I have to admit I am really confused: When dealing with the RF power transfer in a 50 ohm system, I have been told repeatedly that we would want ALL of the amplifier (and filter) stages matched to 50 ohms. That way, I have been told, we would get the most power from, lets say, the output of a radio receiver into its demodulator*. But then how come the power does not decrease when a 50 ohm IF signal is inserted directly into a very high impedance (buffer) op-amp? Why isn't most of that energy reflected right back to the 50 ohm stages and wasted? I just don't get this! And if we can safely mix a low impedance stage with a high impedance stage and still be just fine, then how would I calculate the actual system gain and final output power of such a setup?
Signed;
Bill
In other words, if the stages are not all matched to the same reference impedance, then the output power would be less than what it should be.
Welcome to reality :-) Impedance matching is only necessary when you are working with fixed impedance modules.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
In a radio receiver, what's being passed from stage to stage isn't power, it's information. So unless you're driving the speaker, power transfer isn't a goal; so you do whatever works best.
Impedance matching is rarely done in a radio system if it is designed on one circuit board. Of course, if you have filters or modules that require a certain impedance you have to provide that impedance. Same if there is a long cable from one to the other.
Other than that, impedance matching was taught at most universities including mine. Maybe still is. They even went as far as to say that a large radio transmitter stage must have an output impedance of 50 ohms if it was to be connected to a 50 ohm filter or antenna. Oh man. If somebody really did that you'd see a massive plume of smoke, glass flying all over the place and some power transfer station shutting down. It'll be on the news. Transmitters are much lower in output impedance. If they didn't then their efficiency would be below 50% and not the usual 90+ for a large final amp.
On Sat, 16 Jul 2005 18:46:15 -0700, Jim Thompson wroth:
Another way to look at the situation...
Consider the AC power grid. If the generator at the power plant were matched to the impedance of the grid, then they would have to generate 200 watts for every 100 watts delivered to the grid. Half of their power would go into heating the generator. That's not realistic. And it illustrates the fallicy of the idea that maximum power transfer in a system always depends on matched impedances.
There are other reasons why impedance matching is a good thing, sometimes, but not everywhere.
It is only important to match the antenna impedance, because this signal runs through a long cable and would indeed be reflected or absorbed. But if the distance is much shorter than the wavelength reflections don't occurr.
An opamp doesn't have a high impedance at radio frequencies, because of its high input capacitance. That is why nobody uses opamps for RF stages, only for audio stages. Your view is completely wrong, you should understand complex impedances, it is not a constant scalar number but varies according to the frequency, even if the complex value is constant.
We don't care about power in a receiver, but S/N. Buffers have best values here when input voltage is maximized. We can amplify the audio signal to any convenient level just with a single stage at the end. Forget about impedance matching with audio circuits in a receiver. The wavelengths are too long to make a difference.
On Sat, 16 Jul 2005 20:32:45 -0700, John Larkin wrote:
Hi Bill! Welcome! We're Very Glad you're here!
"I have been told repeatedlly that the greatest thing I can aspire to is to be part of The Revolution."
"I have been told I am NOTHING if I am not part of The Revolution."
You've got to stop listening to the brainwashing, and do some actual thinking for yourself.
"The most dangerous thing I can say is 'I've been thinking.'" Oh, sorry, scratch that, we're trying to get over the programming...
This has to be one of the most confusing concepts in electronics. Perhaps it's seen as something mysterious and only understood by a very few electronics cult gurus, so most people take the "maximum power transfer" concept at its word and believe it to be "the only way to do it" rather than study the actual concept and see where it came from and where it might apply. Suppose you have a very long (several miles, and a few kilometers too) pair of wires over which a voice signal is transmitted. What impedance should one make the receiver (small speaker that goes next to the ear) so that it gives the loudest signal (extracts the most power from the pair of wires)? The question can be (slightly over) simplified to this DC equivalent:
For this example, you have a fixed voltage (the voice source voltage represented by Vbatt) and a fixed resistance (Rsource, representing the wire's impedance). What is the value of the load resistance so that it receives the most power? The answer (derived with simple calculus) is that Rload should be the same as Rsource. Most readers should recognize the 'very long pair of wires' as a telephone line, which has a characteristic impedance of 600 ohms, and so the telephone set should also have an impedance of 600 ohms to get the loudest signal out of the fixed amount of power being sent from the other end (the oversimplification is that for the AC case, Zload needs to be the complex congugate of Zsource, but in most telephone circuits the 600 ohms impedance is assumed to be resistive). Maximum power transfer needs to be done for a phone line because Rsource is rather large and cannot be reduced, and the power of the signal transmitted is rather small. In most situations, within different stages of electronics, Rsource is (or can easily be made) low enough that one essentially does 'voltage transfer' of the signal rather than 'power transfer'. Rload is high relative to Rsource, so very little signal voltage is dropped across Rsource, and the destination voltage is essentially the same as the source. In this case, lowering Rsource causes a higher voltage at the output (or the next stage's/device's input) and thus more power into Rload, whereas in the case of the telephone line, Rsource was fixed, and Rload is changed to be equal to Rsource to get maximum power into Rload. This is a good thing with audio power amplifiers (which with analog Class A and Class B aren't very efficient to begin with - increasing their source resistance will only make them less efficient) and especially with power generators.
The maximum power transfer idea definitely applies to an antenna connected to a receiver, as the receiver's input must be matched to the antenna's impedance (at the received frequency) to get the maximum signal from the antenna.
