180W chinese linear amp woes...how many turns? - Page 1
Forum section: electronics / rf-microwave

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 03:49:23 am
content: I bought the $40 chinese 180W power amplifier from aliexpressI've used it to transmit and its actually worked okay, HOWEVER....The instructions, specs, and schematics are so contradictory and confusing its literally impossible to know the following from them:-how many turns the input xfrmr secondary should have (T1)-how many turns the output xfrmr primary should have (T2)-how many turns the center tap choke should have (L1)BEWARE: there are many schematics drawings and even some websites online talking about this amp. None of them have correct schematics or information that is up to date, as far as I can tell.All xfrmrs have 1 turn for sure on at least one side, because that turn is made out of copper pipes and there is no way for it to be more than one.Here is a schematic that is generally correct, however some of the component values are wrong.So you may ask, how have I been using it with whatever turns I used? Well it seems to work although R13 gets well over 250F after 15 seconds of driving the input with 10W.From what I've seen in the wide array of bad documentation, there is some intent to have different number of turns depending on if you run it at 12V or 24V.Is there a way I can measure the problems caused by the WRONG number of turns and thereby inductively reason I need 1 more or 1 less to improve things? The question is basically 3 turns +/- 2 at most. Its not like 50 or 100 or something.Any help greatly appreciated. I wish I could measure the issue with a digital scope and iteratively figure out which way things should go....

author: pienari
author_rank: Regular Contributor
timestamp: November 06, 2023, 07:39:41 am
content: What frequency you have tested this?.I think IRFP250 will work only low HF.

