This (Ho)ratio detector has me in a half-Nelson - Page 1
Forum section: electronics / beginners

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 27, 2022, 10:27:44 am
content: Spending hours trying to fully get this one. The first thing I'm having difficulty with is a tank circuit above and below resonance. According to 'learn about electronics ' above resonance the circuit is capacitive, but according to NEETS (the old U.S.Navy course) it's inductive. And according to my old ICS manual (page attached) ...you can see their vector diagram of the phase shift is the opposite of the Neets one, in terms of the angles above and below resonance.So that's the first thing. Help and thanks!https://www.rfcafe.com/references/electrical/NEETS-Modules/NEETS-Module-12-3-11-3-20.htm

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 27, 2022, 10:48:35 am
content: Ah...maybe the vector diagrams in the NEETS, for the Foster-Seeley, are different because that tank operates on both halves of the cycle unlike the ratio detector.Still confused about the first bit of my enquiry though.

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 27, 2022, 11:14:34 am
content: Is it true to say a tank circuit above resonance becomes inductive or capacitive depending on your point of view??So inductive in the sense that the coil becomes more 'resistive' and blocking, but capacitive in the sense that as the coil reactance goes up the circuit more and more resembles just a series capacitor of low reactance.(So the current starts to lead the voltage)

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 27, 2022, 11:49:37 am
content: Just to clarify the confusionIn the NEETS page, half way down under ratio detector- " When a tuned circuit operates at a frequency above resonance the circuit is inductive. The secondary coil current i(s) lags the primary voltage e(p). "Makes my head hurt and yours probably...

author: mag_therm
author_rank: Frequent Contributor
timestamp: June 27, 2022, 12:39:49 pm
content: Wow, you are sharp to pick that up.Actually I think the diagram you posted, along with my reference book here, are both incorrect.My book says to disregard the "effect of the diodes and their loads", and then presents vector diagramsshowing the secondary voltage varying in phase. (!)The secondary driving voltage E_sec = -jWM *I_1The secondary capacitor reactance is -j/(WC).There is no R term except for the Q factor of the windings which should be Q= 20 ~ 30, so small R component.Either by Kirchoff or just by parallel impedances,the (-j) terms are added on denominator then reciprocated... ( bring more coffee...)So I think the magnitude of E_sec/2 on each diode will vary, but stay nearly in quadrature.Maybe the quickest way to answer your question would be to set the Foster and Seely up in sim.Note that the FM modulated signal coming in has pre-emphasis both to reduce noise by de-emphasis , and to hold the modulation index down at low audio frequencies.

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 27, 2022, 05:25:05 pm
content: Quote from: mag_therm on June 27, 2022, 12:39:49 pmHow do I set up a sim? I've heard of these but no idea what it really is!Can you explain to me how pre-emphasis keeps the modulation index down please. I know what the index is, but not following how that's so.

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 27, 2022, 05:33:58 pm
content: Here is the complete article (sorry about the order, the running order is bottom top then middle) from my ICS book, which I'm trying to understand. Most of it's ok, the sticky paragraph is at the top of the top page : I don't get how E5 acts from point D, when that point is the lower end of coil L3 which is zero volts on the positive half of the input.

author: mag_therm
author_rank: Frequent Contributor
timestamp: June 28, 2022, 11:18:05 am
content: Hi Qantum. That is the best description I have seen, but I am stillnot understanding the vector diagrams and if they are correct.It is interesting that your article shows the diodes with upper diode cathode right and lower diode cathode left.I can understand with that connection how the deviation above and below resonance will cause rising and falling quadrature voltages and be rectified, but only over a half wave, by both diodes, to get the baseband audio across the audio capacitors..However most references to Fostor Seeley on internet ( and my old reference book) show both diodes cathodes to right, which makes the circuit a just a balanced centre tapped rectifier. Also most references have similar paraphrased text as explanation and I wonder if there is a long term error that hasbeen carried over. Obviously the circuit must work but I don't understand the description given.Anyway to understand further , I will read about the "parallel resonance with resistance only in inductor arm" case, which is what the quadrature side actually is.

