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Basic Adivce/tutoring sought

Discussion in 'Electronic Basics' started by Roy Ingham, Sep 28, 2005.

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  1. Roy Ingham

    Roy Ingham Guest

    Greetings all

    I hope I make sense, as I'm seeking some guidance through this, please.
    I'm trying to get to grips with some basic electronics.
    I know and have learnt the electronic math rules
    W = V x A ; V = IR , etc..

    What I get stuck on is more "why would one use a capacitor
    in a circuit, why not a resistor?"
    I understand the functions that a part plays (sometimes not 100%
    correct) but generally
    capacitor: takes in energy (voltage) and stores it, until it is released.
    Resistor: acts to slow down the voltage as like a form of friction.
    much like one lane of traffic compared to a 4 lane highway.

    But what I don't grasp is why do we place a resistor with a coil to make
    a tuning circuit, why not use a capacitor. I'm trying to grasp what is
    it about the resistor, that makes the two components work and thus form
    a tuning circuit.?

    For example:
    If I take an FM Receiver and Transmitter kit (wireless microphone kit or
    remote control kit), say I connect a 1 Watt power source to the
    Transmitter, (in place of the human voice on the microphone kit)
    what components will effect the Voltage and Amperage (v x a = W) in the
    receiver side, and should it be the same +/- 1 Watt and what would cause
    the decrease should there be a difference?

    I thank you for your assistance up front,

  2. Capacitors remember the past, resistors don't. The voltage across a
    capacitor is proportional to the total charge that has passed through
    it, since it last had zero volts across it. Or mathematically, V=Q/C.
    1/C is just the constant of proportionality between voltage and charge.

    If you look at this as a process, in time, rather than as a result,
    you can say that the rate of change of the voltage across a capacitor
    is proportional to the current passing through it. I=C*(dv/dt), where
    dv/dt means rate of change of voltage with respect to time, with units
    like "volts per second".
    Think of a resistor as a device that enforces a fixed proportionality
    between voltage and current. Another way to look at ohms is to call
    them volts per ampere. Double the volts (electromotive force across
    the resistor) and the amperes through it (amount of charge per second)
    also doubles.
    An inductor also remembers the past, but instead of having a voltage
    in proportion to total charge, it has a current proportional to total
    volt seconds. So it remembers how much voltage for how long has been
    applied to it, and that memory is represented by its current.

    A capacitor and inductor in combination act something like a spring
    and mass act, mechanically to produce a resonance. When the spring is
    at maximum distortion and the mass is changing directions, but
    momentarily at rest, all the energy is stored in the spring. But when
    the mass hits peak velocity and the spring is passing through its
    relaxed state, all the energy is stored in the kinetic energy of the
    mass. The energy sloshes back and forth between spring and mass,
    twice per cycle, and the value of spring stiffness and mass determine
    the resonant frequency.
    I'm not following the question. I think you are trying to combine
    several processes into one, and skipping too many steps. It is sort
    of like saying that if you can run a mile on one hamburger, how fast
    can you run if you eat a gallon of gasoline.

    The FM transmitter combines radio frequency energy with audio by
    varying the frequency of a fixed energy carrier in proportion to the
    amplitude of the audio signal. Replacing the audio with something
    else does not change the power of the carrier.

    The receiver separated the modulated carrier from all other
    frequencies (if it receives enough energy from the transmitter) and
    then discards all information about how strong the carrier is, and
    just responds to the frequency variations to recreate the audio signal
    in proportion to the frequency shifts. Doubling the power of the
    transmitter carrier allows this process to work over a longer
    distance, but does not change the volume of the audio at the receiver.
  3. Roy Ingham

    Roy Ingham Guest

    Firstly thank you for your reply it was helpful.

    So then the an issue of why a capacitor and a inductor / coil are used
    in combination is due the time "remberance" that both components have.

    If a inductor was put with a resistor, then only the inductor would be
    aware of the time factor, while the resistor would not, correct?

    Thus: frequency, (the tuning circuit = inductor with capacitor) which is
    time based, requireds all or both components to be time aware, right?

    Do you know of an online source, that explains the time aspects of a
    capacitor. So far I have only come across the voltage storage principles
    of a capacitor, none mentioning the time rememberance factors?
    I could be skipping steps, not sure?
    Ok, so let me rephrases this another way.
    Using a wireless microphone kit, with a tranmitter and receiver.
    Designed or should be used as follows:
    Human Voice -> Transmitter
    Receiver -> Tape Deck
    Using the wireless system I can now record what is said onto the tape deck.

    Now making an assumption, lets replace the Human Voice with 1 Watt power
    supply and the tape deck with a Light Blub, thus giving:

    1 Watt Power -> Transmitter
    Receive -> Light Bulb

    1) Would the Light blub be able to receive 1 watt of power, if not why?

    2) If the light bulb was to receive power. How much would it receive,
    ans what would cause the loss if the power received is less than the 1
    watt supplied.?

