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Low pass filters

Discussion in 'Electronic Basics' started by zalzon, Jul 30, 2004.

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  1. zalzon

    zalzon Guest

    No matter how much effort I put into understanding these low and high
    pass circuits I always come up clueless. I can't seem to visualise
    what the electrons are doing at the resistor and capacitor. Its been
    slowly driving me insane.

    Why is low frequency AC allowed to pass but not high? Can someone
    explain this to me in easy to understand terms referring to the
    electrons in particular as they move around the circuit.

    Please don't give me equations like Xc = 1/2pifC..etc cos that is just
    crap. I know that already and its not helping me visualise anything.

    I want to know what the electrons do when they reach the resistor in a
    low pass circuit with low frequency AC. What happens then when they
    encounter the capacitor. Why is low freq allowed to pass but not high

    Thank you
  2. Well, keep in mind that electronics is NOT exactly like water... but for this
    explanation, it might work for now. Imagine the following:


    | |
    | |
    flexible | |
    vast ocean of water garden hose | |
    | |
    | |


    Now, imagine that you are somehow able to instantaneously move the ocean of
    water up and down to any level you desire. It's an unlimited supply of water,
    though. No matter how much water you take out or how much water you put back,
    it's level stays right where you put it. Also, the height of the bucket stays
    exactly the same, regardless of the height of the ocean you decide on. This
    means that the garden hose, of course, must be flexible so that the bucket can
    stay at one permanent height while that huge ocean moves up or down.

    Now, the question is, what happens to the level of water in the water bucket
    when you move the ocean up and down, given that there is a garden hose

    If you move the ocean up somewhat and wait, some water will flow into the bucket
    until the levels are equal. And then it just sits there. If you move the ocean
    back down some, the water will flow back out of the bucket and into the ocean
    and the level will once again become stable when the levels are equal.

    In a sense, if you wait long enough, the two levels are always equal. Right?

    The level in the bucket is like a voltage level. The garden hose is like a
    resistor (where the diameter is akin to the resistance -- larger diameter hoses
    are lower resistances, narrower hoses are higher resistances.) The bucket is
    like the capacitor (where the diameter of the base of it is like the capacitance
    -- larger base diameters correspond to larger capacitor values.)

    And the level of the ocean is like the level of some signal at the input of the
    RC low pass filter and the level at the bucket is like the output of the RC low
    pass filter.

    If you move your ocean up and down very slowly, by comparison with the hose
    diameter and the bucket base diameter, then the level at the output will always
    track very closely to the levels at the input. But if you now imagine moving
    your ocean up and down more and more quickly, you can see that the bucket won't
    get a change to completely fill up (on the upswing) to the peak level of the
    ocean, before the ocean's height is already heading back down again. If you
    oscillate the ocean somewhat quickly, the bucket will only get up so high before
    the ocean level goes below that point and the bucket starts draining again. And
    so on. So, if you look at the bottom to top span you are moving the ocean up
    and down (let's say it is a span of 10 meters, from top to bottom) then the
    bucket will probably only go up 2 meters above its average point before the
    ocean has already proceeded below that point and starts draining the bucket; and
    the bucket might only go 2 meters below it's average before the ocean is already
    rising above that point again. And so on. In other words, the bucket level
    would only span 4 meters from top to bottom, while the ocean is moving up and
    down by a total of 10 meters span.

    If you narrowed the hose more, that bucket level would only get up say 1 meter
    above average before proceeding back down, and 1 meter below average before
    going back up. In other words, a narrower hose (higher resistance) means the
    input change is less and less reflected in the output changes. Similarly, if
    you instead simply increased the base diameter of your bucket (increased the
    capacitance), then the same thing would also happen -- the levels wouldn't get a
    change to rise quite as high or fall quite as low. Again, the input change
    would not be as well reflected in the output.

    If you also moved the ocean up and down still faster and faster, but kept the
    same hose and the same bucket, once again the 10 meter changes in the input
    would be even less and less reflected in the level of the bucket (the output.)

