combining multiple solid state RF decks

[kb3ouk]

Is it possible to use a coupler similiar to this: http://w1vd.com/137-500KWTX.html (halfway down the page), modified to work at HF, and have two identical RF decks feeding into it, but have both decks fed from the same RF source, power supply, and modulator? My thinking is that as long as both amplifiers are in phase and the modulation on both is identical, then the two should combine with no problem. So it would be one driver, PS, and modulator driving both final decks, then the two decks being combined in the coupler.

[steve_qix]

In general, it is easier (much eaiser) to combine solid state stages if it is done ahead of any tuned circuits or filtering.  There is quite a bit of information about this on the class E web site http://www.classeradio.com.   Balanced class E amplifiers produce near sinusoidal outputs ahead of the output network, so they are very easy to combine. 

You then use a single output network after everything is combined.  I have successfully combined 12 stages together to make one transmitter.  There are plans showing transmitters that combine 4  6 MOSFET modules to make a 24 MOSFET RF amplifier.  There are a number of these 24 MOSFET amplifiers on the air today.

If you wish to stay in the class D domain, Jay has class D amplifiers for 160 and 75 meters documented on his site.  The 500 kHz transmitter design will not directly translate to 75 meters.  You should use something specific for the band in question, as there are a number of factors that need to be considered when operating relatively large MOSFETs with high element capacitances at high frequencies.

I have combined RF amplifier that use identical but separate modulators and power supplies, operating from a common pulse width source and a common RF source.  Each RF module actually had its own PWM filter, so there were 12 small PWM filters.  The modulators were relatively simple to keep balanced and equal.  Just make everything the same between modules.  The biggest challenge is keeping the RF phase in line.  A difference of even a few degrees between RF amplifiers will cause an imbalance.  The way to achieve good RF phase balance is to use absolutely identical lead lengths, coax lengths (when delivering driving power), components and physical layouts.

I have done combining with and without power combiners, and in fact found a combination of the 2 methods to be the best when combining 6 or more modules.  The power combiners I designed do not use balancing resistors of any kind.  If a module were to fail or otherwise not deliver power, it was not necessary to maintain the combiner impedance, as the transmitter would shut down anyway.  However, for 6 modules and less, it works perfectly fine to put the secondaries of each small RF stage in series to simply add the outputs together, and then run this into a single output network.

I'm sure you could also combine the after-the-filter outputs using a power combiner, as long as you could keep the RF phase shifts ahead of the combiner in line.  Output filters introduce phase shift, of course.

Anyway, it can definitely be done, but I would strongly suggest starting with a basic RF amplifier design that is made to work on the frequency or band(s) in question.

Regards,

Steve

[kb3ouk]

Then I'll focus on the other idea I had, use bigger FETs.

[steve_qix]

Then I'll focus on the other idea I had, use bigger FETs.

Combining definitely works and it's not that difficult - just requires proper planning. 

Bigger MOSFETs mean bigger capacitances.  It depends on the class of operation, but for any switching mode (class D, E, F, etc.), the gate capacitance is the limiting factor.

If the MOSFET has a very high gate C, the internal resistance/inductance withing the MOSFET itself will limit the drive unless the MOSFET is constructed with a metalized gate structure.  Even with this, the package inductance must also be considered.

MOSFETs with metalized gate structures are more expensive for the same size MOSFET, but is generally the best solution when you're up at 14mHz and higher.

I use combining because it gives the best price/performance when considering drivers, MOSFETs and other components.  It is also very reliable and easier to build because there are no very high current paths anywhere in the system, and the impedances are also quite reasonable (not too low), so the layout is much less critical.

