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How can one make a dummyload from a LC circuit ?
It started when the dummyload was too small for a job. It's for 40 mtrs. But making it is barely possible. It should have a Zo of 50 ohms ? And what about the Q factor ? Looking on my swr meter it gets better when i put my hands near it and at a very small C rate but still in the red. Anyone ? |
How can one make a dummyload from a LC circuit ?
On Sat, 21 Nov 2009 18:56:07 +0100, "Ronald" wrote:
It started when the dummyload was too small for a job. It's for 40 mtrs. But making it is barely possible. It should have a Zo of 50 ohms ? And what about the Q factor ? Looking on my swr meter it gets better when i put my hands near it and at a very small C rate but still in the red. Anyone ? Hi Ronald, If you mean by "dummy load" something with a 50 Ohm match and resistive (somewhat redundant to the 50 Ohms) then it has to be a resistor (department of triple redundancy). Finding a resistor that is sufficiently power rated is not a simple (cheap) chore. Making your own will take time and ingenuity. Consult: Caddock Electronics, Inc. 17271 North Umpqua Hwy http://www.caddock.com for resistors that will fill the bill - once you build the heatsink. 73's Richard Clark, KB7QHC |
How can one make a dummyload from a LC circuit ?
It started when the dummyload was too small for a job. It's for 40 mtrs.
But making it is barely possible. It should have a Zo of 50 ohms ? And what about the Q factor ? Looking on my swr meter it gets better when i put my hands near it and at a very small C rate but still in the red. Anyone ? An LC circuit won't make a good dummy load. A "pure" LC circuit is lossless... both the L and the C are pure reactances, and neither will (or can) dissipate any power. In practice, you could make an LC series circuit which would be 50 ohms resistive at a single frequency... you'd use a really lousy inductor (one wound with many turns of small wire) which has a pretty significant DC resistance. The series C would be just enough to tune out (resonate) the inductance at the frequency you're going to be using it with. The inductor's ESR at this frequency is going to be higher than its DC resistance, due to skin effect and other loss phenomena. The resulting circuit would *not* be 50 ohms resistive at other frequencies. It'll have a capacitive reactance at lower frequencies and an inductive reactance at higher frequencies. Seems like an expensive and annoying way to make a dummy load. Unless you've got other requirements you haven't mentioned, I'd suggest just making a halfway-decent dummy load from a bunch of noninductive resistors (e.g. in series-parallel) mounted on some sort of heatsink. -- Dave Platt AE6EO Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
How can one make a dummyload from a LC circuit ?
"Dave Platt" wrote to us It started when the dummyload was too small for a job. It's for 40 mtrs. But making it is barely possible. It should have a Zo of 50 ohms ? And what about the Q factor ? Looking on my swr meter it gets better when i put my hands near it and at a very small C rate but still in the red. Anyone ? A "pure" LC circuit is lossless... both the L and the C are pure reactances, and neither will (or can) dissipate any power. Not dissipate but surely radiate in a closed box! In practice, you could make an LC series circuit which would be 50 ohms resistive at a single frequency... Yes, it's for a single frequency. Since an antenne is also a LC-circuit that doesn't get warm but radiates in fields. I had it all open built so i suspect mantelwaves etc. But on paper a series LC circuit forms a loading of a particular Zo. This statement is still right? Lets say an input on a semiconductor p.a. stage with an LC series circuit.. |
How can one make a dummyload from a LC circuit ?
Suppose we feed 20 watts in a p.a. but don't want to use a dummyload in the p.a. Come on this is done all the time. |
How can one make a dummyload from a LC circuit ?
On Mon, 23 Nov 2009 19:18:37 +0100, "Ronald" wrote:
Suppose we feed 20 watts in a p.a. but don't want to use a dummyload in the p.a. Come on this is done all the time. Ronold, Your discussion gives every impression that you don't know 1. what a dummy load is 2. what it is for 3. why it is principally resistive. Your statement above is true on the face of it, but has no meaning beyond a tautology: you don't use a dummy load IN a p.a. because you don't use a dummy load in a p.a. So what? So, the direct answer to the question in the Subject line: How can one make a dummyload from a LC circuit ? is: You don't You seem to be hedging around another purpose but don't want to tip your hand. Obfuscation does not help find answers. 73's Richard Clark, KB7QHC |
How can one make a dummyload from a LC circuit ?