    If you went fast enough, moving your ocean up and down, you'd probably barely
    see anything more than only very slight changes in the bucket level. Almost
    none of the input change would be reflected in the output.

    It's kind of like that.

  3. John Larkin

    John Larkin Guest

    It's like cars running into bales of hay. They have to slow down and
    fight their way through, because they encounter resistance and can't
    see where they're going.
    This is more like a parking lot. They pile up, hang around a while,
    and then can leave.
    Cars already going slow don't much mind the hay or having to stop for
    a while; the throughput is the same. If you sent cars into this system
    on alternate days, they'd all emerge on those same days. But if you
    sent one convoy every 10 minutes, only a small, smooth trickle would
    come out the other end of the process.

    That's maybe about the best you can do with analogies.

  4. Ian Bell

    Ian Bell Guest

    It's is quite simple really. The electrons cannot cross from one electrode
    of the capacitor to the other. So in the low pass filter with a series
    resistor and a capacitor to ground the rush thru the resistor and bunch up
    on the capacitor plate. The speed the can rush there is determined by the
    resistor. The more of them there are on the capacitor plate the higher the
    voltage across the capacitor. So at low frequencies a lot of them can get
    thru the resistor and onto the plate so the capacitor voltage gets higher.
    At high frequencies far less of the reach the capacitor plate before they
    have to turn round and go back again so the voltage on the capacitor is


  5. Dbowey

    Dbowey Guest

    zalzon posted:
    << No matter how much effort I put into understanding these low and high
    pass circuits I always come up clueless. I can't seem to visualise
    what the electrons are doing at the resistor and capacitor. Its been
    slowly driving me insane.

    Why is low frequency AC allowed to pass but not high? Can someone
    explain this to me in easy to understand terms referring to the
    electrons in particular as they move around the circuit.

    Please don't give me equations like Xc = 1/2pifC..etc cos that is just
    crap. I know that already and its not helping me visualise anything.

    I want to know what the electrons do when they reach the resistor in a
    low pass circuit with low frequency AC. What happens then when they
    encounter the capacitor. Why is low freq allowed to pass but not high
    I will give you something that may help you visualize what you want, but just a
    little math will help you a lot.

    Your question is about an RC filter, so I will confine my answer to just that.

    1. The value of a resistor states the RATIO of the applied voltage to the
    current that will flow in the resistor. Examples: 1 Volt connected to a 1 Ohm
    resistor will cause 1 Amp to flow (1:1). 1000 Ohms and 1 Volt will result in
    ..001 Amps to flow (1:.001).

    At the level of this explanation, a resistor has ONLY resistance and no
    reactance. So, whether the voltage I am applying is AC or DC, the result will
    be the same amount of current for a given resistor.

    2. This should be obvious, but I'll say it anyhow: If two resistors are wired
    in series, the sum of the two is the total resistance (RT) and the RT states
    the E to I ratio for the circuit. The same current (amount of current) flows
    through both resistors in a series circuit rgardless of the resistance of the
    individual resistor.

    3. The voltage drop across each resistor is easily determined by E=IR. E is
    the voltage measured across a resistor, I is the measured or calculated current
    through the series circuit, and R is the value of the resistor for which
    voltage drop is being calculated.

    If the resistors in a series circuit are equal, the voltage drop across each
    one will be the same and the sum of the two will be the voltage magnitude
    applied to the series circuit (see Kirchhoff's law). If the two resistor
    values are different, each will have a different voltage drop, but the sum of
    the two will be the voltage magnitude that is applied to the series circuit.

    4. We're almost there.

    5. The two resistor series circuit that is connected to a voltage source (AC
    or DC) is a voltage divider. Consider any combination of two resistor values
    and determine for yourself the voltage drops across each, so you are
    comfortable with this.

