Merry Christmas! I was watching the video again to digest everything I can and I asked myself a question, where you speak about the UHF transmitter and how to get 100 mW out of 3.3 V by transforming the impedance seen by the output transistor. I was wondering if we could obtain a reasonably high power out of a low voltage source by transforming the impedance to the required value to get that power. So, for example, would it be possible to get 4W of power out of a 3 V supply? Thank you. Miguel
Good question on getting to higher power on low voltage! Yes - that is theoretically possible using a suitable transformation ratio. The trick though would be to achieve a high enough quality factor (low R in inductors, capacitors, etc) to keep the efficiency of the PA usable. But component parasitics could also make it problematic. Let's say we convert from 50 Ohms down to 1 Ohm. We would want to keep series resistances and parasitic inductances in the transistor circuits and matching network components well below 1 Ohm. Even 1 nH of parasitic inductance might prevent achieving a working circuit in reality. 1nH is about 1 mm of PCB trace length - which gets hard to do with high-power circuits. But you got me curious. I had a 70cm Yaesu HT and I think I have it's schematic somewhere. I'll look to see what voltage they ran the output on and what power level it achieved. I'm guessing they ran on 6V and only got a couple watts. Stand by...
I just looked at the radio schematic(s). The radio I was remembering was a Vertex VX-3R and as best I can tell (the schematic is _very_ intense ;-), it uses a L-match like we're discussing. And it's ratings are 1W out when running on 4V, or 2W out running on 6V external supply. I suspect they had trouble going higher for the reasons I mentioned above. And they didn't want to stick in a larger, more expensive FET that would allow higher power. (It's an HT walkie-talkie afterall). Here is a link to the service manual, with the specs shown on page 2. The PA schematic is on page 35 as "SW UNIT", Q3004. But the L match is back on the main board who's schematic is on page 15. SW-UNIT module is at the top of that schematic and it's RF OUT appears to feed a custom inductor L1047 (probably to get needed Q, etc), and the shunt C part of the matching network appears to be C1267 - although that might be the 144 MHz path to the antenna. The rest of the LC stuff is a lowpass filter to clean up the signal harmonics before emission (plus some PIN diodes for transmit/receive switching). www.radioamatore.info/attachments/631_VX-3R_service_manual.pdf
Thanks for a most informative series, I watched all of the videos to get a good overview and will now go back and watch them again bit by bit to really get to grips with the material. Thanks for the time and effort you're putting into this!
Glad you like them! I've been trying to build a set that covers the material from our radio design course. Just uploaded a new one on amplifiers (its long). Next one will maybe be an optional part 2 for that or one on oscillators ...
Hello! Very nice examples of real-life radio receivers to show how impedance matching is performed. With regard to the crystal radio example, I guess the primary and secondary of the transformer will have some associated losses which could be transformed into parallel resistance losses, as we saw in the previous episodes, is that right? The value of those resistances will depend on the Q of the primary and secondary coils... Should those resistances affect somehow the calculation of the impedance transformation we have seen in this example? Thank you
Thanks. Yes - I think so. They have series resistive losses in the coils which can be re-cast as parallel resistive losses when that helps make analysis easier. And figuring out the Q of each winding is perhaps best done by measurement. The setup is a bit complicated also by the limited coupling between coils. I confess I've never actually tried to design a crystal radio set like this. But it's fun to think through how it works ! On a related note - I always wanted to cover series to parallel impedance transformations in the video, but it would have gone a bit long. Here's a nice web page I recently ran across on that area if you're interested: www.rfinsights.com/insights/design/series-to-parallel/
Hello again! Trying to follow the calculations of the L matching network formed by the antenna capacitance and the primary of the antenna coil in the crystal radio set, I have come up with a question: I guess the coil has an internal resistance at the operating frequency, which could be represented by a parallel resistance following the formula: Q=Rp/Xp. If this resistance were comparable to the resulting impedance of the L network (1K, as in the video), how would it affect the circuit? Would the transformed impedance still be 1 K, but now with Rp (internal resistance of the coil) in parallel, generating some power loss? Or what effect do you think it would have? Thank you. Miguel
Yes - it would definitely lower both the signal level in the primary and the Q of the primary resonance and the overall "double-tuned" filtering taking place (double-tuned because the primary is a resonant circuit and the secondary also has one). So it will, in addition to causing some power loss, make the radio a bit less selective in pulling out weak signals in the presence of strong ones. Analysis-wise, maybe try drawing the resulting primary circuit you described, and then moving the 1K loss resistor to the right side of the 17uH inductor. Looking to the left of that resistor to ground, you would see the source converted from 10 to 1K by the capacitor and inductor as before, but now that 1K resistor to the right loads this down, making a V-divider of 1/2. So it seems like we're going to get about a 2X (6dB) hit in voltage, in addition to maybe a 2x'ish broadening of bandwidth. It's late at night, but I think this is a valid analysis. Not sure what the exact effect will be on signal loss and bandwidth though, since the secondary also loads the primary... Good question !
