Finally I found a source that explains impedance matching in depth - and with the equations as a proof - and finally how to meassure that the theory is right. I have searched this for a long time. Thank you very much. This knowledge is vital if you want to design your own HF equipment. Thank you again.

@MegawattKS

2 жыл бұрын

Glad it was helpful!

Lesson learnt from playing around: dirty trick of slightly over valuing inductor(both input and output sides) and then squishing/stretching it in circuit to adjust for the target impedence observed on VNA, seems like easier way than measuring the inductor for its value. As part of final leg of fine tuning, getting down the Reactance component to near zero seems to yield more gain(and SWR) than getting resistive impedence accuracy to target, eg: 45+500mj is better than 49.5+25j.

Thank you for another great video.

Keep up the good work! 73 !

Excellent Explanation! i am also waiting for part 2.

Thanks again, looking forward to part 2!

@MegawattKS

2 жыл бұрын

You're very welcome. Glad it was helpful. I've got part 2 done, but am still trying to refine it. Should be done real soon now I hope :-)

Putting up matching network between NANOVNA and a resistor on NanoVNA testboard hole seats seems like a quick easy way to test matching networks.

I was just thinking about that maximizing the power-transfer from one part of the circuit to the other, and the need to keep the impedance matched (video at 3'14"). As we all know we have almost everything matched to 50 Ohm. I explained it to myself as follows: The 'power sending side' (left, like antenna source) has some power to transfer to the next circuit (rights side, like to the receiver circuit, with some Rin). The 2 resistances (Rant and Rin) are in series and share the same current I, induced by the source. I=Voc/(Rant + Rin). The power on the left side is P1=I^2 x Rant and the power in the right side P2=I^2 x Rin. The (received) power on the right side can never be higher then the (send) power on the left side. That would be a miracle. It can only be lower (due to a mismatch, as we learn). So the best we can get is the same power. Neutral (lossless) power transfer. In that case we can say P1=P2. From that follows: I^2 x Rant = I^2 x Rin. From that follows: Rant = Rin. Please comment if I make some 'thinking' mistakes here 🙂

@MegawattKS

Жыл бұрын

Interesting way to attack the problem/proof. I'm still thinking it through, looking for possible 'thinking mistakes'. Maybe others will weigh-in. The thing I'm considering is the premise that the source has some specific power to deliver. But for an antenna, that's definitely true. It would be Pdensity x EffectiveAreaOfAntenna. I've seen the mathematical proof, but in the classroom we tried to be more 'physical' like you're doing here. I mainly presented a plausibility argument with a set of three cases. I'll give this some more thought after dinner :-)

Amazing set of videos. As a HAM it’s at the perfect level. Been enjoying them thoroughly. Thanks! I do have a question. For the receiver why do we want maximum power in and not maximum voltage? Doesn’t that depend on the device being driven by the receive antenna? For example, if the input amplifier used an FET which is a voltage driven device with a very high input impedance, wouldn’t we just want the maximum voltage developed at its gate? That would imply no matching network needed, right?

@MegawattKS

Жыл бұрын

Thanks! Good question. The short answer is that signals at the receiver's sensitivity level are competing with thermal noise from both the antenna/environment and from the low-noise amp's internally generated noise in the front-end (as well as later stages if the LNA doesn't have high gain). That noise is quantified as Pnoise = kTB (for the antenna/environment and resistive sources), which is based on power. So maximizing the signal power that goes into the LNA optimizes sensitivity. That said, maximizing power and maximizing voltage often turn out to be the same thing. For example, if the LNA has a high input impedance, then when the matching network is used (to maximize power transfer into the amp), the signal undergoes an increase in voltage too (equal to the square-root of the resistance stepup ratio). So intuition is not wrong 🙂

I noticed this playlist series is based on radio receiving. I’m just curious on, are you going to develop a series on radio transmitting. Like how to create an RF transmitter to match a created RF receiver. To be honest in my opinion majority of us aren’t interested in listening to the basic radio stations. We want to control robotics from distances using our own built transmitters and receivers on our own frequencies. I understand it’ll probably be the citizens band. I know I’m asking a lot I apologize. Also, here’s a tip. Nowhere on Google is it giving clear enough information on how to do the things I’m described above. It’s all in expensive books. If you went into the complete depth building transmitter to receiver by how it looks to me, you’d be the only source to go to considering how deep you go with the learning concepts. Thanks for everything by the way.

