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Baisc resistor, electricity, and capacitor questions

Discussion in 'Electronic Basics' started by electronman, Feb 7, 2007.

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  1. electronman

    electronman Guest

    After you read my questions, you will understand why I joined this
    basic electronics group.
    1) Do resistors resist current or voltage or both? Does a resistor
    actually use up voltage or current as it passes through the
    component? or does a resistor just slow down the passage of voltage
    or current? or ...
    2) From my reading it seems that electrons pass from the negative pole
    to the positive pole. Then please explain why current in a circuit
    flows from positive to negative.
    3) I thought the purpose of a capacitor was to hold a charge for a
    brief instance, then release it. So when I connect a resistor then
    capacitor then LED in series with a 9V battery, why doesn't the LED
    blink on and off as the capacitor charges and releases?
    Thanks in advance for your help.
     
  2. Resistors resist the passage of current by using up voltage.
    Remember that ohms (the unit of resistance is just a
    shorthand way to say volts per ampere. 10 ohms of
    resistance uses up (drops) 10 volts per ampere that passes
    through.
    Electrons are repelled by negative voltages and attracted to
    positive voltages.
    Current is not electrons. It is the movement of charge.
    Charges come in positive and negative polarities. When a
    Negative charge (electron) moves one way, the net charge
    moves the other way (a negative going in a negative
    direction is the same as a positive going in the positive
    direction). I know it sounds stupid that hypothetical
    charge is positive when we know that the charge that is
    moving in metal is bits of negative charge. But there can
    also really be positive charges moving in plasmas and
    solutions. The reason that the charge convention does not
    match electrons is because electrons had not been discovered
    yet when the convention was started. Ben Franklin guessed
    that the charge moving through metal was positive charges,
    and had a 50:50 chance of getting it right. But as long as
    you don't get into the actual particle physics (where you
    have to keep track of both the sign of the particle charges
    and the sign of their direction), it really makes no
    difference whether current is made up of moving positive or
    negative charges.
    I think a better way to think of capacitors is that they
    store energy in an electric field by the attraction and
    repulsion of displaced surface charges. The key formula is
    I=C*(dv/dt) or current is proportional to the capacitance
    and the time rate of change of voltage.
    Look at that formula again. When you first connect the
    battery, there is no voltage across the capacitor, so the
    LED has some voltage available and the rest is dropped
    across the resistor. The voltage across the resistor
    determines how much current passes through it and the LED.
    But that current produces a rate of change of voltage across
    the capacitor (charging it up with some of the battery
    voltage). This capacitor voltage bucks against the battery
    voltage, so that the resistor and LED see only the
    difference between them. (Remember that the total voltage
    around a loop must be zero.)

    So with this lower voltage across the resistor, its current
    must also be lower, which produces a slower rate of change
    of voltage across the capacitor. It charges slower and
    slower as its voltage approaches the entire battery voltage.
    So the effect of having an initially discharges capacitor
    in series with a battery, resistor and LED is to produce an
    initial bright light that fades toward zero. After the LED
    fades out, you might remove the capacitor from the circuit,
    and measure the voltage it carries with your volt meter.
    You should just a rather large capacitor for this experiment
    (an electrolytic capacitor of 100 uF to 1000 uF would be
    good) so that this process happens slowly enough for your
    vision to keep up.

    To use the capacitor to time a blinking operation, you need
    something that measures its voltage rise and reacts with a
    switching decision at some voltage, to short out the
    capacitor, and start the charging process over, again and
    again. You might Google the LM555 timer to learn about a
    chip that is often used for this sort of thing.
     


  3. Analogy to current flow is fluid flow. Voltage = Pressure, Mass flow =
    Current. In a pipe we must have mass conserved(assuming ideal fluid), that
    is, the mass entering the pipe = mass leaving. The same olds for flow in a
    conductor. Resistors are like making the pipe smaller. They increase the
    flow rate and decrease the pressure(slow drift velocity due to the increase
    cross sectional area). Since we can't add or remove fluid/electrons they
    either must speed of or slow down.

    This is why you get heat. When the electrons enter a more resistive medium
    they have to move faster to conserve matter and hence there are more
    collisions. This is also why there is a drop in voltage. Similarly why it
    happens in a pipe is because when the fluid is moving faster it doesn't
    spend as much time hitting the walls of the container which reduce pressure.
    When it slows down it then spends more time(on average) colliding with the
    walls of the vessel which means an increase in pressure.

