Transceiver impedance matching - Page 1
Forum section: electronics / rf-microwave

author: andre.lubbock
author_rank: Contributor
timestamp: March 06, 2021, 09:51:20 pm
content: When designing an antenna matching circuit, why do we assume the RF transceiver is 50 ohm? Shouldn’t we measure the transceiver impedance and match the antenna accordingly?I understand how to measure antenna impedance with a VNA/smith chart. Can we use the same tools to measure the transceiver impedance and how is this done in practise? Should the transceiver be power OFF or ON , TX or RX mode etc.?Thanks

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 07, 2021, 12:08:20 am
content: Because if it's not 50 ohm, it's a bad transceiver and you should feel bad.But seriously, it could be anything. It's most convenient, when everything is matched to Zo of whatever the transmission line is between them. So, 50 or 75 ohms for most coax, or 100 for twisted pair, or 300 for twin lead, etc.Put another way: because anything else, that way lies madness.Even then, not actually madness, but precisely tuning the network between antenna, TL and transceiver isn't going to be a pleasant experience, nor is it general (in terms of swapping out different cable lengths, or antennas or transceivers).It's worth noting that that madness sometimes shows itself, anyway. Often, transceivers are made with tuned circuits, and often antennas are resonant. Who knows what the combination of the two will do. It should behave within the common passband -- but the edges of those bands might not be as flat as expected, or you might end up with weird new passbands (producing images, or even overloading), or parasitic oscillation (more for transmitters).Also, note that there are two kinds of impedance we can measure. Consider an audio power amplifier: it has a "8Ω" output, but if you measure its small-signal output impedance, you'll find it's most likely some milliohms. What gives? The 8 ohms is only the expected load. (Conversely, most speakers these days expect a low source impedance, to give strong "damping".) Does that mean you could get more power from it, with a milliohm speaker? Absolutely, at small signals. It will enter current limiting very soon however (or worse..), and the maximum undistorted (or safe operating) power output is highest into the rated load. Output impedance and load impedance are two different things.The same still works at RF. We might have an open-collector output, which has a very high resistance, and capacitance dominates. It still needs a certain load impedance for maximum undistorted power, but this does not, in general, need to be the same as its small-signal impedance.Finally, bringing it all together, networks.We can have impedances mismatched, as long as the response does what we want it to. Generally the receiver will want to be near Zo, so that it both receives maximum power (from a linear source, like an antenna, which obeys the power transfer theorem), and so that the antenna can be somewhat arbitrary (narrow, wide, multiband, etc.), without having to tune the pair together.When the receiver isn't flat Zo, for example if it's a conventional reflective filter, then the reactance at the band edges, and the very mismatched impedance in the stopband(s), may combine with the TL and the antenna's own reactances, creating new and unexpected passbands.Conversely, if the antenna is flat, the receiver doesn't matter so much; but it's somewhat rare that we have a flat antenna. Examples in EMC work include a TEM cell, or ultrawideband antennas like a log periodic or ridged horn.- The TEM cell is little more than an oversized transmission line, with a terminator at one end -- fields inside it are automatically -3dB to the receiver: literally, half is wasted. It's rare that a DUT will couple very strongly with the fields anyway, so the gain is expected to be on the low side, regardless. It's also near-field, so not like it's an exact comparison with antennas in general.- UWB antennas have a scale-independent structure, which makes their impedance and gain flat over a wide range. A receiver with less bandwidth than one of these types, should behave just fine -- the power that isn't received, is simply reflected back out (at some phase shift). This might not be desirable, as it changes the fields, particularly near the antenna -- it's re-radiating out-of-band signals. Between field calibration and well damped environments (anechoic chamber or open-air test site) that's not usually a problem.So, preferably, we have both matched and flat, so we get high gain in the passband, and don't have to worry about out-of-band interactions. Constant resistance filters can help with this, or diplexers, etc.Likewise, the transmitter might not be matched (output) impedance. Here, then, the load should be matched and flat, to avoid surprises. In extreme cases, a bad match might even cause oscillation! This can be measured and analyzed -- an amplifier is "unconditionally stable" if it will not oscillate under any combination of source or load impedance. We might not even need that general case; presumably the transmitter final has a fixed source, so the only thing that's varying is the load. It would have to be pretty badly designed for that to happen, but it is definitely possible. (Another audio example: voice coil speakers are generally well behaved, being mostly inductive at high frequencies (10s kHz). Piezo speakers however, are strongly capacitive, with a strong resonance in there as well. This reactance can trip up poorly designed amplifiers.)Again, transmitter output impedances can be whatever. Some configurations are better than others: for example, a common-source amplifier with broadband neutralization (basically, a shunt feedback resistor from drain to gate) has a more resistive characteristic. An emitter follower might have a very low (or even negative?) resistance.Tim

