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Author Topic: ''The PWM Output Filter''  (Read 3738 times)
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AMLOVER
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« on: January 04, 2024, 07:56:26 PM »

Hi to all,

I read from the class E radio webpage and under the subtitle 'The PWM Output Filter' :

''This particular pulse width modulator implementation is designed for a 5 ohm load. This means that the RF amplifier is adjusted such that it always represents a 5 ohm load to the modulator. As an example, if the modulator (at carrier) is delivering 40 volts DC, the RF amplifier current must be adjusted to be 8 amperes. This yields a load of 5 ohms - [40V / 8A = 5 Ohms]. If the modulator output were 45 Volts, the RF amplifier current would need to be 9 Amperes to maintain the 5 ohm load on the filter.''

I am wondering how can we adjust the RF amplifier current without changing the voltage?

I wish to all a happy and healthy new year.

Stefano
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M0VRF
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« Reply #1 on: January 13, 2024, 05:34:17 AM »

The RF stage has a set impedance so you just modify the pwm filter to match that Z.

The Z if the RF stage can be changed slightly by mismatching, i.e. tuning for maximum power vs nax efficiency.

Not sure of why you want to do what you ask.

Please, some more info?

Regards.

J.
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w9jsw
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« Reply #2 on: January 13, 2024, 07:54:28 AM »

Changing the turns radio of the output transformer is one way. Or using a conjugate matching filter to match the 50 ohm load to the PA. Steve does the later. That is why the actual impedance of the antenna should be close to 50 ohms. If it is not, the efficiency of the PA will be affected. Steve provides an efficiency circuit to help balance the output filter properly to get maximum efficiency, which in this case is to have the PA output impedance be exactly correct with respect to the PWM that was designed for it.
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K9MB
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« Reply #3 on: January 16, 2024, 09:08:30 PM »

Hi to all,

I read from the class E radio webpage and under the subtitle 'The PWM Output Filter' :

''This particular pulse width modulator implementation is designed for a 5 ohm load. This means that the RF amplifier is adjusted such that it always represents a 5 ohm load to the modulator. As an example, if the modulator (at carrier) is delivering 40 volts DC, the RF amplifier current must be adjusted to be 8 amperes. This yields a load of 5 ohms - [40V / 8A = 5 Ohms]. If the modulator output were 45 Volts, the RF amplifier current would need to be 9 Amperes to maintain the 5 ohm load on the filter.''

I am wondering how can we adjust the RF amplifier current without changing the voltage?

I wish to all a happy and healthy new year.

Stefano


As has been stated, the impedance of the amplifier must see s constant load to function properly because the output filter will not work properly if this is maintained.
A modulator could have a dead carrier voltage output of 45 volts at 10 amperes which is 4.5 ohms load. (450 watts)
As modulation reaches peak, it might be 90 volts at 20 amperes (1800 watts peak)
No matter what the voltage is, the current must be V/4.5ohms

If you try to change the driving impedance, it will become very distorted.

To understand now pwm modulators work and do some analysis, read and study Jim Tonne’s monograph from,the link below and download his new version of Elsie, which allows precision design of pwm filters that will not overshoot and cause high IMD on the modulated envelope.

http://tonnesoftware.com/appnotes/pwm/PulsewidthModulators.pdf

http://tonnesoftware.com/

Failure to properly design the output filter and drive it properly will result in high 3rd and 5th order IMD  on your output signal and earn you the scorn of your neighbors on the band and maybe a letter from the FCC if the mess gets too wide…😉

You can design a filter for a given impedance and substitute another input and output impedance and observe the results in tue plots of the respons curves and the transient response curves……
The laws of physics are immutable and uncompromising, so,it,is,best to learn and,observe them…

PS: This subject has been widely discussed innthe class E forum where almost all class E transmitters use class D pwm modulators or equivalent.
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AMLOVER
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« Reply #4 on: January 20, 2024, 10:04:17 AM »

Κ9ΜΒ,
Thank you for suggesting tonnesoftware.com, I got enough more knowledge.

W9JSW and M0VRF ,
What you propose is also my procedure. First to supply voltage to the Rf stage, after to find the best efficiency/power by changing the ratio of output transformer and later to count the impedance (V/I) in order to create the right PWM LPF. In my case it is a 4 pole Butterworth counted for Fc 22Khz and input/output impedances 0 and 6 ohm as far I have 72V/12A dead carrier in the RF stage.

So what I want to say is that the better order is first to adjust the RF stage for best efficiency, then to find its impedance (V/I) and on the end to calculate the PWM's low pass filter. Doing opposite it is very difficult to adjust the RF stage impedance in order to fit it to the PWM's LPF.


