Energy recovery from negative peaks .... and recycle for positive peaks..

[IN3IEX]

 

Really

After a green energy seminar and after observing that common negative peak limiters have the ominously dissipating "resistors" in the circuit.... Let's remove those resistors from our beloved useless circuits:
 
R2 and D5 emulate the class C tube as low frequency load.
R1 is the HV PS + Modulator internal resistance.
C4 is the energy recycling capacitor.

C4*R2 is the discharge time constant, recovered energy will be pushed in the waveform after the end of each negative clipping for a time about C4*R2.

This is a SPICE simulation.
Volunteers required....or report your bad experience if already tested.

Edit:  the optimal value of C4 is the main question here.
First result with a signal made by 300 Hz  +  1000 Hz + 3000 Hz each component with identical amplitude (each 1/3 of the previous example), the capacitance that gives the highest peak is 1.5 uF. In this case the modulator impedance is again 100 ohm and series of three 1N4007 are required because the modulation amplitude has been increased. Final result is wave3. Maximum voltage on capacitor C4 in plot wave3 is 600V.

[Opcom]

I like the idea. Thought about it as well. Does the simulation show that the capacitor's charge will never be more than the below-zero swing of the modulated B+ at the transformer?

I wish you could post the file here, or zip it and post it.

[IN3IEX]

Yes, the capacitor voltage will never be more than the below zero swing of the HV+MOD voltage. Anyway if the capacitor burns, nothing bad happens, the circuit will simply deactivate itself and the modulation transformer will make a different noise... I suppose. The diode in parallel to the capacitor must have a voltage like the capacitor voltage, if it burns open the capacitor will be destroyed, but the transmitter will be safe. The diodes to ground are critical, they must keep the full maximum HV+MOD voltage and if they burn short, it will be a dangerous short.
The capacitor charges through the internal impedance of the modulator and discharges in the RF tube impedance. It seems that it can squeeze the maximum power from the modulator during the charging times. Maybe the modulator will be overloaded when below zero (who cares?), just avoid feedback distortion reduction circuitry in the modulator to avoid saturation of the feedback loop and the resulting transient intermodulation. On the other hand a little feedback could help the charging process, if the modulator is oversize.
The .sch file for a PSPICE student with the old Microsims editor Pspice Schematics is attached.
Why C=1.5uF gives the maximum positive peak with a 10kohm tube and the composite waveform I do not know. With sine signals each frequency has its preferred C value. Obviously C must keep its voltage for about a period....and must charge to maximum voltage through the modulator impedance during the short "below zero" time.
Ok, the simulator says: somebody has to try it.

Please rename the attached file to new.sch

[AB2EZ]

A few questions and observations:


1. When the modulated B+ drops below zero, a large cirrent will flow through the 100 ohm resistor (and the diodes). Dissipated power = i x i x R = v x v /R; where v is the voltage across the resistor. Thus, particularly since R2 >> 100 ohms (+ the series resistance of the diodes), the power dissipated in the 100 ohm resistor (+ the series resistance of the diodes) during periods of negative peak limiting will be much greater than the power dissipated in the matched (to R2) resistor in a "conventional" negative peak limitor during periods of negative peak limiting.  To minimize the power dissipated during periods of negative peak limiting (while still keeping a load on the modulator) you should use the conventional approach.

2. A conventional negative peak limiter maintains (approximately) a constant load on the modulator. This approach will place a very low impedance across the modulator when the negative peak limiting is occurring. Some types of modulators will be very unhappy with this

3. As the time constant T= R2 x C4 gets larger... and if the waveform produced by the modulator produces a negative peak every Ta seconds... and if Ta is less than T... then the capacitor will develop an average voltage that adds to the modulating voltage. This corresponds to an upward carrier shift. This effect may be irrelevant, but it will not be zero.

Stu

[Opcom]

.delete..

[Opcom]

I got about the same results using LTSpice. Sorry I could not get the PSpice to work for me.. Things about the energy harvesting scheme disturb me.