author: TimFox
author_rank: Super Contributor
timestamp: March 07, 2021, 12:26:11 am
content: Short answer: power amplifiers (audio and RF) are designed to work efficiently into a specified load impedance, but the actual source impedance from the amplifier rarely equals that value. Similarly, a receiver has an optimal source impedance for best noise figure that may differ from the impedance you would measure connecting a VNA or other impedance meter to the input.It is true in linear circuit theory that a generator with a specified output impedance will deliver its maximum power into a matched load (complex conjugate is the output impedance is complex), but this would overload an audio power amplifier designed for 8 ohms load with a feedback-produced output impedance of, say, 0.25 ohms. Laboratory signal generators, however, typically have an output impedance (from added resistance and attenuators) that matches 50 or 600 ohm loads.

author: oliviasmithh900
author_rank: Contributor
timestamp: March 07, 2021, 01:15:39 am
content: I think the smaller version is awesome, who has time to read that much

author: andre.lubbock
author_rank: Contributor
timestamp: March 07, 2021, 04:10:06 am
content: Thanks for the reply T3sl4co1lTo clarify, I’m referring to a 50 ohm RF systemAre you saying the impedance of transmit and receive modes could be anything so don’t measure them ?All that’s important is TL and antenna are 50 ohm ?This seems to conflict with the theory that maximum power transfer occurs when source impedance = load impedance.Can you please clarifyThanks

author: andre.lubbock
author_rank: Contributor
timestamp: March 07, 2021, 04:13:01 am
content: Thanks TimFoxThe maximum power transfer theory you mention is the basis of my question. But you also say “the actual source impedance from the amplifier rarely equals load impedance.Are you saying that in my 50 ohm RF system I get less than maximum power transfer ? Doesn’t this impedance mismatch also cause large reflections ?

author: rfclown
author_rank: Frequent Contributor
timestamp: March 07, 2021, 05:38:41 am
content: Quote from: andre.lubbock on March 07, 2021, 04:13:01 amThe issue is a desire to standardize on an impedance (for instance 50 ohms) so that we can connect different things together (transmitter, amplifier, cable, antenna) without hassle. To see why 50 ohms was chosen, Google.When designing an RF power amplifier, you design it for a 50 ohm load (to work with everything else in a 50 ohm system). It is not a "matched" circuit at the device level. The intrisic device doesn't have a 50 ohm source impedance. If it did, it could only be at most 50% efficient (wasting half of it's power in the source resistance). The same term "match" is used ("match" the transistor to 50 ohms), but it isn't maximum power tranfer. It is a circuit designed to give a certain amount of power when loaded with 50 ohms.Say I have a perfect transistor, which behaves like an ideal source. I want to make an amplifier to produce 10 watts into 50 ohms. I want it to use a 10V power supply. Assuming I use an RF choke on the output, the output power will be P=V2/(2R). So for 10 watts and 10 volts, R must be 5 ohms. I want my transitor to "see" 5 ohms at it's output. But my desired load is 50 ohms, so I need to make a circuit which will transform the 50 ohms to 5 ohms. That is the "matching" circuit of an RF amplifier, but it isn't matching the 50 ohms to the impedance of the transistor (which in this ideal case is zero). It is transforming the 50 ohms to a desired load at the device.In a real design case, the transistor datasheet will say something like a 3-j2 load impedance with give 10 watts at 10V. So you design a circuit to transform 50 to 3-j2 at the device. They don't say that the transistor's source impedance is 3-j2 and you therefore design a 3+j2 load for maximum power transfer.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 07, 2021, 06:57:33 am
content: Quote from: andre.lubbock on March 07, 2021, 04:10:06 amNot to not measure them, but to be aware of when they may conflict.A network with two terminations is fine. Your standard LC filter calculators work with these. A transmission line looks like a perfect resistance, at any point along the line, when doubly terminated (hence the use of this type of bus in digital communications: 10Base2, most CAN and RS-485 applications, etc.)A network with one termination can still be fine, with some caveats. LC filters can be tuned for this condition. (Indeed they can be tuned for any ratio of resistances, as long as at least one port is terminated into a known resistance.) A singly terminated transmission line has a characteristic impedance as seen from one end ("load termination"), an R+C characteristic at the other ("source termination"), and a peaky response for connections midway along the line (where the open end acts as a stub with respect to the source). Hence, this configuration is often used for point-to-point communications (TTL and CMOS, LVDS, etc.), but is prohibited for multi-drop configurations.A network with no proper termination, will probably do unexpected things. It's not at all that it's unknowable -- it's perfectly knowable once everything has been measured, just slap together the s-parameters or whatever and see what drops out. It's just not going to be pretty...So, if you're connecting a reactive antenna to a resistive receiver, that's good.Or a resistive antenna to a reactive receiver, that's also good.Or both resistive, doubly good.Both reactive, beware. This is where you need to measure and check. We can't make any assumptions about it, we have to know the particulars.Conversely, connecting a reactive antenna to a resistive transmitter, that's good. (However, the SWR may be poor, and consequently the efficiency too... or worse things happen.)Resistive antenna to reactive transmitter (probably the most common case, again because transmitters tend to be collector/drain output, i.e. high source resistance, with some capacitance), also good.Both resistive, doubly good.Both reactive, same thing as before.Now, for a transceiver, with an rx/tx switch for example, you're probably going to prefer a resistive antenna, because the transmitter might not be well behaved. The receiver seems more likely to behave, but YMMV.Again, we can characterize everything involved in the system -- nothing need be left to chance. But we have to take the time to do that, on the specific hardware. May not be the easiest thing to test anyway (a transmitter's output impedance generally varies with power output, i.e., you have to somehow test its impedancewhile it's transmitting).It's perfectly possible to use sharply resonant antennas, antenna tuners, filters, etc., all inline; it's just not clear, in general, whether a particular combination of antenna, tuner, filter, or transceiver, will behave.QuoteRight -- that's the case for a linear Thevenin/Norton equivalent. But that's not the only possible case.We could for example have a naked voltage source, zero ohms resistance. How much power can be drawn from it? In the linear world, current just keeps going up as resistance goes down, so the answer is infinite power, at \$\lim{R \to 0^+}\$. In the real, nonlinear world, that voltage source only has some rated current output. Maybe its voltage drops, when that current is exceeded; maybe it just blows up. The point is we can't connect a load that low, for a variety of reasons, reasons that a linear theorem simply has no way to consider or express.Conversely, identical reasoning applies to a naked current source, and an infinite load resistance. The maximum output voltage may be limited for similar reasons.Another way to put it: the power transfer theorem is only valid in certain situations. Those situations show up often enough to be useful, so we teach it -- but there are other situations too, and they are more complicated, so they aren't taught until much later in school (hmm, I don't even recall if this case was highlighted in undergrad EE).Tim