I understand the maximum power transfer theorem and I can design complex conjugate matching networks in my sleep but I still don't see why we don't just go for maximum voltage gain (because that is where the signal inteligence is) your beginning example even showed that, sure there is no power to do work but the information is there in the form of a time varying voltage signal, why do we need power at this stage? Once you have a strong voltage signal then follow up that sage with a current amp to increase power. Is this all about the cost of one transistor or is there something that I am missng? BTW I am an extra class Ham llicensee and I hold a EE degree from an ABET accredited school but in fairness my concentraition was in control theorey.
An interesting question for sure. I usually answer that the best signal-to-noise ratio possible is determined by the antenna selection (to bring in the highest signal power), and the kTB antenna noise floor (plus excess input-referred noise added by the LNA). kTB sets how low the signal power can be and still be received The noise is in power terms, so it seems that maximizing signal _power_ into the LNA should get the best S/N we can achieve. I.e. the best noise figure for the front end. [After boosting signal (plus input-referred noise) with the LNA, it doesn't matter as much what we do with impedance interfaces, since the noise floor is now amplified along with the signal, so later noise doesn't degrade the system as much.] But maybe we can come at this from the Z and V domains as well. With a high-Z input LNA, the matching network will increase V as it increases Z to match to the high Zin value. Sounds great, but note that the kTB antenna noise is also boosted. In the end, it's a wash, except when added noise from the LNA is considered also. But what about a common-base amp with low Z input? I think the answer there can come from looking a 2 cases: First consider a 50 Ohm source driving into a, say, 10 Ohm input. In that case we're hit with 15.6 dB attenuation on the way in (20 log (10 / (10+50))). Not good. If we instead match down from 50 to 10 Ohms, there is a voltage step-down. That stepdown is sqrt(10/50) = 0.447 or 7 dB. Of course even after matching, we've got 10 Ohms driving into 10 Ohms, so to be fair, we need to factor in that factor of 0.5 (6 dB) - giving 13 dB hit total. Still, this is better than 15.6 dB. What do you think?
As I recall, there is a complication when noise added by the LNA is considered explicitly. In that case, transistors often spec a best-noise-match impedance value - empirically determined. Then, the goal is to build the matching network to map the antenna Z (say 50 Ohms), to this special value, rather than a perfect conjugate match. Honestly, though, I'm not a big fan of that. Too complex (pun intended) for too little gain in system function. Many modern radio systems are limited by large-signal handling (e.g. intermod behavior) rather than noise figure.
The C and L on the left side have X=100. So if we stand in the middle of the transformer and look left, we see a matching network with Q of 100/10 = 10. So the effective parallel resistance looking to the left (primary) side, is (1 + 10^2)10 = 1K (approximately). The voltage increase is the square root of the impedance ratio, so 10x. (unloaded in this analysis)
The voltage step-up through the transformer is harder to calculate exactly since loading data is missing. Also, the turns ratio is 110:20, which doesn't agree perfectly with the voltage increase I've given. Part of that is because the coupling coefficient is less than one due to lack of a transformer core. Honestly I'm not certain how I arrived at the 11:1 R increase. Still trying to remember my thought process when making this slide with all the unknowns and assumptions... But its likely close to this at least, and at least the 3.3x voltage increase is consistent with the 11:1 R increase which satisfies power conservation... The title of the slide reflects this I suppose 🙂
I don't know if this is already answered somewhere, but I'm wondering about how to use PCB traces when matching is being done. When I have RF on a PCB, I generally design trace widths and spacing to act as transmission lines, and calculate the them for 50 Ohm. But what do you do in the middle of a matching network? Say the output of stage A is 85 Ohm, and input of stage B is 50 Ohm. An impedance match network is created between them to match the impedances. Presumably, you use 85Ohm traces between stage A and the matching network, and 50Ohm traces between the matching network and stage B. But what traces do you use inside the network itself? Or am overthinking this for short traces at typical amateur frequencies? I've mostly been doing PCB design in the 2m band, so the quarter wavelength is around 50cm, and my boards are less than 5cm in overall size, most of the RF traces are 1-2cm....
