Keyboard Shortcuts
ctrl + shift + ? :
Show all keyboard shortcuts
ctrl + g :
Navigate to a group
ctrl + shift + f :
Find
ctrl + / :
Quick actions
esc to dismiss
Likes
Search
Does L502 cause excessive voltages on the QMX finals?
|
The four BS170's and T501 are arranged in a standard voltage mode class D push-pull amp.
The center tap of T501's primary is normally a DC voltage source, as it is on the 50W QCX amp. When one set of BS170's pull one side of T501's primary to 0 volts, the other side pops up to 2xVdc (so 24v if operating from 12v). And then the other set of BS170's pull the second side low, ?creating a square wave at the transmit frequency? with a pk-to-pk voltage of 2xVdc across T501's primary. Since the LPF on the secondary of T501 allows through only the fundamental to the 50 ohm external load, the harmonics are blocked by the LPF on the far side of T501, and the current through T501 is a sine wave.?
The current drawn from Vdc at the T501 primary center tap is a rectified sine wave.
?
Seems to me that adding L502 would enforce a constant current into the T501 primary center tap,
not a rectified sine wave. ?And that the extra current with nowhere to go would be piling up, causing excessive drain voltages that might blow the BS170's. ?
The above paragraphs were originally buried in message ?/g/QRPLabs/message/139736
?
Attached is an LT-Spice simulation schematic of the QMX final at 3.57mhz, with and without L502.
I'll place a copy of that simulation schematic in the files section under KE7ER.
It uses 2n7002 FET's since that is what was handy under LT-Spice.
?
With L502, the Vds on each BS170/2n7002 is very abrupt and spiky, roughly double what it is without L502.
Without L502 the voltage across the drains is close to the expected square wave, but output power is reduced.
Currents from T501 are rather ugly in both cases, but close to a square wave with L502
and close to a sine wave without L502.? I find it curious that a square wave current can exist there?
since the LPF should be blocking it.?
?
The simulation shows the fundamental current going through L508/L1, and the harmonics going down through C515.??
So the LPF behaves differently than a series LC filter would, and perhaps differently than what would work best
with a voltage mode class D amp.
?
It seems that T501 is expected to pass some of the odd harmonics on all bands of interest for this to work well.
A very tall order at 50mhz for a transformer that must also work at 1.8mhz.
?
Comments?
?
Jerry, KE7ER
? |
||||||||||||
|
Hi Jerry,
?
John Z and I did several simulations of the QDX PA.? The most significant issue is the end-of-transmission spike when the energy in L502 is dissipated by the off state of all four BS170s.? You could email John Z (KJ4A) to see if you can get a copy of his last model.? I have attached one of the models I modified for testing.
?
Of note is the power reduction that occurs if L502 is removed.? I did several bench tests on one of my QDX radios to see which solution provided the least power reduction.? The result was that the Zeners provided the lowest power reduction.? John mentioned that the Zeners may not protect the MOSFETs in real-world situations.? I did not have the means to simulate or beach test his points.? The net summation for me is giving up the 1/4 watt to protect the BS170s with the comutation diode solution is the best choice.? Note that there is NOT an end-of-transmission spike for the QMX; it is only the QDX series.? The key shaping circuit allows for a soft landing at the end of the key down.
?
I suggest you look up past posts from John Z (there are many) or PM me if you want more information.
73
Evan
AC9TU
?
?
? |
||||||||||||
|
Hi Jerry, and GM Evan,
?
Evan is correct in all the points he made. Also note that the .asc file he provided includes a model for the BS170 transistor.
?
There is possibly a way to create an L502 inductive spike in QMX, but not in the same end-of-transmission way that QDX is susceptible. In QMX the envelope shaping modulator protects against that possibility, if it is functioning properly.
?
In QMX an inductive voltage spike might result from a rapid change in load impedance , for example when an ATU is searching for best match and undergoes a big transition. The PA current would try to change rapidly and L502 would respond with an Ldi/dt kick. I have not seen that happen physically but it can be seen in simulation.
?
JZ KJ4A?
?
? |
||||||||||||
|
I suspect the topology is wrong on the QMX for a Voltage Mode Class D (VMCD), L502 simply should not be there.