    If you were to look at the basic laws of fluid flow you would see them to be
    exactly analogous to those of current flow.
    Because when electrons were first discovered old Benji got it wrong. Doesn't
    matter though as all equations are self consistant(its like if some other
    culture used -1 for our 1. It just beans its backwards and we can easily
    convert to there system if we needed to).
    Thats now how a capacitor works. It stores charge. When you connect it in
    series with a battery it will store the charge comming from the batter and
    eventually it will fill up. (like closed vessel in the fluid flow analogy.
    This is not exactly true though since the way the battery works)

    The batter just keeps trying to push in electrons onto the capacitor but
    eventually it cannot push any more(because electrons on the capacitor repel
    each other and eventually theres so many of them that no more can fit. Just
    like trying to fill up a gallon jug with water. Eventually you cannot put
    any more and something will break.).

    Now if you remove the battery from the cap you have a small batter yourself
    and you can run the led from it. Depending on the size of the cap, the
    resistance used, and the direction you have the led(it will only work one
    way. You must get the polarity right) you could light up the led for a
    while.

    The cap will not release its charge unless there is a way. If the DC batter
    is pushing in electrons onto the cap then the cap cannot get rid of any
    electrons. There will be a point where niether the battery or the cap will
    be able to get rid of any electrons. You see this when your led turns on for
    a bit and then stops(Actually slowly turns off). The reason is that the
    battery is sending less and less electrons through through because the cap
    is resisting more and more and eventually resisting all the electrons form
    the battery causing no current to flow(there is charge but no flow of
    charge).

    Jon
     
  4. ehsjr

    ehsjr Guest

    As you will see, you have stepped into an unavoidable
    "terminology snakepit". There is no way to avoid it.
    They resist current.
    They convert electrical energy to heat energy.
    That both limits the current and drops the voltage.
    So called "conventional current" is considered from
    positive to negative. It was originally a 50/50
    chance of getting it right, but it was wrong.
    Nevertheless, it has stuck with us. If one views it
    as electrons moving, then it is negative to positive.
    And, as it turns out, that doesn't always hold true,
    either. Thus the "terminology snakepit". Groan.

    We also use "sink" and "source". Sink means
    the current is going into the more negative side;
    source means current is coming from the more positive
    side.

    A capacitor can charge, store and discharge electrical
    energy. If there was such a thing as a perfect capacitor,
    it would store energy forever, or until it was connected
    to a circuit that allowed it to discharge. (Real world
    capacitors always have some internal leakage resistance,
    so they discharge internally when there is no external
    path for that discherge.) The circuit you describe provides
    a path for the capacitor to charge, but does not allow it to
    discharge. Consider:
    (View in fixed font)

    +--[R]--+
    +| |
    - |
    |B| [CAP]
    |A| |
    |T| |
    |T| [LED]
    - |
    -| |
    +-------+

    The top of the cap is always (+) [and the bottom
    is always (-)]. For there to be current, the top
    of the battery needs to "see" a voltage that is
    different from what it has stored inside. The
    circuit provides no mechanism for that. But suppose
    we modify it, as follows by adding a switch:

    S1
    +--[R]--o o----+
    +| \ |
    - | |
    |B| [CAP] |
    |A| | [R]
    |T| | |
    |T| [LED] |
    - | |
    -| | |
    +--------+------+

    With the switch to the left, the cap can charge -
    it has a circuit from the battery, through the
    switch to the top of the cap with the bottom of
    the cap connected through the LED to the (-) of
    the battery.

    Flipping the switch interrupts that charge path,
    and completes a path for the cap to discharge:
    from the top of the cap, through the switch, through
    the LED, to the bottom of the cap.

    (The R in the drawing is to indicate that a resistor
    is often used to limit the current through the LED;
    you always need to ensure that the current through
    an LED is limited somehow, regardless of how you
    do it.)

    Ed
     
  5. Resistors convert voltage and current (or perhaps more accurately,
    electrical energy) into heat. They also limit the current based upon the
    voltage being applied.
    Cuz, Ben Franklin was a scientist. Like all great scientists, he saw
    something intriguing and guessed about it's true nature. Even though he had
    a 50% chance of guessing right, like many scientists he guessed wrong but
    didn't know it. I think it has something to do with a guy named Murphy.
    Hence he (and others to follow) went on assuming Franklin was right about
    the charge polarity and proceeded to come up with much theory based upon it.
    Anyway, like a good many scientific things go, some later scientists were
    able to "look" (check and see whether it was positive or negative) and they
    discovered, much to their dismay I'm sure, that they had indeed been wrong
    all along. Well, to make a long story short, it was too late to go back and
    revise everything so we're stuck with things like they are. They just came
    up with some more theory (as the Johns have explained) to help smooth out
    the wrinkles some. ;-)
    The LED should come on as the capacitor charges up since current is flowing
    during that time period. Depending upon the resistor and capacitor used,
    you may or may not see some light. Once the capacitor is charged up, no
    more current will flow and the light will stop. The capacitor will be
    blocking the DC current flow at this point.