Your conclusion is correct. Usually we just try to keep the circuit small with respect to a wavelength - so that it doesn't matter. At 146 MHz, lambda in free space is 2 meters as you said, so on a typical FR4 PCB it's 1m (velocity of propagation in microstrip on the PCB is about 0.5, depending on trace width and board thickness). As long as we have traces shorter than 1/20th of lambda (about 0.5 cm at 100 MHz), the effect of the transmission line behavior can usually be neglected. If we want to include it however, we can model the trace in this case as a inductance of about 1 nH/mm, so about 5 nH. Then we can think through whether this has any real effect on the matching network design. But the best thing to do is keep all the traces short enough that it's even less than this, use SMD components, and just don't worry about it :-) We usually have to adjust the L (or C) a little anyway in measurement to deal with several other non-idealities...
Amazing set of video, but can you explain more in details how did you found input and output impedance of amp? How did you know that you need to match 50ohms to 10?
Thanks. Yes - it's overviewed on a slide in the video on Amplifiers (Episode 6). Here's a link keyed to the location in that video where the formulas are given: ruclips.net/video/UUlqW-vSq9M/видео.html . See the Common Base amp part. The theory that underlies the formulas given there is covered in "Part 2" of that video, which I decided to call Appendix A, so it would not lead to far away from the main journey. Those details are here: ruclips.net/video/m9X0mfg_8lQ/видео.html After a lot of background, it elaborates the model for a common-emitter amp, from which Rin and Rout can be found. But it doesn't do it for all the other configuations (like Common Base). That is left as an "exercise for the 'reader'" (it would be a homework problem in the university course - or an assigned reading.) But in brief, the input resistance of the common base amp is about 1/gm. With the amp shown in Episode 2, the collector current is about 4.5 mA, so gm is about 0.11 A/V, giving 1/gm of about 10 Ohms :-)
Hello again! There's so much interesting information in this video that I'm still learning after having watched it several times! With regard to the tuned transformer matching network, I have a question: on the secondary side, there's the secondary coil in parallel with a capacitor which, at resonance, will be equivalent to a high resistive value (depending on the Q of the coil). Then, when looking from the primary side, an impedance transformation of that resistive value has been obtained, resulting in another resistive value. But why is this value only resistive? When looking from the primary side of the transformer, there's the primary coil... Shouldn't the impedance seen from the primary be a series combination of the transformed resistive value plus the reactance of the primary coil? Thank you. Miguel
I'm sorry, I said something incorrect, because I forgot to mention the load!: as you say in the video, the impedance transformation would be made from the resistive load (I guess in parallel with the Rp of the secondary coil) to a lower resistive value in the primary, according to the square of the turns ratio. But then, on the primary side, there's no capacitor to resonate or cancel the reactive component of the primary coil. Does this reactance appear or how is it taken away from the transformed impedance? Thank you.
Sorry for the delay in replying. This is an excellent question. I saw your followup info too. Will try to answer both here. On the issue of the primary winding L showing up, I would argue that it would be in parallel, not series. Beyond that, we need schematics and some math to really work it out. But qualitatively, I would offer that the primary and secondary inductances are linked (coupled magnetically), and that there is a mutual inductance involved. So with the secondary resonated, the inductance in the primary is part of that resonance already. This assumes either a good "coupling coefficient" value K, and/or an LC tank circuit Q on one side that is > 1/K. Here's some of the math. Sorry the writing is bad. These are from my class notes and it's a little light on the derivation since I just used these to remind me of what I wanted to talk through. Hopefully some of it will help. ecefiles.org/rf-circuits-course-section-13/
Note that the equations at the bottom of page 1, the circuit model at the top of page 2, and the application circuit in the middle of page 2 are for well-coupled inductors, but without a core. The model includes an "ideal transformer" in it, but the actual construction does not. It's just a simplified model for reasoning through the circuit operation without having to invoke the equations from page 1 again...
Wondering what you recommend for impedance matching between the output of the NE602 mixer at 1.5k ohms and the input to the IF filter at 300 Ohms. I'm coming up with some crazy values so I'm not really sure if I'm doing something wrong, or maybe it's not even needed? And by "crazy", I guess I mean something on the order of a 8 uH inductor, which seems...high? Or maybe it's just my inexperience showing.
That L value sounds right. It's about 540 Ohms reactance, which is somewhere in the range of the geometric mean of 300 and 1500 - which is a quick check. It's also the value I recall from the class 🙂 To get that value, we used a toroid core. Usually T37-2, but sometimes a higher mu core material to decrease the pain of winding too many turns. (Maybe an FT37-77 - I can't recall). The other alternative is to just pad the output R to bring it down to 300 Ohms so the filter sees the right terminating impedance. That helps it have a nice response curve, but of course comes at the cost of losing some gain. That is OK when the IF subsystem has plenty of extra gain - but the IF amp I put in the video series doesn't have a lot of gain, so it will affect the number of stations one can get. Engineering tradeoffs ! 🙂
@@MegawattKS Makes sense, thank you. Curious, what do small FM radio do, then? Like if I open up one of those cheap tiny FM emergency radios ,there won't be giant toroid inductors in there, right? Do they use mixers that are better matched to filters, or like you suggested, make it up with IF gain?