Since it is there, it is forcing a constant current into the final, creating the voltage spikes observed by KB1NLW in post 136435.
The LPF should present a high impedance to harmonics (especially odd harmonics), so the voltage from the transformer can be a square wave.
?
An alternative is Current Mode Class D (CMCD), where we no longer use the T501 center tap and have two L502's.
The LPF should short to ground the harmonics and pass the fundamental, allowing the current from the transformer to be a square wave.
?
My guess is that the LPF does not behave appropriately for either VMCD or CMCD.
Perhaps a hybrid of the two was found to work best given the need to cover so many bands with so few LPF's.
One possible path forward might be to keep the current design for those that want all bands from 160m to 6m,?
but have an alternate build supporting only 6 bands where each LPF can be optimized for the frequency of interest.
?
I'll run a few more simulations, using a series LC instead of the LPF for VMCD, and a parallel LC for CMCD.
Will use Evan's BS170 model, though I suspect the 2n7002 was close enough for the simulation.
?
Thanks to both of you for responding.
?
Jerry, KE7ER |
||||||||||||
|
Attached is a revised LT-Spice simulation of the QMX final, also placed under KE7ER under files.
There are four separate instances running simultaneously allowing waveforms to be easily compared.
T501 is assumed to be perfect, and wired for 9v.? I doubt the 9v T501 matters much in comparing topologies.
?
I am now using Evan's SPICE model for the BS170, have two sets of two in parallel.
Now using L508 and L512 reference designators instead of L1, L2.
Corrected the value of C515 from 580pF to 680pF.
None of that made much difference in the simulation.
?
The four instances are:
1)? QMX Stock Final.??
2)? Same as above, but removed L502
3)? Same as above but replaced the LPF with a series LC filter with a Q of 3, should be a VMCD topology.
4)? Same as above but removed T501 center tap and added two L502's, now an LC parallel filter, should be CMCD topology.
?
There would be tradeoffs in moving to any of these schemes.
Clearly the LC filters would need to be improved on my VMCD and CMCD instances.
Bringing this closer to a true VMCD implementation does reduce Vds-max considerably as expected
though the CMCD implementation has a Vds-max not far from that of the stock implementation.
The CMCD has the most power out of any.
?
We could simply recommend that everyone build for the 9v power supply, that may cure most of the issues
with BS170's getting blown from over-voltage.? However, that would increase BS170 current, and may
cause more failures during high duty cycle operation in digital modes from over-heating.
?
Here's the math (in python) I used in calculating a Q of 3 for the simple serial and parallel LC filters:
from math import pi, sqrt
Q=3; R=50; w=2*pi*3.57e6;? Cs=1/(w*R*Q); Ls=1/(w**2*Cs); print(Cs, Ls)? ? ?# Serial LC values Cp=Q/(w*R); ? Lp=1/(w**2*Cp); print(Cp, Lp)? ? ? # Parallel LC values Qs=sqrt(Ls/Cs)/R; ?Qp=sqrt(Cp/Lp)*R; ? print(Qs, Qp)? # Double check ?
Jerry, KE7ER |
||||||||||||
|
On Thu, Feb 27, 2025, at 06:16 AM, Jerry Gaffke wrote:
?
Jerry,
The model you posted does not include the 100pF capacitor across the inductor that acts as a second harmonic trap.
?
I also got different inductance values from Toroids.info for the transformer windings.? If you use the info from that site and do the 3:2 windings for 12-volt operation, there is almost twice the power difference without L502.
?
73
Evan
AC9TU
|
||||||||||||
|
Evan,
?
You are correct, I missed the 100pF cap.
Thanks.
?
I had noticed that computed values for L508 and L512 were different than the ones Hans had said he simulated with.
I figured since I was simulating may as well go with those, and that final windings got adjusted when he was looking at hardware.
?
I have made both adjustments to the simulation in the attached file QMX_sim03.asc
?
Will try your 3:2 12v transformer, but haven't yet.
In simulation it may not make too much difference, I'd expect that all the voltages after the transformer just go down by that ratio.
From forum posts, it seems people have fewer failures of the BS170's with a 9vdc build, so I figured I'd start with that.
However, perhaps that's only because most people have built for 12vdc.
It does seem that with the higher currents of a 9vdc build we would have more trouble with overheating in digital modes.