Regards,

Steve

[kb3ouk]

Well, I actually went with a third idea. I was looking at a schematic for a 75 meter Class D H-bridge amplifier that was posted on here a while ago, and actually realized that without changing anything, you could get more power from it simply by increasing the power supply voltage. The carrier voltage listed on the schematic was 160v, with the peak being 320v. In an H-bridge, the voltage rating of the fets needs to be 1.5 times the peak voltage. But, the actual voltage that an individual device sees is half the power supply voltage, so in that transmitter, each device is seeing 80v. You could actually double the power supply voltage, to 320v, which would then put 160v on each device (320v at peak). That doesn't leave much of a safety factor, so running something like 200-250v might be a better idea.

[steve_qix]

I believe each device in an H bridge will see the full supplied voltage. 

When the "bottom" MOSFET of a totum pole pair is turned on, the full supply voltage will appear across the "top" MOSFET of that pair.  The reverse is true when the "top" MOSFET is turned on. 

An H bridge is two of these driven out of phase, and the output taken from a transformer between the two totum pole pairs' outputs.

If you can avoid current overlap in the totum pole sections, an H bridge works very well.  The overlap gets worse with large MOSFETs and high frequencies.

Regards,

Steve

[kb3ouk]

Every post I've saw on here concerning an H bridge says otherwise:
http://amfone.net/Amforum/index.php?topic=31087.msg242583#msg242583
http://amfone.net/Amforum/index.php?topic=24512.msg182862#msg182862

The actual schematic of the transmitter I am talking about:
amfone.net/Amforum/index.php?action=dlattach;topic=30798.0;attach=31360

[KF1Z]

I stand corrected.

[kb3ouk]

Well of course, yes the full voltage is gonna appear on the terminals of the top two FETs, but the inside of the FET (even for the top two) is only seeing half the voltage, right?

[steve_qix]

Well, it is easiest to look at the schematic.  For the moment, take away all modulation, and apply a fixed DC voltage to the input - the drains of Q1 and Q3.

Ok, now apply some RF drive to the gates.  When the gate of Q1 goes high, Q1 turns full on, putting the full DC voltage across Q2.  Q4 also turns on at this time, putting the full supply voltage across Q3.

When the cycle reverses, the voltages reverse as well.  So, the full supply voltage is across the MOSFETs.

Regards,

Steve

[kb3ouk]

Ok, so when the lower FETs are on the upper ones get the full voltage across them. But, what is the actual voltage on the ones doing work (the on FETs)? Which brings up a really good  question: If the FETs are seeing the full power supply voltage, then how do the FETs in those broadcast rigs survive 100% (or more) modulation? I'll use the Nautel rig mentioned in the one post I linked to as an example. The devices in that transmitter are APT5010JFLLs, which are rated at 500 volts. The power supply voltage is 400 volts. When that rig peaks out at 100% modulation, the voltage on those FETs is going to be 800 volts. What's keeping the things from blowing up? And one other point I'd like to make, in an H-bridge, Q1 and Q4 would be on at the same time, not Q2 and Q4. If Q2 and Q4 were both on at the same time, how's the voltage getting to them? Think of the FETs as switches. Q1 and Q3 are both off, which means they aren't passing any voltage. Well then if Q2 and Q4 are supposed to be on when Q1 and Q3 are off, how are they getting voltage? They're not. Q1 and Q4 turn on at the same time, the supply voltage goes through Q1 and the horizontal leg before it gets to Q4. That puts those two in series, which is how the voltage across the FETs is half the supply voltage. On the next cycle, Q3 and Q2 turn on. Look closely at how the input transformers are arranged. If Q2 and Q4 were coming on at the same time, then they should be in phase, right? No, Q1 and Q4 are in phase with each other, and Q2 and Q3 are in phase with each other, but they are out of phase with Q1 and Q4. Yes, I will agree with you that the top FET is seeing the full voltage, but in order for the circuit to work, the bottom one has to see some voltage too, which is why the voltage consumed adds up to the full supply voltage. It's like dealing with tube filaments. If you are using a 12v supply and try to run a pair of 12 volt tubes off of it, more than likely the second tube will either not light or not be very bright since the filament of the first tube is taking all of the voltage. If two 6 volt tubes are placed in series across there, then that will work, since the first tube takes its 6 volts, the the second tube only sees 6 volts at its filament pins, where the first tube did have 12 volts on one pin, but its filament only comsumed 6 volts and had 6 volts left over to pass on to the other tube.

