How to do bandwidth calculations using tuned LC with parallel R...

[Al (astral highway)]

Hi folks,

I'm familiar with the physics of resonant conditions in a series or parallel LC circuit, and the lumped circuit equivalents that are often used to characterise these or refer to them.

I'm currently applying some intense focus to the best possible performance of my home-made loop antenna for 1MHz low-powered rebroadcast in a home-internet.

I tuned this to resonance successfully. It has a capacitative voltage divider
but essentially let's say it has 10nf in parallel with 2.5uH.

I know that audio bandwidth needs to be 15KHz for reasonable AM music reproduction right? And I'm guessing 8KHz for voice or something. And that too high a Q in the tank circuit is going to clip things audibly. I don't know much more than this, other than calculating Q, and that I can increase the bandwidth by introducing series or parallel R with the tank circuit. So I need a nice high Q but with very steep shoulders, I imagine, either side of this bandwidth.

In practice, there isn't any clipping just as things are, but that's with a 10K resistor in parallel with my tank circuit. I don't find it satisfactory that I can't quantify why exactly this works (I mean, numerically, not generically, which I understand) or whether it works the best, or what the trade off is in radiated power etc.

So this isn't an 'it's broken, let's fix it!' problem, but an 'ah, it works, but why and could it work better?' inquiry. That's just how my mind works.

Thanks, folks!

[kalee20]

Bandwidth of a LC parallel circuit is given by BW = fc / Q.

But this is only approximate, valid when bandwidth is a small fraction of centre frequency.

If you do the maths, you'll find the response is not symmetric anyway, about the centre frequency, on a linear scale. But it's 'good enough.'

With Al's values, 2.5uH and 10nF, I calculate centre frequency of 1.006MHz. And with parallel shunt resistance of 10k, I calculate a Q of 632. Which gives a bandwidth of 1.58kHz.

To be honest, I would have thought that the 10k resistor would make hardly any difference! If you get good audio, then the bandwidth must be much more than 1.58kHz... which means that Q is lower, so you have much more loss resistance somewhere. And given this, it will be far more than the losses introduced by the 10k resistor.

Are you sure that the resistor is 10k? I've been caught out by multi-coloured bands, only revealed when I've measured the thing!

[Al (astral highway)]

Originally Posted by kalee20

To be honest, I would have thought that the 10k resistor would make hardly any difference! If you get good audio, then the bandwidth must be much more than 1.58kHz...

Ahaha! I snipped it out and there is no audible difference!. Something still isn't right though. There is a lot of bass (more than I need for this purpose) but voice is muffled at times.

So your theory about Q-dropping natural resistive losses is correct. It's hard to see where as I am using very high quality capacitors, but it doesn't matter as the bandwidth is indeed higher. Thanks for your explanation, very clear indeed and I'm sure others will benefit too,.

Wouldn't mind a steer on the muffled voice aspect if anyone has any insights.

[G8HQP Dave]

Much of the loss in a loop is the loop itself.

Lots of bass but muffled sound suggests a limited bandwidth - maybe 2-3kHz or less. Limited bandwidth does not cause clipping. Given the calculations in post 2, you should try a smaller resistor value to get more bandwidth. This will reduce the signal level. You need 3-5kHz for voice and 5-10kHz for AM music. Note that the tuned circuit bandwidth needs to be twice the modulation bandwidth.

You cannot get high Q and 'steep shoulders' from a single tuned circuit. That is why superhets use two or more double-tuned IFTs to get the requires response. For a single tuned circuit if you have enough bandwidth for music then you will have low Q and quite broad shoulders.

[Andrew2]

Wow - is it me just not fully understanding these things (not exactly unknown!) or is that b/w astoundingly narrow for such a low inductance tuned with such a large C?
My own loop antenna (built for the same purpose as Al's) has a -6dB bw of plus and minus 30 KHz, so obviously there is no significant sideband cutting. It would probably be tighter if I'd used nice hefty wire instead of awful Post Office issue thin, single core stuff!
Mine is 3 feet in diameter and measures 23 uH, and it's tuned to resonance on 1575 KHz with (guessing) less than 1000 pf. Match to 50r is done as below. I've also included a plot of the response.

