New 1-FET, 35-Watt Class E "starter" rig

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New 1-FET, 35-Watt Class E "starter" rig

Postby AB2EZ » Fri Jul 15, 2005 1:56 pm

On my web site, I have posted pictures (and some preliminary descriptive material) of my new 1-FET 35-watt Class E transmitter.

http://mysite.verizon.net/sdp2/id11.html

It uses Steve (WA1QIX)'s new prototype Class H modulator board, and a single rail (2 MOSFETs) high voltage / high current Class H modulator output stage (mounted separately on the heat sink).

It is designed to be very easy to build and operate... and more details will be posted soon.

With a single QFET, the problems associated with stabilizing the behavior of a multi-FET class E rig are avoided.

The main power supply is located on the underside of the 12" x 8" x 3" chassis. The "wall plug" power consumption of the rig is about 90 watts. It requires 1 watt of unmodulated RF drive... and I am using it in conjunction with an FT-817 QRP transceiver that easily puts out 1 watt of continuous carrier for this purpose.

The rig occupies a volume of 12" (wide) x 10" (deep), x 8" (high)... including the heat sink. It weights around 20 pounds, and the cost of the components (new) is ~ $450.00 + shipping and tax. Many/most of the expensive components can be reused to build a much larger (4 or 6 FET) rig if one uses this "starter rig" to get some practice with building and operating a class E transmitter.
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Postby JT - W6FO » Fri Jul 15, 2005 3:08 pm

Very nice Stu, thanks for doing this for newbies like me.

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Postby blaine » Sat Jul 16, 2005 12:12 pm

that looks like a large series cap for 35 watts.
could that be shrunk down for a smaller form factor?
looks like the same one on my 10 fet'r
very cool design though :)
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Re: Output tuning capacitor... could it be smaller?

Postby AB2EZ » Sat Jul 16, 2005 1:39 pm

Blaine

Good question! In general, I think the answer is yes! The output air variable tuning capacitor that I used has a larger physical size than is necessary for this power level in this application.

But... please keep in mind:

a. I'm running the output tuning circuit at a Q of 6 (rather than the Q of 3 used in Steve's original 6-FET design) in order to do a better job of suppressing the 2nd harmonic. This is an unbalanced (single FET) design... and, as you know, there is a very large amount of second harmonic in the drain voltage waveform. Although my antenna tuner and my antenna both do a good job of reducing the level of second harmonic that ultimately reaches my antenna / is radiated... I prefer to keep the second harmonic level low at the output of the rig itself. As you know, with a Q of 6, the peak voltage across the capacitor is twice as large as it would be with a Q of 3.


b. A quick calculation of the peak (not rms) voltage... at 150% positive peak modulation... across the tuning capacitor is:

35 (watts) => ~60 volts (peak) across 50 ohms

Multiply by Q=6 to get the peak voltage (at resonance) across the capacitor => 6 x 60 volts = 360 volts

Multiply by 2.5 (1 + 1.5) to get the peak voltage across the capacitor when the modulation peaks up to 150% => 2.5 x 360 volts = 900 volts

So... to be on the safe side, a variable capacitor with around an 1800 volt rating would be fine for this rig. You could get away with a little less, if you don't mind an occasional arc - over.

c. I wanted to select parts that could be reused in a higher power rig... on the assumption that this would be a starter rig to help people get their "feet wet" with class E. Not all of the parts can be reused in a larger class E rig, but most of them can.

Thus I picked a variable tuning capacitor with a 4500 volt rating... which would be good for a rig with 1500 watts of peak output, while retaining the Q of 6 for the output tuning circuit, and retaining a factor of ~2 in safety margin against arc-over.

I also note that the loading capacitor that I selected is, in my opinion, marginally adequate for this application... at least, it has a lot less spare voltage-handling capability than the tuning capacitor has. So... I don't think I would use that loading capacitor for a larger class E rig (although some people have).

Best regards
Stu
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Postby ke1gf » Wed Jul 27, 2005 1:49 pm

Stu, I don't know how you got a Q of 3 for the series resonant network in QIX's 6 FET rig. Here's the calculations that I've been using for a few years now...

From Krauss, Bostian and Raab, Solid State Radio Engineering
page 452.

