Home |
Search |
Today's Posts |
#1
|
|||
|
|||
"Non-dissipative Source Resistance"
It has been theorized that a circuit consisting of a Class C vacuum-
tube r-f amplifier using a tuned tank circuit in its output network provides an operational “non-dissipative source resistance” of 50 ohms for energy present at the output connector of the transmitter. However the information and measured data provided in the text excerpts below is not very supportive of that theory. These excerpts discuss and show the effects of the energy entering a transmitter at its output connector by frequencies offset from the transmitter frequency. There is direct applicability of the conclusions of the paper showing that the operational source impedance of the transmitter near/at the carrier frequency is much different than 50 ohms. If it WAS a functional 50 ohms, then the termination provided to the transmission line for signals entering the transmitter by its output connector (whether on or off frequency) would not be present at the plate of the PA tube to react with the power being generated by the PA tube. Rather the data leads to a logical conclusion that the operational source impedance of this configuration at the carrier frequency will be very low (approaching zero), when it is optimally tuned/adjusted to produce its rated output power. Further discussion or comment is invited. RF From: A STUDY OF RF INTERMODULATION BETWEEN FM BROADCAST TRANSMITTERS SHARING FILTERPLEXED OR CO-LOCATED ANTENNA SYSTEMS, by Geoffrey N. Mendenhall, P.E.* II. INTERMODULATION AS A FUNCTION OF "TURN-AROUND-LOSS". "Turn-Around-Loss" or "Mixing Loss" describes the phenomenon whereby the interfering signal mixes with the fundamental and its harmonics within the non-linear output device. This mixing occurs with a net conversion loss, hence the term "Turn-Around-Loss" has become widely used to quantify the ratio of the interfering level to the resulting IM level. A "Turn-Around-Loss" of 10dB means that the IM product fed back to the antenna system will be 10dB below the interfering signal fed into the transmitter's output stage. "Turn-Around-Loss" will increase if the interfering signal falls outside the passband of the transmitter's output circuit, varying with the frequency separation of the desired signal and the interfering signal. This is because the interfering signal is first attenuated by the selectivity going into the non-linear device and then the IM product is further attenuated as it comes back out through the frequency selective circuit. "Turn-Around-Loss" can actually be broken down into the sum of three individual parts: (1) The basic in-band conversion loss of the non-linear device. (2) The attenuation of the out-of-band interfering signal due to the selectivity of the output stage. (3) The attenuation of the resulting out-of-band IM products due to the selectivity of the output stage. Of course, as the "Turn-Around-Loss" increases, the level of undesirable intermodulation products is reduced and the amount of isolation required between transmitters is also reduced. The small portion of the interfering signal that is not reflected is what causes intermodulation products to be generated. Obviously the lower the output source impedance, the more complete the reflection (lower return loss), with the result being less production of intermodulation products. III. EQUIPMENT PARAMETERS THAT AFFECT INTERMODULATION LEVELS. The interfering signal must be coupled into the transmitter's output stage before the IM products are produced and the output level of the intermodulation products will be related to the interfering signal level. The two parameters (outside of the filterplexing equipment) that most affect the interfering signal level into the transmitter's output circuit are the output loading and the circuit's frequency selectivity (loaded "Q"). These two parameters are interrelated because the degree of output loading will change the loaded "Q" of the output circuit while also affecting the return loss of the interfering signal looking into the output circuit. "Output Return Loss" is a measure of the amount of interfering signal that is coupled into the output circuit versus the amount that is reflected back from the output circuit without interacting with the non linear device. To understand this concept more clearly, we must remember that although the output circuit of the transmitter is designed to work into a fifty ohm load, the output source impedance of the transmitter is not fifty ohms. If the source impedance were equal to the fifty ohm transmission line impedance, half of the transmitter's output power would be dissipated in its internal output source impedance. The transmitter's output source impedance must be low compared to the load impedance in order to achieve good efficiency. The transmitter therefore looks like a voltage source driving a fifty ohm resistive load. While the transmission line is correctly terminated looking toward the antenna (high return loss), THE TRANSMISSION LINE IS GREATLY MISMATCHED LOOKING TOWARD THE OUTPUT CIRCUIT OF THE TRANSMITTER (LOW RETURN LOSS). THIS MEANS THAT POWER COMING OUT OF THE TRANSMITTER IS COMPLETELY ABSORBED BY THE LOAD WHILE INTERFERING SIGNALS FED INTO THE TRANSMITTER ARE ALMOST COMPLETELY REFLECTED BY THE OUTPUT CIRCUIT. VI. CONCLUSIONS 1. "Turn-Around-Loss" is a function of the particular non-linear device and the amount of loading on its output circuit. 2. "Turn-Around-Loss" increases as the interfering signal and the resulting IM products are moved away from the carrier and out of the output circuit passband. 3. "Turn-Around-Loss" will be least when the interfering signal is within the transmitter's passband. The figure posted at the link below shows the measured data supplied with this paper. http://i62.photobucket.com/albums/h8.../TAL_Chart.gif * Geoffrey Mendenhall presently is Vice President, RF Engineering at Harris Corporation Broadcast Division, and a recognized authority on transmitter system design. Harris Broadcast is one of the largest manufacturers in the world of AM/FM/TV broadcast transmitters, rated for power outputs up to 2,000 kW. |