Regards
Stefano


* my D class.jpg (661.56 KB, 1522x1158 - viewed 99 times.)
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K9MB
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« Reply #5 on: January 23, 2024, 12:30:04 AM »

Κ9ΜΒ,
Thank you for suggesting tonnesoftware.com, I got enough more knowledge.

W9JSW and M0VRF ,
What you propose is also my procedure. First to supply voltage to the Rf stage, after to find the best efficiency/power by changing the ratio of output transformer and later to count the impedance (V/I) in order to create the right PWM LPF. In my case it is a 4 pole Butterworth counted for Fc 22Khz and input/output impedances 0 and 6 ohm as far I have 72V/12A dead carrier in the RF stage.

So what I want to say is that the better order is first to adjust the RF stage for best efficiency, then to find its impedance (V/I) and on the end to calculate the PWM's low pass filter. Doing opposite it is very difficult to adjust the RF stage impedance in order to fit it to the PWM's LPF.


Regards
Stefano


I am not very familiar with class D RF Amps, though there are some nice variations that produce good results.
The load impedance is critical, I understand, just as it is with a PWM modulator, which operates in a similar manner.
I have attached a design for a 4th order Butterworth Filter with a 6 ohm output termination and a 0.2ohm input impedance, which is realistic with modern FETs.
Notice that the input inductor is very large in inductance. You cans stack a lot of
T300-2 toroids that can give you this performance and they do not saturate with DC and can take high flux densities.

This will be a very high power modulator for this transmitter-right?
Will the peak voltage be 144 volts and peak current be 24 amperes at 100% modulation? It will take a big supply for this transmitter to have a peak of 3450 watts input….
I also put a design at 3 ohms for your reference in the post. Much smaller inductors


* IMG_7580.png (401.33 KB, 2732x2048 - viewed 67 times.)

* IMG_7579.jpeg (514.18 KB, 2524x1859 - viewed 57 times.)
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AMLOVER
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« Reply #6 on: January 27, 2024, 06:43:37 AM »

          I also design my PWM and RF output stage filters with the RF design tools but a final check with nanoVNA is necessary. My set up can't get 3 ohms but the cores in use with adequate ventilation can handle the power without noticeable saturation.
More cores would of course be much better but I am not interested for very long transmissions.
         
       
 
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W3SLK
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« Reply #7 on: January 27, 2024, 09:46:33 AM »

AMLOVER said:
Quote
More cores would of course be much better but I am not interested for very long transmissions.
What's the sense of doing AM if you are not going to make old-buzzard transmissions? Grin
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Mike(y)/W3SLK
Invisible airwaves crackle with life, bright antenna bristle with the energy. Emotional feedback, on timeless wavelength, bearing a gift beyond lights, almost free.... Spirit of Radio/Rush
K9MB
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« Reply #8 on: January 28, 2024, 10:58:24 AM »

AMLOVER said:
Quote
More cores would of course be much better but I am not interested for very long transmissions.
What's the sense of doing AM if you are not going to make old-buzzard transmissions? Grin


😂😂😂 Exactly!
That is why VOX is an abomination and why that paving brick sits next to the mic switch! 😉
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AMLOVER
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« Reply #9 on: January 28, 2024, 06:10:05 PM »

AMLOVER said:
Quote
More cores would of course be much better but I am not interested for very long transmissions.
What's the sense of doing AM if you are not going to make old-buzzard transmissions? Grin
....... to study and improve the technic ........
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K9MB
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« Reply #10 on: January 30, 2024, 08:42:54 AM »

AMLOVER said:
Quote
More cores would of course be much better but I am not interested for very long transmissions.
What's the sense of doing AM if you are not going to make old-buzzard transmissions? Grin
....... to study and improve the technic ........



Pursuant to that goal of improvement, Jim Tonne’s focus is on low distortion and low IMD numbers. That means linearity and excellent transient response.
A PWM filter is critical in design for impedance match and if you are not using air dielectric inductors, temperature curves for cores is important.
This is great thing about -2 iron cores. They are stable and can handle high flux densities.
Ferrites actually quickly lose permeability at high temperatures.
Good design always has a generous headroom to assure that performance is maximized, in terms of excellence Tom W1JJ has spent decades studying and experimenting with amps to get the cleanest signal possible, not just the loudest..
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steve_qix
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« Reply #11 on: February 01, 2024, 08:04:05 AM »

Interesting discussion...

A class e RF output transformer ratio of 1:1 is usually optimum.  Changing this is typically only done if the circuit values in the RF output network (tuned circuit) become impractical (or impossible) due to:

1) Many modules in series (the impedance of the RF amplifier is too high) and a conventional output network cannot match the load (usually, any value of the loading capacitor will be too high to obtain a proper match).  In this case, a 2:1 step down may be a practical solution.  - or -

2) The impedance of the RF amplifier is very low (high current/low voltage RF amplifier).  In that case, a 1:2 step up may be desirable.