"Richard Clark" wrote in message ... On Mon, 23 Nov 2009 19:18:37 +0100, "Ronald" wrote: Suppose we feed 20 watts in a p.a. but don't want to use a dummyload in the p.a. Come on this is done all the time. Ronold, Your discussion gives every impression that you don't know 1. what a dummy load is 2. what it is for 3. why it is principally resistive. Your statement above is true on the face of it, but has no meaning beyond a tautology: you don't use a dummy load IN a p.a. because you don't use a dummy load in a p.a. So what? So, the direct answer to the question in the Subject line: How can one make a dummyload from a LC circuit ? is: You don't You seem to be hedging around another purpose but don't want to tip your hand. Obfuscation does not help find answers. 73's Richard Clark, KB7QHC I'd like to know more about the 'mantelwaves'! Chris |
How can one make a dummyload from a LC circuit ?
On Mon, 23 Nov 2009 19:18:37 +0100, "Ronald" wrote:
Suppose we feed 20 watts in a p.a. but don't want to use a dummyload in the p.a. Come on this is done all the time. The only reason that works is that the transmitter has a protection circuit that detects high VSWR and either reduces power or shuts down. The idea is to not destroy the output xisitors. Into an infinite VSWR load, you can easily find twice the collector voltage across the output stages, which on many radios, will destroy the xsistors. With other type of loads, the output stages could easily draw twice their rated current, this time destroying them with too much dissipation. Judging by your wording, I don't think you have a clue what a dummy load does or where to install it. It's not "in" the PA stage. It's after the PA, after the low pass filter, after the T/R switch, and in place of the antenna. You use a dummy load for testing and measurement so that you don't pollute the airwaves or fry your radio tuning into an unknown load. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
How can one make a dummyload from a LC circuit ?
A "pure" LC circuit is lossless... both the L and the C are pure
reactances, and neither will (or can) dissipate any power. Not dissipate but surely radiate in a closed box! It will not radiate efficiently. The radiation resistance of a physically-small coil is going to be very small... a small fraction of 1 ohm. What your transmitter is going to "see" will consist of: - The radiation resistance of the circuit (probably less than 1 ohm), in series with - The loss resistance (mostly in the coil) in series with - Whatever reactance remains after the series L and C resonate against one another. If the coil is "perfect" (lossless) and the L and C reactances are equal and opposite, then all you'll be left with is the radiation resistance. If it's in a closed (shielded) box, matters get *worse*. The presence of the reflections from the walls will reduce the radiation resistance even further. Will your transmitter be happy with a load which looks almost exactly like a short circuit to ground. Yes, it's for a single frequency. Since an antenne is also a LC-circuit that doesn't get warm but radiates in fields. And it radiates with reasonable efficiency, precisely because it's physically large - it has a substantial radiation resistance. This will not be the case for a physically-small LC circuit. I had it all open built so i suspect mantelwaves etc. But on paper a series LC circuit forms a loading of a particular Zo. This statement is still right? No, I don't think so... at least, not in the way that you think. If you show us the specific circuit you're thinking of (components and their values, wires, and desired operating frequency) we can analyze it. Lets say an input on a semiconductor p.a. stage with an LC series circuit.. A series LC circuit is not, by itself, going to present a resistive impedance. -- Dave Platt AE6EO Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
How can one make a dummyload from a LC circuit ?
