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Q and crystal radios

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Dr_Doggy

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still get a bit confused with LC circuitry but i get that L & C make up for resonant frequency,

but my question is how is q factor impacted by inductor or capacitor size

ie... does q or voltage go up when inductor value goes up, or which will ring more?
or should i say which of the 3 options will have more volts at output?:
frequency = 100khz:
a)1uH + 2.533 Microfarad
b)10uH + 0.2533 Microfarad
c)100uH + 0.02533 Microfarad
...... how would these values impact my circuit

also if my inductor at resonance is 10uH which is determined by coil diameter, number of turns, and coil length then how should i wrap my coil?
ie
A)a 10uH coil same no matter what?
B)more turns = more volts or watts
C)longer coil = more volts or watts
D)more diameter on coil = more volts or watts

OR ... is it like teslacoil where more turns on secondary = more volts at output?

hope question is clear enough!
 

Tony Stewart

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Q or quality factor is the ratio of reactive to real impedance losses including load.
Therefore choose according to requirements. Tesla chose high Q air coils and 99.99% pure mica with very high Q. You can choose plastic caps rated for ripple current (rms) and voltage you require. He also relied on V= Ldi/dt with very high SRF air coils. with intermediate arccing to create very high di/dt transients and lots of X-Rays
 

dr pepper

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A bigger diameter air core coil exhibits a higher q, I think so does a longer coil.
 

Tony Stewart

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Depending magnetic design if using discontinuous mode with small arc existinguished large voltage arc >>100kV, interwinding capacitance must be minimized with gaps as large conductors for very max self resonant frequency, But if just making a stepup transformer then air core autotransformer just like those used in furnace igniters is Ok when pulsed.

It depends on design goal for output requirements. Start with 3kV/mm for air insulation and increase with better material or decrease with dust and humidity. Choose voltage for trigger voltage to initiate the arc and then follow on current for sustaining the arc at a much lower voltage via a secondary path. Air arcs have negative resistance like an SCR but different thresholds.
 

Dr_Doggy

Well-Known Member
going by the tutorial then, does that mean my equation is for center frequency and the differences between A,B,C is just how we set each band edge
..... but doesnt transformer math say that Vout is all about the *turn ration

also is it implied that i can get 100kv out of crystal radio receiver if tuned right(wont be wearing the earpiece!)!? ... also thats prolly verrry verrry low current

I think* what I want to do is build a crystal radio receiver , and tune it to mains, and replace earpiece with diode and 12v battery....
.... was thinking maybe wrap a coil round a bucket and collect a bunch of scrap capacitors , but now maybe not,
maybe a power supply transformer since it's wound to frequency, and then some caps...
.... what uF and uH should i choose for my crystal radio ... ?

just think it will be an interesting test as i have a huge steel plate about 20ft off the ground(on all corners)... and there was some interesting talk a while back about significant results based on ideal locations
 

Tony Stewart

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My Dad made me a pickle jar AM crystal radio using a quartz earbud hiZ , cat's whisker diode and a magnetwire wrapped jar with a tuning wiper. No batteries included over 50 Yrs ago and reception was great for a couple hundred miles with an antenna wire to the furthest tree on our property.

the interwinding capacitance was such that it resonated with a probably in the range of 100-200 for the load R of the crystal earbud. with a tuning range of LC parallel from 500 to probably >1000 kHz but no channels at that time there.

This is called a passive direct conversion. With freespace impedance transformed by Q to a high Z earbud with voltage gain due to Q of "tank circuit"

Most tiny Schottky diodes today will work in this function as the "hot carrier" diodes of yesteryear with perhaps poor selectivity from adjacent channel interference. with high Q direct conversion being possible but less selectivity or quality than single conversion or "superhet "

Try getting hands on a 7 transistor Japanese radio
 
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MikeMl

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The trick with crystal sets is that the Q of the LC tuner has to be high as possible (bandwidth as narrow as possible, to eliminate hearing several stations at the same time). You need to consider "loaded" Q, as well as "unloaded Q".

Unloaded Q is determined by the losses in the L and C. Capacitors have lower loss than inductors, so high-C/low-L tuned circuits have higher Q. Inductor losses can be minimised by using Litz wire (skin effect), silver-plated wire, big wire, air coils, etc.

Loaded Q is the actual Q of the tuned circuit after you couple an antenna to it, and after you extract energy from it to drive an output load, like hi-Z headphones.
 

Tony Stewart

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yes Q loaded by crystal earbud should have 14kHz BW at ~700kHz for 7 kHz Audio BW which determines L from impedance ratio. and Q = 700k/14k = 50
More gives you better ACI rejection and more sensitivity but narrower signal BW
Q=50 of tightly wound inductance computed online is easy to achieve. Interwinding capacitance calculators also exist, but otherwise guesswork for SRF.
 