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: March 07, 2021, 08:07:09 am
content: To add to T3sl4co1ls point (dammit Tim why do you make your name so damn hard to spell)When we talk about matching an amplifier, there are actually more ways to match. The most common, and the one most people 'know' is a 'small signal' match. Here we match to the thevenin equivalent. The small signal match also happens to be the impedance match where gain (power gain!) is maximal, and where (for small signals) the reflections are minimal. Note that if you match small-signal and then apply a very large signal you might see large reflections popping up after all, because the restive model falls down.For LNAs or other circuits, we might also match for noise - the source impedance where the noise transfer (IE, noise figure of the amplifier) is minimal might not necessaraly be the impedance where the gain is maximal. This is something done a lot at the input of the LNA.For power amplifiers, the load impedance where the output power is maximal might also not be the small signal match. Even more - it might not be the large-signal match either! What we do to figure out this impedance is a 'load pull'. For simplicity sake, I'm going to explain this in pure resistive case (ie, 1D), but in practice we have complex impedances (ie, 2D).We can sweep the load impedance and then look at the power delivered into the load. Think back to the audio amplifier that might have milliohm or lower output impedance. At very high load impedances, the power amplifier can deliver current to the load, but because it has a finite voltage rail, it cannot deliver more power than that voltage swing (well, twice actually). So you are power limited by that swing. As we decrease the load, the power into it will increase (as we are still limited by the maximum voltage swing, but a lower resistance will give more current at the same voltage, so more power). So we decrease it further, until at some point, we hit the current limit of the devices. In theory, we can decrease the impedance further to increase the power, but the amplifier itself might not survive, or, more often in RF cases, the amplifier might become non-linear. This process is called load-pull. The powergainof the amplifier will not be maximized, but the total power will be.We then match this impedance to our antenna or system impedance.

author: TimFox
author_rank: Super Contributor
timestamp: March 07, 2021, 03:20:51 pm
content: One way to set up a transmit/receive system:1. Adjust the transmitter for proper operation into a 50 ohm dummy load through a 50 ohm cable. The 50 ohm cable will be terminated by the load, and there is no reflection back to the transmitter.2. Adjust the receiver input for best noise figure from a 50 ohm source (e.g., switched-diode noise source in -hp- 8970 noise figure meter).3. Tune the antenna system to give 50 ohms resistive at your end of the 50 ohm cable.4. Then you can switch the antenna between the transmitter and receiver for good operation.5. For the transmitter power amplifier, consider a simple circuit with one active device (pentode, NPN BJT, or NFET) transformer-coupled to the load. Look at the characteristic curves: to get good power output, the load line will go from the upper left (high current, low voltage) to the lower right (low current, high voltage), and its slope gives the load resistance to be seen from the device (plate, collector, drain). However, over most of that region of operation, the current vs. voltage curves are relatively horizontal (high resistance), a much higher resistance than the load resistance. In audio, this could be reduced greatly with negative feedback to present a lower damping resistance to the speaker. In both cases (with or without feedback) the actual source resistance into the transformer is unrelated to the load resistance.6. The antenna circuit (including the tuning network and cables) is passive, and Thevenin's theorem applies. In step 2, if you adjust the receiver input circuit to present 50 ohms to the input connector, then the 50 ohm antenna circuit will deliver more signal to the receiver ("power matched"), but the SNR will not be as high as with the proper "noise tuning". This is important at VHF and above, and at low-level audio (microphone or phono pickup), but not so important at HF where environmental noise at the antenna can out-shout the electronic noise of the receiver.

