Peaking 455kHz IF Transformers out of circuit - Page 1
Forum section: electronics / projects

author: Daniel Bingamon
author_rank: Regular Contributor
timestamp: May 22, 2021, 08:12:04 pm
content: I have FeelTech FY3200S signal generator.This is an experiment I've tried with numerous old IF transformers and this has me puzzled.I connected the leads from channel 1 of the signal generator to one of the coils on the IF transformer and the other side to a Fluke 7613 Oscilloscope with a 1X probe.Why is it that these transformers peak around 360-380 kHz when you try to peak them in this manner?I've seen people do this on youtube but there must be something missing.I also have a handmade modulated 455kHz oscillator since the FY3200S doesn't seem to really modulate - a disappointment.

author: TimFox
author_rank: Super Contributor
timestamp: May 22, 2021, 09:25:38 pm
content: If these are vacuum-tube era double-resonant IF transformers, the resonating capacitance inside the can is relatively small (the inductance is correspondingly high). They were designed to work between pentodes, where the plate resistance of the first tube is between 100 k and 1 megohm, and the grid impedance of the second pentode is also very high. The 50-ohm output of the generator will swamp the primary and the CRO probe capacitance will detune the secondary to a lower frequency.

author: Daniel Bingamon
author_rank: Regular Contributor
timestamp: May 22, 2021, 10:19:39 pm
content: What can you do to measure it on a bench top? Is there some kind of resistor network to help match the impedance?

author: DC1MC
author_rank: Super Contributor
timestamp: May 22, 2021, 10:22:05 pm
content: Quote from: Daniel Bingamon on May 22, 2021, 10:19:39 pmTry a capacitve coupling with a VERY small capacitor (like 4.7pF) on both generator and scope.Cheers,DC1MC

author: Daniel Bingamon
author_rank: Regular Contributor
timestamp: May 22, 2021, 10:23:31 pm
content: And yes to the question, these are vacuum tube type. I've been experimenting with building vacuum tube era radios from scratch.As soon as I'm finished with the octal type tubes I'm going to start on the 7 and 9-pin miniature era tubes after that.I'm not a student of any school, just a 59yo Hard Knocks that doesn't stop learning new old things..

author: Daniel Bingamon
author_rank: Regular Contributor
timestamp: May 22, 2021, 10:24:59 pm
content: Would the tiny capacitor alter the frequency? Would a gimmick capacitor work for that?

author: DC1MC
author_rank: Super Contributor
timestamp: May 22, 2021, 10:29:22 pm
content: Quote from: Daniel Bingamon on May 22, 2021, 10:24:59 pmAny type of direct coupling will alter the frequency, but if you put a small cap in series with the generator output and with the scope input you'll alter it less. Ideally you'll want to do an inductive coupling, but is a pain and the capacity of the circuit where you'll install them will anyway change the frequency. So either a small cap for some good approximation, or the best is tuning in the circuit.

author: TimFox
author_rank: Super Contributor
timestamp: May 22, 2021, 10:34:20 pm
content: Gimmick capacitors are useful for sub-pF adjustable capacitors, since < 1 pF caps are hard to find nowadays.Given the parameters of these transformers, it is hard to do your peaking experiment accurately without a vacuum tube at each end.You could take a 6SK7 (or 6SJ7 or 6SG7 or 6SH7) at each end: terminate the grid of the first tube in 50 ohms, connect the generator there, and run the second tube into a low-value load (maybe 1000 ohms). The capacitive load on the 1000 ohms should have negligible effect on the tuning, and the overall gain should be adequate. The actual tube impedances (essentially output capacitance of V1 and input capacitance of V2) will be important, and a bench test would not replicate them. The data-sheet output capacitance of a metal-octal 6SK7 is 7 pF, the input capacitance is 6 pF, but the plate resistance is 120 k (at 100 V) to 820 k (at 250 V). You could try using such a large resistance in series with the primary, but you will still need a high-impedance (and low capacitance) probe on the secondary.

author: Daniel Bingamon
author_rank: Regular Contributor
timestamp: May 23, 2021, 03:26:53 am
content: I tried this technique from this guy on youtube. I've seen it before and I really didn't believe the putting the ground lead on the can would make any difference.I actually got a proper response from a good IF transformer (Miller I3-PC2) and the one that was arcing didn't work very well.I'm sure there's some differences in the circuit but at least now I can tell if the transformer is likely to be "peakable" (is that a word? It is now).