Only if receiver noise is a limitation, as it often is for microwave signals. For AM-HF bands, atmospheric noise is dominant, so you can make a very poor match to the antanna, as many low-frequency receivers do, and it won't matter. Since gain is so cheap nowadays, it's often not worth the trouble to match.
In instrumentation signal conditioning, impedance matching is practically an unknown concept.
Even here, noise match can be more important that power match. They don't (always) happen together.
--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)
Well, it's a theorem. Problem is, people tend to infer all sorts of things from it that aren't necessarily true, and it sticks with them throughout the rest of their career, unless they go on into a specialty where they
*really have to know* how to do it.
They even went as far as to say that a
Very few people appear to appreciate that "designing the output of an amplifier to deliver a specified power into a specified load" does not imply maximum power transfer.
The very last thing you want is for a transmitter output to absorb all the reflected power coming back from a badly mismatched antenna.
--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)
Why is it that the output (or internal) impedance of a practical RF power amplifier is seldom, if ever, anywhere near to 50 ohms or whatever it is supposed to be?
This question should be answered before proceding with the discussion.
Interestingly, the one place where one might think you want to match impedances for sure doesn't work out either. In connecting an antenna to the input of a low noise amplifier, maximum gain occurs for the matched case. BUT, often, if not usually, best signal to noise occurs when there is a slight mismatch.
On Sat, 16 Jul 2005 20:32:45 -0700, John Larkin wroth:
There's at least one point in a receiver at which power must be considered. That's the point between the antenna and the first RF amplifier.
The antenna is a power-limited source. There are only X number of microwatts of power available in the signal.
Any amplifier input requires both voltage and current. The product, watts, specifies the sensitivity of the amplifier.
Impedance matching is critical at the point between the antenna and the first RF stage. Since the antenna is usually a fixed impedance and the amplifier is designed for lowest noise contribution and may have an input impedance higher or lower than the antenna's impedance, passive impedance transformers are very often employed.
Fred had already explained it nicely. Just to illustrate it in a more simple term if you ever have to explain it to a person without RF background: In a class C power transmitter the final stage would ideally be a lossless and infinitely fast switch. Then 100% of the power that this stage takes from the DC rail would be transferred to the antenna. Less tube heater power, filter losses etc.
If this final amp were to have a true output impedance equal to the load impedance it would constantly burn 50% or more of the input power, basically dissipate it. It would slowly melt down because they usually aren't built to stomach that.
If an RF power amp is spec'd 50 Ohms that means it is built for that load. If you load it with less, you may exceed the current limit of the active element in there. A transistor would get hot, a tube would have the plate glowing pretty soon. I have actually seen the glass become soft and being sucked inwards. If you load it with much more than 50 Ohms and there are LC filters in there (class C kind of has to have these) chances are that the transistors die from overvoltage. In the case of a tube you might see some St.Elmo's fire inside or more serious arcing.
It's similar in the power world. A generator at Hoover Dam has an impedance that is a tiny fraction of the impedance of its load. Ideally its impedance would be zero Ohms but that isn't prossible because it uses regular copper wire and not a superconductor for the windings.
Maximum power transfer is not, repeat not, a consideration in the design of an RF power amplifier. What matters is that the amplifier will deliver a *specified* amount of power *into a specified load*. Take an example:
The load for a Class C transistor output stage, operating from a rail voltage Vcc and delivering power Po is given, to a first approximation, by:
Rl = ((Vcc)^2) / 2Po
For the derivation, see any book on transmitter design.
Hence, the correct load for a class C stage delivering 25 watts from a 28 volt DC supply rail is:
((28)^2) / (2 x 25) = 15.6 ohms.
In order to drive a 50 ohm load, it is necessary to interpose a matching network between the amplifier and the load, in order that the transistor is presented with 15.6 ohms resistive. This is usually an L- network of a capacitor in parallel with the 50 ohm load, which transforms the impedance "seen" into a resistive component of 15.6 ohms and a capacitive component which may be tuned out.and an inductor in series with the combination, to tune out the capacitive component thus introduced. It will look more complicated than this because it necessitates adjustable components, but basically it is just an L-network. Its "looking backwards" output impedance will not be 50 ohms, nor does it need to be. Nevertheless the circuit will deliver 25 watts into 50 ohms.
Nothing about *maximum* power transfer here, just the power transfer needed to meet the design specification.
Remember, under true maximum power transfer conditions, half the power gets dissipated in the load and half in the generator. That's not what you want.
It is highly undesirable to have the "looking backwards" output impedance of a transmitter equal to its design load impedance, otherwise, in the case of a mismatched antenna reflected power will be fully reabsorbed by the transmitter, and dissipated there, rather than be re-reflected back and eventually absorbed, partly by radiation, partly in feeder losses.
Remember, *source impedance does not affect VSWR* only load impedance does.
They do - in the context of the maximum power transfer theorem, but that's all it is - a theorem to be learned, then either applied or not, as the case may be. In the case of power amplifiers it is largely irrelevant. Other considerations than *maximum* power transfer apply.
--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)
Am I recalling incorrectly that the inherent resistance of the tube is essentially the slope of the graph (page 14 to which you refer)? Taking an example from the graph, holding VG1 at 50V, adjust the plate voltage from 3 to 4kV and the graph shows less than .1A change in plate current. That works out to about 1000/.1 or more than 10,000 ohms of internal resistance. The tube acts more like a voltage-controlled current source.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.