[ms660]

And me as well, can anyone explain the Q value worked out earlier, 2.5uH seems a low value for resonance at 1MHz for high Q?

Lawrence.

[Al (astral highway)]

Andy, so using your diagram as a reference , my 'inner ' goes to the centre tap of a capacitative voltage divider. On either side is 10nF capacitance, one to the grounded side of the loop and grounded side of the coax. The inductor is a single turn of 3mm wire wound taut over a hexagonal former made of six snooker triangles bonded together. If I recall correctly the outside diameter is around 2metres long.

Lawrence, that is how we get resonance and high Q - about 2.5uH inductance from this single turn and 10nF capacitance.

The intentional single turn is because we want the lowest lumped resistance, the invisible series resistance we would have with multiple turns of thinner wire. We could then use lower capacitance but our Q would be lower. Things could be very different if this was a receiving loop.

[Andrew2]

Thanks for that, Al.
So you are using the same arrangement as me but with 10nF for both caps instead of ~10nF in parallel with the coax and ~ 1nF in series with the loop.
My theory is poor regarding Q etc, but in my experience using a large C and small inductance is 'unfavourable' and leads to a pretty flat response. Quite how you have achieved so much Q that you are muffling the audio is causing a bit of head-scratching here!
A big L and small C leads to much sharper tuning in my experience, but of course there are higher losses due to the amount of wire. Having said that, I think there maybe only 2 or 3 turns on my loop of 23uH (I could open the hula hoop and take a look).
If you have to put a low-ish value of R across your loop to flatten the response, that is also introducing a resistive loss, so we seem to get back to square one.
Really I'm just confused....

[G8HQP Dave]

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.

[Andrew2]

Dave, I'm having trouble with this!
I've just made a loop of about the same dimensions as Al's out of thick coax. Its DC resistance measures as 0.1 ohm, so about twice that of your example.
With two 10n's across the ends and the 50r sig gen across the bottom cap, the resonance came in at 1.2 MHz-ish. Good so far, but the return loss is only 5dB, a poor match.
I then did the bandwidth test in the same way as I did for mine - using the loop to transmit a signal from the sig gen and receiving it on a small pickup loop on the input to my analyser.
The result is shown below. I've had to use a wider span to reveal *any* peak. It's certainly in no danger of affecting the audio response.
I realise my loop has double the resistance of your example, but I can't see the difference affecting things to any great extent.
Any ideas?

[G8HQP Dave]

DC resistance does not tell us much. AC resistance will be higher at 1MHz. However, check my figures for 3mm copper - I could have made a mistake.

If your sig gen has an output impedance of 50 ohms in parallel with 10nF then you have at 1MHz 50R in parallel with about 16R reactive. This transforms to about 4.5 ohms plus 14 ohms reactive in series. The 4.5 ohms will dominate, giving a Q a bit above 3 and very broad bandwidth.

Now what we don't know is the output impedance of his source. I do mean output impedance - not optimum load impedance, which can be quite different. It could be that the muffled modulation comes from his source not his loop. There simply is not enough hard information to go on.

[Andrew2]

Ah, we might be getting somewhere now. I'm so accustomed to working in a 50 ohms environment that I was assuming Al's TX was 50 ohms, but of course it may be nothing like that.
Time for a sit out in the sun while it lasts!

[G8HQP Dave]

Transmitters, even simple ones, very rarely have 50 ohms output impedance. They almost all have 50 ohms optimum load impedance. On the other hand, decent sig gens will often have 50 ohms output impedance as they will finish with an attenuator pad to ensure this.

[Hartley118]

Al: Your original question about working out the Q of a parallel tuned circuit with a parallel resistor is an interesting one because its analysis seems to be usually avoided, as in 'it may be shown that.....'

Instead of looking at voltages and reactances as we do in a series network, the analysis requires a study of the currents in the elements and their susceptances (the reciprocal of reactance). Susceptance is to reactance as conductance is to resistance.

