Ermi Roos wrote: Wenzel's article briefly states that tubes might be used to improve Part 15 AM transmitter designs, but he does not elaborate. He mentioned the interesting fact that 100 mW applied to a 3 meter antenna develops a couple of hundred volts rms across the antenna. (Short antennas are, indeed, "hot.") Wenzel suggests that, since tubes are high-voltage devices, it might be easier to supply the needed high voltage if they were used. He did not, however, propose any tube circuit that can provide high voltage to the antenna.
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However in practical Part 15 AM systems the output power available for radiation by the antenna is a small fraction of the transmitter output power. Most of the transmitter power is dissipated in the resistive losses of the true r-f ground (buried in the earth), and to a lesser extent, the loading coil.
Here are some typical? numbers for a Part 15 AM system on 1600 kHz using a resonant antenna:
- Radiation resistance of 3-m, ground-mounted radiator = 0.1 ohm
- Loading coil resistance = 2 ohms
- Resistance of r-f ground = 25 ohms
- Impedance presented to the transmitter = 27.1 ohms (the sum of the above, if the antenna system is resonant)
Assuming that the transmitter output power was 100 mW, and that the transmitter was capable of matching into a 27.1 ohm load -- then the voltage across the output load impedance would be SQRT(0.1*27.1), which is about 1.65 volts. The current there would be SQRT(0.1/27.1), which is about 60 milliamperes.
So, while much higher voltages may be present at other locations in the antenna system, the voltage (and current) at the loading coil input connection typically should be well within a range suitable for modern solid-state devices and circuits.
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When I wrote about transmitter "efficiency" in the last few posts, I was considering the load to be mostly ground resistance. The 46 ohm load resistor I was using for testing was intended to represent what I consider to be typical ground resistance. I mentioned a couple of posts back that the radiation resistance is orders of magnitude smaller than the ground resistance. So, the "efficiency" I have been writing about is how effectively the limited input power to the transmitter performs the useless task of heating up the earth. However, the more power the ground gets (and wastes), the more power is applied to the tiny radiation resistance. This is "trickle-down" electronics!
While a short antenna hardly radiates at all, it has an intense near field which decreases rapidly as the distance from the antenna increases. The intense near field is produced because the current that is applied to the load (the ground resistance) passes through the high reactance of the antenna capacitance.
In the usual Part 15 AM transmitter circuit, the high voltage that is at the antenna is only at the antenna and across the loading coil, but not at the electronic components. So, in the usual transmitter design, there is no problem with voltage breakdown. However, what Wenzel must have been considering, but did not describe, was a way to directly apply high voltage to the antenna without generating this needed high voltage using an antenna tuner.
Ermi,
Thanks for the effort to experiment and the report.
Just a couple of things:
A series resonant L C circuit with typical dissipative losses can produce very high voltages across the L and the C at or near resonance so it is not surprising that the "antenna" has a high voltage, though as was pointed out, the voltage at the feedpoint of the loading coil can be small. If I touch the end of my antenna with a NE-2 bulb it illuminates. CBers used to attach a NE-2 to the end of the antenna, probably just for fun, and I wounder what that non-linear bulb did to the spectral purity of the signal!
When Wenzel was commenting on the high voltage available from tubes maybe he was considering a non coil loaded antenna and had the idea that raising the feedpoint voltage would increase the antenna current and thus the radiation. But I just speculate.
Your comments about the harmonics are important but it would seem that a near resonant antenna system would suppress these. On the other hand, parasitic effects in the coil could let some of them through. The large loading coil I wound (on a 3 1/2 inch diameter form 17 inches long) on test showed a self resonance at about 3.5 MHz. Since this was about twice my operating frequency, I was not happy, but I have not checked for radiated harmonics so I only assume this may be a problem. This coil is not in service so that will wait.
