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A circuit demonstrating exponential decay....

Discussion in 'Electronic Basics' started by [email protected], Jun 10, 2005.

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  1. Guest

    I'd like to hook up a flashlight lamp, 1.5 v battery, and capacitor in
    series, close the circuit, and watch the lamp come on and slowly go off
    as the capacitor charges up.

    I tried it with a double electric layer capacitor (1 farad), but I now
    think this type of cap has a big internal resistance, so, nothing
    (observable) happened..

    Now I'm thinking an electrolytic cap (0.1 farad) and maybe a 10 ohm
    resistor in the circuit to get the time constant up to > 1 sec (R*C).

    Is this going to work?


  2. John Fields

    John Fields Guest


    Try this:

    +------->\ <------+
    | \ |
    |+ O |
    [BAT] |+ [LAMP]
    | [1F] |
    | | |

    It ought to work with the 0.1F as well.
  3. It might produce a brief flash. There are a couple problems with
    lamps in this experiment. The first is the very low resistance of a
    cold filament. By the time the filament has gotten hot enough to see,
    most of the energy has already been drained out pf the capacitor with
    a much shorter time constant than you are expecting. The second
    problem is that the light output from a filament is at all
    proportional to the current through the lamp.

    I suggest you go back to your original cap, but replace the lamp with
    a red LED. it will produce light very nearly proportional to the
    current through it, and you should see a nice decay each time you
    reverse the capacitor connections. I say a red LED, because it takes
    less voltage to light it than other colors.
  4. Bob Monsen

    Bob Monsen Guest

    I think the cap is doomed to failure. The bulb probably draws more than
    500mA. Thus, the voltage will change at 0.5V/s. If there is even 1 ohm
    of resistance in the cap, it'll suck up 0.5V, probably preventing your
    bulb from lighting at all.

    If you have a darlington transistor, (or two NPN transistors,) you can
    use it to do something similar to what you want, although you will have
    to use a couple of batteries in series.

    Use a 100uF capacitor, and a 100k resistor; connect both to ground, and
    to each other. Connect the positive junction between them to the base of
    the darlington. Then, connect your flashlight battery to the emitter of
    the darlington, and the collector to 3V.

    Now, momentarily touch the junction of the base, 100uF, and 100k
    resistor to 3V, through a normally open pushbutton or something.

    It'll charge up to 3V. When you release the button, it will decay down
    to 0V using a nearly exponential decay. The bulb will thus have an
    exponential decay of the voltage across it, and should dim accordingly,
    over about 5 seconds.

    I just built one using two 2N4401 transistors to simulate the
    darlington. I only had a 3V flashlight, so I set my power supply to
    4.5V, and gave it a try. It starts out quite bright, and decays over a
    period of about 7 seconds to nothing. It is hard to determine if the
    decay is exponential, but I'm guessing that the non-linearities in light
    vs voltage are keeping that from happening.

    The flashlight bulb passes 0.5A at 3V, so the 4401 is getting warm.
    However, it's well within the dissipation limits of the device.

    | | | |
    no switch |/ | |
    o------o----| ' |
    | | |> |/ ---
    | | |-| - 3V minimum
    | | |> |
    '+ .-. | |
    === | | | ---
    /-\ | | .-. -
    | '-' ( X ) |
    | | '-' |
    | | | |
    (created by AACircuit v1.28.5 beta 02/06/05
  5. John Fields

    John Fields Guest


    Or, better yet:

    +-----> |
    | |
    | O
    |+ |
    [BAT] +------+
    | |+ |
    | [1F] [LAMP]
    | | |

    That way, when you turn the switch ON the battery will get the lamp up
    to operating temperature and charge the cap at the same time, then
    when you turn the switch OFF the cap will discharge through the lamp
    without having to use a lot of its charge to heat the filament from a
    cold start.
  6. Guest

    Or, better yet:

    +-----> |
    | |
    | O
    |+ |
    [BAT] +------+
    | |+ |
    | [1F] [LAMP]
    | | |

    I like it, but I think the (assumed) high cap resistance of the 1 f
    double layer electric cap will prevent it from working.

    If I use a 0.1 f electrolytic and the hot bulb has 5 ohms resistance I
    can get a time constant of 0.5 seconds, which isn't what I was hoping
    for but isn't bad. (If the electrolytic has negligible internal
  7. Guest

    I like the LED idea, what is its resistance?
  8. Guest

    | | | |
    no switch |/ | |
    o------o----| ' |
    | | |> |/ ---
    | | |-| - 3V minimum
    | | |> |
    '+ .-. | |
    === | | | ---
    /-\ | | .-. -
    | '-' ( X ) |
    | | '-' |
    | | | |

    Nice work. But.... I'm trying to put this thing together for a
    calculus book for junior high students to illustrate an exponential
    function (starting from the DE characerizing the circuit) in nature, so
    the circuit has to easy for anyone to wire up. Hence, no
    transistors... etc.

    I did take a few EE course in college, and as I recall caps always had
    0 resistance. What are the facts? How about a 0.1 farad electrolytic?
  9. It is very nonlinear, like any forward biased diode. The higher the
    current, the lower the resistance. I was assuming that the high
    internal resistance of the double layer capacitor would dominate the
    time constant.
  10. Bob Monsen

    Bob Monsen Guest

    Caps have both resistance (called ESR) and inductance (called ESL).