author: RFDx
author_rank: Regular Contributor
timestamp: November 06, 2023, 08:32:15 am
content: Quote from: cur8xgo on November 06, 2023, 03:49:23 amDepends on the gain of the transistors. Assuming a gain of 10, the 220 Ohm feedback resistors will reflect to the gate as 20 Ohm shunt resistors. This means, together with R13=10 Ohm, the input transformer would see a load of 8 Ohm. A 6.25:1 (5:2 windings) tranformer would be OK. This is only valid on the lower bands. The internal source inductance of the transistors is very high so the gain will go down fast at higher frequencies.To stabilize the input impedance of the pa you could change the input transformer to a 9:1 version and replace the inductors L4/L5 with apropriate (power) resistors.Quote from: cur8xgo on November 06, 2023, 03:49:23 amAgain, it depends how the output transformer is implemented. You can't deduce that from only a schematic. If the two primary windings are really magnetically coupled and you are looking for linear amplification, you would need a 1:9 transformer and >25V DC supply to get ~180W output power.Quote from: cur8xgo on November 06, 2023, 03:49:23 amWith magnetically coupled primary windings, the middle tap of the output transformers needs to be decoupled for RF with some good ceramic capacitors and not fed directly through a choke as shown.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 06, 2023, 10:04:48 am
content: I think you mean: the $0 power amplifier that you overpaid by $40 for?I don't get why these things... I mean I get why theyexist, because some people -- enough people -- will pay for them; but why dopeople pay for thingsthey literallycan't possibly knowhow to use? There was another thread here a few months ago, some crazy solar / battery charger buck converter, that the guy wanted to use at 10s of kW, a whole house installation, an expensive installation, and paid a couple hundred bucks for this completely undocumented board? Like, what?!Dunno. There doesn't seem to be a huge string of people with such complaints. Which, heh, I guess means either: 1. there's a user group out there that, whether by virtue of the seller/manufacturer, or a bunch of well-enough informed users pulling together, is making these lemons work, and you're justlost*; or 2., most users are silent in their having been swindled, maybe just threw a gamble in the hopes it came with documentation, it didn't, chucked it in the trash and moved on; which leaves the subset, 3.: those who haven't realized either of the above two outcomes, and came here hoping to avoid embarrassment.*And not necessarily for lack of trying! The fragmentation of the internet being what it is these days, there's no shortage of otherwise freely-available information hidden just barely out of sight, in chat rooms, Discords, etc.Well, in the interest of lampshading and immediately moving on... let's treat this as a design exercise, huh?I'm guessing there isn't anything special about the board you have, and maybe we can call it more of a kit, or a little saved labor over tossing parts on copper clad by hand?We can work backward from the power requirement. Suppose we have 180W sine into 50 ohms. That's 94.87V RMS or 134.2V peak. We have a supply of 12 or 24V. To avoid clipping, we need a peak of at most as much at the transistors (and preferably a volt or two less, or further, to avoid Rds(on), and nonlinearity from Coss(Vds) dependence). So we need a 5.6 or 11.2 turns ratio at most. Which, note that gives a load impedance at the drain of 0.8 or 3.2 ohms (actually half this, but the transistors act in parallel so share half and half).BTW, if this is designed the way I think it is, it's probably wrong to begin with. The drains need to be coupled to each other, on a common core. This erroneous omission is frequent enough within ham circles already, so I have a strong suspicion it's happened here, too. The usual design being something like, a copper tube with beads/rings stacked up on it, and a flange at both end for the connections of the "one turn". And then repeat same for the other side, maybe using a common (two hole) flange for the CT. The problem is this:Actually there are two problems. One is unbalanced current: DC bias. Most ferrite rings that size will saturate around 5 or 10A, and you're asking for pretty damn close to that.The other is class AB operation. Namely, it's impossible, with the large series supply inductor (L1 / RFC), and with no coupling between drains, which is the case when it's one turn and one core/stack per drain.Pure class A isn't necessarily a problem, but you better keep those poor transistors pretty goddamn well cooled, and you're going to waste a ton of input power in the process. Maximum efficiency in class A (transformer coupled) is 50%, so you're looking at dissipating 180W at maximum power, 360W input -- that's 180W quiescent dissipation per transistor at no input!!To get proper class B operation, both turns need to be on the same core, and not even like up and down a binocular core (another common, but still erroneous, arrangement!), but one turn all the way around the core,then another same turn all the way around the core, with power to the center tap of those two turns, and the ends to the drains. Power is bypass capacitors, no choke, so that load current can be sunk, dynamically, from the supply to one or the other side; current draw that isn't always there (continuous, quiescent, class A), but additional new current draw that goes directly into output power -- class B efficiency.And since we're talking quite low impedances here (~ohm), those should be low impedance transmission line transformers. A parallel-plate or coaxial connection is feasible, but more likely a stack of multi-pair twist (look up "star quad" arrangement) will do, and then as long as those are terminated correctly, the arrangement of the core itself isn't very important. (I would still recommend a nice squat E core stack, or PQ, RM or P (pot) cores, as these give shorter winding lengths for given inductance and core loss / given maximum power limit and insertion loss.) The secondary can be made up with the primary-pair pairs, and all the primary sets wired in parallel, and all