author: Zero999
author_rank: Super Contributor
timestamp: May 23, 2021, 08:30:01 am
content: Would using a high value resistors be better than capacitors?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 23, 2021, 01:46:51 pm
content: You're missing the effect of impedance on network response. Kind of obvious, but, a network doesn't exist in isolation, it is a part of the system (including sources and loads) -- and you are measuring the system!If you draw the equivalent circuit, this should be more obvious? Consider a typical tube IF interstage:Tube 1: pentode plate. Low capacitance (maybe 3pF?) in parallel with high resistance (plate resistance 200kohms+).Primary: shunt C in parallel with self inductance, resonates at some Zo1 and Fo.Secondary: shunt C in parallel with self inductance, resonates at some Zo2 and Fo. (The tunings should always match: different Fo's "tilts" the double-peaked response.)Tube 2: pentode grid. Higher capacitance (~10pF) in parallel with high resistance (depends; see below).Which, by taking all together (output grid voltage over input grid voltage), we get the total gain for this stage. And so, knowing we need some total amount of gain in our IF strip, we can design for the number of stages this way, alright.Clearly, you have two mismatches in your test:1. the signal generator's 50 ohm output absolutely stomps any behavior of the primary, which is designed for some high impedance2. the 1x probe has considerable capacitance and loss, equivalent to something like uh, probably 1MΩ || (50pF + 50Ω) give or take?Even without taking it apart, we can use some assumptions to figure out what the IF transformer really is:- Since the plate is a high resistance, we can draw more power from it using a higher load resistance. As long as Rp >> R_L, total stage gain goes as sqrt(R_L). (Note that power handling generally goes down, because we might have plate current of some mA and supply of 100-200V, for a maximum-power load of some 10s kohms. Small and large signal gain are different matters! For an IF interstage, we can safely assume small signals.)- Capacitance has a minimum, imposed by the physical circuit. The coil's self capacitance, and plate and wiring capacitance, limit the maximum impedance: Zo(max) = 1 / (2 pi Fo Cmin)- We also want some desired Q factor, since the IF filter should be just tight enough. A bandwidth of 10kHz is 1/45th of the center frequency, so we want a Q around 45.If we take C = 5pF as a guess, we get Zo < 70k. For Q = 45, that means R_L max 3.1M, which is pretty impressive -- and likely higher than expected Rp, so this is actually an overestimate already. That'd be a really big inductor anyway (~24mH), of which we might not be able to get C low enough. So, probably they use L of a few mH, and the lower L requires additional capacitance in parallel, letting us use a varicap for tuning -- which may be easier to work with (whereas a tuning slug will vary coupling as well as self inductance).We use the same design method for the grid coil, with Cmin being higher due to Cin being bigger than Cout, but also the effect of space charge -- a hot, conducting tube has higher capacitance than a cold tube (and guess which one you see in the datasheet, huh?) For good performing types*, the capacitance more than doubles!*Basically to mean, anything since the 1930s, beyond the older, less optimized, big-bulb and grid-cap types. 6SJ7 is probably marginal, it still has kinda low gm and high Cout? Compare to 6CB6 for example, or frame grid or planar types.Cool fact, the capacitance of the space charge is due to the mass of the electrons therein. Makes sense, mass is a conservative (i.e. not lossy) property, so when it's coupled (via electric field), it similarly manifests as a lossless component in the electrical circuit: capacitance. (Indeed it could only ever be L or C, and L would just be... weird.)And what happens if some of that space charge escapes during part of the cycle? Well, it can't give energy back, it's energy lost. Aha, then there must be some component of resistance. In effect, this represents work done to modulate the beam current -- and indeed we measure this. It happens to go something like 1/f^2, so is quite substantial at high frequencies, but should be just about negligible down here. For example 6688/E180F has about 2k input resistance at 100MHz; it has about five times more gm than 6SJ7.Final cool fact, this "beam work" extends to later electrodes -- you can measure gm(g1g3) in a pentode (esp. dual control type), or gm(g2g4) in a heptode -- despite those electrodes being reverse biased. It's not a fluctuation in DC bias like it is at the plate; they feel the beam's space charge as it passes by! Coolest part, this is a nonreciprocal effect: gm(g4g2) ~ 0, give or take mere stray capacitance between them. Makes sense, the electron beam acts like a conveyor belt, allowing electric field and charge to flow one way only!So, for grid resistance, you can probably assume megohms in this application. Which means the grid coil will be pretty much unloaded, and most of the network damping will actually come from the plate, or coil losses. It also suggests we could gain some by increasing the turns ratio, although again with physical and capacitance limitations as they are, we'd probably not get more than 1:2 or so?Either way, this is fine: an ideal (lossless) network needs loss atat leastone port (termination), in order to have a meaningful response. It does have to be designed for it, of course -- consider the standard ladder network for a low-pass filter, where you see the double-terminated case all the time. This isn't for granted, and there are singly-terminated versions as well (one port open/shorted), and, all the coefficients are different; and it's not all that easy to derive from the double-terminated case as far as I know? But definitely possible.So, we should design a test jig, which emulates the impedances expected in the final circuit. We can model a Thevenin source of say 6pF (typical for 6SJ7) as a voltage source in series with a capacitor of the same value, and a large value resistor in parallel with the cap (a few 100k, whatever the plate resistance is).Thevenin doesn't much care what the "cold" end of the resistor returns to, or if we use a current source instead (which might be more physically representative). That is, we can use these source connections without much care (VS = voltage source, '+' = "in series", '||' = "in parallel"):1. VS + (R || C) + tank2. VS + C + (R || tank)3. VS + R + (C || tank) <-- this is probably what you would first think of, in terms of direct Thevenin equivalent4. CS || R || C || tank <-- probably most physically representativeNote that we don't do (4) with a VS because that just stomps over everything. The impedance at the tank must be high, and a low resistance in parallel completely destroys that. There will be some difference between 1-3 in terms of gain and frequency dependence, but we're mainly concerned with observing the peak at all, and as the bandwidth is small, the frequency dependence of this connection will be small (it would be a proportional or inverse relation -- a tiny skew over a ~2% range, and we can easily tune that out in final alignment, which we'll have to do anyway as we're not going to be getting these values exact to the real thing).For the grid side, we have fewer options, because we expect such a high resistance. We might choose (2), where "VS" is instead the probe, and we select C to get the right equivalent grid capacitance. We might ignore R, or take the probe as some equivalent resistance.We can calculate the equivalent as seen by the grid coil, though it's not trivial -- like, I can't just do it in my head, I'd have to plug in some values and do the math. First, we need the RC equivalent of the probe itself, at this frequency; then we connect that in series with a capacitor, and then we solve for the parallel equivalent R and C of that network.This is probably worthwhile to do, partly as an exercise to illustrate what the probe really looks like at AC, and to get experience working with complex impedances. Finally, if the equivalent parallel tank resistance ends up too low (<1M say), we can ballast that out by reducing the series cap to the probe (so it makes up less of the total), and make up the same tank capacitance with a plain cap in parallel with it.So, to measure the probe impedance, you can simply set up an impedance divider between source and probe, and see what its gain and phase shift are for different impedances. You can measure this by resonating it with L and measuring peak gain or bandwidth, and frequency; or dividing with an R of appropriate value and measuring gain and phase shift; etc.Note that if you do this with help from a solderless breadboard, or fly leads, the breadboard adds about 4pF between adjacent positions, and who knows what the leads do. Preferably, just tack everything onto the transformer and measure it like that. It's how tubes would've wanted it... wired point to point.Good luck,Tim