----------------------------------------------------------------------------

Peak Anode Current = 2.86 x Idc

so 45V / 8A = 5.625 Ohms (QIX's optimal tune with 1000pF shunt)

5.625 Ohms x .577 ~= 3.24 (RF Ohms)

Steve uses a 1:2 transformer so 3.24 x 4 ~= 12.9825 (RF Ohms)

---------------------------------------------------------------------------

All the rest is straight AC theory

V = L x ( di / dt )

a bit of algebra

L = V x ( dt / di )

at 4Mhz, t = 250nS

with perfect 50% switch DC

dt = 125nS

di = ( 2.86 x 8A ) ~= 22.88 A (peak)

V = 45V (Across the flyback inductance presented by the output network)

L = 45V x ( 125nS / 22.88A ) ~= .2458uH (on the transistor side of the transformer )

.2458uH (on the transistor side) x 4 ~= .9834uH ( on the output side )

QIX's series inductor, 17 Turns, 2" Dia, 4" Long

((D^2)x(N^2))/((18xD)+(40xL)) = uH

((2^2)x(17^2))/((18x2)+(40x4)) ~= 5.898uH (for the series inductor)

Now, you have to take into account that some of the series inductor is used as part of the L network...

so... real[ 1 / ( ( 1 / 50 ) + ( j x 2 x PI x 4000000Hz x 1343.75pF ) ) ] ~= 12.982 (RF Ohms)

and... imag[ 1 / ( 1 / 50 ) + ( j x 2 x PI x 4000000Hz x 1343.75pF ) ) ] ~=
-j21.92 (RF Ohms)

and thus to cancel the capacitive reactane we need to subtract from the series inductance

Xl = ( 2 x PI x f x L )

algebra

Xl / ( 2 x PI x f ) = L

21.92 / ( 2 x PI x 4000000Hz ) ~= 0.872uH

so...

5.898uH ( the series inductor) - 0.872uH (L network) - 0.9834uH (flyback inductance) ~= 4.0426uH (remaining for the series resonant portion)

Xl = ( 2 x PI x 4000000 x 4.0426uH ) ~= j101.6Ohms

Q = 101.6Ohms / 12.9825Ohms ~= 7.83

So approx a Q of 8...

Stu, how did you get 3 for QIX's 6 FET? Maybe there's something that you'd like to share with us...

These calculations assume 50% DC, it's really more like 30-> 40 On with sinewave drive.

Peace,
-Bill 'GF
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How I calculated the Q

Postby AB2EZ » Wed Jul 27, 2005 5:06 pm

Bill

Perhaps what you are thinking of, when you think about the Q of the output circuit, is something entirely different from what I am thinking of. Therefore, perhaps that is why you come up with a different result.

I am thinking only about how good a job the tuned output circuit does in blocking the 2nd harmonic of the output voltage waveform (i.e., the voltage waveform that appears across the secondary of the output transformer) from reaching the 50 ohm load that I am assuming (for this calculation) is connected at the antenna port.

The output circuit (as I define it for my purpose) consists of: the inductor, the tuning capacitor and the load represented by the antenna ... all in series. For this purpose, I am ignoring the loading capacitor. The load represented by the antenna (for the purpose of this calculation) is 50 ohms. The inductor and the capacitor in Steve's design each have a reactance (one positive and one negative) of 150 ohms at resonance. Thus, as I am defining it here, for this purpose... the Q of the output circuit is 3.

Here is some text from Steve's Web site description of the operation of the 6-FET Class E 75 meter transmitter

"The approximate reactance of the input inductor and the series capacitor are 150 ohms, or about 11 or 12 times the 12.8 ohm load. Now, getting a value for the inductor: Xl = 6.28 x F x L therefore L = Xl/(6.28 x F). Putting in the values we obtained, and assuming a frequency of 3.9mHz (75 meters): L = 150 / (6.28 * 3,900,000) which works out to 6.125uH.
The series capacitor value may be figured in the same way: Xc = 1/(6.28 x F x C), or C = 1/(6.28 * F * Xc). Using our 150 ohm reactance: C = 1/(6.28 x 3,900,000 x 150) = 272pF. "