author: Circlotron
author_rank: Super Contributor
timestamp: June 28, 2022, 11:46:02 am
content: Quote from: mag_therm on June 28, 2022, 11:18:05 amFoster Seeley detector has both diodes pointing right.Ratio detector has diodes pointing opposite directions.https://en.wikipedia.org/wiki/Ratio_detector

author: mag_therm
author_rank: Frequent Contributor
timestamp: June 28, 2022, 12:07:14 pm
content: Circlo, Thanks.That means the description in Quantum's post #6 is for a ratio detector.Edit: And your wiki link has waveforms in time domain showing diode conduction times.This is easier to understand and probably more correct that the usual vector diagrams

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 28, 2022, 03:24:44 pm
content: Quote from: mag_therm on June 28, 2022, 11:18:05 amHi Mag, thanks for chipping in again. I keep thinking I'm nearly there...and then another anomaly pops up. The book is from a 1980 correspondence course - they are pretty good instruction generally.Can I home you in on the top paragraph again on page 2, the vectors page: "again, the combined emf induced in L3 and the lower half of L2..." etc. At this point are they considering the negative half of the primary tank cycle? So arrows reversed? (on the original circuit diagram)The description only makes sense if they are surely?

author: mag_therm
author_rank: Frequent Contributor
timestamp: June 28, 2022, 04:52:03 pm
content: Hi Quantum,That is about what I am questioning too.I was thinking to set up a simplified model in qucs.However I have an old tube FM radio that I use a lot, it needs a tune up.It has the ratio detector with two OA79 diodes and is very similar to your Fig 3.13So I will bring it down, feed a strong signal in from the old Wavetek here which has FM modulation and try to get the voltage, phasing and diode currents. They might be hand sketches or photos from my vintage test gear here.In the mean time look at the time waveforms in link posted by Circlo. I think thay are more useful than vectors, as the diodes only conduct for short periods. They could be expanded with more detail.

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 28, 2022, 06:30:59 pm
content: Quote from: mag_therm on June 28, 2022, 04:52:03 pmWow, that sounds a great idea. We will not be defeated by a ninety year old circuit!Yes, I looked at those wiki waveforms. So Vs2 is the top of the sec. coil and Vs3 is the top of the tertiary coil. It shows Vs3 leading Vs2 by 90°... Which is the opposite way round to my article, and also the Neets description.

author: mag_therm
author_rank: Frequent Contributor
timestamp: June 29, 2022, 07:21:44 pm
content: I did some calcs and measurements white aligning the FM detector in the 1963 Panasonic RE tube FM receiver.The signal generator which has FM modulator was connected to 10.7 MHz IF upsteam of the last IF amp, to align the detector.This detector has 3 resonant frequencies, it would be necessary to destroy the coils to see the connections, so I assume the quadrature coil has 2 coils each with a different inductance or else one coil is wound in opposition ( centre tap soldering under the wax.)The operation can be viewed by connecting 'scope probes to each of the 2 diodes, and observing the amplitude changes while manually stepping the deviation.The positive diode is connected to the coil with highest resonant frequncy.So as the deviation goes positive the voltage rises on the upper resonance , the upper diode conducts, pulling the output high.As the deviation goes negative, the voltage rises on the lower resonance and the lower diode conducts , pulling the output low.I think that both diodes are conducting with zero deviation, however I could not measure that with the limited test points added.I also put some calc amd measurement to support my post #4, that the published descriptions of operation by phase change on the quadrature coil are not valid.FM broadcast has the highest modulation index compared to other communication.Even with high deviation of 75 kHz, it is not possible to get anywhere near usable phase shift at 10.7 MHz, confirmed by today's Q calculation and measurement on page 2 below, showing phase shift of only 4 degrees at full deviation..https://app.box.com/s/v5kvqi88jc2zkkn2w52i8qfd2veyuvrwhttps://app.box.com/s/avi9llwy9ykosw4hvjiwiln3v8sqprv4