author: E Kafeman
author_rank: Regular Contributor
timestamp: March 08, 2021, 06:17:19 pm
content: Many different opinions. For many embedded systems do we always measure impedance for both source and load to be able to optimize filter functions relative impedance, reduce uncontrolled RF-reflections in ground and many other parameters mismatch else could cause problems with. 50+j0 Ohm? that is only a goal for external antenna plug and similar.For precision tools is always 50 Ohm very precise matched for its external ports. Signal generator, its source impedance as well as input impedance in a RF power meter must be very close to ideal impedance with smallest possible reactive load or else would measured precision be bad.There are exceptions when Rx and Tx not is matched to be correct conjugate matched in embedded circuits.A Tx circuit optimal impedance load for best efficiency is seldom exactly same as impedance for best power transfer and in some cases is a certain amount of reflection a part of reducing harmonics.Matching for best Tx efficiency is often a important parameter for wearable units such as headsets as it increases talk-time for a certain amount of battery charge.Situation is similar for Rx as mentioned, best power transfer is for conjugate match but is not always same as impedance giving optimized SnR.Good thing, if you not know which is difference seldom big between these different impedance's so it is often good enough to select conjugate match.It becomes more complicated for transceivers where often Rx and Tx impedance differ. It can often be a weighted decision where often Tx wins due to EMI reasons.It is almost a rule that in a datasheet for an simple radio-chip is its impedance stated "50 Ohm". It not even a complete impedance value.If not else is said in data sheet is 50 Ohm then that ideal load actual chip guarantees its other measurement parameters and stability.Use conjugate match can even cause legal problems if Tx radiated level reaches above allowed levels, due to local rules or FCC as we can assume Tx max power stated in data sheet is measured with a 50 Ohm load.Actual PCB with its transmission lines can also add to result beyond what data sheet says for actual chip so it can be very individual measured value as impedance phase rotates with transmission line length.Situation is much similar in other direction, antenna impedance for embedded handheld units.Antenna impedance is often measured for three different situations: free space, handheld, handheld against head. Ideally resistive 50 Ohm as characteristic impedance does not exist for any of these situations.For these three different impedance's could a matching goal be that mismatch never is above a certain level or antenna efficiency always is above 40% or similar.Mismatch can also cause reduced antenna isolation for MIMO systems which reduces system performance.Similar for antenna polarization for circular polarized antennas, mismatch can increase Polarization Loss Factor, causing elliptical polarization.It can be tricky sometimes to match an antenna for different types of surrounding objects and to best meet all needs.Very common is that antenna impedance in free space must maybe be somewhat miss-tuned in free space with meaning so that it halfway against head passes center of Smith diagram for conjugate matched circuit to avoid too much mismatch when placed close to head.Not doing conjugate matching will affect efficiency and system stability and can case both radiated and none radiated EMI related problems.For some chips is a specific S-parameter curve stated. If it is anything else then general "50 Ohm" is it probably because of good reasons.Often together with shortcuts such as TIS,TRP and SAR.Most of above can be read at this site:https://www.antenna-theory.com/definitions/tis.phpImportant warning, most VNA's have a warning label for max allowed input signal level. If you circuit emits above that level must attenuation be used or else can it be expensive.

author: iMo
author_rank: Super Contributor
timestamp: March 08, 2021, 06:33:07 pm
content: Quote from: andre.lubbock on March 06, 2021, 09:51:20 pmThe output parameters of the final Tx stage (usually a push-pull amplifier with 2 transistors --> output transformer --> output filter) are designed to be as close as possible to 50ohm. The easiest method to measure the actual output impedance is - as has been described above: take two short pieces of 50ohm coax (equipped with 50ohm connectors) with 50+j0 ohm dummy load at the end and do measure the swr with an swr-meter. Something like:TX--------SWR-meter-------50ohm_dummy_loadBased on the measurement's result you may calculate the TX output impedance at the specific band/frequency/output_power..