While Steve talks about a 12.8 ohm load (in the above)... he is referring to the effective output impedance of the 6-FET amplifier ... as it appears at the output of the 1:2 (step up) output transformer. When I speak of "the Q" of the output circuit... what I have in mind is the following question: how much less current will flow through the tuned output circuit (and into the 50 ohm load that is presented by the antenna) at the second harmonic frequency v. the fundamental frequency... for a given amount of voltage across the input to that tuned output circuit. At the fundamental frequency (where the circuit is tuned to resonance) the net load is ~50 ohms. At the second harmonic frequency the load is ~ j150 ohms x (2 - 0.5) + 50 ohms = [jQ (1.5) + 1][50 ohms]... because the impedance of the inductor is 3x the impedance of the antenna load (Q=3) at the fundamental frequency... and (of course) as we move up to the second harmonic frequency, the impedance of the inductor is twice as much as it is at the fundamental frequency. Likewise, the impedance of the capacitor drops from 150 ohms to 75 ohms as we move to twice the freqeuncy. All of this, is based on the assumption that the rig is looking into a constant 50 ohm antenna system. Naturally, when one takes into account the characteristics of the antenna/antenna tuner combination... the whole situation gets more complicated to analyze.

Based on the above (and again, using the simple case of a 50 ohm antenna load that is independent of frequency)...

The tuned circuit attenuates the 2nd harmonic (relative to the fundamental) of the voltage on the secondary of the output transformer by

10 log [(1+ j1.5Q)(1-j1.5Q] dB = 10 log [ 1 + 2.25Q**2]

where Q**2 means QxQ

By using Q =6 instead of Q =3, I increase the attenuation of the 2nd harmonic (relative to the fundamental) from 13.3 dB to 19.3 dB... i.e., a 6 dB increase in second harmonic suppression.

When I look at the actual r.f. output spectrum of my new 1-FET transmitter on a scope that can perform math functions (like Fourier transform), driving a 50 ohm dummy load...
it is 28 DB down from the fundamental.

When I look at the "class E" waveform on the drain of the FET... using the same scope... the secoond harmonic as about 5.5 dB down from the fundamental.

When I look at the class E waveform on the other side (i.e., the secondary side) of the output transformer... the second harmonic is about 5 dB down from the fundamental. This means the the transformer is adding just a bit of distortion... probably due to core saturation effects. As you would expect, the waveform on the secondary of the output transformer is just about twice as big as the waveform on the primary... and looks nearly identical.

Thus, the tuned circuit appears to be producing about 23 dB of selectivity (28 - 5).

Going back to the fomula above... the measurement indicates that the Q of the output tuned circuit is about 9.3. This is (of course) larger than the value of 6 that I calculate based on a 50 ohm load... but the actual load is less than 50 ohms because of the loading capacitor that is in parallel with it. The loading capacitance I am using is about 850 pF... which corresponds to -j48 ohms... so the effect of this on the behavior of the output tuned circuit is to increase the selectivity.

Stu
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On the subject of "the Q": Bill... how about this?

Postby AB2EZ » Wed Jul 27, 2005 5:49 pm

Bill

Perhaps I can bring my definition of the Q and yours into sync...

In my definition, I divide the impedance of the inductor (or the capacitor) at resonance by the nominal 50 ohm antenna load.

In your definition, you (essentially) divide the impedance of the inductor (or the capacitor) at resonance by the effective output resistance of the r.f. output stage (on the secondary side of the output transformer).

If there were no loading capacitor, (which is not the case, of course), and the antenna load were 50 ohms (which is no problem using a dummy load) my definition of Q (150 ohms / 50 ohms) would be a good indicator of the ability of the output tuned circuit to attenuate harmonics.

If there were no loading capacitor (which is not the case, of course), and the antenna load were matched to the effective output resistance of the r.f. output stage (on the secondary side of the output transformer)... which normally is the way to get the maximum power output from a (resistive) signal source ... then I would be using your formula for Q.

In this case, we use the loading capacitor to get a compromise match... that balances the objective of operating in class E (high efficiency) mode against the objective of getting maximum output power.

Therefore, the effective value of Q, for the purpose of computing the selectivity of the output stage, is somewhere between the two values (above).

Using my formula... I get Q~ 300/50 =6 for my 1-FET rig

Using your formula (and the fact that I am using an inductor with a reactance of ~j300 ohms at 3.885 MHz, and only 1 FET)... I obtain Q~ 300/(12 x 6*) ~ 4 (!!!)

*I'm only drawing 1/6 the current that a 6-FET rig would draw

The lesson I've now learned... independent of the issue of how to define Q... is that I need to experiment with a smaller step-up ratio in the 1-FET rig. I'm going to try 1:1 (and maybe 2:3) to see if that has any favorable impact on the behavior of the rig... i.e., either on the overall efficiency or the harmonic attenuation.

