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Capacitor circuit and current flow

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

  1. nicksydney

    nicksydney New Member

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    Hi,

    At the moment I'm reading up about capacitors and resistors and I get the basic part of it is for, but I'm having bit of difficulty in trying to understand the components once it interact together in a circuit. I just need some input and validation to my analysis from the expert here :)




    Here is the circuit that I obtain from the web it is called 'Capacitor Discharge Unit'

    [​IMG]

    The lines and numbering are mine to show you the flow of current that I think happens this is also to refer it easily in my observation below

    ------------------------------------------------------------------------------------
    This is what I think happens when the CDU is connected
    ------------------------------------------------------------------------------------

    1. The current flows through line 1 and it will keep on continue flowing through BD679
    2. The current flows through line 2 and will pass through the resistor flowing through the base of the BD679 (thus allowing step 1 to happen)
    3. The current flows through line 7 and will pass through normally
    4. The current that was flowing pass through BD679 will also lass through line 5 and this in turns will be stored by the capacitor (both 1000u)
    5. The current also flows through line 6 when it will light up the LED

    ---------------------------------------------------------------------------------------
    This is what I think happens when the DCU is disconnected
    ---------------------------------------------------------------------------------------

    The capacitor will be filling up to same 15v like the input source and once it fills up there will be no current flowing from anywhere as the wire are jam packed with electrons and this in turn will reduce the voltage which in turn will "turn off" the signal from the BD679 base. Because the base of the BD679 has not been triggered the gate will not open and this will stops the current to flow.

    I need feedback to validate what I'm thinking is correct in terms on how the current flows through the circuit.

    Thanks for the help
     
  2. duffy

    duffy New Member

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    Ugh. Who designed this thing? The darlington transistor and the four 2k resistors in parallel (which could be replaced with a single 560 Ω resistor) in that configuration form a current regulator. It will charge the caps off that half-wave rectifier, slowly. The 1N4004 marked "7" isn't doing much of anything.

    Looks like you could remove the transistor, remove the diode with the "?", and replace that last 1N4004 with a piece of wire - and the circuit would work almost exactly the same.
     
  3. JimB

    JimB Super Moderator Most Helpful Member

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    As drawn the transistor is doing nothing.
    There is a short circuit from base to emitter, this is done by "line 7" in the annotation.

    A truely awfull circuit.

    JimB
     
  4. dave

    Dave New Member

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  5. alec_t

    alec_t Well-Known Member Most Helpful Member

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    If the input to the circuit is 15VAC RMS then the transistor is pulsed on each half cycle and each pulse supplies charging current to the caps. The caps charge to about 19V. They discharge through the LED and 1k resistor. Line 7 imposes voltage pulses of about 2V amplitude on the + output of the circuit.
    If the circuit input is 15VDC then the transistor is on continuously and charges the caps to about 13VDC. The output voltage, by virtue of line 7, is about 14.3VDC.
    No. The diode on the right prevents line 7 from causing a short-circuit.
     
  6. JimB

    JimB Super Moderator Most Helpful Member

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    JimB said:
    Alec_T said:
    JimB says: Ooops! you are quite right!
     
  7. colin55

    colin55 Well-Known Member

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    I designed the circuit.
    How little you know about electronics.
    The circuit is a Capacitor Discharge Unit for model railways.
    We have sold thousands of kits.
     
  8. DerStrom8

    DerStrom8 Super Moderator Most Helpful Member

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    Hey, no need to be rude! So what if he is a beginner? That doesn't give anyone the right to talk them down for not knowing how it works. That's what we're here for--to help.

    I must ask, why do you have 4 2.2k resistors in parallel? It seems a waste of components. Same with the 2 1000uF capacitors. I'm curious as to your reasoning behind this. At first glance to me, it seems to be a poorly-designed circuit, no offense intended.

    Regards
     
    Last edited: Feb 21, 2012
  9. JimB

    JimB Super Moderator Most Helpful Member

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    Well bully for you!
    But for the rest of us who know nothing, what is it supposed to do?
    Its function may be immediately obvious to a model ralway enthusiast, but my knowledge is about 50 years out of date in that area.

    JimB
     
  10. edeca

    edeca Active Member

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  11. colin55

    colin55 Well-Known Member

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    I am talking about JimB who has claimed 2,830 replies and says: "A truely awfull circuit." (should be truly)
    He has absolutely no idea how the circuit works and I am particularly annoyed by someone criticizing my circuits when they know absolutely nothing about the operation of the circuit.
    He should have said at the beginning of his post: “I have no idea how the circuit works.”
     
  12. duffy

    duffy New Member

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    Oh, I think we can settle this one pretty easy -

    nicksydney: Please remove the transistor, remove the diode with the "?", and replace that last 1N4004 with a piece of wire, and see if it doesn't work exactly the same on your model railroad solenoids.
     
  13. DerStrom8

    DerStrom8 Super Moderator Most Helpful Member

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    Well I'm sorry to say it, but I must agree with him. It does seem to be very poorly designed.
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    What was the original intention of this circuit, what is it supposed to be used for exactly, what was it designed for?

    It says "capacitor discharge circuit" but it appears to be some sort of power supply ??? What capacitor gets 'discharged' ?
     
    Last edited: Feb 21, 2012
  15. nicksydney

    nicksydney New Member

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    Hi all,

    Thanks to all for replying to my question. I didn't state it clearly in my question that the main purpose for the diagram and line is for me to understand the current flow and not to build anything or design anything. I'm very new in electronics as I'm trying to learn the basic and this diagram have combo of resistor and capacitors that will allow me to study how current flows into capacitors and how the different components interact with one another.

    Really appreciate if any of you can tell me from my analysis in my question whether my thinking is correct and if it is not than what is the correct flow.

    Thanks for any feedback.
     
  16. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    I'd be happy to help and im sure others would too, but from the other posts it sounds like we need to know what connects to the output before we can help effectively.

    There are a few questionable areas of this circuit which are not apparent without seeing the type of load that is supposed to be used with this circuit. We have to wait until that becomes clear if you want a really good answer.
    My question to you is what load do you intend to connect to the output?
     
  17. DerStrom8

    DerStrom8 Super Moderator Most Helpful Member

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    I'm going to re-phrase that, since the OP said that he wasn't actually building this circuit. A better way to ask it is what load do you want to assume is connected to the output?
     
  18. colin55

    colin55 Well-Known Member

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    The circuit is designed to connect to a 4 to 6 ohm load.
     
  19. nicksydney

    nicksydney New Member

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    Yes, that's right I'm not building the circuit just trying to understand how the whole current flows as it will give me a strong validation on my electronics learning :)

    Sorry if my question is confusing.
     
  20. nicksydney

    nicksydney New Member

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    Thanks for the reply colin55
     
  21. colin55

    colin55 Well-Known Member

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    This is not a "normal circuit." It is a very clever circuit.
    You will notice no-one is even close to describing its operation.
    It does not matter what load the circuit is supplying, the operation of the circuit can be described when supplying a LOW, MEDIUM or HIGH impedance (or resistance) load and the features of the circuit can be explained.
    This has not been done correctly and no-one has even the slightest idea of how it operates.
    Because of its unusual nature, it is not an ideal circuit to be studying. You will possibly never cme across this type of circuit again. It is "one of a kind." That's why I designed it and that's why it has been so popular. It is designed to prevent a "point motor" burning out.
     
    Last edited: Feb 21, 2012

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