author: Daniel Bingamon
author_rank: Regular Contributor
timestamp: May 23, 2021, 02:28:05 pm
content: I'm still get a grip on impedances - at least for output transformer I have a better understand.Thank you very much for taking the time to write all of this. I'm going to have to read it several times to get a better understanding.So, let me see if I have this right. I noticed that there are "Input IF Transformers" and there are "Output IF Transformers" yet they are both 455kHz - that must expect certain impedances in the two different operational stages. So the primary side had a difference capacitance and inductance yet the tank was still designed for 455kHz at different impedance than the secondary. On schematics it leaves you with the impression that they are both the same --- same but different, okay.

author: TimFox
author_rank: Super Contributor
timestamp: May 23, 2021, 04:00:55 pm
content: The "output transformers" were designed to drive the detector ("second detector" in old-fashioned superheterodyne nomenclature), which is a diode rectifier, and therefore presents a lower impedance than an IF amplifier grid.Yes, the schematic diagram for the IF transformers are the same, only the component values are different for different impedance levels. In 5-tube sets, the third tube combined a triode and two diodes (predating solid-state integrated circuits) in the same vacuum envelope.The reason that many detector-triode tubes (e.g., 6SQ7) have two diode plates (with the cathode common to the triode) is that one diode produces the demodulated audio, while the other produces the AGC voltage (possibly exploiting the cathode bias to "delay" the bias voltage until the signal reaches a given strength).More than you want to know about the loading of the detector can be found in chapter 27 of the classic "Radiotron Designer's Handbook" by F Langford-Smith (1953) , including quantitative design procedures. The book also treats IF transformers in great detail. A 1524-page pdf file of the whole book can be found athttps://worldradiohistory.com/BOOKSHELF-ARH/Handbooks/Radiotron-Designer%27s-Handbook-4th-Edition.pdf, and the book itself is readily found on eBay and book sites.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 23, 2021, 05:57:23 pm
content: Your confusion is understandable -- even manufacturers themselves give scant little data about their IF transformers, if you've seen a contemporary catalog.I guess it's at once, frustrating because of the seemingly intentional ignorance (it's not like they don't know how these things are made), but also the blissful ignorance of simply -- you slap it into the circuit with this and that, and it Just Works(TM).It's not quite as automatic as an integrated circuit for whatever function, but it does happen to be pretty typical that we see tubes with these output and input impedances. So it's kind of a thing that, it doesn't deserve to work, certainly it can't work in general, but in the average case, yeah, it manages to work anyway.Tim