@@jchatterton2113 Good question. You can do either - especially if you do your own IC designs. Over the years, there has been a progression toward full-integration. This was actually my own research area. The goal was to get rid of all the big components - including the passive LC or ceramic filters. It can be done with custom IC design, but there are performance tradeoffs. That said, for FM, most modern radios in the last 10 years use something like the one-chip solution covered in this video. I've keyed it up to the architectures post-superhet which then shows the single-chip FM radio. If you back it up some, then right before that point in the video it shows an earlier radio with less integration - which still used some inductors (probably SMD with low performance). ruclips.net/video/XW3TYQstiuk/видео.html
Hello again! I have a question about the use of a variable capacitor in series with the antenna in the crystal radio set: is it used only to adjust the impedance matching together with the primary of the transformer or could it also be used to tune the resulting series LC circuit with the antenna? If we tune the series LC circuit, would that mean that the desired signal would find no resistance to ground, thus creating maximum voltage in the primary? Thank you. Miguel
Hi. The crystal radio shown at 10:16 doesn't have a variable in the primary - but I think we're talking about the tuned-RF vacuum tube receiver at 12:12 . Assuming so, then yes, that variable C in the primary side is adjusted to resonate the antenna. That provides a sorta zero Ohm path to ground. But remember the antenna resistance (say 10 Ohms like in the crystal radio analysis) is still there. So the current induced is finite. But it does maximize the signal that is linked into the secondary by maximizing the primary signal current (since R goes to a small value as you noted) 🙂 Basically the schematic at 10:12 is what we'd get, if the antenna is Zand = Rant + jXant = 10 + -j100 and the primary winding is +j100. The current induced in the primary would be Voc / 10. And the voltage across the primary winding would be (Voc/10)*j100 - which is 10 times the open circuit voltage Voc 🙂 (Note that there are complexities here that I'm ignoring - like the effect of what's going on in the secondary on what happens in the primary - so some of the numbers are off. But the basic idea remains solid.)
@@MegawattKS Yes, I mixed both circuits up, sorry. I was talking about the tuned-RF vacuum tube receiver :). Thank you for replying, it helped me understand the circuit.
Thanks for the videos, I have a doubt , In episode -1 part-2 when defining Q, the Rp was the resistor value in parallel to the LC ciruit, but here it is the other way around, the formula for Q was shown to be w.r.t Rp and Rs there but now both these reistors places have swapped.
It can be confusing and tricky because of the orientation of components in a schematic. Are we talking about the "Solution for MN Components" slide at about 45 seconds into the video? Assuming so, the 50 Ohm source resistor Rant is actually in parallel with Lm. This is because for the Q circuit analysis, the Voc source is set to zero, which means it becomes a wire. So Rant goes from the top of Lm to the bottom and is labeled Rp for that reason. On the right-hand side, we called the 10 Ohms we're matching to "Rs" because the 10 Ohm load "Rin" is in series with Cm in the matching network. Does that help? Did I interpret correctly which slide you have a doubt on, or is it a different part of the video?
It is common for amateur radio operators to use antenna tuners to match to complex impedance antenna systems. Once tuned, it is believed that power bouces back and forth between the antenna and the tuner until it is all radiated by the antenna. It seems more reasonable that the power is absorbed by the real part of the antenna impedance with nothing bouncing back. Power bouncing back would show up as reflected power to the transmitter, but that is not the case.
Tricky stuff to sort out for sure. I think the resolution is that the meter is on the 50 Ohm side of the tuning circuit where the transmitter is and the LC tuning circuit is downstream from there - between the meter and the antenna. The LC tuning/matching circuit converts the antenna impedance (full complex value) into a 50 Ohm resistive value when properly tuned. So the meter only sees 50 Ohms and does not register any reverse traveling power. All the power that was launched into the line from the transmitter is "absorbed" in the 50 Ohms that the LC tuning circuit (working in conjunction with the antenna impedance) presents at the input to the tuner. What happens on the other side can be considered bouncing back and forth between tuner and antenna, but the transmitter (and meter) never sees that...
Merry Christmas!
I was watching the video again to digest everything I can and I asked myself a question, where you speak about the UHF transmitter and how to get 100 mW out of 3.3 V by transforming the impedance seen by the output transistor. I was wondering if we could obtain a reasonably high power out of a low voltage source by transforming the impedance to the required value to get that power. So, for example, would it be possible to get 4W of power out of a 3 V supply?