    Thank you for your input John, your insight was helpful.

  4. Rich Grise

    Rich Grise Guest

    Not necessarily - only if the receiver is designed to provide that power
    to the load from its own power supply. The actual amount of "power" sent
    from the transmitter to the receiver is minuscule."radio+fundamentals"

    Good Luck!
  5. Roy Ingham

    Roy Ingham Guest

    Thanks Rich

    In the same way that I can put human voice into the transmitter
    (microphone) and get the human voice out of the receiver to the tape
    deck (or house monitors). Can I not "transport or send"
    power/electricity from a transmitter to a receiver (not linked with the
    power required to operate)?

    The 1 watt power source, is not the power source required to operate the
    transmitter, nor was I concerned about the power source(battery) to
    power the receiver. But rather the ability to transmit power from
    transmitter to receiver. I take it this is not possible?

    Thanks again for all your help and input

  6. I read in that Roy Ingham <>
    It's not exactly impossible, but it's so very difficult that it's not
    proved practical. Both Tesla and Yagi (he of the xylophone-like antenna)
    tried. In fact, Yagi invented his antenna for power-transmission

    There has been speculation about transmitting power as microwaves from
    solar-powered satellites to ground, but there are huge problems, not
    least of which is ensuring that the power beam doesn't fry the nearest
    city when the satellite's guidance system is zapped by a solar eruption.

    People have managed to steal a hundred watts or so from a nearby
    high-powered broadcasting station (think 50 kilowatts), but the
    receiving antenna had to be very large, and the people running the
    station could tell it was happening. so the FCC moved in rapidly to stop
  7. Rich Grise

    Rich Grise Guest

    One of these sources can probably explain it better than I can:

    Good Luck!
  8. John Fields

    John Fields Guest

    Not generally, in a practical sense, considering that the energy in
    the transmitted beam spreads geometrically when it leaves the
    transmitting antenna and is, therefore not intercepted greatly by
    the receiving antenna. To a lesser degree, there's also atmospheric
    absorbtion to consider.
  9. John Fields

    John Fields Guest

    Picture this:

    You have a large tank full of water in which is submerged a
    bidirectional pump with its output connected to a hose. The other
    end of the hose is connected to a fitting on the bottom of an
    aquarium, such that the pump can either pump water into, or out of,
    the aquarium.

    The pressure that the pump generates when it's either filling or
    emptying the tank is analogous to voltage, The diameter and length
    of the hose is analogous to resistance, and the volume of water
    which can be pumped either way in a given amount of time is
    analogous to current.

    In the case of the resistor and capacitor (RC) or of the resistor
    and inductor,(RL) they provide a known attenuation of a signal as a
    function of the signal's frequency and either an RC or an RL could
    be used, depending on the application.

    For example, in the water analogy the aquarium is analogous to a
    capacitor, so the corresponding electrical circuit would be: (View
    with a fixed-pitch font like Courier)


    Where E1 is the pump pressure generated, in volts
    R1 is the resistance of the hose, in ohms
    E2 is the pressure on the bottom of the aquarium, in volts
    C1 is the capacitance of the aquarium, in farads, and
    GND is the pressure on the bottom of the empty aquarium, zero

    Now let's say that we start with an empty aquarium and that when we
    start the pump it will generate a pressure of 1 volt and start
    pumping water through the hose, which has a resistance of 1 ohm,
    into the aquarium, which has a capacitance of 1 farad. Initially,
    the pressure at E2 will be 0 volts because the aquarium was empty,
    but as the aquarium fills up the voltage at E2 will rise because of
    the head of water being built up in the aquarium. If the aquarium
    is tall enough to allow the pressure at E2 to rise to 1 volt, then
    the pump will no longer be able to pump water into the aquarium
    because it will be pumping against a 1 volt head with its 1 volt of
    pressure so everything will be in equilibrium.

    What's important to realize is that it took time for the aquarium to
    fill up. There's a certain "time constant" associated with the
    system which is given by:

    t = RC

    Where t is the time it takes, in seconds, for the aquarium to fill
    up to the point where E2 is about 2/3 of E1,

    R is the resistance of the hose in ohms, and

    C is the capacitance of the aquarium in farads.

    If we plug in what we know we'll have:

    t = RC = 1R * 1F = 1s

    So in one second, with a pressure of 1V at E1, we'll have the
    aquarium about 2/3 full and a pressure of ~ 0.67V at E2.

    Now, if at the point where E2 rises to 0.67V we suddenly reverse the
    pump, it will take 0.67s to empty the aquarium.

    So, for an input frequency of 0.5Hz (1s fill, 1s empty) we have a
    peak output amplitude at E2 of 0.67V.