?
Here's the voltage readings from LT-Spice for my original QMX_sim01.asc simulation before and after your changes.
Curious that adding changing the L508 and L512 values brings the output down, but adding the 100pF cap brings it most of the way back up
Output: ?
? ? ? ? ?18.6v as in sim02
? ? ? ? ?15.6v when adjusting L508 and L512 to AL*(t**2) values of 1.875uH and 1.452uH ? ? ? ? ?18.4v when 100pF added ? ? ? ? ?18.4v when both changes made ?
I then looked at that final version with both of your changes, and compared it with the same but without L502,
and also my VMCD and CMCD versions as in the attached file QMX_sim03.asc.
Since Vds-peak values can blow the BS170's, I also looked at peak voltage on one of the drains.
? stock (with both changes) ?vs L1-removed vs VMCD vs CMCD:
? ? Output? ? ?18.4 ?vs ?13.8 ? vs 14.4 ?vs ?18.5 ? ? Vds-max? 33.8 ?vs ?18.5 ? vs 18.5 ?vs ?27.5 ?
? Output with L1 removed is 13.8v as shown, but goes higher to 14.1v if L508=1.83uH,L512=1.55uH)
? For VMCD, FET voltages are square and current is sinewave, for CMCD current is square and voltage is sinewave ?
?
?
Note how much lower Vds-max is for L1-removed and VMCD.?
L1-removed is effectively VMCD, but using the stock LPF which I am still unsure of.
L1-removed and VMCD both also have low output power, which is a function of the square of the output voltages recorded above.
If we were to increase the transformer ratio of the L1-removed version such that Vds-max was equal to that of the stock QMX,
it seems the output voltage should increase to a value of 13.8*(33.8/18.4) = 25.35, which is way beyond the stock output voltage of 18.4
That's the reason I was sucked into this, removing L502 may help with the over-voltage issue at the BS170's
though this may bring on issues due to higher peak currents.
Like all things around RF amps, it's considerably more complicated than one might think at first glance.
?
Anyways, I was up half the night here with a hacking cough, which is why I got that simulation done.
Probably time to get some sleep.
Given my current state, could well be plenty more errors in my attempts at simulation and in the above discussion.
?
Jerry, KE7ER |
||||||||||||
|
Jerry,
?
Not sure if you are aware of two features in LTSpice:
1 - You can use Math expressions for the plots.? Here is one comparing the power as V(n003)*I(R1).? You can plot the current through a device.
?
?
?
2 - you can get the average value with <ctl>left-click on the plot value:
?
?
Unfortunately, you can only get one value at a time.? It is also better to zoom in to a steady state:
?
?
?
There is a good YouTube series that teaches advanced LTSpice commands:
And the online manuals:
?
?
73
Evan
AC9TU
?
?
|
||||||||||||
|
Evan,
?
I had read the LT-Spice docs and used some of the advanced features a few years ago.
Comming at it fresh late yesterday, it was all I could do to edit a schematic and get a transient analysis going.
I had enough head scratching to do without dealing with advanced features.
?
You may have noticed that for the 1:1 version of T501 that I was using, I assumed the inductance
of the two primary windings added together should equal the secondary inductance.
Unfortunately, the two primary windings interact on the same core to give a larger total inductance for the primary.
Instead it is the total turns count of the primary windings that is equal to the secondary winding turns count for a 1:1 transformer.
?
Here's how I now think of it:
?
############
Inductance on a given core is proportional to turns squared.
Lets assume an AL of 1uH per turn for the core, a 16uH secondary, and a 3:2 turns ratio. Then the number of turns for a 16uH secondary winding is sqrt(16) = 4 turns ? (working back we have: ? 1uH/turn * 4turns**2 = 16uH) The total number of turns on the primary is 4 * 3/2 = 6 turns Each half of the primary has 3 turns.The inductance of each half of the primary is: ?1uH/turn * 3**2 = 9uH ?
In general the formula for the inductance of one winding of a center tapped primary given a turns ratio of N:1and a secondary inductance of Ls is:
L_one-primary-winding = ( sqrt(Ls) * N / 2) **2?
For the above example:? ? ( sqrt(16) * (3/2) / 2 ) **2 = 9 ############
?