#2
|
|||
|
|||
"Non-dissipative Source Resistance"
On Jun 12, 1:13*pm, Richard Fry wrote:
It has been theorized that a circuit consisting of a Class C vacuum- tube r-f amplifier using a tuned tank circuit in its output network provides an operational “non-dissipative source resistance” of 50 ohms for energy present at the output connector of the transmitter. However the information and measured data provided in the text excerpts below is not very supportive of that theory. yeah, lots of words that say things kind of related, but more confusing than useful for what is really being discussed. what is so hard about stating: 1. tubes and transistors are non-linear devices 2. tank circuits and other output matching systems TRY to make them appear more linear to the outside world, but are not perfect. any circuit that has non-linear characteristics will have a response that is not like a simple resistance. such systems must be modeled using appropriate non-linear techniques to get the full and proper response characteristics. An APPROXIMATION may be used in many systems that allows an equivalent simple impedance representation, but often only over SMALL ranges of voltage/current or frequency. The problem appears to be in this discussion that some people have used this small signal approximation to derive results in a domain where it is not applicable and are drawing the wrong conclusions. |
#3
|
|||
|
|||
"Non-dissipative Source Resistance"
On 12 jun, 15:13, Richard Fry wrote:
It has been theorized that a circuit consisting of a Class C vacuum- tube r-f amplifier using a tuned tank circuit in its output network provides an operational “non-dissipative source resistance” of 50 ohms for energy present at the output connector of the transmitter. However the information and measured data provided in the text excerpts below is not very supportive of that theory. [deleted] Hello Richard Fry, First, I use your last name also to avoid confusion with Richard Clark. As you may know, I have also doubts about the conjugated match issue. So I am happy with this posting. It is not that I dispute the conjugated match condition itself, but whether or not the conditions are met. In my opinion, required conditions aren't met in many cases, so output impedance of many amplifiers is not equal to the load resistor. To avoid discussions about measurements and traceable measurement instruments/procedures, I did some simulations on a valve amplifier and a class C mosfet circuit (that actually isn't operating in class C mode). Everyone can do these simulations at their own PC en get an opinion. I also described the difference frequency method that is easily implemented in simulation and gives the output VSWR almost directly without complex calculus. The link is: www.tetech.nl/divers/PA_impedance.pdf. Except for the condition of matching to maximum power output given certain drive, all output VSWR of the amplifiers is far from expected load impedance. Just a little voltage saturation (for example to get better efficiency) did drop the plate impedance below the required load impedance (voltage source behavior). When reducing the drive, the plate impedance rises rapidly above the load impedance (current source behavior). So power entering the output of the amplifier will mostly be reflected back into the cable, unless you specially design for output impedance. I know, this is disputed by some members of this group, but therefore I did the simulations and presented the results. I specially took "amateur" examples to avoid comment that I only use "exotic" examples. I did not copy results from professional activities, so I will not run into problems. The circuits are simple to ease reproduction in any spice based simulator. I dispute: "If the source impedance were equal to the fifty ohm transmission line impedance, half of the transmitter's output power would be dissipated in its internal output source impedance." A real class C amplifier with very small conduction angle has efficiency 50% when optimally tuned. When it operates at the transition of current/voltage saturation is can show 50 Ohms output impedance for very small change in load impedance. But as soon as the load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of the amplifier changes rapidly. So practically spoken, most high efficient amplifiers have high output VSWR (or bad Return Loss), but theoretically it can be 50 Ohms (for example). If there are doubts I will add a special simulation example for this. Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me. |
#4
|
|||
|
|||
"Non-dissipative Source Resistance"
On Jun 12, 9:44*am, Wimpie wrote:
A real class C amplifier with very small conduction angle has efficiency 50% when optimally tuned. Would that not be evidence that the amplifier source impedance necessarily is lower than the load impedance? When it operates at the transition of current/voltage saturation is can show 50 Ohms output impedance for very small change in load impedance. But as soon as the load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of the amplifier changes rapidly. For some additional input -- I have taken part in factory tests of high-power Class C single-tube/tuned cavity FM broadcast transmitters driving 50-ohm test loads measured to have 1.03VSWR, showing a DC input to r-f output efficiency of the PA to be in excess of 80% -- including the loss in the harmonic filter. Load power was measured using calorimetric methods. In fact, 80% PA efficiency is the published spec for this transmitter line as long as load VSWR relative to 50 ohms is 1.7:1 or less (any phase angle). Those results don't appear to be fitting very well with the idea that the tx source impedance can be ~50 ohms for loads with low VSWR. RF |
#5
|
|||
|
|||
"Non-dissipative Source Resistance"
On Jun 12, 8:17*am, Richard Fry wrote:
On Jun 12, 9:44*am, Wimpie wrote: A real class C amplifier with very small conduction angle has efficiency 50% when optimally tuned. Would that not be evidence that the amplifier source impedance necessarily is lower than the load impedance? No, not at all. It is important to realize that a real source is NOT necessarily like either the Thevenin or the Norton equivalent. The Thevenin and Norton equivalents, when discussed in any respectable text, will be noted to behave exactly like any other real linear source with respect to the load, but NOT necessarily with respect to the source itself, and in particular with respect to source efficiency and dissipation. For example: I can, at least theoretically, build a switching power supply (a source) which measures its output voltage and current and adjusts the voltage down by 50 volts for each amp that's drawn from it. Because it is always operating as a switching supply with very high efficiency, even when it's loaded so the output voltage is half the zero-current output voltage, it's not behaving internally like a Thevenin equivalent, even though it has a 50 ohm output resistance. When it operates at the transition of current/voltage saturation is can show 50 Ohms output impedance for very small change in load impedance. But as soon as the load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of the amplifier changes rapidly. For some additional input -- I have taken part in factory tests of high-power Class C single-tube/tuned cavity FM broadcast transmitters driving 50-ohm test loads measured to have 1.03VSWR, *showing a DC input to r-f output efficiency of the PA to be in excess of 80% -- including the loss in the harmonic filter. *Load power was measured using calorimetric methods. *In fact, 80% PA efficiency is the published spec for this transmitter line as long as load VSWR relative to 50 ohms is 1.7:1 or less (any phase angle). Those results don't appear to be fitting very well with the idea that the tx source impedance can be ~50 ohms for loads with low VSWR. RF As Wim says, the conjugate-matched output impedance is guaranteed only over a very narrow range of conditions. Specifically, it's for the condition where you can't adjust the load impedance in any direction without lowering the power delivered to the load. For a power output stage with a tank that can tune both resistance and reactance (such as a pi network), that should be equivalent to adjusting the pi network to maximize the power delivered to the load with a particular fixed set of bias/drive conditions for that output stage. But why would I necessarily want to operate the output stage that way? What if I'm providing enough grid drive and plate voltage to my 6146 that it's capable of outputting 150 watts (class C), but if I were to tune up the output matching network to get that much output power, the tube's life would be significantly shortened because the plate dissipation would be too high? What if I'm operating it class AB, with low enough drive so that the plate voltage and current change only by small percentages? Or to me, more to the point: why would I care what the source impedance of my transmitter is? With the power at the wall outlet in my house, I expect a nearly constant voltage, and have no intention of loading it with a "conjugate match." I don't know WHAT its source resistance is, exactly, but I know it's much lower than the loads I put on it. With an electric motor, I have no intention of putting a load in excess of the motor's rating on it, for fear of burning up the motor, even though I know that for short periods the motor can deliver quite a bit more output power than its nameplate rating. With a switching power supply, same thing: I don't load it to the point of significant drop in output voltage. All I want from ANY of these, including the RF power amplifier, is the ability to deliver the expected power to the rated load. In the instrumentation systems I deal with professionally, there are times I care very much about source impedance. With respect to ham transmitters and power amplifiers, though, I just can't get excited about caring that much what the source impedance is. The point of the amplifier or transmitter is to deliver power to my antenna system, and the source impedance it represents is irrelevant to that task. Cheers, Tom |
#6
|
|||
|
|||
"Non-dissipative Source Resistance"
On Jun 12, 11:56*am, K7ITM wrote:
On Jun 12, 8:17*am, Richard Fry wrote: On Jun 12, 9:44*am, Wimpie wrote: A real class C amplifier with very small conduction angle has efficiency 50% when optimally tuned. Would that not be evidence that the amplifier source impedance necessarily is lower than the load impedance? No, not at all. *It is important to realize that a real source is NOT necessarily like either the Thevenin or the Norton equivalent. *The Thevenin and Norton equivalents, when discussed in any respectable text, will be noted to behave exactly like any other real linear source with respect to the load, but NOT necessarily with respect to the source itself, and in particular with respect to source efficiency and dissipation. For example: *I can, at least theoretically, build a switching power supply (a source) which measures its output voltage and current and adjusts the voltage down by 50 volts for each amp that's drawn from it. *Because it is always operating as a switching supply with very high efficiency, even when it's loaded so the output voltage is half the zero-current output voltage, it's not behaving internally like a Thevenin equivalent, even though it has a 50 ohm output resistance. When it operates at the transition of current/voltage saturation is can show 50 Ohms output impedance for very small change in load impedance. But as soon as the load mismatch is above about VSWR = 1.05…1.1, output VSWR/impedance of the amplifier changes rapidly. For some additional input -- I have taken part in factory tests of high-power Class C single-tube/tuned cavity FM broadcast transmitters driving 50-ohm test loads measured to have 1.03VSWR, *showing a DC input to r-f output efficiency of the PA to be in excess of 80% -- including the loss in the harmonic filter. *Load power was measured using calorimetric methods. *In fact, 80% PA efficiency is the published spec for this transmitter line as long as load VSWR relative to 50 ohms is 1.7:1 or less (any phase angle). Those results don't appear to be fitting very well with the idea that the tx source impedance can be ~50 ohms for loads with low VSWR. RF As Wim says, the conjugate-matched output impedance is guaranteed only over a very narrow range of conditions. *Specifically, it's for the condition where you can't adjust the load impedance in any direction without lowering the power delivered to the load. *For a power output stage with a tank that can tune both resistance and reactance (such as a pi network), that should be equivalent to adjusting the pi network to maximize the power delivered to the load with a particular fixed set of bias/drive conditions for that output stage. But why would I necessarily want to operate the output stage that way? *What if I'm providing enough grid drive and plate voltage to my 6146 that it's capable of outputting 150 watts (class C), but if I were to tune up the output matching network to get that much output power, the tube's life would be significantly shortened because the plate dissipation would be too high? *What if I'm operating it class AB, with low enough drive so that the plate voltage and current change only by small percentages? Or to me, more to the point: *why would I care what the source impedance of my transmitter is? *With the power at the wall outlet in my house, I expect a nearly constant voltage, and have no intention of loading it with a "conjugate match." *I don't know WHAT its source resistance is, exactly, but I know it's much lower than the loads I put on it. *With an electric motor, I have no intention of putting a load in excess of the motor's rating on it, for fear of burning up the motor, even though I know that for short periods the motor can deliver quite a bit more output power than its nameplate rating. *With a switching power supply, same thing: *I don't load it to the point of significant drop in output voltage. *All I want from ANY of these, including the RF power amplifier, is the ability to deliver the expected power to the rated load. In the instrumentation systems I deal with professionally, there are times I care very much about source impedance. *With respect to ham transmitters and power amplifiers, though, I just can't get excited about caring that much what the source impedance is. *The point of the amplifier or transmitter is to deliver power to my antenna system, and the source impedance it represents is irrelevant to that task. Cheers, Tom Hello Richard Fry, I quote a sentence from your previous post: "Would that not be evidence that the amplifier source impedance necessarily is lower than the load impedance?" What you have said above is the key to the concern over the output resistance of a Clsss C amplifier being non-dissipative. What seems to be universally misunderstood is that there are really two separate resistances in the operation of these amps; one, the cathode-to-plate resistance, which is the dissipative resistance Rpd that accounts for all the heat, due to the electrons striking the plate; and two, the real output resistance that appears at the output of the pi-network, the resistance comprised of the voltage-current ratio E/R. I am speaking only of tube amps using an adjustable pi-network in the output. Due to the energy storage in the pi-network the output of the amp is inherently isolated from the upstream non-linear portion, thus the output is linear. One of the myths concerning the RF amp is that it cannot have efficiency greater than 50% because half of the power is dissipated in the source resistance. This would be correct if the output source resistance was dissipative, like in a classical generator. But in the real RF xmtr this is