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Caps and Diode setting on meters

Discussion in 'Electronic Basics' started by Uriah, Jun 13, 2007.

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  1. Uriah

    Uriah Guest

    Believe it or not, I have been into electronics for a long time but
    for some reason some things don't sink in. Like what exactly is
    capacitance? I know it is measured in ohms. I know it is like a
    frequency dependent variable resistor. But when the capacitance goes
    up in uF what is going on with the electrons? Does it hold more
    Electrons? Or is the field between the plates holding a higher
    charge? If you have a 1uF cap and a 1pF cap and you send a sign wave
    through it of a particular frequency of 1k. What is happening that is
    different between the two? They both charge to the same potential? Do
    the discharge at a different rate? I can't see the forest for the
    trees, I think, what ever that means.

    Also,

    When you are testing something with your DVM and you have it set on
    the diode setting. You usually get a .400 number and when reversing
    the leads you get an open. What is the meter doing? Is it putting out
    1 volt? And then is it just measuring the difference? Is the.400 in
    VDC? Can someone explain what a meter is doing on this setting.

    Thanks for all of the help
    Uriah
     
  2. Capacitors have impedance, a frequency dependent
    relationship between the voltage across them and the current
    through them. Impedance is two dimensional, and can be
    expressed as magnitude (of ohms) and phase shift (polar
    form), versus frequency, or in artesian form as real and
    imaginary parts of the total.

    Capacitance, itself, is measured in farads. A farad of
    capacitance experiences a 1 volt change across its terminals
    for each coulomb of charge that passes through it. Since a
    coulomb per second is called an ampere, we can say that an
    ampere passing through a Farad of capacitance changes the
    voltage across it by a volt, every second.

    As a formula, this is I=C*(dv/dt), where I is current in
    amperes, C is capacitance in farads, and dv/dt is the rate
    of change of voltage in volts per second.
    Sort of, except that capacitance does not consume energy,
    but stores it and gives it back, depending on whether the
    voltage is rising or falling in magnitude. The energy
    stored in a cap is energy=1/2 * C * V^2, where energy is in
    watt seconds or joules, C is capacitance in farads, and V is
    the volts across the terminals.
    It holds more electrons shifted from one plate to the other
    with less applied voltage. Think of the capacitor as a
    rigid tank with a rubber membrane stretched across the
    middle, and a pipe into the vessel on each side of the
    membrane. In this analogy, the pressure difference
    (voltage) between the two pipes (terminals) is proportional
    to the amount of fluid removed from the volume on one side
    of the membrane and forced into the volume on the other side
    of the membrane. The deformation of the membrane is what
    stores the energy. In a capacitor, it is the electric field
    between the plates that stores energy as the dielectric is
    stressed with voltage.
    The charge is localized to the surface of the plates, but
    the energy is stored in the electric field between the
    plates these surface charges produce.
    If the voltage applied between their terminals is similar,
    as it would be if they were connected in parallel, the 1uF
    cap would be passing a million times more current than the 1
    pF cap. In other words, the impedance of the 1 uF cap is a
    million times lower than the impedance of the 1 pF
    capacitor. The energy stored at any voltage is also a
    million times higher.

    But if you connect these two capacitors in series, and pass
    the same current through each, this is like connecting a
    very low and a very high ohm resistor in series. Most of
    the total voltage drop occurs across the higher value
    resistor, and across the very high impedance capacitor,
    which is the smaller one. The magnitude of capacitive
    impedance is ohms = 1/(2*pi*f*C), where f is frequency in
    hertz, and C is capacitance in farads. Ohms is a word that
    means volts per ampere.
    If they are forced to discharge their voltage at the same
    rate (volts per second) the 1 uF cap will deliver a current
    a million times larger than the 1 pF cap does during this
    discharge. If they are drained at the same current, the 1
    uF cap will have a discharge time a million times longer
    than the 1 pF cap does.
    The main thing that complicates capacitors that is missing
    from the understanding of resistors is time. You cannot
    think clearly about what capacitors do in a circuit or in
    isolation, if you don't include time. The only thing you
    can say without referring to time is the instantaneous
    energy stored in the cap, that is purely a function of
    capacitance and voltage.
    (snip)

    It is delivering a regulated current and displaying the
    millivolts of drop that current produces as it passes
    through the leads and the component they are connected to.
    The reading usually goes off scale if the leads are open
    circuited. You can use a separate volt meter to measure how
    high this voltage goes, open circuit. Something around 2
    volts is common (2000 on the meter). Silicon diodes may
    drop, between .3 and .6 (300 to 600 millivolts) when
    passing this small (and also measurable with a second
    milliamp meter) current.