I've used the above to adjust the transformers in the new simulation schematic QMX_sim04.asc
In addition, I changed the turns ratio of T501 to 1:1 for the "Stock with L502 removed" case.
This increases the output to be slightly more than the stock case, but Vds-max across the BS170's is far less than the stock build.
There may be reasons not to do this, but it is worth investigating since the change in build proceedure would be trivial.
If we use this scheme, then everybody builds the QMX with the simpler 9v version of T501, but 12v builds short across L502.
There should also be a 0.1uF cap to ground at the T501 center tap if L502 gets shorted.
?
Here's some figures for peak voltages using QMX_sim04.asc:
Vds? ? ? Output 38.3? ?21.4 ? ? ? ? ? ?Stock QMX, has 3:2 turns ratio on T501 23.8 ? 17.5 ? ? ? ? ? ?With L502 removed, still has 3:2 turns ratio on T501 23.0? ? ?21.6 ? ? ? ? ? ?With L502 removed, now has 1:1 turns ratio on T501 ?
Note that I am showing 38.3 for Vds-max across the BS170's when simulating a stock QMX at 3.57 Mhz.
Chris reported in a forum post that he measured well over 60 volts using a scope on a stock QMX.
(Unfortunately his graph has been removed.)
?
With the 1:1 turns ratio at T501 and with L502 removed, I see a Vds-max of only 23v
and 21.6v across the load. The power into the 50 ohm resistor is? ?(21.6*0.707)**2/50 = 4.66W
My simulation for a stock QMX 12v builds shows slightly less, but real world tests can give over 5W.
?
That's all just in simulation, should be interesting to try it on some hardware.
And try it with a whole range of complex impedances at the antenna port, not just a 50 ohm dummy load.
Once we figure out operation at 3.57mhz we might find 4.0mhz significantly different.
And then we only have another dozen bands to figure out.
?
It's complicated!
?
Jerry, KE7ER |
||||||||||||
|
Hi Jerry,
?
Thank you for the detailed explanation.? I tried measuring the single windings on BN43-202 cores that I have for another project.? The measured values with an LC meter do not agree with your values or those from toroids.info.? I also get different values when all the windings are on the core.? I am at a loss in determining the correct inductances for the transformers.
?
The inductance value affects the power output compared with and without L502.? I suspect this is due to the flux storage in the higher-inductance cores.
?
I did not want to delay my response any longer, so I am replying before I have compiled all of the measurement data on the transformer winding.? Once that is done, I will post what I measured.
?
73
Evan
AC9TU |
||||||||||||
|
Evan,
?
I wasn't shooting for the correct inductance for each winding of T501.
All that matters is that the ratios of those inductances are correct.
?
> The inductance value affects the power output compared with and without L502.?
> I suspect this is due to the flux storage in the higher-inductance cores.
?
I might expect that to be an issue on 160m and 6m, but not something I'd expect?
on 80m.? What frequency are you testing at???
?
Here's data on the BN43-202:? ??
Says inductance of a winding in uH is:? 2200*(turns**2)/1000
But they don't specify what a turn is.
That dual aperture core works as two independent toroids, so they could figure?
one turn to be one pass through one hole, or one pass through both holes.
?
I got lost in the weeds a bit figuring out how to find the inductance of a center tap winding for LT-Spice
given the inductance of the secondary winding.? Best to just do what you are doing,
take the number of turns for that winding and the AL value for the core and compute the inductance.
You will arrive at the same inductance ratios as I did.?
The AL value is not critical for simulating a transformer in LT-Spice so long as the reactance
is an order of magnitude greater than any loads hanging off of it.
?
I did notice that the inductance ratios you had in the adjusted T501 of QMX_L502_sim01_w3_2.asc
were correct for?a 3:1 turns ratio transformer, not a 3:2 transformer:? ? 2*sqrt(4.2)/sqrt(1.9)=2.97
?
The reactance of T501 windings varies by considerably more than an order of magnitude?
when we try to use the same T501 from 1.8mhz to 54mhz.
What's more, things work best if it can pass the first few harmonics.
?
?
Anyways, the big news is that blowing the BS170's from over-voltage on a 12v build
may be cured by going to the same 1:1 T501 that is used on the 9v build, but removing L502.