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 08, 2021, 09:27:03 pm
content: Quote from: imo on March 08, 2021, 06:33:07 pmInteresting if true; can you show a proof of this method?Tim

author: TimFox
author_rank: Super Contributor
timestamp: March 08, 2021, 09:35:24 pm
content: With a VSWR meter between the generator and the load, the VSWR will be a function of the match between the load and the 50 ohm transmission line. It should not depend on the source impedance of the generator, if the load matches the line. With a perfect match between the load and the characteristic impedance of the cable, there will be no reflected wave in the cable (high return loss) and no standing wave.If the receiver input is noise-matched, and not power-matched, then even with an antenna tuned to 50 ohms, there will be a VSWR > 1.0.

author: radiolistener
author_rank: Super Contributor
timestamp: March 09, 2021, 03:25:22 am
content: you can measure transmitter output impedance by apply two different loads (for example 50 and 75 Ohm) and measure actual amplitude on the load.But note, it can damage transmitter output if you will use it at full power with unmatched load.Use minimum power mode for measurements to avoid damage of the output amplifier in your transmitter!Receiver input impedance can be measured with VNA.Note, be careful! Don't push TX button when VNA is connected to your transceiver, otherwise your VNA will be burned out.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 09, 2021, 05:31:44 am
content: And for the same reason, it's not so easy to test the TX output impedance with the VNA -- note that TX impedance varies with operation, it'll most likely just read open circuit if it's unkeyed. This is least problematic with class A amplifiers (which are probably small, too; there might be some chance of hooking this up, the VNA actually surviving, and getting a representative measurement out of it), and most dramatic with class C, E and other highly nonlinear types (which are also likely more than powerful enough to elicit a "VNA, what VNA?" if attempted!).It would however be possible, with a very good circulator -- the transmit power is directed to the second port (which gets terminated), VNA transmits into the third port (which gets directed into the transmitter). Where does the reflected VNA signal go? Out the second port, with the transmitted power. So we would have to use, not quite a terminator, but a power attenuator, on the second port -- and now we can read the sum of transmitter and reflected VNA signals. The VNA will have to compensate for the weak return (since it's attenuated), and must be selective for its own signal (not just like, reading the magnitude at its input port).So yeah, doing it by classic load shift is much, much easier.Tim

author: andre.lubbock
author_rank: Contributor
timestamp: March 09, 2021, 06:39:34 am
content: Thank you to everyone for the helpful comments. I’m trying to match a 4G cellular chipset to antenna. Trying then to digest all the comments… are these statementsgenerallytrue:1. In TX mode, chipset output impedance is much lower than 50 ohm.2. In RX mode, chipset input impedance is much higher than 50 ohm.3. It is safe to match the antenna to 50 +j0 and don’t worry about the input/output impedances of the chipset.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 09, 2021, 08:17:56 am
content: Oh well that's easy! What does the manufacturer say to do?In general, none of those is true:1. As per above, TX output impedance can be whatever.2. RX input impedance is likely to be whatever system impedance is. Assuming it's designed for 50 ohms, then that. But do check the datasheet. I don't know, maybe it's an oddball impedance, maybe it's differential, how am I supposed to know? There might be some fudge factor due to noise impedance being somewhat different as per above, probably not a huge amount.3. It is safe to match the antenna to whatever the RX/TX is designed to match to, with whatever matching network is specified, if any.Tim

author: radiolistener
author_rank: Super Contributor
timestamp: March 09, 2021, 03:51:07 pm
content: Quote from: andre.lubbock on March 09, 2021, 06:39:34 amNo, all 3 statements are false.Since you're using 4G, it uses very high frequency and impedance matching is very critical. Most of all, the chipset uses 50 Ohm input/output impedance, but you're needs to check it in the chipset datasheet.Impedance mismatch between chipset and the cable leads to a high power loss for TX and a bad sensitivity for RX. Both leads to worse communication quality and shorter distance.In some cases bad performance due to impedance mismatch is acceptable but it still leads to bad performance. It depends on your specification requirements and available budget for design.