Neil
The efficiency tests I performed with the Wenzel circuit output modified to the SSTRAN test circuit were all done without modulation. I looked for maximum efficiency using only the carrier. After posting my results for data obtained only with the carrier, I continued my investigation by checking how well the circuit works when modulating the circuit when it was set for maximum efficency (41%). With this efficiency setting, the top half of the modulation waveform was clipped off, and the circuit was clearly incapable of transmitting intelligence without severe distortion. At the maximum efficiency setting, the 46 ohm load resistance was transformed by the L network to 1060 ohms at the collector of the final stage of the transmitter. 1060 ohms resistance was necessary for maximum efficiency. In order to get the modulation to work, I had to reduce the load resistance, and therefore the efficiency. When I adjusted the L network for a collector load resistance of 725 ohms, I was able to get 74% modulation before the top of the modulation waveform began clipping, but the efficiency was reduced to 28%. By setting the L network for a collector load resistance of 620 ohms, I was able to get 100% modulation, but the circuit efficiency was reduced to 24%.
This particular circuit configuration is unsuitable for being modulated by connecting a modulation transformer to the collector; so nothing much can be done with this circuit to improve its efficiency beyond about 24%.
After thinking about it for a while, it became apparent to me why transconductance modulation does not work when the transmitter is set for maximum efficiency. The Wenzel generates a square wave current waveform in the final stage, nominally 20 mA p-p. Since the voltage from collector to emitter in the final stage is nominally 10 VDC, The maximum RF waveform at the collector of the final stage is nominally 20 V p-p. No more voltage range is available at the collector without clipping. Maximum efficiency occurs when the collector voltage range is maximum. When trying to use transconductance modulation while the transmitter is already set up for a 20 V p-p voltage swing, the top half of the modulation waveform is clipped off. This is exactly what I observed when I tried to modulate the transmitter when it was set up for the maximum 41% efficiency. In order to be able to modulate the carrier, it is necessary to reduce the carrier level so that the positive modulation peaks will not be clipped. Reducing the carrier level reduces efficiency.
If it were possible to modulate the collector, this problem would not occur. The positive peaks of the modulator would increase the collector to emitter voltage to prevent the clipping on positive peaks that occurs with a transconductance modulator. As I already mentioned, however, adding a collector modulator to the Wenzel circuit (after removing the audio input to the transconductance modulator) would not work. To use a collector modulator, the transmitter would have to be completely redesigned, with a class C circuit as the final stage.
Now, we see that, using the Wenzel circuit modified for the SSTRAN output as a baseline, we can get as much as a 6dB improvement in signal strength if we are able to transform nearly all of the allowed 100mW input power to the final stage to RF power. This would make an effort to come up with a really efficient Part 15 AM transmitter design seem more worthwhile.
In my previous post, I reported that the Wenzel circuit, when modified to simulate the SSTRAN output circuit, is capable of 41% efficiency with an umodulated carrier, and only 24% efficiency when it is capable of producing 100% modulation.
Because not much detailed technical information is available to the public for designing Part 15 AM transmitters, I continued using the Wenzel circuit I built as a testbed to discover what more can be learned from it. It is to be expected that a class C output stage should have more efficiency than the overdriven class A output stage used in the Wenzel.
I converted the output stage of my Wenzel circuit to class C operation by making as few wiring changes as possible. I first disconnected the emitter of the output transisistor from the emitter of the oscillator buffer, and connected it directly to ground. I left the oscillator, oscillator buffer, and modulator circuits the same as they were, except for one change to the oscillator to get it to increase its output from the original 1.3 V p-p to 2.3 V p-p. This change was tacking a 5.6k ohm resistor across the 10k ohm resistor in the oscillator circuit. The oscillator buffer and the modulator now served as an emitter follower with an active load to drive the base of the final stage. I no longer used the modulator as a modulator, but as part of the oscillator buffer circuit for driving the base of the final stage. The capacitor that originally provided an RF ground for the base of the final stage was now disconnected from ground and connected to the emitter of the oscillator buffer. There were two resistors that originally provided a bias voltage to the base of the final stage transistor. I disconnected the resistor going to +15 VDC, but left the resistor going to ground. So, the final stage was modified to operate at zero bias. The RF final stage responded to only a portion of the positive half of the waveform from the oscillator buffer. This caused the modified final stage to perform as a class C amplifier.