the secondaries wired in series.That is, consider some triple-twist cable. Let S be the start and E be the end, and the number is the respective conductor. Suppose we have five such cables. The S and E denote the two ends of each cable -- their ports. Thus:Code:1S1 \ / 1E11S2 |__ (twisted) __| 1E21S3 / \ 1E32S1 \ / 2E12S2 |__ (twisted) __| 2E22S3 / \ 2E3...Also suppose Q1 and Q2 are the transistors, and we're connecting to their drains Q1D or Q2D, or the supply +V, or the output OUT or ground GND.For a 1:1:5 transformer, connect:Q1D, 1S1, 2S1, 3S1, 4S1, 5S1Q2D, 1E2, 2E2, 3E2, 4E2, 5E2+V, 1E1, 2E1, 3E1, 4E1, 5E1, 1S2, 2S2, 3S2, 4S2, 5S2OUT, 1S31E3, 2S32E3, 3S33E3, 4S34E3, 5S35E3, GNDThus, all the primary windings (the -1 and -2's) are in parallel, making a 1+1CT total winding; the five parallel pairs gives a transmission line impedance around 20Ω, still not very low but a good start. All the secondary windings are in series, making a 5t total winding; they act as a twisted pair with the primary, giving a transmission line impedance around 60-70 ohms (close enough to the 50 desired).Loop all the cables together as one big omni-cable around whatever cores will do, for as many turns as needed. (Obviously, this must done before making the connections, and the required cable length depends on turns and core dimensions, so this is a bit out of order of actual design and assembly.) Now the voltage drop along any given cable (they're all on the same core, same way around, same drop) is the Q1D-+V or +V-Q2D voltage, so we have a transformer doing what it's supposed to, but notice we only connect to transmission line ports (pairs at a time), so the stray inductance (leakage) and capacitance are minimal in a sense.And obviously, this would apply to 24V, or twice as many cables (and total secondary turns) for 12V, etc.As for other things:The gates themselves don't dissipate much power; the input impedance has an R+C characteristic, plus whatever feedback due to Crss, and Ls (source inductance: common to both gate and drain circuits), but aside from matching to that impedance, the rest just has to be burned entirely. So, you can add some series resistance to tweak that RC characteristic, but some shunt R is required to get a matching input impedance at low frequencies (<< 1/(RC)). This is what R13 is doing, and obviously it needs to be rated for full input. If it's not a ~20W part, it should definitely be increased to one. 250F seems reasonable enough for a resistor, but 15s also isn't very long. Or it isn't mounted correctly (e.g. heatsink types must be heatsink mounted, obviously enough, but this too is an all-too-common oversight!).Note that R13 and the gate equivalent R+C act in parallel, meaning the input won't be very well matched at either frequency extreme. Maybe you don't care about mismatch, but also it can potentially be an R+L, with L/R13 = Rgate*Cgate, so that they cross over from R13 at LF, to gate R+C at HF. (This also needs to take into account stability and parasitics, and maybe some R+C -- perhaps from each gate to GND, or each drain to GND, and yes not just differential between them, but both ways are options to consider and it depends -- will be found useful to optimize Zin, Zout and bandwidth/flatness.)Keep source inductance as low as possible. Mount the transistors flush, as far down in the board as they can go (vertically), or along the edge (horizontally) so that a wide plane connection can be made as close to the body as possible. The heatsink tab has output/AC on it, so should be insulated (yet another thermal limitation!), and the heatsink grounded to return displacement currents to circuit ground plane; alternately, large enough heatsinks must be used to dissipate the required power, but also insulated from everything else so they can be hot (thermal greased joint only, no insulator), and then shields placed around them to shunt the displacement current (E-field between sink and surroundings) to circuit GND all the same. Which will be a large enclosure so I suspect the former will be preferred. (A hybrid approach, using a heat spreader, then a wider insulator area to a grounded heatsink, is also possible, but usually more bother than it's worth.)There should be a thermistor or BJT on the heatsink, to sense temperature and allow Vgs tracking. Fixed Vgs (absolutely fixed, regulated source into a pot) means bias increases as temperature increases (Vgs(th) drops), a potential runaway situation. How this should be arranged, depends, but there are many ways to compensate bias.And obviously, R7 and R8 will get hot, depending on value (negative feedback / "neutralization"*), and R9-R12 for the aforementioned reason (dissipating in the R+C gate equivalent).*I always thought this a funny name, because it's not neutralizing anything, it's swamping the effect of reactances (mostly Crss) with resistance. Neutralization per se is canceling it out, e.g. an inductor resonating with Crss, to get higher total gain in a narrow (tuned) band. Or perhaps used to peak response at Fmax (it works for wideband amps too), in which case the R and X_L (at Fmax) will be comparable. But just R, I'd rather just call that shunt NFB.There should probably be a shunt sense resistor on VCC_PA (where this named connection is shown), perhaps using a INA1xx current sense amp or the like, to get a low-side referenced voltage corresponding to supply current. This can be used for servo biasing (but keep in mind it doesn't read just quiescent current, but additional draw due to class B operation as well), or just setup and testing (in which case just as well, you might use the bare shunt resistor and provide test points; there's very little AC at this point, so attaching DMM probe leads will (read: is supposed to) have no effect on the RF path.Oh, C9, C10, C18 should probably be some ~uF. Particularly if chip type. (They shouldn't even be leaded type...) You may want to move them so they're in the same place, indeed the transformer (both drain ends and the CT), these bypass caps, and the MOSFETs should all be in very close to the same place as is possible, since length between these components adds to stray inductance in the loop between them.Tim

