Are impedence matching networks necessary in regular AM recievers. - Page 1
Forum section: electronics / rf-microwave

author: Dan Moos
author_rank: Frequent Contributor
timestamp: February 25, 2024, 05:53:16 pm
content: I'm trying to teach myself RF stuff. I should thank W2AEW for his videos btw. Huge resource. Currently I'm just fooling around with the the basic building blocks of a AM superhet receiver.It seems most books and YouTube videos are kinda geared towards ham stuff. And it seems matching networks are important at those frequencies.But for AM broadcast, the frequencies have really long wave lengths. I'd think reflections aren't a concern when coupling circuits. Yet old transistor radios are full of impedence matching transformers.And I don't see why maxing power transfer would matter. (Although apparently it does). In audio stuff, it's the information of the signal in passing around (basically the voltage), and you amplify the current at the end. Why isn't that the same for circuit coupling in radios?

author: radiolistener
author_rank: Super Contributor
timestamp: February 25, 2024, 08:52:40 pm
content: Yes, you can amplify signal which is attenuated due to bad power transfer. But the issue here is that you're limited with noise floor of your circuit. When you use amplifier it amplify noise the same as the signal. So, if you lose signal power due to bad power transfer you will not be able to restore it anymore, your amplifier will apply gain to signal together with noise.In other words you will always lose SNR (signal to noise ratio). If you're receive very strong signal, that SNR loss may not be critical, but if you want to receive weak signal that SNR loss can lead to unable to listen the station, because it will be flooded in noise and your amplifier will amplify noise.As result, for receiver impedance matching is less critical than for transmitter, because it's more easy to compensate lose with amplifier, but if you're want to receive weak signal, it still critical, because allows to keep max available SNR for the signal. While not matched transmission can lead to SNR lose and unable to receive weak stations.

author: Andy Chee
author_rank: Super Contributor
timestamp: February 25, 2024, 09:28:49 pm
content: Quote from: Dan Moos on February 25, 2024, 05:53:16 pmThose transformers in the IF stages of a superhet receiver, aren't really impedance matching transformers per se. Many are simply playing the role of isolation transformers, isolating each amplifier stage and bias voltages from the next stage. I suppose the standout feature of these transformers is their Q, which is designed to peak at the IF frequency (e.g. 455kHz)