author: E Kafeman
author_rank: Regular Contributor
timestamp: March 09, 2021, 11:45:42 pm
content: Ham radio does still leave a lot of space for tools such as VSWR meters and narrow band tuning by trial and error method until VSWR needle goes down.Modern wearable small RF chips for communication normally work over a frequency band, not just a single frequency.Relative wide BT 2400-2480 MHz or real wide as for some LTE circuits, 700-2700MHz.A matching network must match for whole freq. range, which is a complex curve for an embedded antenna. Radio impedance is also complex varying over a wide frequency range.Function of a reactive matching netwok is to match between these two curves.Rx, Tx, different Tx levels, different surrounding such as handheld or free space does also add to curve variations but it is in most cases just a part of the job to find a good enough weighted solution that fit all cases without to much losses or reflections.Impedance matching did become a necessary thing when cellphone antennas did become both embedded and dual band, 900-1800 or 850-1900MHz, which did happen more then 20 years ago.Without matching circuit was it else not possible to design an antenna that had high efficiency at two different bands which still is hard and more band have since then been added and available antenna space have been reduced.To keep SAR low is it important to avoid RF in ground loops which else will cause hot spots were it shouldn't.These RF currents are caused by not proper impedance match of Tx circuit, resulting in reflective currents.I have designed such antennas and matching circuits whole this period, since it all did begin with embedded antennas.No destroyed VNA yet.Any TX power above 50mW and a attenuation is added before calibration, typical 10-15 dB if max TX power is 30dBm for DUT.Noise floor will be lifted due to the attenuation but remaining dynamic range of 60-100 dB is more then enough to evaluate best matching circuit with a quality VNA.Proper impedance tuning is not an option, it is a must if good distance coverage and good battery economy are factors for wearable things.Even if a BT antenna is an external ideal 50 Ohm antenna and if BT-chip output impedance is ideal 50 Ohm, how many are able to design 20 mm transmission line such that it not have become something very else from impedance view at external port without proper matching design?Tx impedance is normally measured in by FCC-specified constant carrier mode, no matter if it is BT, WiFi, Lora or nb-IoT, no worries.Class E is a special case but we all know that, it is no big deal, except for its harmonics which often are at the limit for what can be attenuated as a part of the matching circuit.Many BT/WiFi transceivers have balanced Tx and have an impedance around 100-150 Ohm (vector value!). No problem, I am used with it and no problem to measure with a VNA calibrated for 50 Ohm. A VNA does not stop working because measured impedance not is ideal or balanced and center of Smith chart is just a point among others.From practical view, if Tx impedance is 100+j100 Ohm, is it perfectly fine if matching circuit in serial with antenna measure 100-j100 Ohm, measured at same location for both directions. No need to temporary jump down to 50 Ohm in between as it then will need two matching circuits to no use.Double-step matching just adds to cost and losses and makes tuning unnecessary complex.Tx and Rx and if needed, TIS and TRP, are verified at least as a final measurement in an anechoic chamber to check that implemented matching network does what it is expected to do and for the different specified surroundings and positions.VSWR-meter or coaxial cable is ok to use as measurement element as long as they not extend transmission line at PCB and as long as resulting value is a frequency curve. It is a must to be able to do a wide band tuning, covering radio full frequency range.Matching for a single frequency is more of use for narrow band radios, such as 144MHz.433MHz is also narrow-band but must be tuned wide band to be able to cope with different type of surrounding, free space and handheld for example, with reasonable well tuned antennaMeasured impedance is a result of chip impedance in serial with your PCB-design transmission lines and your connecting measurement tools.Include measurement tools in circuit matching is not right way to do it.That is why most VNA's support proper port forwarding. Port forwarding can not be done with a VSWR meter.For similar reasons are ferrit tubes often used to avoid that VNA measurement cable extend ground properties for DUT.Using an VNA and be able to measure in a correct way can be demanding compared to hook up a VSWR-meter but a VSWR meter is not an usable tool for wide band matching and gives no reactive information. It is not enough information to be able to design a wideband reactive network.A conventional VNA is an expansive instrument for good reasons but for personal use or just for learning can a tool such as nanoVNA be of high value at a low cost. My first VSWR meter did cost more then a nanoVNA and it was then barely usable at as highest 30MHz.

author: rfclown
author_rank: Frequent Contributor
timestamp: March 10, 2021, 01:15:41 am
content: Quote from: T3sl4co1l on March 09, 2021, 08:17:56 amThat is your answer in a nutshell. Even if you have the equipment to measure the antenna, and the software tools to design a matching circuit, you still have to start with the manufacturer's datasheet and app notes. Otherwise, you are in the dark and no one can suggest what to do.

author: David Hess
author_rank: Super Contributor
timestamp: March 10, 2021, 12:26:28 pm
content: Maximum power transfer occurs if the output impedance of the power amplifier matches the load however this also results in half of the power being dissipated in the amplifier so for higher efficiency, the power amplifier's output impedance is much lower.

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 02:32:46 pm
content: Yes. In general, with power amplifiers (audio and RF), the actual output (source) impedance differs from the optimal load impedance. The terms "output" and "load" are often misused.

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 05:05:15 pm
content: Quote from: TimFox on March 10, 2021, 02:32:46 pmNo, any impedance mismatch leads to lowerefficiencypower transfer. The bestefficiencypower transfer is possible for good impedance match only.Low impedance amplifier cannot keep max power on a high impedance load, due to Ohms law (it cannot keep high enough voltage on the load to put the max power into the load).For RF applications almost any power amplifier has the same impedance as feeding line, the standard impedance is 50 Ohm. This is because RF frequency can make significant standing wave even on a short transmission line with unmatched load on the end.Low impedance amplifiers are used for ultra low frequency, where wave length is too long against transmission line and where bad power transfer efficiency is not an issue. For example audio amplifier, where max power depends on a speaker impedance, but this is not a big issue.