not true. Returning now to your quote above, your reference to the source impedance (resistance Rpd) is actually the plate-to-cathode resistance, the dissipative resistance, which is less than the output resistance R = E/I appearing at the terminals of the pi-network. I must add that when the pi-network has been adjusted to deliver all the available power at any particular grid-drive level, the output resistance of the network equals the load resistance, but not necessarily 50 ohms, i.e., whatever the load resistance might be. If you have doubts concerning the conditions I stated above, I invite you to review Chapter 19 in Reflections 3, which is available in two parts on my web page at www.w2du.com, as the first part appears in Reflections 2 and the second part appears in 'Preview of Chapters in Reflections 3'. In that chapter I use an example taken directly from Terman's 'Radio Engineers Handbook'. You will note that the dissipative resistance Rpd (3315.6 ohms) is less than the output resistance (6405 ohms) obtained from the E/I relationship appearing at the output This example verifies the explanation in the text. In addition, I also invite you to review the portion of the chapter appearing in the 'Preview Chapters in Reflections 3', in which I report additional measurements that offer additional proof that the output source resistance is non-dissipative. The numbers prove it to be true. Walt, W2DU |
#7
|
|||
|
|||
"Non-dissipative Source Resistance"
walt wrote in
: .... What you have said above is the key to the concern over the output resistance of a Clsss C amplifier being non-dissipative. What seems to be universally misunderstood is that there are really two separate resistances in the operation of these amps; one, the cathode-to-plate resistance, which is the dissipative resistance Rpd that accounts for all the heat, due to the electrons striking the plate; and two, the It is my understanding that the average power (heat) generated at the anode of a triode can be found by averaging the product of the instantaneous anode current and anode-cathode voltage over time. In a Class C amplifier, the voltage and current are not linearly related to each other, ie there is no constant of proportionality, no constant or fixed resistance. I don't understand why then, that people try to explain the anode dissipation in terms of some value of resistance. Owen |
#8
|
|||
|
|||
"Non-dissipative Source Resistance"
On Sat, 12 Jun 2010 06:13:15 -0700 (PDT), Richard Fry
wrote: It has been theorized that a circuit consisting of a Class C vacuum- tube r-f amplifier using a tuned tank circuit in its output network provides an operational “non-dissipative source resistance” of 50 ohms for energy present at the output connector of the transmitter. However the information and measured data provided in the text excerpts below is not very supportive of that theory. Hi Richard, Deja Vu all over again. I suppose you posted this to inspire me to, once again, remind you from Mendenhall's own notes about Class C amplifier construction - and so I will: "It was thus necessary to determine the plate load impedance = (Eb - Emin) / I1 ~= Eb/ Idc ~= 1000 Ohms [Not very far from Walt's determination of 1400 Ohms from his own bench.] "Since this was to be coupled into a Z output of 50 Ohms, a impedance transformation of 20:1 was needed. This was accomplished by using a voltage divider of two series capacitors to series tune the plate circuit." The tube is an EIMAC 4CX300A running no where near capacity. All details (and more) written in Mendenhall's own hand. 73's Richard Clark, KB7QHC |
#9
|
|||
|
|||
"Non-dissipative Source Resistance"
On Sat, 12 Jun 2010 06:13:15 -0700 (PDT), Richard Fry
wrote: It has been theorized that a circuit consisting of a Class C vacuum- tube r-f amplifier using a tuned tank circuit in its output network provides an operational “non-dissipative source resistance” of 50 ohms for energy present at the output connector of the transmitter. From Mendenhall: "VHF amplifiers often exhibit a somewhat unusual characteristic when tuning for maximum efficiency. ... If the amplifier is tuned exactly to resonance, the plate load impedance will be purely resistive and teh load line will be linear." 73's Richard Clark, KB7QHC |
#10
|
|||
|
|||
"Non-dissipative Source Resistance"
On Sat, 12 Jun 2010 06:13:15 -0700 (PDT), Richard Fry
wrote: It has been theorized that a circuit consisting of a Class C vacuum- tube r-f amplifier using a tuned tank circuit in its output network provides an operational “non-dissipative source resistance” of 50 ohms for energy present at the output connector of the transmitter. From "Care and Feeding of Power Grid Tubes:" "The first step in designing the output circuit is to specify the resonant load impedance of the tube, the loaded Q of the circuit and the desired output impedance of the network. The resonant plate load impedance for the 4065A is determined by dividing the plate peak RF voltage swing by the plate peak fundamental RF current. [work shows Resonant load impedance = 7600 Ohms] "if it is assumed that the output impedance of the network is to be 50 Ohms, and the loaded Q is to be 15, the output tuned circuit may now be designed. The output impedance of 50 Ohms will match into a properly terminated 50 Ohm transmission line." This document provides plenty of math and charts for the design of efficient output networks (finals' plate load). 73's Richard Clark, KB7QHC |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|