author: vk6zgo
author_rank: Super Contributor
timestamp: February 25, 2024, 11:58:40 pm
content: Quote from: Andy Chee on February 25, 2024, 09:28:49 pmTheir main function is supplying the major part of the receiver's selectivity, hence the high Q.If you look at schematics for tube superhets, you will find that both the primary & secondary of IF transformers are tuned.This is reasonably easy to achieve, as both the anode & grid circuits of tubes are relatively high impedance.The collector impedance of BJTs is "high-ish" & the following base, low so to have both primary & secondary tuned, it is necessary to tap the base circuit down the winding so it occupies a small part of the secondary, in other words, acts like a auto-transformer, indeed being an impedance matching device.The alternative, which in practice, was the nearly universal choice, as it was cheaper & less difficult to tune, was to use less turns on the secondary to more closely match the low collector impedance, & leave it untuned, concentrating all the selectivity in the primary.So, in fact, with BJT circuitry, the IF transformers do, apart from their primary purpose, perform impedance matching.At this point, someone will ask. "Why not use less turns on the primary & just tune it using a larger capacitor?"The answer to this is, that it is then perilously easy to end up with an oscillator instead of an amplifier."Miller capacitance" between collector & base offers a (normally negative) feedback path from the output to the input of a stage.A collector or base tuned circuit with a preponderance of one or the other reactance can easily cause phase rotation, turning that feedback from negative into positive.This was actually done on purpose with tubes, where similar LC networks in the anode & grid circuits were deliberately tuned to obtain this positive feedback. (Google "tuned plate, tuned grid oscillator".)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2024, 01:56:44 am
content: Because audio signals are such low bandwidth, and such high level, that noise margins are essentially irrelevant.You need to pay some attention for analog transducers, where mV, µV even, signal levels are common, and so microphone designers, phonos back in the day, various optical systems nowadays, etc., need careful consideration; a good microphone might have >100dB+ of SNR or dynamic range. Or mastering equipment, where you want to maximize the DR of each channel before summing them all together (while minimizing noise gained by processing steps, effects loops, etc.). But fully mastered media, where at most you're reducing its volume to average listening level? Hardly a problem.Ultimately what you're doing is impedance-matching the noise of the amplifier. This minimizes the joint effect of voltage and current noise, giving whatever the noise figure is for the amplifier. If the impedance doesn't match, one or the other dominates.Noise match is generally close to impedance match, though they typically differ by a bit, for unclear reasons.Now, AM radio specifically, has a lot of tolerance, as the SNR is quite high to begin with. It has to be, because the atmospheric noise background is quite high. We're all bottled in here by ionospheric reflection, both noise from lightning and the transmitters. This means radio stations might have to be 50kW or more, but can also broadcast over a huge area (100s miles radius), even with a typical reduction in power level at night. The power levels mean, despite the long wavelength, quite short antennas can be used; you throw away all your noise margin on antenna gain, and still need a pretty sensitive / low-noise receiver, and come out better on the whole because improving the antenna into a bulky ferrite or loop let alone dipole, gets more expensive way faster than a single JFET or whatever in the receiver, plus an IC doing pretty much all the tuning, IF and detection in one. That is, a "pretty good" receiver plus an awful antenna, is cheaper than a mediocre antenna and mediocre receiver, let alone a "good" antenna.If you want to throw away some of your noise margin as impedance mismatch rather than antenna factor alone, you can get away with that, to as many dB total between the two (mismatch + antenna) as you would have otherwise. This may make the impedance matching less critical -- but it also means strictly speaking you need a bigger, better antenna, to have optimal reception.Tim

author: David Hess
author_rank: Super Contributor
timestamp: February 26, 2024, 02:37:32 am
content: The impedance matching maximized the gain, or really signal-to-noise, of each stage. Without it, more stages would be necessary. As T3sl4co1l says, it optimizes the ratio between voltage and current into each stage.The impedance matching could also be done by varying the collector current of each stage, however this would sacrifice bandwidth when low collector currents are required.

