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Wireless Mains Adapter-Frequency etc.?
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11-06-2013, 08:25 AM
Your circuit sounds like a FET version of the one I built when i was a teenager.
The output was a TV overwind. It never welded aluminum as intended. What is the finished article going to do?
11-06-2013, 08:46 AM
It will be a wireless mains adapter for a Hacker radio. it will charge the two batteries ( 6 x AA rechargeables in PP9 cases and run the radio.
As for the heating I referred to, it was a side effect of a high power wireless power Txn system which used a single turn of 8mm copper pipe as the tuning coil. If I remember correctly this also ran in the 100s of KHz range. With my little system I'm looking to get 2 x 3-5Watts across about a 1.5 cm distance. Mike
11-06-2013, 09:11 AM
""The problem is that at resonance the inductive reactance is zero""
I'm confused. Lawrence.
Maybe I'm getting it all wrong, but I thought that at resonance the current in the cct saw only the resistive "real" part.
http://www.electronics-tutorials.ws/acci...nance.html, Mike
11-06-2013, 11:04 AM
""The problem is that at resonance the inductive reactance is zero""
I'm confused. Mike, I'm not disputing the purely resistive element, but I am confused in that you say that the inductive reactance is zero at resonance. I always thought that the reactances of the components concerned (Xl & Xc) at resonance had to be of equal magnitude for resonance to occur thus suggesting they must have a value at resonance. Lawrence.
They both have equal value at resonance.
A parallel tuned circuit has its highest impedance at resonance while a series circuit has its lowest. Just think of a capacitor being almost open circuit at low frequency and the inductor being almost open circuit at high frequency and you will see why this it true.
Hi Gents,
You're both right, the reactance is equal ie. XL=Xc at resonance in both parallel and series arrangements, I should have said that the 2 reactances appeared to cancel leaving only the resistive element. Heat dissipation being a function of Vsq/Rint. This I guess leads on to a bigger can of worms with Q involved too. I made a new Tx coil with the same wire and 18 Turns which I will try out tomorrow. What about the Fets, is there any way to bias them so that the drive to the coil will be low until a load is present? Mike
No one has pointed out the one significant aspect of this 'reactance at resonance' issue: it's all a question of phase.
Inductive reactance and capacitive reactance are 180° apart. The magnitudes of those two reactances, at all frequencies except one, will be different: hence, a net reactance will be produced: either capacitive or inductive, depending on which one has the greater magnitude. But when the magnitude of those two reactances are equal, since their phase difference is 180°, the net reactance becomes zero: this occurs at one frequency. At that one critical frequency, the only limiting 'opposition' to the current flowing is the effective total resistive component of that circuit. This concept is easy to imagine in a series L/C cct. that has resistance in series (real or equivalent) with the capacitance and the inductance. In a parallel L/C cct., the concept is not quite so easy to comprehend. The two reactances are now connected in parallel, not in series, and because of this type of connection, the net reactance at resonance now theoretically approaches infinity. However, since a perfect reactance is impossible to realise in practice (for example, the coil has a real physical resistance), a very high 'opposition' to current flow results. This 'opposition' has all the properties of a real physical resistance, since the supply voltage to, and the supply current in, the L/C cct. are exactly in phase. And the ratio of that voltage and current gives the resultant 'equivalent' resistance: the dynamic resistance of the L/C cct. at resonance. This will automatically include the real 'physical ' resistance of the cct. That dynamic resistance = L/CR, where R = that real resistance. The 'sharpness' of that impedance curve (at resonance only) is assessed by the Q of the cct.: Q = (1/R)*√(L/C). All of that can be explained much more simply by recourse to some elementary mathematics involving the j operator . . . . but that's another topic for another day. So there we are. I hope that that little summary of reactance and resonance clears up any confusion here. Al. / June 12, 2013 //
11-06-2013, 10:27 PM
My post #48 crossed with post #47 by boiss: it appears that a supplementary comment is now necessary.
When we have an L/C cct., operating at resonance, which is electro-magnetically coupled to another impedance, the behaviour of that L/C cct. will become modified depending on the nature of that coupled impedance and the degree of coupling it has to that L/C cct. The effect of that impedance on the L/C cct. is called the 'reflected impedance'. That impedance could be reactive or resistive: if it is another L/C cct. and if it is also at resonance, it will reflect a resistance into the original L/C cct. that acts as a parallel resistance with the dynamic resistance of that original L/C cct, thus reducing its effective dynamic resistance and, consequently, the resultant Q. That is why, under those circumstances, the supply current to the original L/C cct. increases as the coupling is increased: energy is being transferred from the L/C cct. to the new, additional L/C cct. Should the new L/C cct. have a resonant freq. that is slightly different to that of the original L/C cct., we obtain the traditional band-pass tuning response curve: two resonant peaks, the separation of which depends on several factors, the degree of coupling between them being the most significant. Al. / June 12, 2013 // |
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