Above about 98% modulation, ugly things start happening with the modulator's current. It starts because of the diodes. If the modulator were a voltage source with no care about current that would be fine, but a push pull tube modulator will be unhappy, and I think it can't hold regulation well enough to keep an undistorted sine wave above 98%. By the time the modulation percentage is 150% it is a very serious problem.

Some of these things may happen in an ultramodulation setup, but none so severely as the effect of the energy recovery capacitor.

The green trace is a differential measurement of the voltage across the capacitor, showing it charging and discharging. At higher frequencies it stays charged better but the high current charging spikes (red) are still present, just narrower. This charging current is demanded from the modulator.

The modulator called for by the 98% modulation parameters on this RF amp running 1KV/100mA (near as I can figure from the original simulation) would expect a 10K Ohm load an be at least a 50W modulator. A pair of 811's coasting or two 807's or something. What is the typical output impedance or damping factor of such a modulator?

It was mentioned that no feedback would be used, if I read right. That means a high output impedance no matter if triodes or beam tubes are used. It's typical of a pair of triodes to use no feedback in ham gear but in BC gear they do. Beam tubes should always use it. Even so, the impedance of the modulator as a source would be too high.

So a normal modulator is much too small to make a 1.2A charging pulse without severe distortion. The simulation I did does not show the distortion but common sense says it would be there in the form of clipping. It might even cause instability and pop something.

I'm not saying this can't be done, only that there are challenges to be addressed.

A large solid state amp driving an oversized transformer such as a toroid power transformer might work. Since the 1.2A charging pulse is drawn at the peak of a half-cycle (of which the normal peak is 200mA), the amplifier would have to provide six times the power (current) if maintaining a regulated output voltage. In this case of a 1KV/100mA RF amplifier having 100W input, the usual 50W modulator would have to be able to provide 300W during that pulse. SS would be a good choice.

To be fair i also tried with a keep-alive supply and examined the area where the modulated B+ nears zero volts. My attempts at this simulation show that a usual keep-alive would still be advisable to prevent the RF stage from seeing 0 volts.

I don't know how to simulate a real modulator and abuse it in cyberspace and I do not have a typical 100W plate modulated rig here, so that's all for now..

[W1RKW]

Neat concept.  Would like to apply that to motors and generators.  Capture unused inductive energy and store it for later use.

[IN3IEX]

Please note that the modulator might be overloaded but only when its output is NOT needed for modulating the RF tube.
1.5 uF is suggested for 10000ohm load. 3 uF for 5000 ohm, etc.
For an audio amplifier, the transient that appears while exiting the overload state is minimized if NO feedback is used.
That is why this method may work for recycling lost audio power and convert to useful RF power.
This method may not provide the best audio quality, for which it is not designed.

Just another concept to try if the TX is open for experimentation...

[Opcom]

It would be great to see this tried for real just to see how it works.

[AB2EZ]

Again...

Calculate (or estimate) the energy dissipated in the resistors that represent the series resistance of the modulator. This energy is not stored in the capacitor. It is converted to heat.

If you make this calculation, you will conclude that this circuit wastes (converts to heat) much more energy during periods of negative peak limiting than either of the following approaches to negative peak limiting:

a) switching the modulated B+ into a load that is matched to the modulation resistance of the RF output stage (good for traditional modulators)

b) removing the load on the modulated B+ during negative peak limiting (good for a solid state modulator that acts like a voltage source... and which doesn't care if its load is removed).

SDP

[Opcom]

Again...

Calculate (or estimate) the energy dissipated in the resistors that represent the series resistance of the modulator. This energy is not stored in the capacitor. It is converted to heat.

If you make this calculation, you will conclude that this circuit wastes (converts to heat) much more energy during periods of negative peak limiting than either of the following approaches to negative peak limiting:

a) switching the modulated B+ into a load that is matched to the modulation resistance of the RF output stage (good for traditional modulators)

b) removing the load on the modulated B+ during negative peak limiting (good for a solid state modulator that acts like a voltage source... and which doesn't care if its load is removed).

SDP

No argument there. Each time I tried making the modulator realistic, it didn't work as first shown. dang it, nothing's free. Where's all the free green energy?