author: Dan Moos
author_rank: Frequent Contributor
timestamp: February 26, 2024, 03:35:05 am
content: Great answers everyone. I hadn't considered noise as a reason.Some one glanced on something I am also curious about. I see in a lot of schematics, the first mixer stage (which seems often to double as the LO oscillator) has the primary of a transformer as the collector load, but it is taken from a tap on the primary rather than the whole thing. I can almost just see what is happening here (you can use a primary winding that is tuned for the IF, but tap it so the collector load is different maybe?), but would love some clarification.I have a pile of books, and a bunch of web resources I'm using, but what I can't seem to find is a detailed component by component tutorial on a typical AM superhet. Details like the reasoning behind all the transformer coupling never really gets mentioned.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2024, 05:31:12 am
content: The motif of a tapped resonator helps matching source/load/system impedance with resonator impedance.You might need too high or too low of an LC impedance (Zo = sqrt(L/C)) to get the desired Q directly, but by tapping it down, a more reasonable ratio can be used.For AM radio stuff, a load impedance of 100kohm might not even be unusual; very high impedances are hard to make at AC, but at low frequencies and narrow bandwidths, quite high values are possible. But good luck making a resonator directly at that -- 455kHz and a q of even 10 would require 1Mohm reactance or 0.35pF and 0.35H. Such an inductance, with less stray capacitance than this, isn't possible to manufacture! But a much more reasonable say 1k ish resonator is, well 1000 times other than that or 350pF and 350uH -- very ordinary values.There's also the coupled-resonators motif, which isn't really relevant in AM radio design beyond the usually double-tuned IF stages, but this can be used when higher selectivity is needed to make a multi-peaked filter. A typical example is the chain of crystals in an SSB radio filter: desired is a flat-topped (or modestly rippled) passband of a few kHz, with sharp skirts outside. Crystals are normally modeled as a series resonance with parallel capacitance; by making a pi network with capacitors to GND, the series resonance is effectively transformed into a parallel resonance, with coupling between resonators determined by the loading capacitors between crystals.That is, equivalent to a circuit of parallel LC resonators, joined from top to top with small coupling capacitors. Which you can imagine, if the capacitors are very small, the resonators mostly act alone, and you get a narrow peak for each tuned frequency; as you crank up the value, and the peaks start to overlap and interact; at critical coupling, insertion loss (for a given input/output coupling or impedance match) reaches minimum, and going up, and the response starts breaking into multiple peaks.The cleanest way to couple resonators is mutual inductance, which doesn't affect the resonant frequency of the resonator(s) and gives symmetrical asymptotes. Coupling capacitors tilt the asymptotes, giving a shallower upper stopband, and load the resonators, shifting resonance downward.You could also use resistors to couple between resonators, but, obviously, losses; or inductors, but they're expensive, and perform poorer than capacitors.Tapping the inductors also affords some coupling coefficient to the tap, as well as being kind of just in series with the inductor. So there's a spectrum between tapped and coupled inductors, whether direct main-inductor-to-main-inductor, or with coupling links (a few turns here or there), or taps anywhere from near 0 to 100%.Anyway, about the most you see is N=2, or "double tuned" stages. The coupled resonators motif is effective for high Q (over 5 or 10, often much more), and at lower Q (including most regular LP/HP filters) a conventional ladder structure (as you see from a calculator tool) is more effective.Tim

author: szoftveres
author_rank: Regular Contributor
timestamp: February 26, 2024, 05:37:59 am
content: Speaking of audio, one instance of impedance matching is the power supply (power source) matching to the optimum speaker(s) (load) impedance at full power.

author: RoGeorge
author_rank: Super Contributor
timestamp: February 26, 2024, 07:28:58 am
content: Quote from: Dan Moos on February 25, 2024, 05:53:16 pmFrequency is irrelevant.Even at DC, if you want maximum power transfer from a battery to a load, ideally would be to have a load with the same resistance as the internal resistance of your battery. Impedance matching is required at DC, too, if you want maximum power transfer. If the DC load and the battery internal resistance are the same, then half the power is transferred to the load [and that is maximum possible], half is dissipated as heat inside the battery. It is not possible to transfer more power than that.If the load is either bigger or smaller than the internal resistance of your battery, even less than half of the power will be transferred to the load, and more power will be dissipated as heat by the internal resistance inside the battery.Power "splashes back" when the impedance between the source and the load does not match. Matched means complex conjugate for the AC impedance, Z, or equal R for DC. It's an inescapable physics law of this Universe.Reflections at an impedance mismatch happen no matter the frequency, DC or AC, and no matter the type of power that has to be transferred: mechanical, optical, thermal, electrical, etc.

author: Solder_Junkie
author_rank: Frequent Contributor
timestamp: February 26, 2024, 02:41:17 pm
content: There are advantages to make receivers modular with 50 Ohm in/out RF modules, you can substitute modules and test them without much difficulty. Using crystal roofing filters avoids the need to use IF transformers.The Hycas IF module makes building such a receiver relatively straightforward. This is one example of a high performance modular receiver based on the Hycas IF.https://www.qsl.net/g4aon/g4aon_rx/SJ

author: jwet
author_rank: Frequent Contributor
timestamp: March 18, 2024, 05:14:27 pm
content: Just an FYI for your library (no special or financial interest...)There is a great., fun book called "Build Your Own Transistor Radios" by Ronald Quan that is very good and beginner friendly. Its not a fluffy little project book like its title might suggest. Its kind of an exploration and goes through a lot of circuits of varying complexity with pretty detailed analysis and design. The author is a senior engineer with a lot of patents and papers. Great book on this somewhat unusual topic.

