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Old 23rd May 2017, 4:51 pm   #21
John_BS
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Default 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.

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Old 23rd May 2017, 5:04 pm   #22
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Default Re: How to do bandwidth calculations using tuned LC with parallel R...

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Originally Posted by Hartley118 View Post
That would be right when R is in series with L, but as I read Al's question, his R is in parallel with the inductor+capacitor parallel combination.

In that configuration, we'd expect Q to increase as R is increased because damping is reduced. So we'd expect R to be in the numerator, not the denominator.

Hence, Q = 2pifCR feels right to me when L, C, and R are in parallel.
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?

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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Old 23rd May 2017, 5:46 pm   #23
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Default 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:
Originally Posted by G8HQP Dave View Post
3mm diameter copper has a resistance of 0.028 ohms per meter at 1MHz. 2m is thus 0.056 ohms. A 2.5uH inductance with then have a Q of 285, which means bandwidth 3.5kHz so audio bandwidth of 1.75kHz. This will sound muffled.
That's really helpful, it indicates that the 10k resistor is having far less effect than the series coil resistance.

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.

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Originally Posted by astral highway View Post
So we seem to be midway between doing a thought experiment and pure empiricism.

In this case, I'm saying, (and looks like other are saying...) 'Wait up, theory gets us so far, but other things are happening here that we [I, certainly] can't account for.
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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Old 23rd May 2017, 5:51 pm   #24
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Default 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.

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Old 23rd May 2017, 5:51 pm   #25
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Default 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.
 
Old 23rd May 2017, 5:53 pm   #26
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Default Re: How to do bandwidth calculations using tuned LC with parallel R...

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Originally Posted by kalee20 View Post
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!)
I know, I know. I was semi-serious.

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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Old 23rd May 2017, 5:54 pm   #27
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Default Re: How to do bandwidth calculations using tuned LC with parallel R...

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I've been watching this thread form the beginning, drawn a few Smith charts trying to solve, but I am still to shy to post to Dekatron or Octodes,
Hi Jacek,

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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Old 23rd May 2017, 5:55 pm   #28
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Default Re: How to do bandwidth calculations using tuned LC with parallel R...

Kalee20: Our posts crossed, but I think our messages are similar.

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Old 23rd May 2017, 7:12 pm   #29
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Default Re: How to do bandwidth calculations using tuned LC with parallel R...

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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.
Exactly!
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Old 23rd May 2017, 7:57 pm   #30
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Default 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:
Originally Posted by Hartley118
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.
Calculate one Q assuming the other loss mechanism is absent. Do it the other way round too. You have two Q values: Q1 and Q2. Then total Q is given by 1/Q = 1/Q1 + 1/Q2 -- just like combining parallel resistors.

Alternatively, do a series-parallel transform (in whichever direction you like) and add the resistances/conductances (as appropriate).
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Old 24th May 2017, 7:36 pm   #31
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Default Re: How to do bandwidth calculations using tuned LC with parallel R...

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Originally Posted by G8HQP Dave View Post

Alternatively, do a series-parallel transform (in whichever direction you like) and add the resistances/conductances (as appropriate).

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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Last edited by Al (astral highway); 24th May 2017 at 7:59 pm.
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Old 25th May 2017, 12:22 pm   #32
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Default 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.
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