Since I was using the original modulator circuit only as an active load for the oscillator buffer, and no longer as a modulator, I added a collector modulator. I also no longer used the 15 VDC supply that powers the Wenzel to power the final stage, and I used a separate adjustable voltage DC supply instead. I disconnected the 2.6 mH RFC that goes to the collector of the final stage from the +15 VDC supply, and soldered a 100 ohm resistor in series with it. At the junction of the RFC and the 100 ohm resistor, I soldered a .01 uF capacitor to ground for RF bypassing, but I wanted a fairly high capacitive reactance for audio frequencies. I connected one terminal of the secondary of an audio transformer to the 100 ohm resistor in series with the RFC. The other terminal of the secondary of the audio transformer was connected to the output of the variable DC supply, with a bypass capacitor connected to its output. The primary of the audio transformer was connected to an audio source.
I had set the variable DC supply to 8.6 VDC. The voltage between the 100 ohm resistor and the RFC was 7.23 VDC. The collector current was 13.7 mA. The DC input power to the final stage was 99 mW. Of course, I did not get this result by chance. I had to adjust the oscillator buffer output and the variable power supply to get nearly 100 mW DC input power to the final stage. I also had to be sure that, when the power supply voltage was doubled, to simulate the positive peak for 100% modulation, the RF voltage across the load also doubled. Transistors are peak current limited for a given drive power, and I had to be sure that the upper end of the modulation envelope did not clip. I did not get peak current limiting until the supply voltage was 19 VDC.
Peak current limiting is not as big a problem with tubes as with transistors. Transistors have an absolute limit in the amplitude of the current pulse, while tubes do not. The space charge in tubes provides a current reserve, and the tube current pulse is limited by the plate power dissipation. My current pulses were "M" shaped, as they should have been, and had a 103 degree conduction angle.
I got 80% efficiency when using the output stage at the peak current limit. The need for 100% modulation caused me to operate the carrier well below the peak current limiting level, and this caused the efficiency with modulation to be only 57%.
Ermi,
Great report as usual from you, but as I try to visualize the test circuit in my mind a question arises.
Did you use the 100 ohm resistor as the output load for this circuit or did you couple the signal from the collector through the ell network to the load as you previously described? From your description the RF signal was bypassed to gnd. before reaching the 100 ohm resistor and the resistor was not the load, but I wanted to be certain I got this right.
Thanks again for your work and for your report.
Neil
Neil,
Thank you for carefully reviewing my last post. I know that it is not easy to figure out a circuit from only a written description.
When altering the Wenzel to class C operation, I did not start with the original circuit designed by Charles Wenzel, but with my modified circuit that I used to simulate the SSTRAN output. I described this modified circuit in earlier posts.
Before my changes for class C operation, the modified circuit had a radio frequency choke coil (2.6 mH) going from the collector of the final stage to the +15 VDC supply. I disconnected the choke coil from the +15 VDC supply, but left the choke connected to the collector. I soldered a 100 ohm resistor in series with the choke coil where I had disconnected the coil from the +15 VDC supply. The other end of the 100 ohm resistor was connected to one terminal of the the secondary winding of a modulation transformer. The other terminal of the secondary of the modulation transformer was connected to a separate, variable, DC power supply. This supply had a bypass capacitor at its output. At the junction of the choke coil and the 100 ohm resistor, I soldered a .01 uF capacitor. I connected the other end of the .01 uF bypass capacitor to ground.
The 100 ohm resistor and the bypass capacitor you referred to are not at the collector of the final stage, but, instead, at the other end of the RF choke connected to the collector. So, the 100 ohm resistor is not the load of the final stage, and the bypass capacitor does not bypass the collector to ground.
Sorry for the redundant, jaw-breaking, language I used to describe the circuit modification. I tried to be as precise as possible.
This is the original subject of this thread. Wilcom Labs was interested in creating a new Part 15 AM transmitter product design using tubes, and he asked the Forum for opinions. In my first post in this thread, I pointed out that tubes should be used only if they have significant technical advantages over other devices, and that new tubes manufactured today are inferior to those manufactured years ago. I also stated that tubes will probably work better for Part 15 AM than other devices.