    Think of water flowing in the circuit instead of electrons. When current is
    first flowing, you can think of the capacitor as having a sort of flexible
    membrane that allows water to flow out of the capacitor on one side, and to
    flow into the capacitor on the other. At some point, the membrane is
    stretched until the tension on it matches the pressure of the water and the
    flow stops. You could say the capacitor is charged up to some amount of
    potential (voltage in the case of electicity or pressure in the case of
    water). As you can imagine when the capacitor is charged, it is very
    willing to give up the energy stored inside. The problem is that it can
    only flow back where it came from, not thru the membrane in the original
    direction of flow. This is not a great analogy, but it should get the basic
    idea accross.
     
  6. Both. If you try to push through a crowded room trying to go faster (more
    voltage) only helps a little.
    As heat.
    Edison guessed and got it wrong. It actually doesn't matter that much.
    There is no 'release' in this case.
     
  7. Rich Grise

    Rich Grise Guest

    Everyone seems to have answered your questions quite adequately. But,
    while looking for some stuff about the positive/negative thing, I
    happened on a web page that doesn't necessarily teach you much, but
    it's kind of a fun read:
    http://www.sciencemuseum.org.uk/on-line/treasure/objects/1925-814.asp
    It's especially amusing if you've heard of the "audiophool" set. ;-)

    We old fart techies learned about current flow in the age of vacuum
    tubes ("valves"), where the charge carriers are obviously electrons.

    Here's the bottom line:

    College boys use positron flow.
    Techs, the guys who do the actual work, use electron flow. ;-)

    It's really irrelevant which convention you follow, as long as
    you're consistent throughout.

    Cheers!
    Rich
     
  8. Ratch

    Ratch Guest

    If the total resistance in a circuit increases, the current will be
    less. If a single resistor increases with respect to other resistors in a
    series circuit, the voltage will increase across it, and decease across the
    other resistors. Does that mean the resistor in "resisting"?
    Since voltage is the energy density of the charge, and the resistor
    dissipates the energy as heat, then there is less energy density of the
    charge after passing through the resistor. Therefore the voltage is less.
    There is the same amount of current entering as leaving the resistor, so
    there is no loss there.


    The current moves through a conductor very slowly. The voltage field
    establishes itself across the resistor at light speed. A good analogy is a
    water hose filled with marbles. Force a marble into the hose at one end and
    almost immediately another pops out the othe end. It will be a long time
    before the same marble comes out the other end. A conductor is filled with
    a sea of charges, so the current responds very fast to the voltage, but the
    actual charge movement is very slow.

    So many people get hung up on that one. And a lot of folks blame it on
    Franklin. Here is their problem and yours. You all are trying to determine
    charge flow by looking at the movement of the electrical charges. At first
    it seems logical, but it gives rise to confusion. Why? Because there are
    just as many positive charges in a balanced universe as a negative ones.
    They include, but are not limited to positive ions in electrochemistry,
    semiconductor "holes", positrons, etc. Get the picture? So Franklin was
    wrong about defining current as electrons going from positive to negative,
    but he was right about positive charges like positrons and ions going from
    positive to negative. If he said that current was defined by charges going
    from negative to positive, then he would be right about electrons, and wrong
    about positive charges going from negative to positive. To avoid having to
    keep track of what polarity each charge is, and having to change directions
    and signs in loop equations, the scientists and engineers, not Franklin,
    came up with a convention that says charges flow from positive to negative,
    no matter what their polarity is. This gives consistent values that are
    correct signwise for positive charges, and opposite for negative charges.


    I would have to see your circuit. Ratch
     
  9. jasen

    jasen Guest

    yes, both
    "as it passes through" that means it comes out the other side,
    the voltage that was pushing the current through comes out reduced though.

    this is described by "Ohms law"
    current,charge, and or voltage was discovered before electrons. The
    discoverer guessed wrong.
    What capacitance is your capacitor (somewhere between 10 and 100uF
    would be about right)

    Bye.
    Jasen
     
  10. jasen

    jasen Guest

    That won't work LEDs are Diodes.