    6. Now we will consider a voltage divider where one of the parts is not a
    resistor, but is a capacitor: We will look at a high-pass filter and a
    low-pass filter using the same components.

    This is where Xc=1/2pi*F*C would be very helpful, but since you don't want to
    get into that we will gloss over it. A capacitor *does* have reactance (Xc)
    and we cannot ignore it for a filter (or anything else, actually). For this,
    consider reactance in Ohms to be the same as AC resistance in Ohms.

    7. On to our filter - Low Pass:
    Place a resistor and capacitor in series. Let's assume the Resistor is 180
    Ohms and the capacitor is 1mF. Connect a 1 kHz. signal to the free end of the
    R. Connect a AC voltmeter or scope to the midpoint of the R and C, which is
    the output of this filter. Connect the common (ground) of the signal
    generator, the common of the meter, and the free end of the capacitor,

    At 100 Hz the reactance of the capacitor is about 1800 Ohms. At 1 kHz the
    Reactance is about 180 Ohms. At 10 kHz it is 18 Ohms. As the frequency
    changes, the voltage drops will change, but the sum of the two will be the
    applied voltage.

    Assume the applied signal is 1 Volt, at 1 kHz there will be 0.5 volts drop
    across the resistor and 0.5 volts drop across the capacitor.

    You should be able to calculate, from Ohms law, the current in the series
    circuit and the voltage drops of the two components.

    You will see it is a low-pass filer, because the reactance of the capacitor is
    lower at 10 kHz than at 100 Hz. so it shunts more signal to ground at 10 kHz.
    than at 100 Hz. (passes more Lo than Hi).

    8. High Pass:
    Place a resistor and capacitor in series. Let's assume the Resistor is 180
    Ohms and the capacitor is 1mF. Connect a 1 kHz. signal to the free end of the
    C. Connect a AC voltmeter or scope to the midpoint of the R and C, which is
    the output of this filter. Connect the common (ground) of the signal
    generator, the common of the meter, and the free end of the resistor, together.

    Using the same process as above, you will see this is a High pass
    filter,because the capacitor has less voltage drop at 10 kHz than at 100 Hz.

    Reactance isn't crap if you want use capacitors or inductors when you learn
    enough to actually design something - like a good filter, a transmitter, a
    receiver.... you name it.

  6. Here is a try:

    First, try to figure out how a capacitor works. If you can visualize
    how a cap works, you can visualize an RC filter. A cap is just two
    conducting plates separated by some small distance. If you add a
    charge to one side, that pushes a charge on the other side away.

    Now, for a DC charge on the capacitor, its easy to see that no charges
    are coming into or going out. There isn't any current. The voltage
    across the capacitor doesn't really matter. However, as you add
    charges (ie, if there is a current flowing into one side, and out the
    other,) the voltage will change by some amount. Its possible to show
    that this amount is related to the geometry of the device. The
    mathematical relationship is

    change in V = a constant * the current.

    That is, the change in voltage is proportional to the current (which
    is change in charges per second). If you feed a capacitor with a
    constant current, you get a ramp voltage, the slope of which is
    related to the size of the capacitor.

    Well, because of the fact that the change in voltage is proportional
    to the current, then the faster we change the voltage, the more
    charges are moved per second, right? That means that the current is
    proportional to how fast the voltage is changing. For an AC voltage,
    the voltage is changing faster if the frequency is higher, obviously.
    Thus, as the frequency increases, more charges are moving into and out
    of the capacitor in a given second; the 'ac current' is higher.

    Now, for a resistor, this is not true. For a resistor, the number of
    charges that are passing through it per second is only dependent on
    the voltage at any time. At any frequency, the relationship between
    current and voltage is fixed by the resistance.

    Thus, if you connect a capacitor and a resistor in series to an AC
    voltage source, you can see that while the number of charges moving
    through the resistor is independent of frequency, the number moving
    'through' the capacitor is higher when the frequency is higher, ie,
    the resistance to the movement of charges is lower as the frequency

    So, if you hook up a capacitor to a resistor to ground, and put an AC
    voltage source across it, you can see that as the frequency increases,
    the amplitude of the AC measured at the junction will increase.