author: uer166
author_rank: Super Contributor
timestamp: March 18, 2024, 05:50:27 pm
content: Quote from: RoGeorge on February 26, 2024, 07:28:58 amNot true if the electrical length of the link is "short". If the battery has very low resistance, and load has high resistance, then most of the power will be dissipated in the load. Think about it: an EV battery can push out 300kW, do you think it also dissipates >300kW internally? No, in fact it dissipates <1-2%, and the rest goes to load.Having a low-impedance source with a higher impedance load is a good way to maximize efficiency if you don't need to maximize power transfer to absolute maximum that is possible.

author: uer166
author_rank: Super Contributor
timestamp: March 18, 2024, 05:52:03 pm
content: Quote from: RoGeorge on February 26, 2024, 07:28:58 amHeh? How can you have a reflection if you don't have a wave? This makes no intuitive nor physical sense.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 18, 2024, 07:13:16 pm
content: It's always waves.More particularly: for LF/DC purposes where nodal analysis is effective, we can equivalently analyze the circuit, by dividing a given cross-section of it into ports, viewing the instantaneous V,I at those ports as incident plus reflected waves at the port's nominal impedance.That is, a wave going one way or the other, carries some V/I = Zo, with v(t) or i(t) (whichever equivalent description we like) being the waveform. The sum of incident and reflected waves gives the total V,I on the port.The port impedance is a matter of definition, an arbitrary number bearing no relation to the surrounding circuit; we've basically added extra complication for no benefit. It's an equivalent description of the system, but we've drawn no insight from it, and only made more work.You can think about a DC battery sitting on a shelf, sending out a continuous DC wave of I = V(oc) / Zo, at whatever Zo you care to define, which is transmitted into an open circuit, 100% reflecting in phase, and thus reflected right back at the same current, for a total of zero current flow. It doesn't mean anything, really, it's just a diversion.Where it shines, is at high frequencies where LC networks and transmission lines are unavoidable, and analysis is simplified when setting the analytic port impedance equal to characteristic impedance of those elements. Hence we use 50 ohm transmission lines in 50 ohm systems, with 50 ohm terminators, and filter networks and etc. Now the perspective shift is not only useful, but in terms of wave mechanics, it's actively descriptive, and as you go further into transmission line and free-wave regimes, the instantaneous-field and transient perspective does less and less (or takes more and more effort to extend; wave simulation and stuff).Tim

author: RoGeorge
author_rank: Super Contributor
timestamp: March 18, 2024, 08:13:15 pm
content: Take a battery. Say it can give 1V when open (so 1V at 0A) and 1A when shorted (so 0V at 1A). This also means we have a 1V battery with an internal resistance of 1 ohm.Now, attach an external load resistor to that battery. Pick any value you want for the external resistor, but the condition is to get the maximum possible power into that load resistor.What value should the external resistor be?To get maximum possible power into the load, than it should be of the same value as the internal resistance of the battery (in this example 1 ohm, and the power into a 1 ohm load will be 0.25W). Any other value for the load resistor, either bigger or smaller than the internal resistance of the battery will mean less power into the load resistor.Let's take an example, a load resistorbiggerthan the internal resistance. For example, for a 9 ohms load resistor, the power dissipated by that 9 ohms will be I2*R = 0.1A*0.1A*9ohms = 0.09W, which is less than 0.25W.Let's take another example, a load resistorsmallerthan the internal resistance of the battery. For example, for a 0.1 ohms load resistor, the current will be 1V/1.1ohms, about 0.909A. The power dissipated by the 0.1 ohms load is 0.909A*0.909A*0.1ohms = 0.08W, which again is also smaller than 0.25W.0.25W is the max power one can get out from that battery, and it happens when the load resistor is equal with the internal resistance of the battery, here 1 ohm. Anything else other than 1 ohm as a load, either bigger or smaller, and you'll get less than 0.25W into the load resistor.