[steve_qix]

Ok, so when the lower FETs are on the upper ones get the full voltage across them. But, what is the actual voltage on the ones doing work (the on FETs)?

0 Volts (in theory - probably some small voltage - maybe a volt or two.  They are full on.

Which brings up a really good  question: If the FETs are seeing the full power supply voltage, then how do the FETs in those broadcast rigs survive 100% (or more) modulation? I'll use the Nautel rig mentioned in the one post I linked to as an example. The devices in that transmitter are APT5010JFLLs, which are rated at 500 volts. The power supply voltage is 400 volts. When that rig peaks out at 100% modulation, the voltage on those FETs is going to be 800 volts. What's keeping the things from blowing up?

These rigs use pulse width modulation if they have modulators, and pulse width modulators are series modulators.  The voltage does not ever rise above the total power supply voltage in such a system.

And one other point I'd like to make, in an H-bridge, Q1 and Q4 would be on at the same time, not Q2 and Q4.

That is correct.  Q1 and Q4, which is what I said in my post.  The mention of Q2 in my post refers to the fact the full voltage appears across the Q2 device.  It is not turned on when Q1/Q4 are on.

If Q2 and Q4 were both on at the same time, how's the voltage getting to them? Think of the FETs as switches. Q1 and Q3 are both off, which means they aren't passing any voltage. Well then if Q2 and Q4 are supposed to be on when Q1 and Q3 are off, how are they getting voltage? They're not. Q1 and Q4 turn on at the same time, the supply voltage goes through Q1 and the horizontal leg before it gets to Q4. That puts those two in series, which is how the voltage across the FETs is half the supply voltage.

That is not correct about half the voltage   Q1 and Q4 are on.  Q2 and Q3 are off.  There is (in theory) 0 (very little - a few volts or less) voltage drop across Q1 and Q4 when they are on.  The power supply voltage is conducted through Q1, through the load, and through Q4 which is also turned on and has 0 voltage drop (ok, a very small drop - just single volts).  The full voltage being supplied to the H bridge appears across the load.  It also appears, it turns out, across the MOSFETs that are currently turned off.

The opposite occurs on the next half cycle.  Q2 and Q3 are on; Q1 and Q4 are off.  The voltage across the load REVERSES, so you get a full AC voltage across the load, in theory, equal to minus the power supply voltage on one half cycle, and then plus the power supply voltage on the next half cycle.

On the next cycle, Q3 and Q2 turn on. Look closely at how the input transformers are arranged. If Q2 and Q4 were coming on at the same time, then they should be in phase, right? No, Q1 and Q4 are in phase with each other, and Q2 and Q3 are in phase with each other, but they are out of phase with Q1 and Q4.

 Yes, I will agree with you that the top FET is seeing the full voltage, but in order for the circuit to work, the bottom one has to see some voltage too, which is why the voltage consumed adds up to the full supply voltage.

The turned-on FETs have a VERY small voltage across them (single volts, if even that).  Every effort is made to reduce the turn-on voltage to 0, as the more voltage drop across the turned-on FETs, the lower the efficiency of the RF amplifier.  The full input voltage appears across the turned off MOSFETs.  At no time are the MOSFETs sharing the voltage, and there is no voltage division between the MOSFETs, save a very very small voltage caused by the R D-S on of the MOSFET.

It's like dealing with tube filaments. If you are using a 12v supply and try to run a pair of 12 volt tubes off of it, more than likely the second tube will either not light or not be very bright since the filament of the first tube is taking all of the voltage. If two 6 volt tubes are placed in series across there, then that will work, since the first tube takes its 6 volts, the the second tube only sees 6 volts at its filament pins, where the first tube did have 12 volts on one pin, but its filament only comsumed 6 volts and had 6 volts left over to pass on to the other tube.