Suppose we feed 20 watts in a p.a. but don't want to use a dummyload
in the p.a. Come on this is done all the time. Sure. If one does not want to have a dummy load in the PA, one uses an *external* dummy load. A dummy load, almost by definition, is something which accepts RF, and dissipates it as heat. An ideal dummy load doesn't radiate at all. If it does radiate significantly, we call it an "antenna". One *cannot* magically make those 20 watts disappear. They either have to be radiated (out into space) or converted to heat and dissipated. A small LC circuit will not do a good job of either of these things... won't radiate efficiently and won't have enough resistance to act as a comfortable dummy load (e.g. Zo near 50+0j). -- Dave Platt AE6EO Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
How can one make a dummyload from a LC circuit ?
Ronald wrote:
Suppose we feed 20 watts in a p.a. but don't want to use a dummyload in the p.a. Come on this is done all the time. Done all the time, but usually with resistive loads. Could you elucidate exactly why you want to do this? I can't think of any good reason to use LC. I suspect you might have some terms and principles a little mixed up. A dummy load is used to absorb RF power at the end of the transmit circuitry. The transmitter does everything it's supposed to do, but the power goes into the dummy load, and is dissipated as heat. Ideally no RF makes it out of the load. The ideal impedance for a transmitter is resistive. For that nothing can beat a resistor. - 73 de Mike N3LI - |
How can one make a dummyload from a LC circuit ?
"Dave Platt" wrote to our discintent: A small LC circuit will not do a good job of either of these things... won't radiate efficiently and won't have enough resistance to act as a comfortable dummy load (e.g. Zo near 50+0j). All amateurs here who i suppose 'believe' to much. Do you go to church every sunday ? - sometimes dummyloads are used in input stages of a valve or mosfet p.a. because valves and mosfet only use the rf voltage NOT the power. So the driver stage (xmttr) has to be dummyloaded and tapped. Understand ? -Then the remark : " a LC circuit won't radiate efficiently " . So what? Thats exactly what i need here. - so every LC series circuit you see in schematics, ìf you see them, you call shortcuts ?? - If you use an antennatuner for a 'nittingneedle' you will get a 1:1, and we don't care the bad radiation pff, and that antennatuner has a series LC circuit inside !! So no shortcut at all, dummy. So whats your point ? I only asked about the practice of the consept. I see its easier the become doctor in the usa then farmer in Belgium ... |
How can one make a dummyload from a LC circuit ?
On Tue, 24 Nov 2009 18:00:42 +0100, "Ronald" wrote:
I see its easier the become doctor in the usa then farmer in Belgium ... Classic trolling. |
How can one make a dummyload from a LC circuit ?
In article ,
Ronald wrote: - sometimes dummyloads are used in input stages of a valve or mosfet p.a. because valves and mosfet only use the rf voltage NOT the power. So the driver stage (xmttr) has to be dummyloaded and tapped. Understand ? Sure. And, if those are truly "dummy loads", they are *not* simply LC circuits. They have a resistor of some sort incorporated into them, which is responsible for providing the correct termination impedance and which serves to dissipate the power from the driver stage. In other cases, what you call "dummy loads" for driving a MOSFET are *not* simply "dummy loads". Rather, they are impedance transformers, to step up the RF voltage to the correct level needed to switch the MOSFET or tube properly. In many cases, an LCR input stage to a final amplifier may have both of these functions... impedance transformation *and* load termination. -Then the remark : " a LC circuit won't radiate efficiently " . So what? Thats exactly what i need here. In standard radio technology, a "dummy load" is a device which dissipates most of the power fed to it. An LC circuit isn't one. You *need* something with resistance in it, if you want to terminate the driving signal properly (i.e. dissipate its power). - so every LC series circuit you see in schematics, ìf you see them, you call shortcuts ?? How in the world did you come to that ridiculous conclusion? Please don't put words into my mouth. LC circuits have *many* useful functions in a radio schematic. Two of the primary ones are impedance transformation (matching), and filtering (bandpass or notch). Acting as a dummy load (in the accepted sense of the term) is *not* one of the good functions of an LC. - If you use an antennatuner for a 'nittingneedle' you will get a 1:1, and we don't care the bad radiation pff, and