    AC --||--X--/\/\/-- GND

    Measured at the X, the AC voltage (ie, the amplitude) will increase as
    the frequency of the AC increases. Thats because more charges pile up
    at the X when the frequency increases; its easier to push through the
    capacitor, but its just as hard to push through the resistor. At lower
    frequencies, its harder for the charges to push through the cap, so
    the pile is smaller. Now, a pile of charge is voltage, and the bigger
    the pile, the higher the voltage.

    Now, imagine that the AC isn't just a single sine wave, but is a
    combination of lots of different sine waves. If you could separate all
    of these sine waves out and feed them into the circuit one by one, you
    could see that the higher frequency sine waves will all have bigger AC
    voltage at the X than the lower frequency sine waves. Now, its an
    amazing fact that if you combinine all the waves together and feed
    them into the thing, its just the same as if you fed them in
    individually and then added the result. Thus, for this circuit above,
    if you measure at the X, the high frequency components are bigger than
    the low frequency components. This is the definition of a high pass

    Now, imagine the inverse circuit:

    AC --/\/\/--X--||-- GND

    You can see that all the same things hold true, except that the
    voltage at the X will decrease for higher frequencies, and increase
    for lower frequencies. This is the low pass RC filter.

    Bob Monsen
  7. Rich Grise

    Rich Grise Guest

    OK, I'm going to fall back on the water model. The resistors are just
    pipes, but pipes with friction, so the pressure at the outlet isn't
    the same as the pressure at the inlet. That loss in pressure corresponds
    to "voltage drop," which isn't like dropping volts down a well, it's
    just the difference in potential, or voltage, from one end of the resistor
    to the other.

    For the capacitor, imagine a tank with a port at either end, and
    in the middle, a big rubber baffle. It's completely full of water,
    so when you force water into one end, water comes out the other
    end and the rubber stretches.

    So take your restrictive pipe, and start pushing water into the
    capacitor/tank. There will be a certain flow, and your pressure
    gage at the junction will start at 0 and slowly increase,
    as the rate of flow decreases, because of the back pressure.

    So the pressure lags the input. A capacitor opposes a change
    in voltage, an inductor opposes a change in current. So
    it takes time to move water through the pipe to fill the one
    side of the tank; and as the rubber stretches, the opposing force
    increases, until it either reaches the source pressure or
    some kind of feedback changes the input flow rate.

    So the resistor-capacitor network slows down the ability of the
    circuit to keep up with the signal, which means it doesn't
    pass high frequencies as well as low.

    Hope This Helps!
  8. One more thought crossed my mind. When I was first learning about capacitors,
    one of the things that really bugged me is how can it be that current can flow
    through a capacitor? I'd actually built a capacitor out of a sheet of glass and
    huge sheets of aluminum foil. And I knew darned well that there was NO WAY that
    electrons were going to flow through that glass. So I was simply stuck on this
    impedance thing, based on frequency. I had it firmly in mind that there was no
    way electrons could flow through a sheet of glass (a good insulator) and it
    didn't matter to my mind whether there was AC or DC there. It simply couldn't
    flow. Period.

    But if you realize that electrons don't exactly flow through the glass, you can
    at least imagine that they pile up and accumulate on one side, while many free
    electrons on the other side are removed, too. And, for a short time as these
    electrons are pushed into a pile-up situation on one side and while other
    electrons are sucked away from the other side, that in some sense this short
    duration flow looks just the same as real current. Of course, eventually, the
    flow stops because as more and more electrons pile up, the resist more of them
    arriving and eventually that resistance overcomes the voltage trying to push
    them there and maintains a counter-force to neutralize it.