@MegawattKS

6 ай бұрын

Thanks for the inquiry and the ideas. The short answer is that I stayed away from transmitters intentionally - for a number of reasons including licensing and special issues with respect to equipment and safety. BUT... there is a playlist on Antennas and Propagation here. kzread.info/head/PL9Ox3wpnB0kqNLnCdCtWcjGq7nRVHYxKA Arguably this is very relevant to transmitters, as it covers decisions about frequencies vs antenna size, and how the signals propagate with and without obstructions. I hope its helpful !

I wonder if impedence matching could be explained with the example of resonance sound tuning forks(or playground swing), where transferring power from vibrating fork to an idle fork becomes optimal and easy at resonant frequency of the target. Matched frequency makes idle fork resonate, and matched impedence also resonates(its an AC signal afterall).

@MegawattKS

2 жыл бұрын

That actually sounds like a 2-pole bandpass filter :-)

i figure additional gain/input matching network are needed to compensate for the losses from BFP filter stage.

@MegawattKS

2 жыл бұрын

Yes, but there's about 100 dB of gain in the overall receiver chain of circuits (filter, lna, mixer, IF-subsystem amps, etc), and the FM receiver has to work over a dynamic range of about 80 dB typically. So losing a few dB doesn't make a big impact - unless it's in the first filter where it will affect the overall receiver's noise figure. But even if it does, the only thing that will do is maybe prevent receiving the very weakest station on the band.

Very useful video series - thanks. However, I have a (perhaps naive) question. At 2:17 , you say that you want to maximise antenna-front end *power* transfer (which you can do by impedance matching). However, it's not entirely clear to me why this is the optimum strategy when using an antenna, when, for example, with a microphone (whose low-power signal also competes with noise at the front end amp. input), we typically drive an amplifier whose input impedance is much higher than mic. impedance (and hence we maximise *voltage* transfer). The two situations seem to be pretty analogous, as far as I can see. Am I missing something obvious here?

@MegawattKS

2 жыл бұрын

Actually a great question. The answer is that we need to maximize signal to noise ratio if we want to receive the weakest possible signal - which is always competing with noise - both from the environment the antenna is 'seeing', and from the amplifier itself. Without getting in too deep, here's one way to think about it: If we match up from 50 Ohms to, say, a high-Z amplifier input impedance like 10K (as opposed to just driving directly into it), then we get a voltage boost in the process. So the signal level getting into the amp is higher. Since the amp has a somewhat fixed level of internal noise (shot and thermal noise from semiconductors, etc), the SNR is higher since we increased the input signal voltage it now sees relative to its so-called "input-referred" internal noise. I think the other part of the answer is that in audio, signal levels are often so high relative to the noise that we don't care about this issue (except if we're doing a low-noise microphone amp). For the record, that's often true in RF design in stages other than the antenna input. Once you get the signal level up sufficiently with an LNA, downstream circuit matching can sometimes be dispensed with. Also, audio is inherently ultra-wide band and impedance matching is hard to do. (In old tube amps, transformers were used sometimes - especially in the power amp output section, but they're big, expensive, and somewhat lossy/noisy themselves...)

@scollyer.tuition

2 жыл бұрын

@@MegawattKS Thanks for the response, which I shall have to ponder carefully. I suspect that I need to revise noise figure calculations, amongst other things (there's that formula with the Gs in the denominator - my memory is hazy with age :-( ) However, one point occurred to me after posting my question: with an antenna, if we don't transfer all the received power to the next stage, we have to worry about re-radiating it when it reflects off the unmatched stage that it is feeding. That is probably undesirable. When we have a microphone at the front end, reflected energy can be dissipated by lossy components, and merely ends up as heat - that is much less of a problem. I shall have to consider the noise and SNR issues carefully though.

@MegawattKS

2 жыл бұрын

@@scollyer.tuition I think some amp topologies don't improve that much with matching, although it will help their gain and therefore the NF of the overall receiver (as calculated with the Friis equation with the Gs). Indeed, it depends on how much noise they generate internally and how much the mismatch is. With a hypothetical amp with no internal noise, the NF hit would be nil since input signal and input noise are amplified the same, and noise factor is the ratio of the SNR at the input to the SNR at the output. On the issue of reflections, I don't think there's any issue there. The signal that is reflected from the LNA input is the same as the one that was coming into the antenna anyway. Typically the more problematic issue is radiation of the LO signal if the LNA doesn't have good 'isolation' (S12).

