55

    colin55 Well-Known Member

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    The current gradually dies to 15mA.
    We are not switching off the inducutor. The current drops to 15mA.
    And secondly the energy produced (if the inductor is switched off) is determined by the voltage and current flowing though the inductor. These values are very small and so any back energy will be very small.
    It's not even a value worth considering.
    On top of this you have two diodes that will clip any back voltage so I don't know why you even brought up the point.
     
    Last edited: Feb 22, 2012
  2. kubeek

    kubeek Well-Known Member

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    use limit switches
     
  3. colin55

    colin55 Well-Known Member

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    If that were possible, thousands of constructors would not have bothered to buy the kit.
     
  4. dave

    Dave New Member

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  5. ljcox

    ljcox Well-Known Member

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    I assume you mean the average current is 15 mA.

    As I see it, when the solenoid is across the output & the caps have discharged, the transistor is off & there is a half wave rectified current through the solenoid.

    Thus the solenoid current is going on/off at the mains frequency, so there would be a significant back EMF if D2 was not there.
     
    Last edited: Feb 22, 2012
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  6. colin55

    colin55 Well-Known Member

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    Totally incorrect
     
  7. Roff

    Roff Well-Known Member

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    The half wave rectified sine wave's slew rate is low, and the inductance is low, so the flyback voltage will probably be on the order of millivolts, or maybe tens of millivolts.

    I think it's a clever circuit.
     
    Last edited: Feb 22, 2012
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Colin,

    You're saying that there is no back emf, then you're saying that you have two diodes that will clip any back voltage. So which is it then?
    The fact is that depending on the capacitance in the circuit and the inductance in the circuit we may see the back emf develop a current in the opposite direction. Some or all of that energy may be absorbed by the capacitors safely if they still contain some energy.

    The other thing is near the input of the schematic we see "15v AC or DC". With 15v DC we'll see a one time only response, but with 15v DC we may see repeated responses similar to if we connected a 15v source, disconnected it, then connected it back up again. This means the load (solenoid) would be pulsed rather than turned on once and then turned off slowly. If the 'switch' you talked about is on the output, that means the switch may open during a time of relatively high current anyway.

    In an more obscure design like this it is often harder to critique the design itself rather than the designer mostly because there are so many undocumented variables that only the original designer might know. This means we can question the designer and see if he is assuming something unreasonable. If we find something it doesnt mean that the design is flawed, but it does mean there's a good chance it is if he assumed something unreasonable that was more important. What would be done then is he could be asked to explain the reason behind the choice or spec of a component, or a circuit connection. If he cant explain it, then the design falls into a questionable area, but if he can explain it then perhaps it works to some degree or else he finds out that he has to improve the design.
    Let me give a simple little illustration here...


    In a certain design we find a resistor with a power rating of 10 watts, but in simulating the circuit as well as we can we find that the most likely power dissipation is 12 watts in that one resistor. We are also told that there are several units in operation in the field. What to do?
    Well, we think there's a design flaw, so we look at the schematic and in the lower (or upper) right hand corner we see a block of text with the date and designers name. So we go back to the designer and ask him/her if something could be wrong.
    He/she said they dont think so, but they will go over that part of the design again and see if they can find anything. We also ask him/her to back up the calculated data with some real life measurements and if necessary some scope pictures.

    So it is a little harder sometimes to get to the bottom of it all, but one thing that helps a lot is some real life test data which in the case of a transient operating circuit like the one we are talking about in this thread some scope pictures could be the ultimate turning point, where we either verify the design or improve it.
     
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  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,


    Oh yeah one more little thing.

    If you replace the transistor with a 1k resistor and keep the caps you get almost the same kind of response. The cap charges up when the solenoid is off, and when the external switch is closed the solenoid can only get it's turn on energy from the caps because the 1k resistor limits the current flow to the caps from the power input.

    So this brings the circuit count down to 2 or 3 components: a resistor and a capacitor (one or more sized). If you're worried about back emf then use a reverse diode across the cap too.
     
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  10. kubeek

    kubeek Well-Known Member

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  11. colin55

    colin55 Well-Known Member

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    The kit costs $8.00
    Try to sell limit switches to someone who has 25 point motors on his layout.
    How are you going to fit them. How are you wire them?
    We have already gone though this discussion 25 years ago.
    That's why we designed the circuit.

    The condition of back EMF doesn't exist.
    However if you think it does exist, the two diode will quench the voltage. So it's a pointless discussion.
    The extra diode has been added to make the design "bullet proof" - just in case someone abuses the design by pulsing the point motor and creating a back EMF. Under normal conditions, no back EMF is produced.
    In fact I have used the BD679 in a purpose-breakdown (zener) circuit and although it is designed for 80v, it zeners at over 180v.

    Yes, you can replace the transistor with a resistor but the transistor decreases the recharge time to less than 1 second.
    This allow the operator to connect to multiple points.
    A 1k resistor would make the unit useless.

    <mod edit: self promotion deleted>
     
    Last edited by a moderator: Sep 16, 2014
  12. lebevti

    lebevti Member

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    LOL

    really? somebody posts a random circuit, and it happens to be your design? what a coincidence
     
  13. colin55

    colin55 Well-Known Member

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    Coincidences do occur when you get 1,696,330 pages viewed per month on my site with hundreds of circuits designed by me.
    Although I know 0.001% of electronics, I have studied the basics of electronics and can impart this to anyone who wants to learn.
    With 332.64 GB of traffic each month, I must be providing something.
    Even this simple circuit has taken a lot of delving before an acceptable description has been delivered.
    As I said at the beginning, you could have considered a HIGH, MEDIUM and LOW impedance load and produced an outcome.
    From this you can discuss the suitability of the circuit for the three cases.
    Many of the circuits discussed on this forum come from my site. This is not the first and won't be the last.
    All my circuits have bee built and tested - a far cry fom many of the circuits preseneted on the web.
    <mod edit: self promotion deleted>
     
    Last edited by a moderator: Sep 16, 2014
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    "Yes, you can replace the transistor with a resistor but the transistor decreases the recharge time to less than 1 second.
    This allow the operator to connect to multiple points.
    A 1k resistor would make the unit useless."