    However, if we only fill the aquarium for 1/2 second and empty it
    for 1/2 second, the voltage at E2 won't have enough time to rise to
    0.67V, so you can see that the higher the input frequency, the lower
    the amplitude of the output signal.
  10. This is why, together this pair is called a second order system.
    Correct, and this pair makes up a first order system.
    Frequency selective circuits can be made with just an RC or RL pair,
    but this kind of selectivity is just low pass (everything lower than
    some cut off frequency is passed through with minimal attenuation) or
    high pass (everything higher than some cut off frequency is passed
    with minimal attenuation) but it takes at least a second order system
    to produce resonance that act as a band pass filter.
    The search key words might be [capacitor capacitance tutorial].

    So far I have only come across the voltage storage principles
    And you may not find such a description. The mathematical way to say
    "remembrance factor" is to refer to a function of time. The voltage
    across a capacitor is a function of current passing through it and
    time. The rate of change of voltage across a capacitor is
    proportional to the current passing through it. Current is charge per
    time. Rate of change of voltage with respect to time is a function of
    time. If the present value involves time, there is some sense of
    history in the present value.
    In the voice receiver combination, the amplitude of the voice signal
    produces a proportional receiver output signal, even though the actual
    power in the receiver's output comes from its power supply, not from
    the voice. The voice provides a controlling signal that determines
    how the receiver's power supply is converted to a proportional copy of
    the original voice signal.

    So the conversion of 1 watt going into the transmitter has no
    particular form implied to make it an appropriate control signal at
    the receiver. If the transmitter is designed to modulate its carrier
    energy with a few millivolts from a microphone, and you replace the
    microphone output with a 1 watt audio tone (say, a speaker drive
    signal) at the very least, the transmitter would produce a distorted
    modulation and be received as a distorted (like a fuzzed electric
    guitar) version of the original tone. At worst, the large audio power
    to the transmitter input would destroy the modulation circuit.
    Power leaves the transmitter in all directions into space. Only a
    tiny fraction of that energy is picked up by the receiver. That tiny
    signal is amplified (a more powerful copy is created that matches the
    instant by instant variations of the radiated wave) with power from
    the receiver's DC supply. In a similar way, power steering copies the
    movements you make with the steering wheel, only with much more force
    capability, using power from the engine via the power steering pump.

    The modulation (audio) information is recovered from that amplified
    copy of the radiated wave, and then that signal is probably again
    amplifier with more power from the receiver's supply before it comes
    out of the audio jack of the receiver.
  11. You are probably biting off more than you can chew at this stage. The
    concept of tuned circuits are not something that a beginner should
    start with, esp if you don't fully understand how a capacitor works
    Start off with basic DC theory which involves resistors and voltages,
    and then move on to basic AC theory with Resistor and Capacitor (RC)
    circuits. Stuff like RC time constants, how a capacitor lets AC through
    but not DC, that kind of stuff. Then learn about inductors, and
    Resistor and Inductor (RL) circuits, and then you might be ready to
    grasp LC tuned circuits and other more complex AC theory.
    All of these are basic "building blocks" which you combine to make a
    circuit that does something useful.

    I can recommend "Circuit Theory & Techniques" by Hans Goodman. Vol 1
    and Vol 2.

    Dave :)
  12. Ban

    Ban Guest

    Actually it's done all the time with RFID systems, but maybe not in the
    ballpark that the OP imagined.
    You need to be quite close to the transmitter, as the field falls off
  13. Jasen Betts

    Jasen Betts Guest

    capacitors behave like springs
    they don't

    a typical tuning circuit uses a capacitor and an a coil

    with a resistor and a coil you can make a filter circuit.

    1 watt is way too much input to the microphone terminals, more like 1

    1 watt into the power source (battery terminals) is quite a lot too,
    the volume knob :)

    or if you mean the strength of the radio signal received the antenna,
    is probably the main consideration.
    there will be a decrease, as the 1 watt radio signal spreads out to cover
    more area it gets weaker like the way a distant lamp gives less illumination.

  14. Jasen Betts

    Jasen Betts Guest

    In the same way that I can put human voice into the transmitter
    it's possible, just not practical. a "crystal set" am receiver is totally
    powered by the received signal, they typically need a large antenna, and
    only output into a earpiece.
    energy is sent, but the energy sent is one millionth or less of the energy
    used to power a typical receiver... the energy sent is just used to carry
    the information - the voice.
    if you put a louder input into the transmiiter you'll get a louder output
    from the receiver (within limits), but still the receiver is responsible
    for the energy in the output, the transmitter only for the content
    (information part)

  15. John Fields

    John Fields Guest

    Oops... 1s
  16. Roy Ingham

    Roy Ingham Guest

    Wow!! so much input, thank you

    I didn't even think about the radiation of the signal from the
    transmitter, thus making it obvious that the receiver can only get a
    part of the signal.

    For everyones input thank you.

  17. Don Kelly

    Don Kelly Guest

    David Jones gave very good advice- follow it and learn the basics. There is
    a hell of a lot between understanding of V=IR and radiation of signals and
    the intermediate steps are important.
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