This is shown in the "Stock without L502" simulation in QMX_sim04.asc that I posted yesterday.
I'll try that out in hardware when I get a chance, see what blows up.
?
There are lots of other things we could try, but that's the most painless solution
that I see possible.
?
A clean push-pull RF amp would be either VMCD or CMCD as demonstrated in the last two
simulations in QMX_sim04.asc.? What we have now is a VMCD getting force-fed a constant
current through L502, which will indeed cause Vds voltages to get out of hand.
But there are all sorts of compromises to be made in getting something to work at all
from 1.8mhz to 54mhz, and I can believe that the current design was found to work better
on the bench than a clean VMCD.? The current design may well turn out to work better than
what I am suggesting, will be interesting to see how this sorts out.
?
Jerry, KE7ER
?
?
On Fri, Feb 28, 2025 at 12:00 PM, Evan Hand wrote:
|
||||||||||||
|
Evan,
?
Here's a datasheet for the BN43-202:
?
They do seem to measure stuff with one turn going through both holes.
But these things are not designed to give accurate inductance values:
?
> Multi- aperture cores in 73 and 43 materials are controlled for impedance only.
> The 61 NiZn material is controlled for both impedance and AL value.
> The high frequency 67 material is controlled for AL value.
> Minimum impedance values are specified for the + marked frequencies.
> The minimum impedance is listed on our catalog drawing.
?
Jerry |
||||||||||||
|
Jerry,
?
Thank you for all of the information and for taking the time to explain it to me.? It prompted me to dig into binocular cores and how they work.? I found that one turn is through both apertures and 1/2 turns through just one.? Using the info from Toroid.info and plugging in the number of turns, I came up with values of :
For the BN43-202.
?
Plugging these values into an LTSpice simulator, I confirmed that three windings on the binocular core equal two 1.5 windings.? So, the 12-volt output transformer has two 4.95uH in series linked to a 19.8uH secondary.
?
When I think about the amount of core material involved, it does make sense.? Two turns are next to one core of material in one tunnel, and only one turn is in the other tunnel.? That is equal to 1.5 turns through both.? A half-turn is only possible with the binocular core.
?
This link also helped me understand the following:
?
73
Evan
AC9TU
? |
||||||||||||
|
Evan,
?
Looks good except this:? "a 19.8uH secondary"
?
The 19.8uH was had from the calculation for a 3 turn winding, which is the full primary.
The secondary is 2 turns, so is 8.8uH, not 19.8uH
?
The AL value doesn't matter much, let's arbitrarily assume it is 1uH per turn, not the toroids.info figure of 2.2uH per turn.
Now we can take the square root of the inductance figures you gave to find turns for a double check.
2*sqrt(4.95) / sqrt(8.8) = 1.5? ? ?use the number of windings for one side of the primary and multiply by two
sqrt(4.95) / sqrt(8.8) = 1.5? ? ? ? ?use the total number of windings for the primary
?
Still not quite right though.? I said previously that the two haves of a binocular core work like two different toroids.
This is evident in the tandem match SWR detector on the QMX, many web designs for a tandem match
show that as being built with two toroids instead of one binocular core.??
So that extra half turn on each half of the primary does not couple with all of the rest of the primary, only some of it.
This means that the figures of 4.95uH and 19.8uH are somewhat high, and the effective turns ratio
is somewhat less than 3:2 for the 12v version of T501.? Like the addition of L502, this makes the 12v final
much more difficult to analyze.? I'd much prefer a final with a 1:1 T501 and no L502.
?
Another good thing about a 1:1 transformer is that it is easy to make using 50 ohm coax.
We're using it as a balun, all we need is a toroid with some RG174 wrapped through it a few times.
Using transmission line makes the transformer behave much better at high frequencies,
RG174 works just as well when passed through a toroid as it does when lying flat on the desk.?
This is important for the QMX because the harmonics should pass through the transformer
and be reflected back from the LPF to keep the voltage waveforms? at the drains of this VMCD
amplifier (I'm assuming L502 is removed) sharp and square.
The 3:2 transformer we have now will likely have trouble doing this on the upper bands.
?
Jerry, KE7ER
?
?
?