@scollyer.tuition

2 жыл бұрын

@@MegawattKS I shall go away and study this (I have a big antenna theory book that deserves some attention) but my final comment/query re: signals reflected from the front end: surely it would be at least somewhat out of phase with the received signal, and will cause some amount of re-radiation? Particularly if we had a long transmission line between the two? And that's an interesting point re: the LO. I think I've seen videos of hams poking around with tinySAs trying to detect it.

Hey there, Im curious if you made these double sided pcbs yourself? And if so, is that something the average bloke can attempt? Any info on cheap one-off pcb at home is appreciated.

@MegawattKS

Жыл бұрын

Great question. I'm actually building out a website to contain more information, including that. Just finished a first attempt at what you're asking about. Go to the 4th bullet here ecefiles.org/ (Sadly I haven't figure out how to put the actual layout file on the site yet - but all the other info is there)

At 10:17, I'm a little confused, as from the antenna's perspective, the output impedance ideally should be 50 Ohm for max power transfer, but if you add Xp + Xs +R as a series RLC circuit, the impedance is 55 Ohms, not 50?

@MegawattKS

Жыл бұрын

The answer to this is tricky, but the main problem is that we cannot add X and R values. The inductor and capacitors have "complex" impedances (j Xp and -j Xs), so complex number math (including the j values) has to be used. Also, we can't just look at the three components in series in the absence of the loading on the left side. But - let me try this... Stand at the antenna and look to the right of the 50 Ohm antenna resistance. We see jXp in parallel with the series impedance -jXs+Rs. If we do a "series to parallel" conversion on -jXs + Rs, we can replace the two series components Cm and Rin with a _parallel_ Cp and Rp. See the webpage below for how this is done. When the series to parallel conversion is done, the Rp value is 50 Ohms and the reactance of the C value in the converted circuit is Xp = Xs ((1+Q^2)/Q^2) = 20((1+2^2)/2^2) = 25 Ohms - and becomes the same as that of Lm (25 Ohms). So Lm resonates with the parallel-equivalent capacitance Cp at fo and they cancel each other. So the only thing left is the 50 Ohm resistance (Rp). Sorry - this is hard to relate in a paragraph. Hope some of it is helpful. aaronscher.com/Circuit_a_Day/Impedance_matching/series_parallel/series_parallel.html

For the inductors on the output side of the real physical circuit, it shouldn't be a problem converting parallel inductors to one smaller inductor right(630 || 160 = 130nh) right ? Fewer inductors(with even fewer turns) on a PCB seems like less of headache. Probably similar for input side, but realizing such tiny one would be more of trouble than 2 inductors there.

@MegawattKS

2 жыл бұрын

Absolutely correct. It's nice when it works out like that. As you said, a single 130 nH is much nicer than two - especially since one is 630 nH. 130n makes for a reasonable number of turns and size :-)

Hi, Thanks for the videos, I have a silly doubt, at 12:00, in video, if the resistance from emitter to ground is 1K, then how come Rin was taken to be 50 before MN is done

@MegawattKS

Жыл бұрын

Hi. The amplifier shown is a "common-base" design - so it's input resistance is much lower than the physical 1K used in biasing. The details are in Part 3 of the series here: kzread.info/dash/bejne/h4mg07lmprbXabA.html (sorry about that forward-reference). Much more detail can be found in the associated website associated with this video series. Specifically in the lecture on RF amplifier design here: ecefiles.org/rf-circuits-course-section-2/ . The third class-handout on that webpage shows the common-base design with the associated formula for Rin. It's approximately Rin = 1/gm since 1/gm is much lower than the 1K bias resistor or the internal BJT device resistance "r_pi". Thanks for the question! Hope this helps.

What is the name of the connectors at 13:22? I found some SMD connectors of various types available, but I could not find cables with complementary connectors. Do you calibrate the VNA at the end of these cables? If yes, how, considering the calibration kit has SMA connectors?

@MegawattKS

7 ай бұрын

They have a few names, but I call them U.fl connectors. There is a really nice board that includes short, open, load references that uses this connector, and it comes with cables. It also has lots of other interesting examples of Smith Charts and components. I got mine from Amazon for about US$15. Search for "rf demo kit nanovna". I also bought a set of 50 connectors for US$12 from Amazon :-)

@MegawattKS

7 ай бұрын

Pictures here :-) kzread.info/dash/bejne/hGShtLail7zgk9I.html

@nick1f

7 ай бұрын

@@MegawattKS Thank you for all the info. I already came across the RF demo kit, I just didn't realize it can be used to calibrate the VNA and that it has these cables. I am going to buy a NanoVNA for myself and a demo kit and a set of connectors. I already have a spectrum analyzer, I hope it still works since it is about 30 years old.