    Then, if you reverse the voltage that was pushing those electrons and make it
    start to suck them away, and cause the other side to start supplying electrons
    to the side that already had them removed, earlier, then you'd see another flow
    of current for a short time -- until, once again, the counter-force rose up
    enough to stop the flow again. So, for a time, it looks just like current
    flowing. Then it stops.

    Now, if you oscillate that voltage across the capacitor so that it is ALWAYS
    changing, instead of just flipping it this way and that, there will always be
    some flow going on -- sometimes one way, sometimes the other, but never stopping
    because the capacitor never gets a chance to completely counteract it and thus
    stop it. The more rapidly you try and change that voltage, the higher this
    apparent oscillating current will be.

    Of course, the current still doesn't actually flow through the glass. It just
    flows off of one side of the capacitor and through components attached there and
    also onto the other side, again through components attached to that end. But to
    other components, like resistors that are hooked up, they cannot tell the
    difference between current that actually flows through a device or current that
    flows from some reservoir on the device. As far as they are concerned, some
    charge is flowing through them and that means current. So, even though nothing
    actually goes through the glass itself, from the outside world it looks just as
    though it did.

    Folks heavy in math have worked out the details of this flow rate based on rates
    of voltage change that match up with sinusoidal shapes to give you a reactance
    equation that provides a kind of equivalent to Ohms which can be used in simple
    equations to compute the effective current. The reason they keep it called
    reactance is that there is a delay between current and voltage in the case of
    capacitors that doesn't exist with regular resistors (where a change in voltage
    is immediately met by a change in current fitting the I=E/R equation.)
    Capacitors have their highest rate of current flowing exactly at the point where
    the voltage is changing the fastest and this is, in the case of sine waves,
    right at the point where the voltage crosses the midpoint between the top and
    bottom voltages (which is when the voltage difference is zero.) In other words,
    when the voltage is zero, the current is at a max. This is so different from
    the way it looks with a resistor that they really needed a different word for

    Other ideas, such as the fact that any complex wave can be thought of as being
    made up of some combination of sine waves, allows designers to analyze a circuit
    based only on sine wave analysis (using simplified rules, in other words) and by
    also breaking up the expected complex signals into combinations of sine waves,
    each of which can be analyzed individually. This allows one idea about the
    circuit to be applied repeatedly to see how a complex shaped wave might pass
    through it, by treating it as a combined set of sine waves, each with their own
    factors, and then recombining the results at the other end to see what comes
    out. It works.

  9. CFoley1064

    CFoley1064 Guest

    Subject: Low pass filters

    You're not going to get this without the math. It isn't crap, it's the way to
    describe what's happening. And the problem isn't with the description. Sorry.

    Good luck
  10. Just accept it at face value? If the capacitor is between the source
    and the load, higher frequencies get through while lower ones don't.
    If the capacitor is shunted to ground, the lower frequencies can't go
    through the capacitor, so they go through, while the higher
    frequencies find their way to ground through the capacitor. It's the
    path of least resistance (in this case impedance).

  11. John Fields

    John Fields Guest

    Let's say that you've got an automotive shock absorber with one end
    anchored to something so that it can't move and the other end
    connected to one end of a spring that you can push and pull back and

    Now let's say that you've got a magic marker (MM2)attached to the
    place where the spring and shock are connected together and another
    magic marker (MM1) attached to the other end of the spring, and that
    you've got a lo-o-o-o-o-ng piece of paper moving under the magic
    markers so that there are two lines being drawn on the paper at right
    angles to an imaginary line drawn axially through the center of the
    shock and the spring.

    Now, very slowly, push and pull your end of the spring back and forth
    say, six inches, and you'll notice that the magic markers also move
    back and forth, and that the line being drawn on the paper by MM2
    follows the line being drawn by MM1 and that if you're going slowly
    enough, it will also go back and forth pretty close to six inches.
    You may also notice a curious thing, and that's that MM2 doesn't
    follow the six inch movement of MM1 instantaneously, but is always
    behind. That's because you're not pushing on the shock directly, but
    storing energy in the spring which gets transferred to the shock at a
    rate depending on how much the shock resists it and how strongly the
    spring is pushing/pulling on the shock's piston.