Thank you.
Miguel
Good question on getting to higher power on low voltage! Yes - that is theoretically possible using a suitable transformation ratio. The trick though would be to achieve a high enough quality factor (low R in inductors, capacitors, etc) to keep the efficiency of the PA usable. But component parasitics could also make it problematic. Let's say we convert from 50 Ohms down to 1 Ohm. We would want to keep series resistances and parasitic inductances in the transistor circuits and matching network components well below 1 Ohm. Even 1 nH of parasitic inductance might prevent achieving a working circuit in reality. 1nH is about 1 mm of PCB trace length - which gets hard to do with high-power circuits. But you got me curious. I had a 70cm Yaesu HT and I think I have it's schematic somewhere. I'll look to see what voltage they ran the output on and what power level it achieved. I'm guessing they ran on 6V and only got a couple watts. Stand by...
I just looked at the radio schematic(s). The radio I was remembering was a Vertex VX-3R and as best I can tell (the schematic is _very_ intense ;-), it uses a L-match like we're discussing. And it's ratings are 1W out when running on 4V, or 2W out running on 6V external supply. I suspect they had trouble going higher for the reasons I mentioned above. And they didn't want to stick in a larger, more expensive FET that would allow higher power. (It's an HT walkie-talkie afterall). Here is a link to the service manual, with the specs shown on page 2. The PA schematic is on page 35 as "SW UNIT", Q3004. But the L match is back on the main board who's schematic is on page 15. SW-UNIT module is at the top of that schematic and it's RF OUT appears to feed a custom inductor L1047 (probably to get needed Q, etc), and the shunt C part of the matching network appears to be C1267 - although that might be the 144 MHz path to the antenna. The rest of the LC stuff is a lowpass filter to clean up the signal harmonics before emission (plus some PIN diodes for transmit/receive switching). www.radioamatore.info/attachments/631_VX-3R_service_manual.pdf
@@MegawattKSThank you for looking into it and sharing the schematic. I'll have a close look at it ;)
Miguel
Thanks for a most informative series, I watched all of the videos to get a good overview and will now go back and watch them again bit by bit to really get to grips with the material. Thanks for the time and effort you're putting into this!
Glad you like them! I've been trying to build a set that covers the material from our radio design course. Just uploaded a new one on amplifiers (its long). Next one will maybe be an optional part 2 for that or one on oscillators ...
Hello!
Very nice examples of real-life radio receivers to show how impedance matching is performed.
With regard to the crystal radio example, I guess the primary and secondary of the transformer will have some associated losses which could be transformed into parallel resistance losses, as we saw in the previous episodes, is that right? The value of those resistances will depend on the Q of the primary and secondary coils... Should those resistances affect somehow the calculation of the impedance transformation we have seen in this example?
Thank you
Thanks. Yes - I think so. They have series resistive losses in the coils which can be re-cast as parallel resistive losses when that helps make analysis easier. And figuring out the Q of each winding is perhaps best done by measurement. The setup is a bit complicated also by the limited coupling between coils. I confess I've never actually tried to design a crystal radio set like this. But it's fun to think through how it works ! On a related note - I always wanted to cover series to parallel impedance transformations in the video, but it would have gone a bit long. Here's a nice web page I recently ran across on that area if you're interested: www.rfinsights.com/insights/design/series-to-parallel/
Thanks for teaching matching networks. Always learn in this hobby.
Thanks for the nice feedback !
Nice set of examples, thanks!
Hello again!
Trying to follow the calculations of the L matching network formed by the antenna capacitance and the primary of the antenna coil in the crystal radio set, I have come up with a question: I guess the coil has an internal resistance at the operating frequency, which could be represented by a parallel resistance following the formula: Q=Rp/Xp. If this resistance were comparable to the resulting impedance of the L network (1K, as in the video), how would it affect the circuit? Would the transformed impedance still be 1 K, but now with Rp (internal resistance of the coil) in parallel, generating some power loss? Or what effect do you think it would have?
Thank you.
Miguel
Yes - it would definitely lower both the signal level in the primary and the Q of the primary resonance and the overall "double-tuned" filtering taking place (double-tuned because the primary is a resonant circuit and the secondary also has one). So it will, in addition to causing some power loss, make the radio a bit less selective in pulling out weak signals in the presence of strong ones. Analysis-wise, maybe try drawing the resulting primary circuit you described, and then moving the 1K loss resistor to the right side of the 17uH inductor. Looking to the left of that resistor to ground, you would see the source converted from 10 to 1K by the capacitor and inductor as before, but now that 1K resistor to the right loads this down, making a V-divider of 1/2. So it seems like we're going to get about a 2X (6dB) hit in voltage, in addition to maybe a 2x'ish broadening of bandwidth. It's late at night, but I think this is a valid analysis. Not sure what the exact effect will be on signal loss and bandwidth though, since the secondary also loads the primary... Good question !