There's a neat analysis in the electronics tutorials series at http://www.electronics-tutorials.ws/...resonance.html

In summary, the equation for Q is in the picture (because I don't know how to insert a square root). Please click on the thumbnail:

[Al (astral highway)]

Originally Posted by G8HQP Dave

Now what we don't know is the output impedance of [Al's] source. I do mean output impedance - not optimum load impedance, which can be quite different. It could be that the muffled modulation comes from his source not his loop.

Here is the source impedance, diagram below.

Two notes;

1) There is a mod. of mine in the output transistor's collector, which will have an indeterminate effect on the bias, I realise.

I have put a 220uH choke in the collector instead of the 22uH shown. The source text says this isn't critical and suggested even 100uH, but I realise this choke forms part of the output impedance together with the loop's lumped characteristics.

Capacitor values in the source text are for resonance closer to the top of the AM band, so mine is on point for 1Mhz exactly. I peaked this on a signal generator and 'scope.


2 ) Although my inductor isn't copper tubing, as shown, and isn't a square, the as measured inductance of my version is very similar.

3) I have an 18R resistor in series with the 470uF emitter bypass electrolytic, instead of the 22R shown. I know this will have some effect on linearity but I don't think it's the source of the muffling.

I can bring the whole thing to high enough bandwidth with a 3 metre long wire and series matching inductance/ pi network, but I don't want this, I want the resonant loop.

Originally Posted by G8HQP Dave

There simply is not enough hard information to go on.

Now there is, Dave! It would be great to get to the heart of this.
Cheers!

Originally Posted by Andrew2

I realise my loop has double the resistance of your example, but I can't see the difference affecting things to any great extent.
Any ideas?

Andy, I appreciate your head-scratching and parallel brain-power and experimentation. Does this diagram make it easier?

[Al (astral highway)]

Originally Posted by Andrew2

Thanks for that, Al.
So you are using the same arrangement as me but with 10nF for both caps instead of ~10nF in parallel with the coax and ~ 1nF in series with the loop.

...exactly, Andy. Mine is just configured as a capacitative voltage divider as the swing would be too large across the collector otherwise. (Diagram shows what
I mean...)

[ms660]

Originally Posted by Hartley118

Al: Your original question about working out the Q of a parallel tuned circuit with a parallel resistor is an interesting one because its analysis seems to be usually avoided, as in 'it may be shown that.....'

Instead of looking at voltages and reactances as we do in a series network, the analysis requires a study of the currents in the elements and their susceptances (the reciprocal of reactance). Susceptance is to reactance as conductance is to resistance.

There's a neat analysis in the electronics tutorials series at http://www.electronics-tutorials.ws/...resonance.html

In summary, the equation for Q is in the picture (because I don't know how to insert a square root). Please click on the thumbnail:

In my reference book it gives the formula for Q in a series circuit as 2piFL/R
it also gives the same 2piFL/R formula for a parallel circuit.

Lawrence.

[Al (astral highway)]

In my reference book it gives the formula for Q in a series circuit as 2piFL/R
it also gives the same 2piFL/R formula for a parallel circuit.

Hi Lawrence, this inquiry isn't into the theoretical factors determining Q - we know that with R as the denominator, it's critically important - either because we want Q all high or we want it quite high but not so high that it affects bandwidth. I wanted my R as low as possible, so went with 3mm copper wire instead of multiple strands of thinner wire and more capacitance. But I can't theorise the lumped resistance of the circuit because it obviously isn't just a fractional part of an Ohm. There's some elsewhere, but we just can't exactly determine it.

But now it looks like I've overcooked it and we need some damping, some intentional R for de-Q-ing and to increase bandwidth. Or not.

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. And one of those may just be the source impedance of the circuit.

[JacKam_]

Hi All,

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, my screening grid is still at very negative voltage.

Could someone write an abstract, on what happened till this point ?

Best regards,

Jacek

[Hartley118]

Originally Posted by ms660

In my reference book it gives the formula for Q in a series circuit as 2piFL/R
it also gives the same 2piFL/R formula for a parallel circuit.

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.

Martin