    Bye.
    Jasen
     
  11. electronman

    electronman Guest

    - -- - - - - - - - -
    Rich,
    This Email is intended as a huge thanks to all of you who replied so
    thoughtfully yesterday to my first post to this group. What a wealth
    of information and explanation I received! I've definitely found my
    electronics Internet home here. As I work through my introductory
    electronics books, I now know where I can get my questions answered.
    The two books I'm studying so far are "Electronics for Dummies" and
    "Electronic Circuits for the Evil Genius." Any other suggestions?
    I'm also breadboarding each project in the "Evil Genius" book as I
    move along, so I'll have lots more questions. Thanks again. -
    "electronman"
     
  12. Rich Grise

    Rich Grise Guest

    .
    Doing your own studying and actually building stuff is highly regarded
    around here - I'm sure you'll fit right in. :)

    As far as the books, If I had "Electronic Circuits for the Evil Genius."
    and "Electronics for Dummies", I'd stick with the "Evil Genius" book -
    after all, which would you rather identify with? ;-) and give the Dummies
    book to the dog or something. ;-) (unless it's actually useful to you,
    which, how would I know? in that case, put this in the "90%" bin. ;-) )

    Cheers!
    Rich
     
  13. ehsjr

    ehsjr Guest

    Yup, you're right. Easily fixed:
    Move the LED below the resistor
    on the right.

    S1
    +--[R]--o o----+
    +| \ |
    - | |
    |B| [CAP] |
    |A| | [R]
    |T| | |
    |T| | [LED]
    - | |
    -| | |
    +--------+------+

    Cap charges through left hand R when switch is
    left, and discharges through the LED and R on
    the right when switch is changed to right.

    Ed
     
  14. Mike_in_SD

    Mike_in_SD Guest

    (Rich Grise) wrote in <pan.2007.02.08.18.30.05.670569
    @example.net>:
    Look on Amazon for a book by Forest Mimms called
    " GETTING STARTED IN ELECTRONICS" IMHO it is the
    best book you could possible read to learn electronics.

    I have personally bought it for about 15 people throughout
    the years and every person was floored.

    good luck
    mike
     
  15. Robin

    Robin Guest

    Think of voltage as "pressure". Then it makes sense e.g. if you double
    the pressure then the current doubles (in the same value resistor).
    Likewise if you halve the voltage then the current halves.

    Your Capacitor is too small. Capacitors are a lot smaller than first-
    impressions. If e.g. buy a 3.3 Farad capacitor from e.g.

    http://www.mouser.com/search/refine.aspx?Ntt=555-DZ2R5D335

    you can do wonderful things like charge it up from a 3Volt battery
    (for five minutes) and then put an LED across it (no resistor
    required*) and watch it glow for minutes.

    Cheers
    Robin

    *This is abnormal for capacitors - they are generally viscious. These
    "super capacitors" have a "high internal resistance" and that makes
    them safer than the usual type capacitor which would, for this size of
    several Farads, be rather dangerous.
     
  16. ehsjr

    ehsjr Guest

    Robin wrote:
    (much snipped)
    Quite possible - he didn't mention the cap size.
    I figure he needs at least 2.2 uF to get a readily
    visible blink. He could charge it to say 6 volts,
    then discharge it through the LED and a 1K resistor.
     
  17. MassiveProng

    MassiveProng Guest


    Excellent method of defining Ohm's Law in yet one more
    understandable way.
     
  18. Jamie

    Jamie Guest

    I just wanted to say, interesting name you have Mr MassiveProng.

    You are referring to components of a plug correct?, i just
    can't for the life of me think of it being anything else!
     
  19. A capacitor does not "hold a charge then release it" as you seem to
    expect.

    I suggest that you consider a capacitor as a (very small) rechargeable
    battery - when you apply a voltage to a capacitor, it will charge up
    to that voltage, and retain that charge. If you disconnect the
    charging voltage and connect a load (resistor and LED, of example) the
    capacitor will discharge through that load. With a suitable voltage
    and a large enough capacitor, the LED will blink once. You would have
    to recharge the capacitor, then connect the load again to see the LED
    blink again.

    To make the LED blink on and off, you need some active devices -
    transistors or an IC - in a suitable circuit to switch the LED current
    on and off. Part of the circuit will use a capacitor to set the
    on/off timing, but a capacitor alone won't make anything blink.


    --
    Peter Bennett, VE7CEI
    peterbb4 (at) interchange.ubc.ca
    new newsgroup users info : http://vancouver-webpages.com/nnq
    GPS and NMEA info: http://vancouver-webpages.com/peter
    Vancouver Power Squadron: http://vancouver.powersquadron.ca
     
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