I understand the maximum power transfer theorem and I can design complex conjugate matching networks in my sleep but I still don't see why we don't just go for maximum voltage gain (because that is where the signal inteligence is) your beginning example even showed that, sure there is no power to do work but the information is there in the form of a time varying voltage signal, why do we need power at this stage? Once you have a strong voltage signal then follow up that sage with a current amp to increase power. Is this all about the cost of one transistor or is there something that I am missng? BTW I am an extra class Ham llicensee and I hold a EE degree from an ABET accredited school but in fairness my concentraition was in control theorey.
An interesting question for sure. I usually answer that the best signal-to-noise ratio possible is determined by the antenna selection (to bring in the highest signal power), and the kTB antenna noise floor (plus excess input-referred noise added by the LNA). kTB sets how low the signal power can be and still be received The noise is in power terms, so it seems that maximizing signal _power_ into the LNA should get the best S/N we can achieve. I.e. the best noise figure for the front end. [After boosting signal (plus input-referred noise) with the LNA, it doesn't matter as much what we do with impedance interfaces, since the noise floor is now amplified along with the signal, so later noise doesn't degrade the system as much.] But maybe we can come at this from the Z and V domains as well. With a high-Z input LNA, the matching network will increase V as it increases Z to match to the high Zin value. Sounds great, but note that the kTB antenna noise is also boosted. In the end, it's a wash, except when added noise from the LNA is considered also. But what about a common-base amp with low Z input? I think the answer there can come from looking a 2 cases: First consider a 50 Ohm source driving into a, say, 10 Ohm input. In that case we're hit with 15.6 dB attenuation on the way in (20 log (10 / (10+50))). Not good. If we instead match down from 50 to 10 Ohms, there is a voltage step-down. That stepdown is sqrt(10/50) = 0.447 or 7 dB. Of course even after matching, we've got 10 Ohms driving into 10 Ohms, so to be fair, we need to factor in that factor of 0.5 (6 dB) - giving 13 dB hit total. Still, this is better than 15.6 dB. What do you think?
As I recall, there is a complication when noise added by the LNA is considered explicitly. In that case, transistors often spec a best-noise-match impedance value - empirically determined. Then, the goal is to build the matching network to map the antenna Z (say 50 Ohms), to this special value, rather than a perfect conjugate match. Honestly, though, I'm not a big fan of that. Too complex (pun intended) for too little gain in system function. Many modern radio systems are limited by large-signal handling (e.g. intermod behavior) rather than noise figure.
Hello. How did you calculate the V=10Voc and R=1K in 11:30? and the R=11K and V==33OC
Thank you
The C and L on the left side have X=100. So if we stand in the middle of the transformer and look left, we see a matching network with Q of 100/10 = 10. So the effective parallel resistance looking to the left (primary) side, is (1 + 10^2)10 = 1K (approximately). The voltage increase is the square root of the impedance ratio, so 10x. (unloaded in this analysis)
The voltage step-up through the transformer is harder to calculate exactly since loading data is missing. Also, the turns ratio is 110:20, which doesn't agree perfectly with the voltage increase I've given. Part of that is because the coupling coefficient is less than one due to lack of a transformer core. Honestly I'm not certain how I arrived at the 11:1 R increase. Still trying to remember my thought process when making this slide with all the unknowns and assumptions... But its likely close to this at least, and at least the 3.3x voltage increase is consistent with the 11:1 R increase which satisfies power conservation... The title of the slide reflects this I suppose 🙂
I don't know if this is already answered somewhere, but I'm wondering about how to use PCB traces when matching is being done. When I have RF on a PCB, I generally design trace widths and spacing to act as transmission lines, and calculate the them for 50 Ohm. But what do you do in the middle of a matching network? Say the output of stage A is 85 Ohm, and input of stage B is 50 Ohm. An impedance match network is created between them to match the impedances. Presumably, you use 85Ohm traces between stage A and the matching network, and 50Ohm traces between the matching network and stage B. But what traces do you use inside the network itself?
Or am overthinking this for short traces at typical amateur frequencies? I've mostly been doing PCB design in the 2m band, so the quarter wavelength is around 50cm, and my boards are less than 5cm in overall size, most of the RF traces are 1-2cm....