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 02:12:42 pm
content: Quote from: pienari on November 06, 2023, 07:39:41 amI've only used it at 7MHz (40m). The "specs" say 5MHz 5o 15MHz.

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 02:28:37 pm
content: Quote from: RFDx on November 06, 2023, 08:32:15 amOkay that schematic is not entirely correct value wise, here is my hand-drawn simplified version with correct values, as built (see below).So to update things according to my understanding of what you said:"Assuming a gain of 10"...-how do I figure out the gain of the transistors?"the 220 Ohm feedback resistors will reflect to the gate as 20 Ohm shunt resistors."...-sch had error so correct val is 560 so 56 ohm shunt resistors. btw do you mean 22 ohm? just so I understand whats happening"This means, together with R13=10 Ohm, the input transformer would see a load of 8 Ohm."...-the input resistor is actually 22 ohm so...instead the input xfrmr sees 22 ohms in parallel with 56*2 = 18.4 ohms..."A 6.25:1 (5:2 windings) tranformer would be OK."-sooooo the goal is to make the input xfrmr look like 50 ohms right? so in your explanation it was seeing 8 ohms and 8 x 6.25 would be 50 ohms. But with the corrected values it would be more like 50 / 18.4 = 2.7 turns or 3 turns?EDIT: Wait my description is wrong. The impedance transformation is the square of the turns ratio. So if the secondary sees 18.4 ohms then the input sees 50 ohms with a turns ratio of sqrt(50/18.4) = 1.64. If the secondary is 1 turn (its pipes) then the primary needs to be 1.64 turns? So 2? That would yield 73 ohms on the input? But now I dont understand what you said anymore. If the original version was 8 ohms on the secondary, and you wanted 50 on the primary, you would want 2.5 turns ratio (or 2.5 turns since primary is 1). Yes?"To stabilize the input impedance of the pa you could change the input transformer to a 9:1 version and replace the inductors L4/L5 with apropriate (power) resistors."But here I'm lost! Is this to address high frequency gain rolloff or to address something that happens even in a single band? Trying to understand...L4 and L5 conduct less at high frequency and therefore the bias voltage influences the gate less at high frequency. So by putting resistors there it would resolve that. I guess I would feel uncomfortable doing that because I don't know why those inductors are there in the first place (why wasn't it resistors to begin with?)

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 03:59:01 pm
content: Quote from: T3sl4co1l on November 06, 2023, 10:04:48 amQuoteI think you just answered your own question!And should add one more to your list: 4. Someone looking for a project and to learn.There were other amp projects I could have chosen which have a much stronger user base/knowledge base and were more realistic as far as their specs, with accurate build instructions, at the same price. But I intentionally chose this one because I am very low on RF design concepts and I want a steep learning project. Even in just the couple days I've been working on it its $ per value rating is through the roof. I think its because it struck the perfect combination of functionality and malfunctionality. It actually does work and I got on the air at 77W (12V) and made FT8 contacts (15s!) at much higher SNR's than with 10W. But if I want the tantalizing 180W (or even close to that) I'm forced to understand whats going on here. The essence of ham radio!QuoteOkay so trying to understand:(lets just stay with 12V)134V peak across output xfrmr secondary, with 12V peak across input.So turns ratio 11.2Impedance ratio is 11.2^2 = 125. So 50 / 125 = 0.4.So I follow so far. But it appears that I do not actually understand how the push pull works. I dont get how the transistors act in parallel. I would think when one is on the other is off. I also dont understand why coupling the drains and class AB operation are related. I thought this design had 180 degree out of phase gates and therefore class B (edit) conduction.QuoteI understand the DC bias problem, but I dont understand the class AB operation issue being related to coupling between drains.Can I prove the ferrites are saturating with a measurement of some kind? Perhaps a current measurement?Is it possible that the drains couple somehow via capacitive coupling through the heat sink?QuoteI think I can actually do this because I've cooled TO-247's in the 120 watt range with the particular heat sinks I'm currently using at long duty cycles. If this thing actually ends up putting out 180W continuous and the only problem left is heat dissipation thats an achievement!QuoteThanks Tim, I'll have to study this for awhile its definitely further up the hill than I currently am.Regarding the heatsink though:I'm currently using 1 heatsink with the insulators for both TO247s. But I have lots of these heat sinks. What if I use a separate heat sink for each TO247? This seems like an obvious solution to me but I would think it would have been mentioned if it really was. Is there some reason I can't do that? Then I could get rid of the thermal insulators (and double the heatsink area!). Or is the capacitive coupling between the fet drains needed?If I cant do that and need to keep them insulated and on one heat sink, then are you saying I can ground the heatsink in the existing design?I have a cooling fan attached to the heatsink and powered separately by a linear isolated supply. When I transmit it shuts off the supply! So yeah some radiated energy there! (presumably)