    Now, if you push and pull the spring more and more quickly (while
    keeping the same 6" stroke) you'll notice that the faster you go the
    smaller the squiggles MM2 writes will be, because the shock just
    doesn't have the time it needs to move back and forth as much as MM1
    does. That's a lowpass filter, and if you want to find out how the
    amplitude and phase of the squiggles written by MM2 compare to those
    of MM1, you'll either have to get a shock, a spring, some magic
    markers and a lo-o-o-o-o-ng piece of graph paper with some way to pull
    it past the markers at a known, fixed speed, or do the math!-)
  12. Do you understand how a series RR voltage divider can reduce the
    voltage applied across it to something less across one of the

    There is a resistor, say R1 between input and output, and a second
    resistor, say R2, between output and ground. Lets also say the source
    of the input signal has zero impedance (current from this source does
    not affect its voltage) and the output signal feeds an infinite
    impedance load (the output voltage causes no current to pass through
    the load). So the only currents are those through R1 and R2 from
    source to ground. The way voltage across a resistor is related to the
    current through a resistor is proportionality. The voltage is always
    proportional to the current. Ohms are just a short hand way of saying
    "volts across per ampere through". Since there are no other paths for
    current in this circuit except through R1 and R2, those currents must
    be equal. So the voltages across those resistors must be proportional
    to their resistances by the same factor (the common current). So the
    input voltage is used up partly by R1 and partly by R2. If R1 is 4
    times the resistance of R2, then 4/5 of the input signal voltage must
    be dropped across R1 and 1/5 must be dropped across R2. The frequency
    (or wave shape, which can be thought of as a combination of
    frequencies) of the signal voltage does not matter, since this voltage
    division is instantaneous. All frequencies are treated equally.

    RC high and low pass filters are similar voltage dividers, except that
    one of the elements (the capacitor) has different impedances (AC
    voltage across per ampere through) at different frequencies.

    For example a resistor R is between the input signal and the output
    signal, and capacitor C is between the output signal and ground. The
    source and load impedances are as in the first example.

    The voltage drop across the resistor is instantaneously proportional
    to the current through it (its resistance is the proportionality
    factor), but knowing the voltage across a capacitor does not tell you
    anything about the instantaneous current through it, and vice versa.
    The relation between voltage and current for a capacitor is that the
    current through the capacitor is proportional to the time rate of
    change of the voltage across the capacitor, regardless of what the
    voltage happens to be. Capacitor voltage rises as current passes one
    way, and falls as current goes the other way.

    You might picture a capacitor with a fluid analogy as a rigid tank
    that has two pipe connections and an elastic membrane stretched across
    the tank between the two pipes, separating the tank into two volumes.
    The voltage across the capacitor is equivalent to the difference of
    pressures in the two halves of the tank. You might imagine that this
    difference has nothing directly to do with how much fluid is passing
    into one pipe and out of the other, but has a lot to do with how much
    the membrane is stretched to one side or the other at the moment.
    flow changes the pressure difference by changing how much the membrane
    is stretched. If you think of the resistor as a capillary tube you
    have all that you need to visualize the RC low pass filter. The input
    end of the capillary is driven by a pressure that varies sinusoidally
    at some frequency. The electrical ground (zero volts) is replaced
    with a vent to zero (atmospheric) pressure. Now, as the input
    pressure swings positive and negative, fluid flows into and out of the
    side of the tank that is connected to the capillary tube, and the
    membrane is pushed and pulled by various amounts and the pressure in
    the half of the tank connected to the capillary varies positive and
    negative, but never quite catches up with the input pressure wave.
    The pressure in that side of the tank is also the output signal.

    The faster the input wave changes, the further the output signal falls
    behind and the smaller the relative pressure swing it achieves.