Your conclusion is correct. Usually we just try to keep the circuit small with respect to a wavelength - so that it doesn't matter. At 146 MHz, lambda in free space is 2 meters as you said, so on a typical FR4 PCB it's 1m (velocity of propagation in microstrip on the PCB is about 0.5, depending on trace width and board thickness). As long as we have traces shorter than 1/20th of lambda (about 0.5 cm at 100 MHz), the effect of the transmission line behavior can usually be neglected. If we want to include it however, we can model the trace in this case as a inductance of about 1 nH/mm, so about 5 nH. Then we can think through whether this has any real effect on the matching network design. But the best thing to do is keep all the traces short enough that it's even less than this, use SMD components, and just don't worry about it :-) We usually have to adjust the L (or C) a little anyway in measurement to deal with several other non-idealities...
@@MegawattKS - thanks!
Amazing set of video, but can you explain more in details how did you found input and output impedance of amp? How did you know that you need to match 50ohms to 10?
Thanks. Yes - it's overviewed on a slide in the video on Amplifiers (Episode 6). Here's a link keyed to the location in that video where the formulas are given: ruclips.net/video/UUlqW-vSq9M/видео.html . See the Common Base amp part. The theory that underlies the formulas given there is covered in "Part 2" of that video, which I decided to call Appendix A, so it would not lead to far away from the main journey. Those details are here: ruclips.net/video/m9X0mfg_8lQ/видео.html After a lot of background, it elaborates the model for a common-emitter amp, from which Rin and Rout can be found. But it doesn't do it for all the other configuations (like Common Base). That is left as an "exercise for the 'reader'" (it would be a homework problem in the university course - or an assigned reading.) But in brief, the input resistance of the common base amp is about 1/gm. With the amp shown in Episode 2, the collector current is about 4.5 mA, so gm is about 0.11 A/V, giving 1/gm of about 10 Ohms :-)
thank you so much @@MegawattKS
Hello again!
There's so much interesting information in this video that I'm still learning after having watched it several times!
With regard to the tuned transformer matching network, I have a question: on the secondary side, there's the secondary coil in parallel with a capacitor which, at resonance, will be equivalent to a high resistive value (depending on the Q of the coil). Then, when looking from the primary side, an impedance transformation of that resistive value has been obtained, resulting in another resistive value. But why is this value only resistive? When looking from the primary side of the transformer, there's the primary coil... Shouldn't the impedance seen from the primary be a series combination of the transformed resistive value plus the reactance of the primary coil?
Thank you.
Miguel
I'm sorry, I said something incorrect, because I forgot to mention the load!: as you say in the video, the impedance transformation would be made from the resistive load (I guess in parallel with the Rp of the secondary coil) to a lower resistive value in the primary, according to the square of the turns ratio. But then, on the primary side, there's no capacitor to resonate or cancel the reactive component of the primary coil. Does this reactance appear or how is it taken away from the transformed impedance?
Thank you.
Sorry for the delay in replying. This is an excellent question. I saw your followup info too. Will try to answer both here. On the issue of the primary winding L showing up, I would argue that it would be in parallel, not series. Beyond that, we need schematics and some math to really work it out. But qualitatively, I would offer that the primary and secondary inductances are linked (coupled magnetically), and that there is a mutual inductance involved. So with the secondary resonated, the inductance in the primary is part of that resonance already. This assumes either a good "coupling coefficient" value K, and/or an LC tank circuit Q on one side that is > 1/K. Here's some of the math. Sorry the writing is bad. These are from my class notes and it's a little light on the derivation since I just used these to remind me of what I wanted to talk through. Hopefully some of it will help. ecefiles.org/rf-circuits-course-section-13/
Note that the equations at the bottom of page 1, the circuit model at the top of page 2, and the application circuit in the middle of page 2 are for well-coupled inductors, but without a core. The model includes an "ideal transformer" in it, but the actual construction does not. It's just a simplified model for reasoning through the circuit operation without having to invoke the equations from page 1 again...
Wondering what you recommend for impedance matching between the output of the NE602 mixer at 1.5k ohms and the input to the IF filter at 300 Ohms. I'm coming up with some crazy values so I'm not really sure if I'm doing something wrong, or maybe it's not even needed? And by "crazy", I guess I mean something on the order of a 8 uH inductor, which seems...high? Or maybe it's just my inexperience showing.
That L value sounds right. It's about 540 Ohms reactance, which is somewhere in the range of the geometric mean of 300 and 1500 - which is a quick check. It's also the value I recall from the class 🙂 To get that value, we used a toroid core. Usually T37-2, but sometimes a higher mu core material to decrease the pain of winding too many turns. (Maybe an FT37-77 - I can't recall). The other alternative is to just pad the output R to bring it down to 300 Ohms so the filter sees the right terminating impedance. That helps it have a nice response curve, but of course comes at the cost of losing some gain. That is OK when the IF subsystem has plenty of extra gain - but the IF amp I put in the video series doesn't have a lot of gain, so it will affect the number of stations one can get. Engineering tradeoffs ! 🙂
@@MegawattKS Makes sense, thank you. Curious, what do small FM radio do, then? Like if I open up one of those cheap tiny FM emergency radios ,there won't be giant toroid inductors in there, right? Do they use mixers that are better matched to filters, or like you suggested, make it up with IF gain?