Dr_Doggy

Well-Known Member
it still seems to me that reducing turns in L (or moving the whisker down )would also decrease volts ? , not just resonant frequency?

also doesn't this all mean that when a induction tuner dial is adjusted the capacitor needs adjustment too, to keep Q up?

if high Q means tighter bands but more amps then what happens to the noise, is it energy lost in to eddy currents, maybe it would be more ideal for me specifically to keep a smaller Q and allow for the noise to pass through, although i doubt there is much else out in the air there so maybe not? .

or maybe if i wanted to capture a second frequency simultaneously it would be best to build second receiver... so they both have high Q?
 

spec

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Hope this has not already been said, but the resonant frequency of a circuit, parallel and series, is given by the formula, f0= 1/(2* Pi * root[L *C])...........(F1)
Where:
f0 = frequency in Hertz
Pi= 3.142
L = inductance in Henrys
C = capacitance in Farads

To answer your question about different values of L and C producing the same resonant frequency, that is true- they will all have the same resonant frequency but the impedances of the circuits will differ; in general, the lower the L and the greater the C values, the lower the impedances of the resonant circuit.

The reactance of an inductor is given by the formula XL= 2 * Pi * f * L .........(F2)
Where:
XL = inductive reactance in Ohms
f = frequency in Hertz
Pi = 3.142
L = inductance in Henrys

The reactance of a capacitor is given by the formula XC= 1/(2 * Pi * f * C) .......(F3)
Where:
XC = capacitive reactance in Ohms
f = frequency in Hertz
Pi = 3.142
C = capacitance in Farads

Notice that the formulas for XL and XC are reciprocals so that when the inductive reactance and capactive reactance are the same they cancel each other out in a series tuned circuit and add to each other in a parallel tuned circuit; this is the essence of resonance. This is a simple explanation which omits the finer points, but if you had perfect inductors and capacitors the impedance of a series tuned circuit would be zero Ohms and the impedance of a parallel tuned circuit would be infinity Ohms.

A Xtal radio set is all about power transfer. You need to extract the maximum power from the antenna and feed that into the detector diode (Xtal) and then you need to get the maximum power into the headphone from the detector. To get maximum power transfer the power source and power sink should have the same impedance. This is known as the maximum power transfer theorem (don't confuse this with matching most transistor audio power amplifiers, bench power supplies, or batteries to a load which is a completely different matter).

The only problem is that the more you load a resonant circuit, the more you lower the Q, so the design of a Xtal radio is a balancing act.

There is quite a bit about Xtal radios and detectors on the 'Transistor Equivalent' thread: http://www.electro-tech-online.com/threads/transistor-equivalent.146091/

spec

Links
http://www.electronics-tutorials.ws/accircuits/parallel-resonance.html
 
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Tony Stewart

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I told u how to define the optimal Q, optimal L from a crystal earbud at least 50kohm.

Any questions? You will have more than adequate signal to Noise ratio with a LONG antenna wire , but to hear it you need a strong signal maybe 1mV, then It will work.
 

dr pepper

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Hmm is this a free energy project?
Your right to an extent a reactance tuned receiver will to a degree vary its q as its tuned, this is taken care of by the front end tracking filter to prevent images.
Transformer action isnt realy how its works though, as you know a capacitance and inductance will resonate at a certain freq, this property is used to tune to a particular radio freq, parallel resonance can give you voltage magnification, not power magnification though.
With perfect components yes its possible to get kv's from off air signals, however components are not perfect and the parasitic leakage in a circuit will prevent you getting high voltage.
Tuning to the mains even if you got it very effective is only going to register on the power company's meter, unless you abscond it from somewhere else.
Sorry to be a killjoy,.
 

Dr_Doggy

Well-Known Member
busted, but its not free energy as it is generated at a source , almost literally right over my head... so no closing of the threading!
that is very interesting that it could actually reach a kv high voltage, and I realize prolly less efficient than the kelvin water dropper, but my expectations are not high, a 12v trickle charger may still provide useful, really this is just for curiosity sake of how much...

also thanks for patience and repetition, i always have had those 2pifc and xc-xl equations memorized from school, but never understood them enough for visualization/application,
sorry i drag out this too, i think i understand now how to tune C & L, and that longer antenna is better,.... but when i went to wind the induction coil i was not sure how many turns to aim for, in my head it seems that turns (ratio) is a important factor..... i realize that there is X amount of power coming from my antenna, and a good Q will mean a better power transfer, but how to set my vout with that, ie, how would I establish if I have [email protected] output or [email protected] .....
I think this was mentioned about in regards to impedance ... which should be matched to load impedance...?

also dont be sorry!, Its unbelievable how many times a day i need to be told to stop chasing my tail, but it helps to wrap my head around things!
 

spec

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but when i went to wind the induction coil i was not sure how many turns to aim for, in my head it seems that turns (ratio) is a important factor..... i realize that there is X amount of power coming from my antenna, and a good Q will mean a better power transfer, but how to set my vout with that, ie, how would I establish if I have [email protected] output or [email protected] .....
I think this was mentioned about in regards to impedance ... which should be matched to load impedance...?
There are instructions on the net about winding the coil and the value and type of tuning capacitor.