I constructed a Part 15 AM test circuit using an old, slightly used, Sylvania 12AX7. The 12AX7 is a dual, high-mu, triode, which is still popular today in retro audio equipment that uses tubes. I originally envisioned using one triode as a Pierce crystal oscillator and the second triode as a power amplifier. It quickly became apparent that the crystal oscillator did not produce enough grid drive for class C operation of the power amplifier. It is not safe to drive a crystal with RF current more than about .5 mA. So, I used the first triode as an amplifier for an external crystal oscillator and buffer. What I used as the external oscillator and buffer was the same Wenzel oscillator, buffer, and modulator (used as an active load) that I used to drive the transistor I used as a class C amplifier, which I described in a previous post. I used separate, adjustable, high voltage, plate power supplies for the driver stage and the final stage so that I could set the RF drive to the grid, and the DC power input to the final stage. Quite a lot of RF drive to the grid was needed. To get good efficiency, it is necessary for the peak grid voltage to be higher than the minimum RF voltage swing at the plate. My, grid drive was about 75 V p-p, but the waveform was not symmetrical. The positive peak was somewhat clipped because, with this much drive, the grid drew an appreciable amount of current, and loaded down the driver. I had set the output stage with 100 VDC plate voltage and 1 mA plate current, giving 100 mW input power. I got 66.7 mW of output power, making the efficiency 66.7%.
Here is some information about my test circuit: I, at first, used a small 25 pF piston cap between the plate of the output stage and ground for tuning. This capacitor was obviously not made for significant voltage, since I soon heard crackling that indicated that the cap had broken down. I dug up a multiple-plate, air-variable, capacitor in my junk box that was able to do the job. A rough rule of thumb is that air dielectric can withstand a field strength of about 80 Vrms/mil. I measured the spacing between the plates of the cap I used to be 32 mils.
I used a 47 k ohm grid-leak resistor in the final stage. I tried other resistor values, but this value worked as well as, or better, than any other. I also added a variable power supply to provide additional grid bias. I found no benefit to providing additional grid bias, so I left the bias supply in the test circuit, but set to minimum voltage.
To avoid the need for adding a neutralization capacitor to the final stage, I used a tuned circuit at the output, but not at the input. Even so, I had some trouble. The output of the driver stage used a .5 mH RF choke coil. I had to add a 1 k ohm resistor in series with the choke coil to reduce its Q in order to eliminate squegging. It looks like the choke coil and stray circuit capacitances formed a kind of tuned circuit at the input of the final stage which caused parasitics to be generated.
From what I saw in working with this test circuit, further improvements can be made. At 1.5 MHz, the dummy antenna I used, which was a 30 pF capacitor in series with a 46 ohm resistor, has an equivalent parallel resistance of about 270 k ohms. However, because I used only 100 VDC of plate voltage to the final stage, Its optimum output impedance was only 54 k ohms. So, the dummy antenna could not be connected directly to the plate, making the tuning circuit complex. However, to get the correct impedance match right at the plate, a higher plate voltage is needed. This higher plate voltage has to be accompanied by a corespondingly smaller plate current so that the input power to the final stage would be 100 mW. (Wenzel mentioned driving the antenna directly at the plate of a tube.) With my test circuit, I was not able to use higher plate voltage that did not correspond to excessively high plate current. There was nothing I could do by adjusting grid excitation amplitude and grid bias voltage to avoid the excessive current at higher voltages. I think I know how to solve this problem, but it requires a new clever circuit design, or maybe even an invention.
Instead of driving the grid of the final stage with a quasi-sinusoidal signal, pulses with widths much smaller than the period of the carrier frequency could be applied to the grid. This will cause the conduction angle of the current in the final stage to be small, which reduces the DC current. I don't know how to generate this train of narrow pulses. The pulses have to have tens of volts in amplitude and have enough energy to supply significant grid current. Of course, digital one-shots are out of the question because they supply too little energy. I thought of using a triggered blocking oscillator, but the transformers used in blocking oscillators are such that they can't produce really narrow pulse widths. I think, however, that the resolution to this pulse generation problem is well within the realm of possibility. What is required is ingenuity and lots of hard work. Really high efficiency should be possible.
If you have a good oscilloscope with a good probe, you may be able to measure the efficiency of your transmitter yourself. I say "may" be able to, because, if your transmitter is of the SSTRAN-type, the efficiency measurement is tricky, since the antenna loading coil forms part of the "L" network that performs the impedance transformation between resistive portion of the antenna circuit and the final stage. Measuring the efficiency of an SSTRAN-type transmitter has been discussed in other posts in this thread.
However, if your transmitter is of the Rangemaster-type, the efficiency measurement is appreciably easier, since all of the impedance transformation components are in the transmitter circuit.