that antennatuner has a series LC circuit inside !! That antenna tuner has an LCR in it. The "R" is the loss in the coil... as I pointed out several postings ago, a coil does have a loss resistance. In the case you're talking about, "matching" a nitting-needle load with a typical T-configured antenna tuner may result in an apparent match. What's happening, is that you're creating a rather high-Q resonance inside the tuner... and all of the power is being dissipated in the coil's loss resistance. Very little of the power is actually going into the load - it's not really "matched". The coil heats up, and (if you've got a cheaply-made tuner) you melt the coil form and burn up the tuner. The problem with depending on this sort of circuit (e.g. using a coil, and a couple of capacitors) to dissipate power (acting as a very crude sort of "dummy load") is that it's unpredictable, dangerous, and inefficient. It's unpredictable because the series resistance of the coil is typically not specified by the manufacturer. In fact, the better the coil (heavier wire, open-spaced windings) the lower the series resistance and the losses, which means that you have to use a higher-Q tuning (more critical adjustment) to establish the match. It's dangerous, because the sort of small coils you'd normally find in the driver-to-final network of an amplifier can't dissipate the amount of power you're talking about (20 watts). They'll burn up. It's inefficient, because for the price of one adjustable coil which can be tuned in this way (with a couple of capacitors), will give the correct termination impedance (typically 50 ohms), and not melt down or catch fire or otherwise self-destruct, you can buy *dozens* of very well made 50-ohm noninductive resistors, which will give a good flat termination impedance over a *much* wider frequency range than a high-Q resonant circuit. Such resistors will be cheaper, smaller, and more reliable than a lossy-LC-circuit kluge, and won't require any tuning at all to work properly. So no shortcut at all, dummy. So whats your point ? My point is, you're attempting to choose the *wrong* technical solution for your problem. You'll waste money, and end up with a device which is bigger, tricker to tune, and less reliable than is necessary. I only asked about the practice of the consept. And, the accepted engineering practice of this concept is "This is a bad idea, we don't do it that way. We use resistors." -- Dave Platt AE6EO Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
How can one make a dummyload from a LC circuit ?
Dave Platt wrote:
In article , Ronald wrote: So no shortcut at all, dummy. So whats your point ? My point is, you're attempting to choose the *wrong* technical solution for your problem. You'll waste money, and end up with a device which is bigger, tricker to tune, and less reliable than is necessary. Your self control is admirable. I would have lost my temper, at least a bit. He is just a troll, after all. tom K0TAR |
How can one make a dummyload from a LC circuit ?
tom wrote:
Dave Platt wrote: In article , Ronald wrote: So no shortcut at all, dummy. So whats your point ? My point is, you're attempting to choose the *wrong* technical solution for your problem. You'll waste money, and end up with a device which is bigger, tricker to tune, and less reliable than is necessary. Your self control is admirable. I would have lost my temper, at least a bit. He is just a troll, after all. I dunno. Not to call Richard or yourself incorrect, but it could just be a language problem coupled with a short fuse. - 73 de Mike N3LI - |
How can one make a dummyload from a LC circuit ?
On Wed, 25 Nov 2009 10:08:09 -0500, Michael Coslo
wrote: I dunno. Not to call Richard or yourself incorrect, but it could just be a language problem coupled with a short fuse. Hi Mike, The best evidence is from his postings. If there is a language difficulty (we've been assaulted with paisano fruit seller dialect from Italy when this is obviously Hollywood cliché) then Ronald's re-characterizations expressed in understood English are not miscomprehensions of language. Ronold: "A dummy load is an LC circuit." anyone else: "A dummy load is NOT an LC circuit." Such an interchange is more seriously logic limited if "misunderstood." I think Google translate would cope easily here. Negating the original phrase does not lead to the sophism of doctors and farmers - trolling does. If you want to be gracious, we have a Belgian CBer who is trying to work 40M. That level of technical ability is NOT masked by translation. Reminds me of the dwarf's argument in the movie "In Bruges." 73's Richard Clark, KB7QHC |
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