author: uer166
author_rank: Super Contributor
timestamp: March 18, 2024, 08:14:55 pm
content: Quote from: T3sl4co1l on March 18, 2024, 07:13:16 pmThis was my main issue with the previous post claiming that *not* perfectly matching source/load impedance means more is dissipated in the source. I personally don't find this helpful for larger understanding of practical impedance matching.

author: uer166
author_rank: Super Contributor
timestamp: March 18, 2024, 08:17:09 pm
content: Quote from: RoGeorge on March 18, 2024, 08:13:15 pmThis is true because of themaximum possible qualifier. I'm not arguing about this statement, but rather the one claiming that you can never get more than 50% of effective power transfer to load in all cases. If you remove the "maximum" qualifier, you can obviously reach higher than 50% of power transfer.

author: baldurn
author_rank: Regular Contributor
timestamp: March 18, 2024, 08:24:40 pm
content: Most batteries can not actually survive being almost shorted. The real max power output is decided by thermal considerations except for very short durations. The internal battery resistance might not be linear either and will change depending on load and temperature.Also in most cases it is the amount of energy delivered not max power that is important. To get the maximum energy you want the voltage drop to be minimum, which means you need the current to be as low as possible.

author: RoGeorge
author_rank: Super Contributor
timestamp: March 18, 2024, 10:31:22 pm
content: Quote from: uer166 on March 18, 2024, 05:50:27 pmI've read again, including the context. That phrase of mine is wrong, indeed, I've strike out that text. Thank you for pointing it out.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: March 20, 2024, 10:02:17 pm
content: Getting back to the original question.... Yes the stages have to be somewhat matched in impedance and also to the frequency being transferred from one amplifier stage to another. The first stages are RF and have to be broad enough to let the entire band of the radio thru and also somewhat match the output of one stage to the input impedance of the next. Sometimes this can be just a capacitor link to keep DC from one stage out of the input of the next. The links between the stages have to be resonant for the frequencies involved and also match the impedances of the stages they connect. If they do not pass the frequency desired they would not receive anything. Also not passing out of band frequencies is desirable.After the mixer, In the IF stages, the adjustable cans are essentially LC circuits tuned to the ONE IF frequency. Sometimes there are multiple IF stages with different IF frequencies. Sharper filters can be used such as crystal filters. These are not necessary for the broadband AM radio reception since the signal is very wide and getting the full signal thru sounds better. FM broadcast signals are also very wide.Also every component adds noise, even resistors, and every mismatch in stages looses part of the desired signal.The choice of the transformer (or LC circuit) has to take into consideration other capacitances and inductances present in the circuit. This includes the capacitances of the transistors and or the valves involved. I think this can get rather complicated and is where an Engineering Degree would help. Maybe I will go back to Engineering School.Sometimes the tuning cans you see are filters that just get rid of an unwanted freq (pass it to ground for example) after it has already mixed and made the frequency you need, so not all the cans are passing a signal thru, but sometimes getting rid of something.Lots of old Ham radios and Short Wave radios also used a Pre selector tuning circuit to match the antenna input to the frequency that is to be received. Some of them use the transmitter Pi filter to limit the received signal frequencies.

author: vk6zgo
author_rank: Super Contributor
timestamp: March 20, 2024, 11:03:14 pm
content: Quote from: Wallace Gasiewicz on March 20, 2024, 10:02:17 pmThe RF stages ofconventionalSuperhet MF AM receivers arenot"broadband" to prevent the twin problems of "image interference" & "double spotting".Without pre-mixer tuning, in the former case, (assuming a normal 455kHz IF for convenience) two signals, one at 455kHz above the local oscillator frequency, & another at 455kHz below the LO frequency will both be received, whilst in the latter, the same signal will be received when the LO is 455kHz aboveor belowthat signal.Normal tuned circuits at the input of the mixer stage will usually offer adequate protection.