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I'm fairly new to electronics, but have some idea,BUT how do these work?

Mark Simms

Apr 6, 2016
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I'm new to electronics and the forum so Hi all, I hope to be a useful member. I have questions for you high I.Q fellas here . please excuse my noob problems, I'm trying real hard.

A resistor voltage divider can step down a voltage for say 12v DC to 6v DC, but what resister values do I use? the current for the load say a 1Watt 6v bulb seems to generate a much higher current in the voltage divider series resistors than that of the bulb load circuit.

do resistors generate heat when reducing current, what I'm asking is am I draining a lot of power from the battery with resistors or are they just mostly dropping voltage?

how do coils work in lay-mans terms, here is what I know so far, they don't restrict DC, but oppose AC at higher frequencies, also they build an EMF field, I'm not bothered about formulas at the moment I'm looking at how they do what they do, what puzzles me is if a coil is in parallel with a capacitor what does the resonating achieve? does the DC just go straight through the coil?

I'm looking at a crystal AM radio set and the capacitor and coil resonate back and forth, the antenna acts as a capacitor , and the diode demodulates, but why is the coil actually needed? the antenna resonates with the radio waves, what use is the magnetic field in the coil? would the variable capacitor no work with the antenna without the coil? I can sort of think of the resonating as tuning to a frequency, but I cant see why ?
 

duke37

Jan 9, 2011
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RMS is root mean squared. It is a measure of an AC (alternating current) waveform to give the same energy as a DC (direct current) supply.

As you have found, a resistive voltage divider is very wasteful of energy. The wasted energy shows up as heat.
If you are running a 6V 1W bulb from 12V then a simple series resistor will drop the voltage, this will need to dissipate 1W, and be the same (hot) resistance as the bulb.
You need equations, start with Ohms law. V = I * R where V = voltage across the resistor, I is current through the resistor. As you can see, voltage and current are interrelated.
To go one step further, power (W) = V * I or using Ohms law, W = V * V /R or I *I *R.

An oscillating circuit works by sending energy from an inductance to a capacitor and then back again. A crystal set may be able to run without a coil but it will not be possible to tune to a specific frequency. The antenna (US), aerial (UK) will be unlikely to resonate at the frequency you want.
Find out how a pendulum works, with the energy being interchanged from potential energy and kinetic energy.
 

Mark Simms

Apr 6, 2016
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duke that's great, you truly solved the problem I had with stepping down a voltage.

I still am confused about the resonating , I can visualise the back and forth action with it, but if the coil is resonating how does it select the particular frequency of the aerial , would they not be resonating at different frequencies ? the aerial sends the AC current into the coil then through to ground, the coil is in-partnership with the capacitor, I don't see the handshake with the resonating coil and the flowing AC current, they seem to be different to me. I hope I'm being clear here I can see myself rambling.
 

Minder

Apr 24, 2015
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An aerial can be tuned to a particular selective frequency, as in tuned to a particular TV ch freq for e.g. look up dipole antenna.
A transistor radio has a fairly broadband antenna and is not tuned to one particular frequency, it receives a wide spectrum of frequencies and feeds the tuned circuit which selectively detects the station.
M.
 
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Arouse1973

Adam
Dec 18, 2013
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duke that's great, you truly solved the problem I had with stepping down a voltage.

I still am confused about the resonating , I can visualise the back and forth action with it, but if the coil is resonating how does it select the particular frequency of the aerial , would they not be resonating at different frequencies ? the aerial sends the AC current into the coil then through to ground, the coil is in-partnership with the capacitor, I don't see the handshake with the resonating coil and the flowing AC current, they seem to be different to me. I hope I'm being clear here I can see myself rambling.

The tuned circuit, in this case would most likely be a parallel type with the capacitor connected across the inductor as the simplest form.

220px-LC_parallel_simple_svg.png

images.png
The above formula is the frequency at which the tuned circuit is resonant and also has the highest impedance (parallel circuit only). This allows that particular frequency to have virtually no attenuation from the inductor and capacitor. All other frequencies above and below the resonant frequency will be attenuated. The amount of attenuation of all other frequencies is dependant on how close they are to the actual resonant frequency. Frequencies much higher or much lower will be attenuated the most. It gets a bit more complicated than that in the real world. If you want to know more then just ask, but you did seem to not want any formula at this time.

Adam
 

duke37

Jan 9, 2011
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The aerial will have a series of resonant frequencies, depending on its length and proximity to other objects. The set will have a tuned circuit which will be modified when the aerial is connected. The mathematics can get quite complicated.
If the coupling between aerial and tuned circuit is low, the energy transfer is not great but the tuned circuit will behave as expected. A crystal set needs tight coupling since all the energy comes from the aerial and you need all you can get. In this case the aerial and tuned circuit behave as one and the tuning is lousy.
 