First, you need to construct a dummy antenna made by connecting a resistor and capacitor in series. The capacitor represents the capacitive reactance of the antenna, and the resistance represents the ground resistance of the antenna. The difficulty here is that you can only guess at the values of the resistor and capacitor, because it is difficult to measure them with any accuracy. The dummy antenna I use is 30 pF in series with 46 ohms. The capacitor is a 30 pF +/- 5% mica capacitor. The resistor is a 47 ohm (nominal) 1/4 watt carbon composition (can be carbon film) resistor that measures to be actually 46 ohms. If you are optimistic, and think that your ground resistance is better represented by a lower standard resistor, like 27 ohms or 12 ohms, then use that value in your dummy antenna. As for the capacitor in the dummy antenna, if your antenna is 3 meters long, 30 pF is a reasonable approximation of the antenna capacitance.
Tune up your transmitter with your dummy antenna connected to the antenna terminals of your transmitter, your "real" antenna disconnected, and your oscilloscope probe connected across the resistor (R) in your dummy antenna. Your probe should be of the 10/1 type, with about 10 pF of capacitance. Don't use a 1/1 probe! Set the transmitter to exactly 100 mW input and measure the peak-to-peak amplitude (Vp-p) of the unmodulated carrier waveform.The waveform should be a sine wave with very little distortion. The output power is P= [(Vp-p)^2]/(8R). Vp-p is in volts. R is in ohms, and P is in watts. Dividing P by the input power of .1 watts gives the efficiency.
For example, suppose R is 46 ohms and Vp-p measures to be 5 volts. In that case, the output power is P= 0.068 watts. Dividing by the input power of .1 watts gives an efficiency of .68, which is 68 %.
Ermi and all,
If you are interested in bench testing a transmitter, Ermi's advice is excellent. He mentioned the importance of using a 10X probe and this is an often overlooked and misunderstood aspect of measurements using a scope.
He is correct and all I want to do here is to offer a bit of explanation. A 1X probe presents a load to the test point which includes the scope input resistance and capacitance plus the cable capacitance of the probe. Typically the input R of a scope is 1 Mohm with a capacitance of 15 pF. This is significant at even medium RF. A 10X probe presents primarily a high resistance load which will minimally affect a circuit.
RF measurement with a scope or a DVM requires caution since the measurements depend to a great extent on the characteristics of the measurement device and the probe. Ermi's suggestion on using a 10X scope probe is dead on! Also, make sure the probe is properly compensated and that the bandwidth of the scope is adequate, though most scopes will suffice for the AM BCB frequencies. DVMs, however, are mostly not satisfactory for AC measurements at RF.
Moral of this story is that measurements need to be interpreted with knowledge of the methods and equipment used.
Thanks, Ermi, for the great tutorial and I hope I added some good information. I assume that you, as will I, will answer any further questions.
Neil
In my previous post about the use of tubes for Part 15 AM, I stated that improved efficiency can be obtained if the RF drive to the final stage is obtained from a generator of narrow pulses rather than a quasi-sinusoidal waveform. At the time of my previous post, I did not know how to make the grid pulser.
As I stated in another thread ("Great technical thread," initiated by Neil), I am investigating the loss in efficiency due to a parasitic diode from drain to source of transmitters using power MOSFET output stages. This is called the "Turn-off behavior " of a power MOSFET. When the MOSFET is on, and is then turned off, ringing is produced. The ringing waveform is a train of positive pulses at the drain with the negative half cycles clipped by the parasitic diode. It occurred to me that I can make a good grid pulser by keeping a MOSFET on for most of a cycle, but turning it off long enough to generate a single positive-going pulse. The turn-off time is established by logic circuits. In my circuit, I had set the turn-off time to be 0.1 the period at 1.6 MHz. The inductance and capacitance at the drain should be set to be so that the turn-off time is pi[SQRT(LC)]. The inductor should be wound on a large core so that the energy storage in the inductor would be large. I used an IRF 510 MOSFET with a .95 uH inductor wound on a 3" diameter low-permeability ferrite core. The LC product is smaller than would be needed for a half cycle of the ringing to be a tenth cycle at 1.6 MHz, but the pulser works well, nevertheless.