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 05:55:19 pm
content: A 50 ohm cable terminated with a 50 ohm load (resistor or antenna matched to 50 ohms through an appropriate circuit) at one end will present an impedance of 50 ohms at the other end.A transmitter with good efficiency from its power source into that load will, in general, not have a source impedance equal to that load impedance. The loss of power in the transmission line (with finite loss per unit length) itself from the power incident into one end to the power into a matched impedance at the other end does not depend on the source impedance.

author: David Hess
author_rank: Super Contributor
timestamp: March 10, 2021, 06:15:38 pm
content: Quote from: radiolistener on March 10, 2021, 05:05:15 pmThat is completely wrong in practice for the reason I gave; outside of special and low power applications, nobody operates an RF power amplifier into its characteristic impedance because this immediately results in half of the power being dissipated in the amplifier. You can get more output power with a proper source match, but modify an existing RF power amplifier to do this will destroy it. An feedback audio amplifier would also be destroyed for the same reason; you sure do not see them using 0.1 ohm speakers although this would be briefly great for high power.When designing RF power amplifiers, no consideration is given to matching output impedance. Instead with the output voltage being known, the output matching network is selected to produce the desired output power without exceeding the current and power dissipation available. A matching network which matches the actual output impedance to the transmission line will produce more output power, but also very likely destroy the amplifier by exceeding current and power limits.QuoteAnd that is wrong also; the transmission line only requires a load termination which then prevents a reflection whether the source is properly terminated or not. The situation you are describing is double termination with proper source and load termination which is not required.This subject came up in old RF test equipment all of the time. While a double termination is desirable for many reasons, it also sacrifices signal level so more gain is required because efficiency is *lower*, with half of the power being dissipated in the source. The high impedance of series feedback stage or the low impedance of a shunt feedback stage will happily directly drive a transmission line with a proper load termination and see only a resistive load with no reflections. The advantage of this is that for a given gain-bandwidth product, the frequency response is several times higher or less gain is required so fewer stages are needed.Outside of power amplifier design, I first really encountered it with oscilloscope vertical outputs where Tektronix used a source impedance of 950 ohms to drive a 50 ohms transmission line. Bandwidth depended *only* on the load termination.Update:This is also why source termination is *not* required, or even desired, for digital line driving. A special gate is required to meet the output current requirements but only a load termination is required to prevent reflections. Source termination is actually undesired because combined with the load termination, the logic levels would be divided in half. This is why parallel termination of digital logic works at all.Another thing about digital line driving is that often the source impedance is wildly different between logic levels, with the extreme case being open collector/drain drivers like with coaxial Ethernet and CAN bus.

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 06:32:28 pm
content: Quote from: TimFox on March 10, 2021, 05:55:19 pmYes, the loss of power in the transmission line doesn't depends on the source impedance. With one exception, when we don't have double reflection from both sides of transmission line.But the loss of power on the transmitter output amplifier depends on the load impedance applied to the transmitter output.Just forgot for transmission line. We're talking about amplifier and it's load.The output amplifier is working from a fixed voltage, so it cannot produce higher voltage for case when the amplifier is loaded to a high impedance load. It requires to use impedance match between low impedance amplifier and high impedance load.Also, the current across the output transistors is also limited, so they cannot produce higher current for the case when the amplifier is loaded to a low impedance load. It requires to use impedance match between high impedance amplifier and low impedance load.So, the best case when amplifier is loaded with optimal impedance which is equals to the output impedance of the amplifier. Otherwise you will need to use impedance match circuit.

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 06:54:50 pm
content: The power out of the transmitter into the load depends on the load impedance seen by the transmitter output connector.That does not mean that the optimal source impedance for power efficiency from the power source (Vbb for tubes, Vcc for BJTs, etc.) to the load equals the load impedance. David Hess gives a detailed explanation.As a simple example, consider a 12 V storage battery with 0.1 ohm internal resistance. It can deliver 120 A into a dead short (but no power, since the dead short has no voltage across it), or 0% efficiency; 60 A at 6 V into a 0.1 ohm load (360 W, maximal power for 50% efficiency; or 10.9 V at 10.9 A into a 1 ohm load (119 W) for 91% efficiency. Into a very high load resistance, with zero current, the voltage is maximum but there is no power delivered and the efficiency is not defined (0/0 is indeterminate).In the maximum power case (0.1 ohm load), half of the energy stored in the battery is wasted. In the shorted case, all of it is wasted. In the high load resistance case here, only 9% is wasted. This parable relates to efficiency.

