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Components and Circuits For discussions about component types, alternatives and availability, circuit configurations and modifications etc. Discussions here should be of a general nature and not about specific sets. |
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23rd May 2017, 4:51 pm | #21 |
Octode
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Re: How to do bandwidth calculations using tuned LC with parallel R...
The "Q" of a system like this is simply the ratio of energy stored to energy dissipated per cycle: so you can take either the L or C impedance (as they're the same at resonance), and
for a series R: Q = ZL/R (or Zc/R) for parallel R: Q = R/ZL etc. If you happen to want good selectivity /Q whilst preserving a wider pass-band, you need to add a second tuned element and couple them to form a two-pole filter. You're then into filter design. John |
23rd May 2017, 5:04 pm | #22 | |
Dekatron
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Quote:
http://www.electronics-tutorials.ws/...resonance.html Lawrence. Last edited by ms660; 23rd May 2017 at 5:14 pm. Reason: alteration |
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23rd May 2017, 5:46 pm | #23 | |
Dekatron
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Re: How to do bandwidth calculations using tuned LC with parallel R...
I usually dump the frequency!
Q = R / (√(L/C) ) if L, C, R are all in parallel (like the OP's first post). Easy to remember, as R goes to infinity, so does Q. Q = √(L/C) / R if L, C, R are all in series. Again easy to remember, if R goes to zero, Q goes to infinity. If the resistance is in series with L, in a parallel tuned circuit, then the resonant frequency is affected very slightly (depending how it is defined!) but the second formula is still approximately correct. Making the coil with 3mm copper, 2 metres of it, it's instructive to see: Quote:
However, are you sure that the conductor resistance is the only loss? With a coil as large as that, at 1MHz I can see there being a bit of radiation loss, which will drop the Q even further... but I don't know how to calculate it! There is also loss in the capacitor - an ESR of as much as 0.05 ohms is not unreasonable. All this adds to broaden the bandwidth and improve the audio frequency response. It also acts to vastly reduce the sensitivity at the peak frequency, of course. Theory will get you as far as you want to go! When you take into account EVERYTHING that could POSSIBLY have an effect on the result, theory predicts EXACTLY the result you measure! (The problem of course is quantifying it and having sufficiently advanced maths!) |
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23rd May 2017, 5:51 pm | #24 |
Nonode
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Location: Cambridge, Cambs. UK.
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Re: How to do bandwidth calculations using tuned LC with parallel R...
I hear what you're saying but I'm still a bit confused by that, here's a link below showing that Q = 2piFCR = R/2piFL, so using Q = R/2piFL (I find that easier) and suppose the parallel resistance of 10k in Al's circuit was replaced with a resistance of 100k, what's the Q value then? I make it stupidly high unless I've missed something?
As with most of these things, the formula is a simplification and assumes zero series resistance in the inductor, so if the parallel resistance is very high, Q will tend to infinity. In practice, the inductor series resistance Rs will limit the Q because it can't go higher than 2pifL/Rs. I guess that a good challenge would be to work out the Q formula incorporating both parallel resistor R and inductor resistance Rs, but I think it's beyond the scope of this thread. Easier to look at them separately because one or the other will usually dominate the picture. Martin
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23rd May 2017, 5:51 pm | #25 |
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Re: How to do bandwidth calculations using tuned LC with parallel R...
There is still is one undefined variable, the resistance (yes the real term, the one that sogs Q) of the collector of the output transistor.
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23rd May 2017, 5:53 pm | #26 | |
Dekatron
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Quote:
At my stage of the journey, theory is the thing which, when I look back, explained I didn't know I didn't know! (Which is, still, to arrange a formula such that a parallel R's affect on bandwidth can be modelled/ quantified.) Also, I'd be interested to hear how much of a mismatch there is between the source impedance in the output collector circuit and the loop? Cheers folks!
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Al |
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23rd May 2017, 5:54 pm | #27 | |
Dekatron
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Quote:
Dekatron, Octode or the next thing may or may not mean very much. In my case it means I've held an opinion more times than I can remember and also asked quite a lot of questions! Nobody bites round here, we're all a friendly bunch. Welcome!
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23rd May 2017, 5:55 pm | #28 |
Nonode
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Kalee20: Our posts crossed, but I think our messages are similar.
Martin
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23rd May 2017, 7:12 pm | #29 |
Dekatron
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Exactly!
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Al |
23rd May 2017, 7:57 pm | #30 | |
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Re: How to do bandwidth calculations using tuned LC with parallel R...
My guess is that the output collector will have a high impedance, the exact value being nontrivial to calculate and possibly varying with modulation. It could easily be 10's of k or more. Hence not loading the loop very much. The main loss mechanism will be loop resistance. At 1MHz the loop will not have much radiation resistance - which is good as it means the loop will not radiate much beyond the 'pantry'. Nearby receivers will actually see the local induction field, not the radiation field.
There is a somewhat counterintuitive behaviour of capacitive taps, which is that a very low resistance or very high resistance applied will mean high Q but a middling resistance means low Q. Quote:
Alternatively, do a series-parallel transform (in whichever direction you like) and add the resistances/conductances (as appropriate). |
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24th May 2017, 7:36 pm | #31 | |
Dekatron
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Quote:
So, I make it that ... When Q=250, 3dB bandwidth =4KHz When Q=175, 3dB bandwidth =5.7KHz Which would be perfect... I think at the moment, Q is between 250 and 500. At Q=500, 3dB bandwidth is only 2KHz! I don't think it's (Q I mean) this high. Just somewhere in between. Voice is perfectly clear but tuning is critically difficult but just acceptable to peak, on an 81 year old set and on a 35 year old-ish tuner. I have a FFT (fast Fourier Transform) function on my 'scope but it is a pain to use. It incorrectly suggests that the carrier is frequency modulated, which it isn't. Also I don't know if the sampling rate is high enough (I think the bandwidth of my scope is only 25MHz). I was hoping to see when the spectral lines first peaked to measure the sideband. The centre value is almost off the scale at the left of the screen...
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Al Last edited by Al (astral highway); 24th May 2017 at 7:59 pm. |
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25th May 2017, 12:22 pm | #32 |
Rest in Peace
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Re: How to do bandwidth calculations using tuned LC with parallel R...
Remember that to get music bandwidth of f you need the loop to have bandwidth 2f. You need half the Q figures.
Producing some FM when you meant to produce just AM is quite normal for simple circuits. |