The thing is, the MOSFET circuit shown is not a lossy or resistive circuit.  It is a switching circuit, and there is no voltage division as occurs in a resistive (analog) circuit.

The distinct advantage of an H bridge is that you can get more power for a given MOSFET because you can run more voltage.  The hard thing with an H bridge is to prevent the MOSFETs connected in the totum-poles (the ones in series) from turning on at the same time.  MOSFETs have a rather long turn-off delay, and the turn-off delay is longer than the turn-on delay by a significant amount.  So, some pulse shaping is generally required to prevent overlap.  The other issue is the body diode.  The body diode recovery time is very long, so if it comes into conduction, say, from a mismatched load or inductive load, there can be trouble.

These issues can be mitigated, but some amount of planning and thought is required in the design to prevent such occurrences.  A properly designed solid state transmitter can, assuming no capacitors or mechanical components fail, last almost forever - certainly forever as measured in a human lifetime.

I have designed a number of H bridge output solid state RF amplifiers in the past, and will probably use this design again at some point in the future.

[kb3ouk]

Then that must be where everyone is getting that they only see half the voltage, becuase the modulator would be passing 200 volts at carrier, then at peaks be passing the full power supply voltage.

[steve_qix]

Then that must be where everyone is getting that they only see half the voltage, becuase the modulator would be passing 200 volts at carrier, then at peaks be passing the full power supply voltage.

You know, it may be - or it is (more likely) just a mistake.

Everything in solid state has to be built to the "edge" conditions - those absolutes of current and voltage that occur quickly and seldomly, but do nevertheless take place.  One such event can easily kill the transmitter and ruin your day! 

So, I know when I design something like this, everything gets 100% headroom over and above the absolute maximums that I calculate.  If I were designing this RF amplifier, and I were going to use a 400 volt power supply, then a minimum of an 800V MOSFET would be required to give 100% headroom in the voltage domain.

That's just me.  I don't like things to break under any conditions of nature, chance, or stupidity (like running full power into no load which I do more often than I'd like to admit!).

Regards,

Steve

[kb3ouk]

And that's what makes figuring this stuff out confusing. Nautel is using 500v 41A FETs, 20 modules (each of which is a seperate H bridge), so there are 80 devices total in the transmitter. Running at 50kw, each one of those modules would be putting out 2.5kw each. They claim 90% efficiency, so each module is drawing 2.8kw at carrier. With a 400v PS, that means each module is drawing 14A at 200v. On peaks, that would be drawing 28A. Now, in this case, would the current be divided between the two FETs? So on peaks each turned on FET would be handling 14A? that's only a third of what they are rated at. Which going by your safety factors for class E, is acceptable, but barely.

[steve_qix]

Yes and no    The world is REALLY different down below 1.6mHz.  Switching waveforms are easy to keep clean, capacitances are reasonable for the frequency, and stray inductances don't cause a lot of problems.

Start getting up to 3.5mHz and higher, with reasonably priced MOSFETs, and now things that worked perfectly at 1mHz won't work at all !!  Waveform ringing, parasitics, overlapping turn-ons, switching times, energy to charge and discharge device capacitances, etc. are normal problems that must be solved or at least well understood.

When I started doing solid state transmitter designs (30 years ago!!!!), I started out with class D and H bridge designs.  I had a 200 watt carrier output transmitter (on 75 meters) using 48 IRF610s in a totem pole arrangement and a pulse width modulator using IRF620s.  Somewhere, parts of that transmitter are kicking around.  It's funny to look at!  It worked, but reliability was an issue.