I have encountered both situations over the years.  One of my transmitters used a 1:.5 (one to point 5 ratio) RF output transformer.  That particular RF amplifier consisted of 8 modules of 2 MOSFETs each.

The EFFICIENCY of the RF amplifier is determined largely by the duty cycle of the driver RF waveform, proper values of RF circuit components (shunt capacitors, L/C ratio in the output network), and of course the power.  Usually, the efficiency will be better (up to a point) at lower power levels, however, the goal is to achieve the highest efficiency AT THE DESIRED POWER LEVEL.

The RF output network will usually work into a fairly wide range of antenna load impedances.  I have found that a mismatch of up to a 3:1 SWR is not a big problem (assuming your RF output network can be properly tuned to the RF load). 

My "backup antenna" which is a coax fed dipole presents a 2:1 SWR to the transmitter, which will happily work into that load, and I do it all the time. 

Sometimes, with the main antenna (open wire fed dipole), I will move frequency by 50 kHz or so, and I am too lazy to retune the antenna tuner (and I have to adjust the transmitter anyway), so I just adjust the transmitter to the proper drain current, and all is well, even though the SWR is 2:1 or even more.

That is the probably the main reason why I prefer class E over class D, because it will work into RF loads other than 50 ohms.    But, this is a personal preference, and class D is good technology as well - just a slightly different application.
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AMLOVER
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« Reply #12 on: February 06, 2024, 09:37:20 AM »

Steve,

My set up is in class D. Four half bridged modulators are modulating eight H bridges. Each output impedance transformation is 1:1 and all eight secondaries are connected in series in order to make 48 ohms plus some leak inductance which is reduced in the input of the output LPF. The efficiency is 88% and the duty cycle is 45/55. My antenna is as your back up, a dipole center fed with 50 ohms unbalanced line. I use a tuner L type to move about 20 khz left or right and i always get a perfect match.
No problem there even if the power is considerable.

I understand that the current in class E could be adjusted sacrificing a little efficiency by the output tuning network.
In class D however this procedure is not easily possible because of the standard wideband output LPF.

As far I don't wish to change the rf output impedance transformation from 1:1, as you also suggest it as the proper one, i am wondering if there is another top secret procedure in order to adjust the current of a class D H bridge.

Is it possible that the parameters (RDSon, Ciss, Coss, Qg, ect) of each individual mosfet type is the main responsible for the current level of the D class H bridge?

In this case i could change the Si mosfets to Sic as far they have much better parameters (RDSon, Ciss, Coss, Qb) but this will be my last step.

When i get the opportunity I'll make an entertainment video with my D class set up in duty.


* D CLASS SET UP.png (42.57 KB, 2912x1280 - viewed 62 times.)
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AMLOVER
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« Reply #13 on: February 06, 2024, 11:16:55 AM »

K9MB,

Inductors with material -2 or even better air cored are very good choice but not for PWM LP filters or rf output transformers. I think that each part of the set up demands a different core material.

In this moment i need to improve some issues about current adjustment at the class D rf amplifier and then i'll check the possibilities for the improvements you suggested me.

If you could suggest me a solution about that it would be very appreciated.

Thank you for good advising.
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vk3alk
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« Reply #14 on: February 06, 2024, 09:40:09 PM »

Hi Stefano…

A bit late with a reply but hope that my comments are helpful for you…
What band is your Transmitter operating on and also the expected power levels too etc:
It’s a nice looking build you should be pleased with your work…
Also it’s an unusual design ….. 8 Modules in an adder circuit is not common …
Was there a reason for this ?

There is a difference between Class E and H Bridge in terms of obtaining efficiency…..
Basically a H Bridge is a non resonant switch really and when you connect a CRO on the output FETs before the transformer you should be able to trigger on the waveform displaying some kind of squarewave at the same frequency as the input……

I understand why your TX modules need to see 5R in your adder design…..
From my experience building these things have never been able to achieve high efficiency under appox 10R load…..
For me 88% efficiency is rather low …. did you try for max efficiency or were you just wanting them to see 5R….

When I make my TX modules my aim is always max efficiency….
You can experiment with the output transformer winding ratios to some extent  but this was always governed at the end of the day by efficiency ( have said that word a lot ) ….

That’s all for the moment…


Wayne
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vk3alk
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« Reply #15 on: February 07, 2024, 05:36:31 PM »

Hi Stefano

Just one final thing….
If it was my transmitter which it isn’t of course I would disconnect 4 of your H Bridge modules.
The remaining 4 are still active…
You would have to ground the secondary wire on the last Module though…
This would create a load to each of the TX modules of 12R appox…
You would probably find your efficiency will increase into the 90s hopefully…
Apply a low voltage and test to see what happens…

If all good and your happy with everything you then redesign your PWM LPFs to match 12R….