Ratch

Mar 10, 2013
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duke that's great, you truly solved the problem I had with stepping down a voltage.

I still am confused about the resonating , I can visualise the back and forth action with it, but if the coil is resonating how does it select the particular frequency of the aerial , would they not be resonating at different frequencies ? the aerial sends the AC current into the coil then through to ground, the coil is in-partnership with the capacitor, I don't see the handshake with the resonating coil and the flowing AC current, they seem to be different to me. I hope I'm being clear here I can see myself rambling.

Resonance occurs when two energy storage elements, either mechanical or electrical, swap energy back and forth at a sinusoidal rate at a particular frequency. In electrical circuits, this happens when the reactance of the two elements are equal. Another requirement is that the current-voltage phase of each element has to be 180° different from each other so that one element is absorbing energy while the other element is dispensing energy. Coils and capacitors work well at this because the voltage of a coil is proportional to the derivative of the sinusoidal current while the voltage of a capacitor is proportional to the integral of the sinusoidal current. This assures a 180° current-phase relationship. This condition also means that one coil or two coils can never resonate by themselves. Same thing for capacitors. In a series circuit, if one element has more reactance that the other, it will also drop more voltage than the other element so that the voltages can never cancel, and current will not be at a maximum. In a parallel tank circuit, if one element has more reactance that the other, the current in the branches will never cancel out and some current will leak past the tank circuit so that the parallel circuit will not be at a minimum current. So the equal reactance of each element is determined by the resonant frequency.
 

Mark Simms

Apr 6, 2016
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That's all good information to think on, thankyou. I've just ordered the parts to make the set.
 

Mark Simms

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so my understanding so far is, the antenna oscillates AC signal, the coil receives the AC signal and reacts to the resonate frequency by only! oscillating at the tuned frequency, and ignoring the above and below frequencies. then the circuit can be peeked at by the diode and earphone.
 

Mark Simms

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so I think I'm getting it, the coil keeps boosting the frequency , but dose not help the frequencies higher or lower that's the handshake I was looking for, wow this I think is getting better to understand thankyou
 
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Mark Simms

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I have seen a circuit with a coil and capacitor in parallel feeding the collector of a NPN transistor from a DC voltage, how dose the coil and capacitor deal with DC like this?
 

Mark Simms

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am I correct in thinking the wire thickness does not matter, does a current make the same magnetic field with thin or thick wire?
 

duke37

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Another equation

X = 2*pi*f
The reactnce of an inductance is proportional to frequency so with DC (frequency = zero), the reactance is zero and the DC goes through uninhibited. This does not take into account any wire resistance.

Thin and thick wire will have similar magnetic effects but thin wires have a higher resistance and so higher power loss. In the case of radio signals, a higher resistance in inductors degrades the resonant peak so it will not be so easy to peek at the signal.:)
 

Mark Simms

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Thankyou duke, I have a good grounding now, and I will have a better time of it after these answers I've got
 

Mark Simms

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Is it possible to put a transistor preamp circuit in a crystal radio set before the diode to boost the signal , then allowing a normal 0.7v diode to be used instead of a low powerd germainium diode? I know you can amp it up after the diode. or is there an easier way? how do you boost the aerial signal? I know you can get signal boosters but what is it?

also is it possible to easily step up a small DC voltage, eg 9v to 12v ? I suspect you can put a battery cell in seres, but is there a way with componants?
 
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duke37

Jan 9, 2011
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A transistor could be used either before or after the diode and you could go much further with lots of components but this takes away from the idea of a crystal set. Good for experiments though. You could try adding a slight voltage to help the diode to turn on but this means the set will be powered.:)
It is necessary to use high sensitivity, high impedance headphones.
A Schottky diode will have a very similar action to a germanium diode but might be easier to obtain.

Why do you want to raise 9V to 12V? I can think of very few applications. The way to do it is with an oscillating circuit which can give lots of electrical noise if not made properly. These voltage convertors are available very cheaply if you want one ready built.
 

Mark Simms

Apr 6, 2016
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I'm trying to learn radio electronics at the moment as it seems an interesting start point. I'm hopeing to build radios , but my only understanding of one at the moment are crystal sets, so amping a crystal set was the next step, I would love to see a very basic circuit of a transistor radio, but as simple as possible because I'm not the sharpest tool in the box, but I get there with help. adding a slight voltage is great advice it was what I was hopeing to learn thankyou.

I want to know how it is possible in a simple way to step up a voltage without a transformer, I just cant figure it out how.
 

Mark Simms

Apr 6, 2016
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forgive me, but do I need a base biasing setup on the transistor to boost the signal in a crystal set, or can you somehow just add the transistor, Ive just learned a common emitter preamp,

when I add a voltage to the crystal set to bias the diode, does adding a voltage drown-out an AC signal, because if it dosnt that's great
 
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