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Basic Adivce/tutoring sought

R

Roy Ingham

Jan 1, 1970
0
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
eg:
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,

Thanks
Barry
 
J

John Popelish

Jan 1, 1970
0
Roy said:
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?"

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".
I understand the functions that a part plays (sometimes not 100%
correct) but generally
eg:
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.

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.
But what I don't grasp is why do we place (replace?)

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.?

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.
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'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.
 
R

Roy Ingham

Jan 1, 1970
0
Firstly thank you for your reply it was helpful.

John said:
....
Capacitors remember the past, resistors don't.
....
An inductor also remembers the past,

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'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.

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

Questions:
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.

Barry
 
R

Rich Grise

Jan 1, 1970
0
.
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

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

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.

http://www.google.com/search?q="radio+fundamentals"

Good Luck!
Rich
 
R

Roy Ingham

Jan 1, 1970
0
Thanks Rich

Rich said:
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.

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

Barry
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Roy Ingham <[email protected]>
But rather the ability to transmit power from transmitter to receiver.
I take it this is not possible?

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
experiments.

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
it.
 
R

Rich Grise

Jan 1, 1970
0
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?

One of these sources can probably explain it better than I can:
http://www.google.com/search?&q=wireless+power+transmission

Good Luck!
Rich
 
J

John Fields

Jan 1, 1970
0
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)?

---
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.
 
J

John Fields

Jan 1, 1970
0
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
eg:
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.

---
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.

---
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.?

---
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)


E1
|
[R1]
|
+---->E2
|
[C1]
|
GND

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
volts.


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.
 
J

John Popelish

Jan 1, 1970
0
Roy said:
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.

This is why, together this pair is called a second order system.
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?

Correct, and this pair makes up a first order system.
Thus: frequency, (the tuning circuit = inductor with capacitor) which is
time based, requireds all or both components to be time aware, right?

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.
Do you know of an online source, that explains the time aspects of a
capacitor.

The search key words might be [capacitor capacitance tutorial].
E.G.
http://www.electronics-tutorials.com/basics/capacitance.htm

So far I have only come across the voltage storage principles
of a capacitor, none mentioning the time rememberance factors?

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.
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

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

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.
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.?

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.
 
D

David L. Jones

Jan 1, 1970
0
Roy said:
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
eg:
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,

Thanks
Barry

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
yet.
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 :)
 
B

Ban

Jan 1, 1970
0
John said:
I read in sci.electronics.design that Roy Ingham <[email protected]>


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 experiments.

Actually it's done all the time with RFID systems, but maybe not in the
ballpark that the OP imagined.
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 it.

You need to be quite close to the transmitter, as the field falls off
quickly.
 
J

Jasen Betts

Jan 1, 1970
0
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
eg:
capacitor: takes in energy (voltage) and stores it, until it is released.

capacitors behave like springs
Resistor: acts to slow down the voltage as like a form of friction.
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.?

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.


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)

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

1 watt into the power source (battery terminals) is quite a lot too,
what components will effect the Voltage and Amperage (v x a = W) in the
receiver side,

the volume knob :)

or if you mean the strength of the radio signal received the antenna,
is probably the main consideration.
and should it be the same +/- 1 Watt and what would cause
the decrease should there be a difference?

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.

Bye.
Jasen
 
J

Jasen Betts

Jan 1, 1970
0
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?

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.
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)?

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.
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?

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)

Bye.
Jasen
 
J

John Fields

Jan 1, 1970
0
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.
^^^^^
Oops... 1s
 
R

Roy Ingham

Jan 1, 1970
0
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.

Barry
 
D

Don Kelly

Jan 1, 1970
0
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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