Stu
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Postby Tom WA3KLR » Wed Jul 27, 2005 8:52 pm

O.k. I haven't built any Class E transmitter yet as you fellows have. But with all of the simulations I have done, my opinion is that the Q should be kept at 5 or less and then a multi-section low-pass filter should be added to the output of the transmitter. From what I gather now, no one is adding the additional low-pass filters! The additional low-pass filters were always standard practice with the Class C transmitters. (Why the reluctance?) It doesn't seem to be a good practice to me to use a high Q PA in pursuit of suppressing the second harmonic. There is too much additional inductor loss and the tuning becomes much more critical. One advantage of the high Q design with low values of 'R' is the lowered variable capacitor value however.

My simulations of the single-ended PA with a Q of 5 had the second harmonic down about 15 dB as I recall. So you do need additional suppresssion of the second harmonic (and higher harmonics too). I use resonant dipoles here on 75 and 40 meters and no antenna tuners, so I do plan to make low-pass filters for my Class E PAs.

The Q I am referring to is the Q value as plugged in to the formulas of Nathan Sokal's article "Class-E RF Power Amplifiers" published in the Jan./Feb. 2001 issue of QEX.

As I have done in the past here on the Class E forum, I again offer my Excel spreadsheet which incorporates Nathan's formulas for Class E design (and more). To get the latest spreadsheet, send me a PM with your e-mail address stated.

A quick shot at duplicating the QIX circuit situation as stated by Bill with my spreadsheet, I came up with a design Q of 9 rather than 3.
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Postby ke1gf » Wed Jul 27, 2005 10:49 pm

Stu, I see what you are saying. I just have never seen Q discussed in this way before with a class-E amp.

Tom, with SEPP class-E you can see -65dBc or better on the second harmonic supression, I've even heard people quote -100dBc, but I'll believe it when I see it. With only an L network as a filter/impedence match.

When I used to run my single-ended E rig, I was heard loud and clear on 40M.

Peace,
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I now agree with Bill and Tom

Postby AB2EZ » Thu Jul 28, 2005 6:35 am

I now agree with Bill and Tom that when one takes into account the effect of the loading capacitor on Steve's 6-FET design ... the effective resistive load associated with the output tuning circuit is much closer to the ~12 ohm output impedance of the FET (as measured across the secondary of the output transformer) than it is to the nominal 50 ohm impedance of the antenna (or dummy load). Thus the Q of the output circuit is closer to 150/12 than it is to 150/50.

In my 1-FET "starter rig" I have now tried a 1:1 output transformer (instead of 1:2) in order to make the effective output impedance at the the secondary of the output transformer ~ 12 ohms (instead of ~50 ohms). In concert with that, I increased the value of the loading capacitor.

The measured result was a reduction in the 2nd harmonic (into a 50 ohm dummy load) from about -28 dBc to about -45dBc, with essentially no change in output power [because the loading capacitor allows me to match the antenna to the new (lower) value of effective output impedance of the amplifier]. Part of that improvement has to do with a resulting further increase in Q ... and the rest has to do with the details of how the r.f. stage is interacting with the load it is looking into.

I agree with Tom and Bill that with a single-ended rig, some additional low pass filtering is advisable. In my case, it is provided by the selectivity of the antenna tuner I am using, and the fact that my dipole has a high SWR at the second harmonic frequency.

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Postby Tom WA3KLR » Thu Jul 28, 2005 8:04 am

Since I rambled my post off of the top of my head last night, I recall now that the 15 dB down 2nd harmonic is with a load resistor equal to the design R value; in other words - just the series tank network connected to the load resistor.

When the shunt "loading" capacitor is added to the output network, then the output network becomes 2 networks; the series tank network plus an additional L-C matching network which steps up from the 3 - 12 Ohms value to 50 Ohms. Part of the series tank inductor is the L for this later L-C step-up impedance matching network. In this case, due to this second matching network which is in a low-pass configuration, additional harmonic filtering is obtained. My simulations of the output circuit with the loading capacitor incorporated gives about 30 dB attenuation of the second harmonic. So here again 25 - 50 dB of additional attenuation of the second harmonic is desired.