The coil is connected between earth and the antenna. First you match the coil impedance to the antenna impedance for maximum transfer of power from the antenna to the coil. Then you tap the coil near to the earth end to match the characteristics of the detector diode and detector capacitor. The coil with the tap acts like an auto transformer

The voltage across the capacitor will indicate the received power. You could wire a 50uA moving coil meter (in series with a resistor) across the detector capacitor to have a permanent indication of received power. Suitable meters are available quite cheaply from eBay for example.

spec


 
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dr pepper

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It isnt down to me to close a thread I'm not a mod, besides I find your approach interesting.

I have seen a crystal rx light an led, so that would have been 2v at around 2ma, 12v might be doable with a good rx and a strong signal, but a ma is probably the most you'd get.

If your thinking am then your stuck with tuned l/c, however fm you could make a 1/4 wave whip, which has a low impedance, voltage is going to be low, but maybe you could drive a 'joule thief' with it, fm transmissions usuially dont go as far as am so rx power is going to be very low.
I dont think Q is going to make a big diffrence if you use a voltage step up circuit, if your not using a step up and want to get the voltage up high enough to charge a battery then maybe try an adjustable q tank circuit, you can put a resistor across a tank circuit to change its q, the resistor should be around the same impedance as the tank circuit at resonance, certainly no more than 2 to 3x.

I dont really want to tell you to do anything, however I will share my knowledge.
 

Dr_Doggy

Well-Known Member
just need to be careful using the F word, never know whos watching!
ok I think i am getting it now!
the earbud is to match the impedance of the LC circuit, which should also match the series antenna impedance, but i wonder , since impedance is like resistance then does that mean i can look at the antenna coil and tank coil as a volt divider circuit much like if i had 2 resistors there, ie if antenna z =50 ohm, and coil z=50 ohm, then Volts going in to tank circuit = 1/2 volts found at antenna, doesn't that mean that if i bring tank coil up to 100 ohm voltage would follow ohms law:
vout = (vin * Xtank) / (Xtank + Xantenna)

or maybe:
also it is mentioned about series tuned and parallel tuned, where series tuned antenna coil has ~0 ohm impedance and parallel tank coil had ~infinity ohms, is this what makes the "wave" roll off of the antenna in to rest of circuit since it is pure dampened, and in to the tank part that is the pure inverse(absorbent), causing max vout at tank...?
 

dr pepper

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Making me think now.
The best way I can explain that is maximum energy transfer happens at Rf when source and destination are the same impedance, if they are not matched you get reflections, ie not all of the energy from the source gets to the destination or load, so really if you made the load higher impedance transfer would reduce, what you say is true for audio but not for Rf.

The second part of your post I dont get, your correct about parallel and series circuits, not sure what the rest means, however if you understand matching impedance is important that might clarify things a little.

If I was to try something like this I'd contruct a 'flashing joule thief', this circuit charges up a cap untill the boost converter has enough voltage to run, then it empties the capacitor into the Led resulting in a flash, you could use something on these lines to charge a battery, allbeit very slowly.
 

Tony Stewart

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Most Helpful Member
just need to be careful using the F word, never know whos watching!
ok I think i am getting it now!
the earbud is to match the impedance of the LC circuit, which should also match the series antenna impedance, but i wonder , since impedance is like resistance then does that mean i can look at the antenna coil and tank coil as a volt divider circuit much like if i had 2 resistors there, ie if antenna z =50 ohm, and coil z=50 ohm, then Volts going in to tank circuit = 1/2 volts found at antenna, doesn't that mean that if i bring tank coil up to 100 ohm voltage would follow ohms law:
vout = (vin * Xtank) / (Xtank + Xantenna)

or maybe:
also it is mentioned about series tuned and parallel tuned, where series tuned antenna coil has ~0 ohm impedance and parallel tank coil had ~infinity ohms, is this what makes the "wave" roll off of the antenna in to rest of circuit since it is pure dampened, and in to the tank part that is the pure inverse(absorbent), causing max vout at tank...?
The antenna impedance is not 50 Ohms since you don't have a tuned length. Antenna impedance rises as X length reduces < Lambda/10

So you need to specify/compute antenna length/impedance at desired frequency F then choose LC from there to match then choose Load Q. But in your case a radio with "no batteries included" requires pre-engineered design or you have to compute LC impedance in order to shape bandwidth correctly wiothout being too far off antenna real impedance in your f band.

So find your crystal earbud first (>>50k) then specify f then design antenna length and then LC.

That being said your impedance will rise well above free space impedance of 377 Ohm. I have no idea what band you are are using.
 
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Colin

Active Member
Most of your requirement has been answered above.
But I make 2 points: When the cap reactance Ind reactance are the same, the max amount of energy from the coil is transferred to the cap and from the cap to the coil.
I have covered all this in my Crystal Set article but I will repeat it here.
The incoming signal from the radio station may only be a few micro volts, but the output of the tuned circuit may be a few volts. This represents an enormous Q.
What happens is this.
It's like pushing "big momma" on a swing. All you have to do is push a little bit each time and eventually she is swinging over the roof.
The same thing with a crystal set. It takes a lot of cycles to get the 2v and only one signal is pushing at exactly the right time. All the others are pushing at the wrong time and many are cancelling each other.
You can't take much current because the input from the right station is very low. When you load the circuit with a LED, the Q drops enormously and you get the benefit of all the nearly stations and that's why the LED lights.
 
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