    Here is a tutorial containing more that you ever wanted to know about

    BTW, if the students have access to multimeters, it's easier to just
    have them take time measurements of voltage across a resistor in an RC
    setup. 100uF + 1MEG ohm gives a time constant of 100, so measurements
    are likely to end up looking exponential on a graph in a lab. Two
    students, one saying 5, 10, 15, 20... the other recording voltage at
    that time.

    Also, the parts are much cheaper (except the multimeter, of course,
    which they could share).
  11. John Fields

    John Fields Guest

    Take a look at this:

    Note that the 1F unit has an internal resistance of less than or equal
    to 0.12 ohms.

    If you look up the spec's for a 1728 lamp (1.35V, 0.06A) it'll present
    a resistance of about

    E 1.35V
    R = --- = ------- = 22.5 ohms,
    I 0.06A

    when it's hot, so that 0.12 ohm ESR in the cap is only going to "kick
    in" and decrease the initial current through the lamp by about 5% when
    the switch is opened.
    See above. With a 1F cap discharging into a lit bulb with a filament
    resistance of 22.5 ohms in series with the cap's 0.12 ohm ESR one
    would expect a time constant of about 22.5 seconds, but since the
    lamp's filament resistance will be dropping as the cap's charge
    becomes depleted and its output voltage decreases, the time constant
    will be considerably shorter. How much shorter? I don't know, but if
    I took a WAG I'd say by about a factor of two.

    IMO, that would be an interesting experiment/ exercise for you or your
    students to perform. That is, calculating the filament resistance by
    measuring the current through it with different voltages across it,
    and then determining the time constant of the system as the resistance
    falls because of the decay of the capacitor voltage across the falling
    filament resistance.
  12. Rich Grise

    Rich Grise Guest

    Be sure and use _analog_ multimeters, you know, the kind with a
    meter needle? DVMs are almost useless for this sort of thing.

    And I've done this very thing, except with a fairly large aluminum
    electrolytic cap, maybe 1000 - 10000 uF (I don't remember the exact
    value), ~24 VDC, and an LED + 1K resistor. The voltage across the
    resistor was 1V per mA, of course.

    I was amazed - from 20 mA down to 10 mA, the brightness seemed not
    to change _at all_, at least to the naked eye (shame on you! Put
    some clothes on!) over a period of about a second, and then decreased
    practically linearly from 10 mA to zero. I _saw_ the knee at 10 mA! I
    even turned off the room lights, where the LED was the only light in the
    room, and the LED just kept dimming, and dimming, and dimming, until there
    was zero light at zero current.

  13. Guest

    Here is a tutorial containing more that you ever wanted to know about

    Nothing on double electric layer caps that I can find :).
  14. Guest

    And I've done this very thing, except with a fairly large aluminum
    electrolytic cap, maybe 1000 - 10000 uF (I don't remember the exact
    value), ~24 VDC, and an LED + 1K resistor. The voltage across the
    resistor was 1V per mA, of course.

    Excellent. I see that 100000 uF electorlytics are available cheap, so
    that's good.... and I suppose I could go with a 9 volt battery.... (not
    having a power supply)

    I'm not really concerned that the LED brightness doesn't accurately
    represent an exponential, I just want the thing to slowly turn
    off.....that will be enough.

    I'll give this one a go.

  15. Guest

    Take a look at this:­F/DZ-N.PDF

    Excellent. I didn't find the internal resistance spec for my Panasonic
    double layer 1 farad ... but according to the spec for the ELNA the
    internal resistance is 10% the usual, and is what .12 ohms which means
    the Panasonic should be about an ohm which means my original circuit
    should have worked.....

    In any case..... I'd like to get one of these caps if they're cheap.

    I did hook mine up the wrong way .......once.... maybe I cooked it.

    Hmmmmm, in any case, we're getting there.

  16. Bob Monsen

    Bob Monsen Guest

    Hmm, I guess it isn't the be all and end all site. Drat.
  17. John Fields

    John Fields Guest

    Not necessarily. Because of the low cold resistance of the lamp
    filament in series with the cap's one ohm ESR,

    you might not have been able to get enough energy into the filament to
    heat it up to incandescence before the charge in the cap was depleted.
    It also depends on the thermal time constant of the filament.

    Note that the ESR of their 10F offering is <0.1 ohm
    Mouser's got them:
    Easy enough to find out.

    Charge it up through a resistor and measure the charge current. If
    the current into the cap doesn't eventually decay to the specified
    leakage current of the cap, then you've hurt it. If it looks good,
    then discharge it through the resistor. If T = RC is out of spec,
    then you've hurt it.
  18. Guest

    Thanks to all for the assistance. I finally got something to work
    using an LED. I found the spec for the Panasonic cap, it has an
    internal resistance of 30 ohm at 1000 Hz...this doomed my original
    circuit.... I do have the Elna cap on order and will hopefully get
    something going with a bulb.....

  19. mike

    mike Guest

    If it's just a demo of exponential decay, why not make it easier on
    Use a LED, higher voltage and a MUCH smaller cap.
    If you must have an incandescent, use an emitter follower to drive
    it from a much smaller cap. Ain't nobody gonna be able to tell by
    looking whether it's perfectly exponential anyway.

    The key phrase is, "much smaller cap" ;-)


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