At that time, the devices really weren't good enough to do class D at any sort of a reasonable power level (more than 500 watts).  So, made the switch to class E, which works much better at higher frequencies, and the waveforms are MUCH easier to deal with.  That's more or less the history of why, at least for these designs, class E was better.

Fast forward to 2012 - we have better MOSFETs.  Now, Class D and class E will give similar performance.  Both have their advantages and problems.

Class D amplifiers are smaller to build, and less expensive because there aren't any tuning components in the transmitter - but that function must be provided elsewhere (antenna tuner) because class D transmitters need to be properly terminated in their design impedance.   

Class E stages can be combined effortlessly at high frequencies  (20 meters, 40 meters - works great), so it is very easy to attain high power.  Because there is a tuned circuit, you can operate the transmitters into loads that vary by 150% of the design value - no problem, but the components in the tuned circuit add cost and size.

[steve_qix]

And that's what makes figuring this stuff out confusing. Nautel is using 500v 41A FETs, 20 modules (each of which is a seperate H bridge), so there are 80 devices total in the transmitter. Running at 50kw, each one of those modules would be putting out 2.5kw each. They claim 90% efficiency, so each module is drawing 2.8kw at carrier. With a 400v PS, that means each module is drawing 14A at 200v. On peaks, that would be drawing 28A. Now, in this case, would the current be divided between the two FETs? So on peaks each turned on FET would be handling 14A? that's only a third of what they are rated at. Which going by your safety factors for class E, is acceptable, but barely.

So, if each module is pulling 14 A, each MOSFET pair is pulling 7A (average).  At 100% positive modulation, each pair will be pulling 14A average, which is a good safety factor.  Current is not NEARLY the problem as is voltage.  You can overcurrent a MOSFET by a LOT as long as it's for a small period of time.  The same is not true for overvoltage.  A sufficient overvoltage will puncture the device and ruin it instantly.

[kb3ouk]

So mosfets are more forgiving when it comes to current than tubes are, but don't like the voltage. But running 400v on a 500v fet seems like they are pushing things to the edge for sure. The only thing that is limiting them as far as current is the filter impedence. All the H bridges I've looked at seem to like running the filter at 56-57 ohms. If that filter was a lower impedence, then you could push more current through, which is where class E does have the advantage, you can pretty much run whatever current you want, and the output network will match to the load. But then you are limiting yourself as to how much voltage you can safety run. What might work would be to replace the lowpass filter with either a tuned network, or a filter that presents a lower impedence at the input than it does at the output.

EDIT: I reread the schematic for the H bridge transmitter. You might be able to run a higher amount of current through by changing the turns ratio of the output transformer. It's a 1:1 transformer in the schematic. If you made it step up withsomething like a 2:1 ratio, then that should cause the current to double, which would double the output power, and should be safe for the fets. Now the question becomes do you really want to be stepping up the output of the amplifier, I've never saw a schematic that's showed it being done before. (rated at 14A, in the schematic as it is currently written, the carrier current is 2.82 for the entire transmitter, which works out to 1.41A per fet at carrier and 2.82A on peaks. If you doubled the current, then each one would see around 2.82A at carrier and 5.64A on peaks, which is reasonable)

EDIT AGAIN: I have saw 4:1 transmission line transformers in the output though.

[steve_qix]

To be fair, there are most definitely a lot of components that would need to be added to create a commercial quality, reliable RF amplifier.

It is pretty easy to do at 1mHz, in the broadcast band.  Harder to apply at HF.

Overcurrent protection is VERY easy to apply, no matter what.  I have it in all of my class E designs.  This identical circuit will work with any RF amplifier.  You cannot damage the transmitter in any way with an overcurrent as long as the circuit is there and working properly.

Overvoltage is much harder to manage. Most commercial transmitters have ALL KINDS of protection devices on gates, drains, networks, etc.  TransZorbs, arc-back protectors, those sorts of things.

Load mismatches are the biggest real danger to any transmitter because of the potential for very high voltages to be created.