    If you test a large capacitor, you should see something near
    zero (uncharged) that rises smoothly to the meter overload
    (cutoff) as the current ramps up the capacitor voltage, as
    charge is stored in the cap. Not a good idea to test the
    charge capacitor in reverse, because the meter is not
    expecting its voltage to go negative. Touch the capacitor
    leads together to dump the stored charge, to repeat the test.

    Once you calibrate this capacitor test (zero to cut off time
    for a given capacitance) you can roughly estimate other
    capacitances by their charge time, relative to this one.
    For instance, if it takes 2 seconds for a 100 uF cap to
    charge to meter cut off, it should take about 20 seconds for
    a 1000 uF cap to do the same thing.
     
  3. Charles

    Charles Guest

    Darn, John you wrote a mini-textbook here ... nice answer!
     
  4. Thank you.

    I just remember how long I was frustrated with not finding
    explanations of simple electronic principles when I was
    trying to teach myself electronics. So many sources were
    either so simplified that they didn't really explain
    anything, or they dove directly into differential equations
    and left me behind.

    I tried to keep the differential equations to a minimum.
     
  5. DJ Delorie

    DJ Delorie Guest

    Without going into too many details...

    Capacitance is the ability to store electric charge, much like a
    balloon stores air.
    Capacitance is measured in Farads. Impedance is measured in ohms.
    Which unit you use depends on what you're measuring. Impedance can be
    calculated from capacitance and frequency, if the signal is a sine
    wave. For other shapes, you have to consider each frequency component
    (harmonics etc) of the wave separately; each sees a different
    impedance.
    It can be, but not always. When dealing with AC signals, where the
    capacitor is part of a filter, it makes sense to think of it that way,
    because each component (frequency) of the signal will see a different
    impedance. For DC or DC-like signals (long-duration square waves, for
    example, like the 555, or non-periodic signals), it's better to think
    of them as storage devices.
    Both, sort of. Capacitance depends on the surface area of the plates
    and the distance between them.
    Vin ------| |------* Vout
    Ic ---> |
    |
    Gnd ---------------* Gnd

    Let's say Vin is increasing. The voltage across the capacitor won't
    change immediately, so the voltage across the load does. This voltage
    drop causes Ic. Ic causes the cap to charge, which increases the
    difference between Vin and Vout - effectively reducing Vout.

    If the capacitance is smaller, the charge reacts faster (takes less
    time to charge), thus Vout stays closer to zero. A larger capacitance
    takes longer to charge, letting Vout go further from zero.

    The neat part about sine waves is that the integrals and derivatives
    (math formulas that relate charge to current) are also sine waves
    (although of various amplitudes and phases). If you put waves of
    other shapes through the above circuit, you won't get the same shape
    out. So when Vin is a sine wave, Ic and Vout are also sine waves,
    although at different amplitudes and phases. When looking at the sine
    wave amplitudes, one can think of the capacitor and resistor above
    like a voltage divider, hence one can calculate the "impedance" (or
    effective resistance at that frequency) of the capacitor.
    If the applied voltage is the same, yes. However, it takes more or
    less total current to do so, depending on the capacitance.
    Assuming all else is the same, yes. That is, if two capacitors have
    the same charge-induced-voltage and load resistance, they'll both
    start out with the same discharge current. However, the larger
    capacitor will be able to sustain that current longer before it's
    discharged as much as the smaller one. Example: two caps at 10v with
    1k load resistors. One is a higher capacitance than the other. The
    larger one will take longer to discharge to 5v than the smaller one.
    I think it's just putting a small fixed current through the diode, and
    measuring the voltage drop across it. When forward biased, the diode
    passes the current with a drop of Vf. When reverse biased, the diode
    does not pass the current, and the whole power supply's voltage is
    across it.

    So, if you test the diode forward-like, the meter tells you the
    voltage drop across it. If the number makes sense, the diode is good.
     
  6. The meter puts a fixed current (1mA is common) through the diode and
    the display is actually showing the voltage drop across that diode. So
    yes, the display is in VDC.

    Dave.
     
  7. Fred Abse

    Fred Abse Guest

    Well, well !!

    ;-)
     
  8. Damn spell checker!
     
  9. John Fields

    John Fields Guest

     
  10. Uriah

    Uriah Guest

    Thanks John, DJ and David for taking the time to answer my question.
    I will let it sink in.
    Uriah
     
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