@@jchatterton2113 Good question. You can do either - especially if you do your own IC designs. Over the years, there has been a progression toward full-integration. This was actually my own research area. The goal was to get rid of all the big components - including the passive LC or ceramic filters. It can be done with custom IC design, but there are performance tradeoffs. That said, for FM, most modern radios in the last 10 years use something like the one-chip solution covered in this video. I've keyed it up to the architectures post-superhet which then shows the single-chip FM radio. If you back it up some, then right before that point in the video it shows an earlier radio with less integration - which still used some inductors (probably SMD with low performance). ruclips.net/video/XW3TYQstiuk/видео.html
Hello again! I have a question about the use of a variable capacitor in series with the antenna in the crystal radio set: is it used only to adjust the impedance matching together with the primary of the transformer or could it also be used to tune the resulting series LC circuit with the antenna? If we tune the series LC circuit, would that mean that the desired signal would find no resistance to ground, thus creating maximum voltage in the primary?
Thank you.
Miguel
Hi. The crystal radio shown at 10:16 doesn't have a variable in the primary - but I think we're talking about the tuned-RF vacuum tube receiver at 12:12 . Assuming so, then yes, that variable C in the primary side is adjusted to resonate the antenna. That provides a sorta zero Ohm path to ground. But remember the antenna resistance (say 10 Ohms like in the crystal radio analysis) is still there. So the current induced is finite. But it does maximize the signal that is linked into the secondary by maximizing the primary signal current (since R goes to a small value as you noted) 🙂 Basically the schematic at 10:12 is what we'd get, if the antenna is Zand = Rant + jXant = 10 + -j100 and the primary winding is +j100. The current induced in the primary would be Voc / 10. And the voltage across the primary winding would be (Voc/10)*j100 - which is 10 times the open circuit voltage Voc 🙂 (Note that there are complexities here that I'm ignoring - like the effect of what's going on in the secondary on what happens in the primary - so some of the numbers are off. But the basic idea remains solid.)
@@MegawattKS Yes, I mixed both circuits up, sorry. I was talking about the tuned-RF vacuum tube receiver :).
Thank you for replying, it helped me understand the circuit.
Excellent. Thank you.
You are welcome!
Great! thank you.
Thanks for the videos, I have a doubt , In episode -1 part-2 when defining Q, the Rp was the resistor value in parallel to the LC ciruit, but here it is the other way around, the formula for Q was shown to be w.r.t Rp and Rs there but now both these reistors places have swapped.
It can be confusing and tricky because of the orientation of components in a schematic. Are we talking about the "Solution for MN Components" slide at about 45 seconds into the video? Assuming so, the 50 Ohm source resistor Rant is actually in parallel with Lm. This is because for the Q circuit analysis, the Voc source is set to zero, which means it becomes a wire. So Rant goes from the top of Lm to the bottom and is labeled Rp for that reason. On the right-hand side, we called the 10 Ohms we're matching to "Rs" because the 10 Ohm load "Rin" is in series with Cm in the matching network. Does that help? Did I interpret correctly which slide you have a doubt on, or is it a different part of the video?
@@MegawattKS Yes it was the question, i got it now, thanks🙂
It is common for amateur radio operators to use antenna tuners to match to complex impedance antenna systems. Once tuned, it is believed that power bouces back and forth between the antenna and the tuner until it is all radiated by the antenna. It seems more reasonable that the power is absorbed by the real part of the antenna impedance with nothing bouncing back. Power bouncing back would show up as reflected power to the transmitter, but that is not the case.
Tricky stuff to sort out for sure. I think the resolution is that the meter is on the 50 Ohm side of the tuning circuit where the transmitter is and the LC tuning circuit is downstream from there - between the meter and the antenna. The LC tuning/matching circuit converts the antenna impedance (full complex value) into a 50 Ohm resistive value when properly tuned. So the meter only sees 50 Ohms and does not register any reverse traveling power. All the power that was launched into the line from the transmitter is "absorbed" in the 50 Ohms that the LC tuning circuit (working in conjunction with the antenna impedance) presents at the input to the tuner. What happens on the other side can be considered bouncing back and forth between tuner and antenna, but the transmitter (and meter) never sees that...