author: Quantumplate
author_rank: Regular Contributor
timestamp: June 30, 2022, 04:42:22 pm
content: Quote from: mag_therm on June 29, 2022, 07:21:44 pmAppreciate your work, thanks! A lot for my beginner brain to chew over.So your radio (which I see has a Wiki entry - nice old thing you have there) has the Foster-Seeley variation, working and conducting on both halves of the input. If your results show that phase shift alone is too small to produce enough voltage variation, then does that mean the main voltage function is produced from the Slope Detector effect of tuned tanks off resonance?I've attached a crappy drawing of most of the ratio detector from the NEETS page. The low pass pi filter is just out of view, but I'm trying to determine whether indeed this circuit also works on both halves of the input, despite online info to the contrary (including the NEETS description). If you look at the NEETS circuit again you'll see at the O/P an a.c. audio waveform, NOT an alternating d.c.I hope it's viewable as I rather stupidly chose two dark colours to indicate the current directions (proposed) for both halves of the input wave. On the negative half cycle in blue we have two loops of current going on, and on the +half cycle I see a loop indicated in grey pencil...but , it would need E2 to be less than Ep...But is that so?They've offset coil L3 from L1, which maybe indicates that E2 is greater than Ep, but then again L2 and L1 are meant to be loosely coupled magnetically speaking, so not much emf should be produced in L2. Any thoughts?To get the audio waveform they've shown you would need an FM signal with AM modulation at the point I've written on my drawing in blue. Ergo, it works on both halves of the input. It's interesting that in my ICS article they also describe the voltages on each of the diodes in such a way that means they must be considering both halves of the I/P in turn.Finally, can I try and answer my question to you from earlier in the thread: so does pre-emphasis hold down the modulation index by reducing the dynamic range of the modulating signal? Is that anything to do with it?

author: mag_therm
author_rank: Frequent Contributor
timestamp: June 30, 2022, 07:00:53 pm
content: Hi Quantum, I finally today found the papar by Foster and Seeley which has a detailed description.Search ( it is not pay walled) :"Proceedings of the Institute of Radio Engineers volume 25 March 1937"Page 290 ~ 291 show the idea; that when the two inductively coupled tanks are tuned to exactly the same centre frequency,The SCALAR AMPLITUDEs of the sum of the voltages will have two peaks, above and below the centre frequency in Fig 1 page 291.That is analogous to "over-critical " coupling of two resonant circuits where 2 humps appear each side of the centre resonance.Then the coil connection circuit as we know it today is drawn, at Fig 2, which shows both resonance peaks on the same set of axis,and the names E1 and E2( Later shown connected to the diodes) are having the two peaks E1 : lower side below centre frequency, and E2 upper side above centre frequency )The rate of change of scalar amplitude as the frequency swings about the centre frequency is described verbally on page 292and math at (7) etc page 296.The above generally coincides with what I measured yesterday (post #13 link 1) on the old Panasonic .Centre = 10.66 MHZ set by slugE1 ( my TPB) =10.55 MHzE2 (my TPR) =10.84 MHzOn Page 294 the math is not so different to today except the vacuum tube transconductance is included in calc.just treat that as FET driving the circuit.The Complex voltages are stated for E1 and E2, at (3), then the Scalar magnitude for E1 is given at (4) and E2 will be the same.So there you have it. It is fairly simple when the inventors have explained it so well.The old Panasonic here is happy now all the dust vacuumed out and is aligned better than ever,but not so happy about the slot cut in his discriminator can to get the test points out.

author: Quantumplate
author_rank: Regular Contributor
timestamp: July 01, 2022, 01:13:04 pm
content: Quote from: mag_therm on June 30, 2022, 07:00:53 pmah that's great - I can't get inside it however. I pasted the above into google, came up google books, but all I have there is a contents page.How do i get inside?

author: mag_therm
author_rank: Frequent Contributor
timestamp: July 01, 2022, 02:03:07 pm
content: https://worldradiohistory.com/Archive-IRE/30s/IRE-1937-03.pdfIt is a scan.If you want to go back to the derivation of function (3) which I think is where answer to your first question is,you will have to unravel some of the parametric functions they used.At least, they explain most of them at top of page 293.Edit to Add: The reason I set my centre frequecy to 10.66 MHz and not the conventional 10.7 MHz IF,was that the primary slug vas either jammed or glued and I did not try to loosen it for fear of cracking it.The secondary slug was free, so I set it 10.66 to coincide as nearly as possible with the primary (scalar difference of 0 at deviation of 0 as shown in Fig 2)