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 04:04:10 pm
content: UPDATE:The aliexpress seller got back to me with info from the "supplier":"When 24V power supply, other unchanged. Only need to adjust the output transformer from 12V power supply 4 turns to 24V power supply 3 turns "Output transformer secondary 3 turns?It currently has 5 turns.How can 3 turns possibly make sense? So a voltage transformation of 3 and an impedance transformation of 9? 50/9 = 6 ohm load for the fets? Why would that be the intent? 15V peak on the drains at 24V supply?? Why would that happen?If I wanted to give them the benefit of the doubt and try this, what could l monitor to see how its going?I really want to make some measurements while the circuit is operating to see the horrors up close. I almost know enough to do that .I think I could start by observing the peak drain voltages and seeing if things make sense.

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 10:57:43 pm
content: Okay I took some measurements.Setup:UBITx to 180W amp to LPF to 50 ohm dummy loadSupply voltage ~13V (notes in pics say 12V but its really ~13V)Modulating with 1KHz sine sent via 600:600 to UBITx MIC input, keying MIC manually very briefly.Tuned to 7.074MHz upper side bandVpp measurements:INPUT XFMR: 3:1 (PRI:SEC)INPUT XFMR PRI: 76VppINPUT XFMR SEC: 26VppOUTPUT XFMR: 1:5 (PRI:SEC)OUTPUT XFMR PRI: 93VppOUTPUT XFMR SEC: 410Vpp? (exceeds max scope measurement!)LPF OUTPUT: into 50 ohm dummy loadLPF OUTPUT: 170VppSo from this seems like the voltage ratios are as expected. 76/26 = 2.9, 410/93 = 4.4 (close enough?)And it appears R13 (input xfmr sec resistor) is dissipating only ((26Vpp / 2) * .707 / 22 ohms)^2 * 22 = 4 watts..it looks like a 5W resistor to me so guess thats fine.EDIT:But the output of the LPF is 170Vpp....that doesn't make any sense. Thats like 240W across 50 ohm. My 50 ohm dummy load is a jar of salt water that I verified is 50 ohms with a nanovna. But maybe I did something wrong?Output is 85Vpeak (170Vpp) which is 81W at 50 ohm load. That seems about right.Some dummy load measurements (salt water)SWR @ 7MHz = 1.1Resistance = 48 ohmReactance = +2.8 ohmLPF:SWR @ 7.07MHz (where I was testing) 1.6 <-- hmmLOGMAG @ -3dB 7.8MHz <-- hmmBEST SWR: 1.05 @ 7.46MHz

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 06, 2023, 11:19:15 pm
content: UPDATE:Okay I replaced the salt water dummy load with a bona fide RF 50 ohm 100W dummy load resistor. The voltage at the resistor went up to 180Vpp!Then I tried transmitting at 7.4MHz instead of 7.074MHz because thats where the SWR was lowest. Went down to 170Vpp!Ahh but I get it now..the calculation is peak not pp...So in that case I'm seeing 81W at the load which makes sense. (180/2)^2 / 100 * 1000 = 81Wokay so back to the measurements to see if this makes sense.....

author: ahbushnell
author_rank: Frequent Contributor
timestamp: November 07, 2023, 12:02:20 am
content: Is this for induction heating or a transmitter?