    At very low frequencies, the tank pressure almost reaches the applied
    pressure before the direction reverses, and there is little difference
    in output amplitude compared to input amplitude. Small changes in
    either the resistance of the capillary or tank membrane elasticity
    (resistance or capacitance) have little effect on the output signal.
    This shows frequencies that are in the pass band. But at some
    characteristic frequency, the pressure swing in the tank reaches only
    70.7 % or the applied pressure in each direction and falls behind the
    input wave by 45 degrees. This is the filter corner frequency and
    represents the transition between pass band ands stop band. Above
    that frequency, the tank pressure swing drops rapidly with increasing
    frequency. At much higher frequencies the ratio of the pressure
    swings in the capillary side of the tank to the applied pressure
    swings fall almost in proportion to the source frequency. Double the
    frequency and the output swing falls by almost half.

    The characteristic or corner frequency where the the filter response
    changes is related to the product of the resistance and capacitance.
    When 2*pi*f=R*C, f is at the corner frequency.
  13. I like this analogy. If I gather it correctly, though, I'd want the membrane to
    become increasingly difficult to stretch further as it stretches, so that the
    counter pressure increases in opposition as more water is retained on one side
    as opposed to the other side. Common rubber membranes, for example like balloon
    material, has the opposite sense -- being much harder to "start" than to
    continue inflating, once started.

    Alveoli sacs in the lungs would be just like balloons in this sense, too, and
    would some would overinflate -- except for the fact that there are tiny cells
    that release a surfactant into the interior surface to break up the surface
    tension of water and make them easier to inflate at first; but not so much
    surfactant that it can fully cover the interior surface area as the sac gets
    larger. In that way, the surfactant dominates when the sac isn't inflated
    allowing an easy start but the surface tension of water begins to dominate more
    and more as the sac inflates, making it increasingly difficult to inflate them
    without limit. This is how the air we breathe in gets uniformly distributed
    across many thousands of sacs.

    With that addition, I like the analogy.

  14. Agreed. This analogy requires a very distinct membrane character. A
    better analogy might be a floating piston sandwiched between half
    compressed springs. Spring constants are closer to the dielectric
    springiness of the electric field in the dielectric of capacitors.
    Keep in mind that real capacitors have all sorts of non ideal
    dielectric behavior that limits our ability to describe the details of
    their operation with simple, linear mathematics.
  15. I've been spending some free time rereading the Feynman basic physics
    series, and recently came across a fun section on resonance, where he
    makes an analogy between a physical mass and spring system and and
    electrical system. The table is on page 23-6 in volume I.

    One interesting thing is that, in this model, capacitance is the
    reciprocal of the stiffness of the spring. Thus, a smaller capacitor
    is like a stiffer spring.

    Here is the whole table:

    characteristic = mechanical = electrical
    independent var = time = time
    dependent var = position = charge
    inertia = mass = inductance
    resistance = drag coeff = resistance
    stiffness = spring stiffness = 1/capacitance
    resonant freq = omega naught squared = stiffness/mass = 1/(inductance
    * capacitance)
    period = 2pi.sqrt(mass/stiffness) = 2pi.sqrt(inductance * capacitance)
    figure of merit (Q) = resonant freq * mass / drag = resonant freq *
    inductance / resistance

    Bob Monsen
  16. BobGardner

    BobGardner Guest

    No matter how much effort I put into understanding these low and high
    Can you grok how a resistor works? Like it lets a certain amt of current thru
    depending on the voltage and resistance. Got that? Now how about a pot or
    voltage divider? Simple concept? Small resistor on bottom, big resistor on top,
    small voltage out? Now can you dig that a cap is a freq dependent resistor? And
    the r and c form a voltage divider. When the 'resistance' of the cap (geeks
    call it 'impedance') is the same as the resistance of the r, you have half the
    voltage (Harold Bode who invented the frequency response graph call that the
    '3dB point'). Did you like this explanation?
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