    Then so would be a lower current wall wart or other input power source as that would be the limiting factor for the charge time then.

    As i said before, you wont get ANYONE to understand this circuit if you dont tell them what the load is going to be and also how it is going to be operated. And every time someone provides an explanation or improvement you change the design specifications. Lets look at the history of your load specification in this thread:
    1. No particular load specified.
    2. Then, 4 to 6 ohms load.
    3. Then, a solenoid with external switch.
    4. Now finally (?), multiple solenoids with multiple external switches.

    Now if you want to keep changing the spec's you'll never get anyone to understand this circuit or offer up an improved solution.

    The spec's as they stand now:
    1. 15v AC or DC input (no particular current rating supplied yet).
    2. Multiple externally switched solenoids (AKA 'points').

    Care to make any more changes before we go on?

    So i guess i can connect a 15v DC 50ma wall wart to the input and this would work fine?
     
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  15. colin55

    colin55 Well-Known Member

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    The circuit will only activate one set of points at a time. It has a recovery-time of about 1-2 seconds.
    The supply is 12v to 15v AC or DC with a capability of 1 amp.
     
  16. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Colin,

    Ok thanks for pointing that out. I think that 1 amp spec should also be on the schematic.

    So your circuit is starting to look better now, now that we know what the load is and how that load is to be operated, and also the input spec's. That makes a really big difference on how we evaluate this circuit as you now know. And switches in the output load changes everything quite drastically too. That makes this a load controlled power supply, which is very different than a regular common power supply and something we dont see too often unless it is built into something else.
     
    Last edited: Feb 23, 2012
  17. ljcox

    ljcox Well-Known Member

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    This shows that you do not understand inductance!

    The back EMF is due to the changing flux. The back EMF is given by:-

    v = L di/dt.

    Which is the definition of inductance & it reflects Lenz's Law.

    Lenz's Law is very general & applies to many electric/magnetic situations.

    In this case it could be stated thus:-

    When a current is applied to an inductor, the back EMF opposes the rise of flux.

    When the current is switched off, the back EMF opposes the fall of flux.

    The inductor resists changes in current (therefore changes of the flux).

    If there is no path for this current when the source is switched off the back EMF can be very high. It is only limited by what is connected across the coil (if anything).

    For example, say there is a 3 Amp current flowing through an inductor that has a resistance of 20 Ohm & there is a 100 Ohm resistor connected across it.

    When the source is opened, the inductor (according to Lenz's Law) "wants" to oppose the decay of flux.

    So, in this example, the initial back EMF will be 3 * 120 = 360 Volt.
     
    Last edited: Feb 23, 2012
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  18. Roff

    Roff Well-Known Member

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    Post deleted.
     
    Last edited: Feb 23, 2012
  19. ljcox

    ljcox Well-Known Member

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    nicksydney,
    With all the discussion, we did not fully answer your questions.

    I'll describe the case where the supply is DC since this is simpler than, and a good introduction to, the AC case.

    I'll describe the AC case in another post if you wish.

    When the power is first applied, the 1000 uF caps are discharged, there is no resistance across the output & so base current flows through the base/emitter junction & through the capacitors to ground.

    The gain of the Darlington transistor (hFE) ensures that the emitter current will be much larger and therefore the capacitors will charge at a fairly rapid rate.

    Note that D3 prevents a short between the base & emitter at this point since the base is more positive than the emitter.

    Thus the Caps charge via the transistor until a point of the equilibrium is reached where the base/emitter voltage falls to the point where the emitter current is just sufficient to provide the LED current.

    It is difficult to calculate this point since the collector current of the transistor is an exponential function of the base/emitter voltage and the LED current is an exponential function of the voltage across the LED.

    Now, when the soleniod is applied across the output, the caps discharge into it and cause it to operate briefly (ie. until the current through its coil falls below its hold level).

    The situation regarding the back EMF generated by the solenoid depends upon the energy stored in the magnetic field around the solenoid which is a function of the inductance & the current through it.

    If the inductance is sufficiently large, there will be enough energy stored in the magnetic field that the current, due to the back EMF, will discharge the capacitors and charge them in the reverse direction until diode D2 starts to conduct and thereby clamp the voltage to about 0.8 Volt; thus protecting the capacitor & the Darlington from possible damage.

    But if the inductance of the solenoid is smaller, there will be insufficient energy stored in the magnetic field to fully discharge the capacitors so they will discharge to the voltage determined by the voltage divider effect of the resistors and the resistance of the solenoid.

    When the transient currents have decayed to zero, the base/emitter voltage will not be sufficient to turn the transistor on due to the voltage divider effect I mentioned above. So the transistor will remain off until the solenoid is disconnected.

    Then the transistor will turn on and the capacitors will charge has described above.
     
  20. colin55

    colin55 Well-Known Member

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    The back EMF will NEVER rise above the delivery voltage to the inductor because the delivery voltage NEVER turns OFF and it certainly does not turn off rapidly and thus back EMF does not occur. I have mentioned this previously but my statement was not understood.

    I don't see the relevance of this statement.
    The two diodes on the output can be removed and the circuit will function exactly as before.
     
  21. Roff

    Roff Well-Known Member

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    You can't remove the diode between base and emitter. This would short the emitter to the base. It will still work after a fashion, but you won't get the quick recharge from the transistor.
     
    Last edited: Feb 25, 2012

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