Using the pulser, I was able to get the load impedance at the plate to match the parallel-equivalent resistance of the ground resistance. The smaller the ground resistance is, the larger the parallel-equivalent resistance will be. A larger load resistance requires a greater DC plate voltage and a smaller plate current. The use of high voltage can be a problem. I zapped a couple of 12AX7 tubes when working with more than 500 VDC.
A vacuum tube with a grid pulser is probably the only Part 15 AM transmitter design that can give very high efficiency over a wide range of output impedances.
Ermi,
As I digest your latest report I wonder why, though you have posted much about this, that solid state circuits cannot perform as do tube circuits in terms of efficiency.
Your reference to the parasitic diode effects in MOSFETS was very helpful and I did my homework and now appreciate the difficulty.
It sure would be easier if the 100 mW was the the output rather than the input power.
On VHF home brew PAs I have attained efficiencies of 60 to 70% range with about 40 watts output power using bipolar transistors. Do you think this range of efficiency would be realistic for part 15 AM devices with bipolars?
My measurements of the Ramsey AM-25 peak efficiency gave numbers in the 30% range with a variable resistive load.
My measurements of the SSTRAN efficiency is much lower but I haven't published the numbers since I have not made these measurements using a suitable load where I could determine the peak. This will be a winter project.
Just some points to ponder.
Neil
Neil,
Thank you for your evaluation of my posts in this thread.
One of my posts describes my tests on the Wenzel circuit in which I converted the output transistor circuit for class C operation. I got in the vicinity of 60% efficiency. So, bipolar transistors can give reasonable efficiency. This may be good enough. Increasing the efficiency further will not give much of an improvement in dBs. As I had previously pointed out, the efficiency of the bipolar output stage is slightly worse if the stage has to be collector modulated (which is the case for Part 15 AM) than if it were made for FM or CW. This is because the modulated stage has to have greater RF drive at the base than a CW stage in order for the output RF voltage to double when the DC voltage at the collector (or, more exactly, DC + AF peak voltage) is doubled.
Tubes work efficiently with Part 15 AM because a short grounded monopole can provide a high-impedance load to the transmitter. A tube can work well with a high impedance load because it can be operated with high plate voltage and low plate current. Grid drive is a little difficult, because a lot of drive is needed for good efficiency, but overdriving the grid so that the grid voltage exceeds the minimum plate voltage causes secondary emission, which is harmless if it is not too high, but it can result in the destruction of the tube.
As for the power MOSFET, in preliminary tests, I got 25% when the parasitic diode was repeatedly forward-biased during each cycle. By capacitively loading down the drain so that only a single pulse occurred during a cycle, I got the efficiency to increase to 66%. When there was only a single drain pulse during each cycle, the parasitic diode did not conduct. Using additional capacitance at the drain converts an RF transformer with an untuned primary into a double-tuned circuit. This considerably complicates output tuning.
I am continuing to do further work studying the power MOSFET stage, and will report on my efforts once I am finished.
Hi Ermi -
Re your post "Tubes work efficiently with Part 15 AM because a short grounded monopole can provide a high-impedance load to the transmitter. A tube can work well with a high impedance load because it can be operated with high plate voltage and low plate current."
But the reason that the short antenna has very high feedpoint impedance is due to its capacitive reactance. When that reactance is offset by the proper inductance in a loading coil, then the transmitter sees an external load consisting of the ~ 1/10th of an ohm of antenna radiation resistance, plus the r-f resistance of the coil and ground system. Probably the net impedance seen by a typical Part 15 AM transmitter ranges from 15 to maybe 50 ohms, j 0 for a resonant 3-meter antenna system, which isn't really "high impedance."
Are you thinking that a Part 15 tube transmitter can work into a 3 meter monopole without the use of a loading coil, because the tube tx can operate with high plate voltage and low plate current? Does it not still require a non-reactive net load for efficient power transfer?
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Ermi wrote:
As I had previously pointed out, the efficiency of the bipolar output stage is slightly worse if the stage has to be collector modulated (which is the case for Part 15 AM) than if it were made for FM or CW.
Thanks for the reminder about this. It has been a while since I read back in this thread and this point had dimmed in my memory.
Indeed, the VHF amp. I referenced was for FM and your explanation of the situation with AM makes sense.
Neil