I wonder how well the push-pull configuration performs for the 3rd, 4th, 5th, 6th, and 7th hamonics, compared to the single-ended/LPF approach?
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Postby frank carcia » Thu Jul 28, 2005 10:08 am

Bill,
People who claim 100 dB of second harmonic suppression may be using a rice box S meter as a calibration standard. I'm amazed when I hear performance claims that only NBS standards can measure. gfz
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Postby ke1gf » Thu Jul 28, 2005 1:25 pm

Tom WA3KLR wrote:Since I rambled my post off of the top of my head last night, I recall now that the 15 dB down 2nd harmonic is with a load resistor equal to the design R value; in other words - just the series tank network connected to the load resistor.

When the shunt "loading" capacitor is added to the output network, then the output network becomes 2 networks; the series tank network plus an additional L-C matching network which steps up from the 3 - 12 Ohms value to 50 Ohms. Part of the series tank inductor is the L for this later L-C step-up impedance matching network. In this case, due to this second matching network which is in a low-pass configuration, additional harmonic filtering is obtained. My simulations of the output circuit with the loading capacitor incorporated gives about 30 dB attenuation of the second harmonic. So here again 25 - 50 dB of additional attenuation of the second harmonic is desired.

I wonder how well the push-pull configuration performs for the 3rd, 4th, 5th, 6th, and 7th hamonics, compared to the single-ended/LPF approach?


Tom, I don't actually know how good the odd order harmonic supression is(not as good as the even obviously), but I can tell you that you will be compliant.

I've made a nice SWCAD 100W PEP SEPP Class-E with IRF610's and totem-pole drive model (for 75M), if you'd like to tinker I'd be willing to share it with you.

Gonna submit my publication for IQP tomorrow. Man they weren't kidding that AI is a gold mine in the computer industry... Did some work with FIMLOF (Full Information Maximal Likelihood via Optimal Filtering), the stuff's worth it's weight in gold for strategic planning software.

Frank you work in MIL/Aero industry, ever had a taste of it?

FIMLOF would be kickass in SPICE, I'd just guess on the shunt cap for a Class-E and it would optimize for me...

Peace,
-Bill 'GF
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Postby ke1gf » Thu Jul 28, 2005 3:07 pm

The lesson I've now learned... independent of the issue of how to define Q... is that I need to experiment with a smaller step-up ratio in the 1-FET rig. I'm going to try 1:1 (and maybe 2:3) to see if that has any favorable impact on the behavior of the rig... i.e., either on the overall efficiency or the harmonic attenuation.


Stu, from what I've learned from experience is that BB transformers do not like reactive loads and will do funky stuff (frank brow-beated it into me). So I wouldn't go too crasy with turns ratios, just remember that the load is reactive on a class-E amp... Also I noticed that your series inductor is huge and edge-wound with insulated wire, have you checked it for paracitic resonances? (tina, and QIX are running copper tubing, I'm running #4 AWG), all with wide turns spacing, is your series coil getting smoking hot? Just remember that the flyback is ringing at probably around 180% of the fundamental operating frequency. I'll blaber all night and mine is cold to the touch... Also there is the issue with paracitics on the drive waveform from the reverse transfer capacitance from harmonic currents on the drain waveform, my rig is borderline with this problem but I have no time to fiddle and fart around with the Q of the output to solve this problem, it works but it's not perfect. I just throw a lossy circut at the problem to glue the driver to the gates of the mosfets to solve this at the current moment... Some of this stuff is PFM in my book. With a lot of my experiementation last summer into totem-pole drivers my next rig will have completely elimiated the TP and the BB transformers, each gate with get it's own driver, slamming the gates on and off with trapazoid drive at 50% DC, I'm working on it, but my pockets are turned out for capital, I couldn't take on a job this summer because of my IQP.

Peace,
-Bill 'GF
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Postby frank carcia » Thu Jul 28, 2005 3:17 pm

My efficiency improved about 4% improved by going from 1:4 to 1:2 turns ratio. Also reducing the Q from 9 to 7 helped. Circulating current
is prety high 1/8 tubing pretty hot at 1 KW out. I think I was at 88%
the last time I checked. It still wants lower Q. I'm at around 400 pf series and 1800 load. Broadband transformers get interesting at higher turns ratios. It takes more than a couple wires to get it to work properly.
But then I'm busy banging nails myself. Hopefully this winter there will be time to get serious again. fc
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