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 07, 2023, 12:16:20 am
content: I left out an important measurement. Output XFMR primary center tap to drain for one of the fets.Looks like its hitting 20V (the supply is 13V)This is CT - DRAIN so negative going means the fet is on.If I wanted to get this down to a volt or two below the supply voltage, I'd need a turns ratio of about 1:10 I think right?I think maybe thats not whats going on here though....

author: joeqsmith
author_rank: Super Contributor
timestamp: November 07, 2023, 12:18:52 am
content: Quote from: cur8xgo on November 06, 2023, 11:19:15 pmBeen following and seeing your posts about salt water. No idea what you were going on about then finally did a search and now believe you really are using salt water for a load. I would have never guessed that.https://kv5r.com/ham-radio/2018-projects/salt-water-dummy-load/

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 07, 2023, 12:20:42 am
content: Quote from: joeqsmith on November 07, 2023, 12:18:52 amYep thats the one. I wanted something in the few days before my actual RF dummy loads got here. It seems to work. But I'm going to retire it because I have the real loads now and although the salt water one seems to perform okay with the nanovna, tipping it over in its glass jar would be a disaster.

author: joeqsmith
author_rank: Super Contributor
timestamp: November 07, 2023, 12:28:03 am
content: He posted how the performance was poor over the 160 to 10 meter bands. Decent on 80. What did you find with yours? I've never heard of anyone doing this until now and wonder how it behaves with frequency, temperature, age, different electrode geometries?

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 07, 2023, 12:39:15 am
content: Quote from: joeqsmith on November 07, 2023, 12:28:03 amI just did some nanovna measurements of it..this is all I know regarding performance.SWR @ 7MHz = 1.1Resistance = 48 ohmReactance = +2.8 ohmIts so easy to make I would suggest just making one and playing with it. Go easy on the salt! A little goes a LONG way and then you overshoot minimum SWR

author: joeqsmith
author_rank: Super Contributor
timestamp: November 07, 2023, 01:47:00 am
content: You only use the 40M band? With the Nano, you could try a sweep from 3 to 50MHz like the author and post the return loss or VSWR. Maybe then post a few photos of yours.From the link, there was that second comment where poster suggests mostly flat for their setup.I saw someone post about using a home hot water tank heater for a load once. They found a sweet spot on one band where I guess it sort of worked but it couldn't be used over a wide range (meaning the HF bands).

author: cur8xgo
author_rank: Regular Contributor
timestamp: November 07, 2023, 03:12:59 am
content: Quote from: joeqsmith on November 07, 2023, 01:47:00 amDisplay shows 3MHz to 50MHz, 0.1 SWR per division, bottom is 1 SWR.Looks like around 1 at 3MHz and around 1.5 at 50MHz. I have no idea how that might be impacted by the wire placement inside the jar or other things. Its very easy to modify. It had a SWR of 1.05 I think when I first made it a couple days ago. Not sure why it changes. It was very handy at the moment though.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 07, 2023, 03:16:36 am
content: How did you probe "CT to drain"?Oh yikes, that filter really doesn't have ground plane... does it...Tim

author: joeqsmith
author_rank: Super Contributor
timestamp: November 07, 2023, 03:53:33 am
content: Thanks for running that. Yours is much much better than the person shows in that article. Very surprised.

author: RFDx
author_rank: Regular Contributor
timestamp: November 13, 2023, 06:37:20 am
content: Quote from: cur8xgo on November 06, 2023, 02:28:37 pmThe 220 Ohm feedback resistors reflect back to the input as 220/(1+G). With a gain G=10 that would be 20 Ohm single ended or 40 Ohm gate-to-gate. Together with R13=10 Ohm and the 6.25:1 transformer the input would show 50 Ohm. With the input transformer being 9:1, R13=22 Ohm and with the feedback resistors of 560 Ohm, this isn't the case.Quote from: cur8xgo on November 06, 2023, 04:04:10 pmWith a 24V supply voltage and 1:9 output transformer, an ideal power amplifier in class-AB would deliver theoretically ~207W. Taken into account the saturation voltage of the MOSFETs, you are left with ~170...180W. Unfortunately your amp doesn't work in class-AB. Compared to class-AB, under identical conditions, you can expect about half the output power from your amplifier.