One of the really good things with PWM and H Bridge or Class D setups is you can vary the voltage to the transmitter causing output power levels to vary and you do not have to change anything else….
The Audio input remains the same … the Mark to Space Ratio as well …. the loading to the TX Modules and the load to the PWM LPF….
Everybody is happy….

Hope you don’t mine these comments…


Wayne
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AMLOVER
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« Reply #16 on: February 08, 2024, 09:10:12 AM »

Wayne,

I can't understand exactly what you mean so a little schematic would help a lot.
The Rdc of each module (H-bridge) is 73V:6.1A=12 ohm so each one's Rload with 1:1 OT ratio is 6 ohm no depended the way their OT secondaries are later connected to each other.
As far i feed two modules from one PWM, the LPF is designed for 6 ohm output.
The target is to get the DESIRED POWER LEVEL and in the same time with the OT secondaries connected in series to get as closest as possible to 50 ohm in order to get the best adjustment with the rf output LPF.
Efficiency can become of course a little better with some trimming and here comes my question.

Is there a way to adjust the current of a H-bridge in order to get a desirable Rload?
 
About efficiency in relation with power level i agree with Steve:
'Usually, the efficiency will be better (up to a point) at lower power levels, however, the goal is to achieve the highest efficiency AT THE DESIRED POWER LEVEL.'
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vk3alk
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« Reply #17 on: February 08, 2024, 06:16:54 PM »

Hi Stefano...

Is there a way to adjust the current of a H-bridge in order to get a desirable Rload?

It all comes down to the OT winding ratio to obtain the loading you want ....
I feel there is some misunderstanding going on here though....

The efficiency of the H Bridges I use are constant thoughout the power range of say 1 watt to maximum ....

The FETs you use play an important roll in this too ..... greater losses as the power input increases effects efficiency...
By power input I mean increasing the voltage level to the TX Modules which in turn increases the current though the devices..

As a matter of interest what FETs are you using ?


Wayne
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vk3alk
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« Reply #18 on: February 08, 2024, 11:43:05 PM »

Hi Stefano....

I had better stop making posts as I'm getting carried away here....
Also not trying to prove my point at all just forums are interesting places to discuss information etc:

Have uploaded a photo.
It shows input waveform at the bottom and top in yellow is the output waveform...
The waveforms show the output as larger in voltage amplitude as it is just a non resonant switch ....
If I increase the voltage on the drain the higher the amplitude of the waveform...
That waveform cannot go higher than the voltage on the Drain..
There is no Miller Capacitance to cancel out no adjustments at all...
Of cause the input level has to be able to turn the FETs fully on and have a duty cycle of 50%...

Efficiency is mainly dependant on 2 things...
The correct loading usually over 10R for reasons I cannot explain...ugg
Losses in the FET...

The efficiency is a same from low power to higher powers levels and there is no point of maximum efficiency to achieve....

Think I had better go and sit on the fence  Grin


Wayne






* 80M inout waveform.jpg (112.41 KB, 800x480 - viewed 50 times.)
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AMLOVER
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« Reply #19 on: February 10, 2024, 08:41:39 AM »

Hi Stefano...

Is there a way to adjust the current of a H-bridge in order to get a desirable Rload?

It all comes down to the OT winding ratio to obtain the loading you want ....
I feel there is some misunderstanding going on here though....

The efficiency of the H Bridges I use are constant thoughout the power range of say 1 watt to maximum ....

The FETs you use play an important roll in this too ..... greater losses as the power input increases effects efficiency...
By power input I mean increasing the voltage level to the TX Modules which in turn increases the current though the devices..

As a matter of interest what FETs are you using ?


Wayne

Dear Wayne,

I use FDA38N30s
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vk3alk
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« Reply #20 on: February 10, 2024, 05:39:00 PM »

Hi Stefano...

Ok on the FDA38N30 Fets your using...
To be honest it's not one that I would order in and try...
The Gate Capacitance and Total Gate Charge are too high for me but you never know until you try etc:

There is no special circuit I know that will change the current though the TX FETs with the applied Voltage been constant...
The only way maybe is to use an ATU on the output to change the loading on the Transmitter...
A small change there will not effect the overall performance of your transmitter etc: but if it achieves what you want then  Grin
When we make our PWM LPF and use them in circuit are they really the same as when we design them on the Computer ... probably not...

If using an ATU on the output of your TX to achieve what you want ... then I would do it...
Its your Transmitter  Smiley and nobody would care...

Also if it passes all the health checks and have good heatsinking you are ready to go.....



Wayne

I see all the fuses there...
Are they your overcurrent protection ?



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