On Sat, Mar 1, 2025 at 06:13 AM, Evan Hand wrote:
|
||||||||||||
|
Correction:
Where it says? "sqrt(4.95) / sqrt(8.8) = 1.5? ? ? ? ?use the total number of windings for the primary"
should say "sqrt(19.8) / sqrt(8.8) = 1.5? ? ? ? ?use the total number of windings for the primary"
?
Jerry, KE7ER
?
?
On Sat, Mar 1, 2025 at 09:35 AM, Jerry Gaffke wrote:
sqrt(4.95) / sqrt(8.8) = 1.5? ? ? ? ?use the total number of windings for the primary |
||||||||||||
|
Jerry,
?
I think you misinterpreted what the split transformer simulation did for me.? I wanted to see how LTSpice XVII calculated the transformer with a split winding.??
?
?
Has the same output across the single primary.
?
?
When I run the simulation, the voltage across the resistor is the same.
?
I used this to validate my understanding of the center-tapped transformer operation.
?
73
Evan
AC9TU
?
?
?
|
||||||||||||
|
Your simulation is as expected.
My primary objection was that you had stated?
"So, the 12-volt output transformer has two 4.95uH in series linked to a 19.8uH secondary." ?
To which I wrote:
"Looks good except this:? "a 19.8uH secondary"
The 19.8uH was had from the calculation for a 3 turn winding, which is the full primary.
The secondary is 2 turns, so is 8.8uH, not 19.8uH"
?
And that is what you simulated.
?
?
I still object to how the 3:2 transformer is wound on the binocular core.
Those extra half turns on each end of the primary do not get coupled to both sides of the binocular core.
?
And I object to stuffing L502 into a VMCD amp to get more power.
It also gives bigger voltage peaks on the drains when current from L502 has nowhere to go.
The correct way to get more power is to change the winding ratio to 1:1 on 12v builds.
On 9v builds, including L502 is not so much an issue because voltages are naturally lower.
?
My objections are "theoretical".
It may well turn out that the design as is performs better on the bench
than something built the way I think it should be.
I would have no objection to that.
?
Jerry, KE7ER
?
?
|
||||||||||||
|
Jerry,
You had said:
?
"Still not quite right though.? I said previously that the two haves of a binocular core work like two different toroids.
This is evident in the tandem match SWR detector on the QMX, many web designs for a tandem match
show that as being built with two toroids instead of one binocular core.??
So that extra half turn on each half of the primary does not couple with all of the rest of the primary, only some of it."
?
and:
?
"I still object to how the 3:2 transformer is wound on the binocular core.
Those extra half turns on each end of the primary do not get coupled to both sides of the binocular core."
?
...
?
OM, the idea that the two apertures of the binocular operate as independent toroids is correct for cases like the Stockton SWR bridge T507 in QMX. Turns of a given winding are confined to one aperture, completing by wrapping around its exterior. They are not wound around the central post. Any flux lines entering the central post from one cylinder "core" must also exit the central post to return to the starting cylinder, leaving no flux to couple into the other cylinder. Two such windings, each on their own cylinder are effectively isolated from each other.
?
This is not the case for transformers where all the windings wrap around the central post, as in the QMX PA transformer T501. Here the the flux from each segment of every winding links to all other windings, Half turns count as such. It is a very different situation from the SWR bridge application.
?
Keep up the good work experimenting and simulating, Jerry!
Regards, JZ KJ4A
?
?
|
||||||||||||
|
John,
?
You're right, thanks for setting me straight.
?
I was triggered by discussions of how several published designs of linear amps
30 yrs ago would not work well because they used two side-by-side choke cores.
Primary was a single turn through both cores with a center tap at the far end.
It took me a couple minutes and a sip of coffee this morning to conclude that
the two cores are a totally different case, the magnetic fields are separate.
?
It's good having you and Allison contributing here to keep this stuff on the right track.
?
Thanks,
Jerry, KE7ER? ? Retired, digital design
?
?
?
On Sun, Mar 2, 2025 at 05:23 AM, John Z wrote:
This is not the case for transformers where all the windings wrap around the central post, as in the QMX PA transformer T501. Here the the flux from each segment of every winding links to all other windings, Half turns count as such. It is a very different situation from the SWR bridge application. |
to navigate to use esc to dismiss