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 07:48:54 pm
content: Quote from: TimFox on March 10, 2021, 06:54:50 pmLet's see another example, more close to power amplifier.Assume that the load is const 50 Ω and power supply is 12 V.So, let's check efficiency for different amplifier impedance:- 10 Ω: I = 0.2 Amp, Pout = 2 W, efficiency = 83%- 50 Ω: I = 0.12 Amp, Pout = 0.72 W, efficiency = 50%- 100 Ω: I = 0.08 Amp, Pout = 0.32W, efficiency = 33%10 Ohm leads to a high current. The current is limited by transistors. So there is no reason to make low impedance amplifier, because it will require more expensive transistors and more distortions. Isn't it?The same there is no reason to make high impedance amplifier, because it will needs power supply with higher voltage to put the same power into the load. Higher voltage needs more expensive transistors. Isn't it?

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 08:11:29 pm
content: Quote from: David Hess on March 10, 2021, 06:15:38 pmThis is true when transmission line is properly terminated at the end of line (don't have reflected wave). Otherwise we can catch second reflection from the input of transmission line. Even in a digital logic the small reflected waves leads to a jitter or unwanted phase delay. Isn't it?

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: March 10, 2021, 08:25:09 pm
content: Quote from: David Hess on March 10, 2021, 06:15:38 pmIt's not entirely true that no consideration is given to matching output impedance. It depends on the application, but in cases like arrays (which are really becoming quite common) you do need to take this into account, since mismatch will result in apparent phase shifts (because you have reflections and coupling into the antenna, which will reflect at a mismatched PA output again, with a phase offset to the original signal)Also, in higher frequency bands, given modern CMOS technologies, they have such low voltages that they really do match for the highest output power.

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 09:44:22 pm
content: In your example with constant load impedance and variable source impedance, you obtained higher output power (into 50 ohms) and higher efficiency with the lower source resistance. How does that demonstrate your point?There is a tendency to confuse source or output impedance with load impedance and transmission-line characteristic impedance.Note that test equipment, e.g., signal generators, typically have a source impedance that matches the load impedance, but that usually requires adding a resistor after a lower-source-impedance output and an attenuator after the resistor to obtain accurate voltage outputs, but that reduces the power efficiency (not generally important for test equipment).

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 09:49:11 pm
content: Matching the load or antenna to the transmission line is a good thing, to avoid standing waves on the line that will increase the power loss in the line. That is a separate question from1. What load impedance is appropriate for the transmitter output.2. What is the actual source impedance of the transmitter output. A poster above discussed measuring the source impedance by varying the load resistance (best done at reduced power).

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 10:09:40 pm
content: Quote from: TimFox on March 10, 2021, 09:44:22 pmI know that reflection from transmission line output doesn't depends on load impedance applied at transmission line input. But we don't talking about transmission line after amplifier connector and don't talking about standing wave in the transmission line.We're talking about amplifier and it's load (transmission line length = zero).All RF power amplifiers that I tested have output impedance close to 50 Ω. The only exception is a power amplifier designed to use load different than 50 Ω.

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 10:14:00 pm
content: That simplifies the situation further. It is still true that when you connect the appropriate load resistor or load impedance directly to the transmitter output terminal for efficient operation, so that there is only one relevantload impedance, then thesource impedancefeeding that load is not, in general, equal to that load impedance.It is difficult to measure the source impedance. I suggest that you measured the load impedance.

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 10:27:33 pm
content: Quote from: TimFox on March 10, 2021, 10:14:00 pmit's not so difficult. Just use two different load with known values and measure actual amplitude on the load. Then you can calculate output impedance.Let's assume we have two test loads:R1 = 50 ΩR2 = 75 ΩJust put fixed power on the amplifier input and measure amplitude on the each test load:U1 for R1U2 for R2Then calculate output impedance:R = (R1 - R1*(U1/U2)) / (U1/U2 - R1/R2)The possible issue here is to not burn your amplifier when testing with 75 Ω load. To avoid that, just do not use full power, use low power for measurements to leave some margins for the amplifier output transistors

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 10, 2021, 10:55:46 pm
content: Quote from: radiolistener on March 10, 2021, 07:48:54 pmNo, the problem is still more fundamental than that: amplifier impedance can be set independently of overall voltage and current range.QuoteIf I use an op-amp with mixed voltage and current feedback, I can adjust its output impedance by the ratio of those feedback signals. At one extreme, the circuit is the standard voltage follower (inverting, noninverting, whatever). At the other extreme, it resembles the Howland current pump.In all cases, the output voltage and current range are fixed at whatever the op-amp ratings are.I can adjust the output impedance to 1mohm or 1Mohm, and still get the same e.g. +/-6V 20mA range, with maximum undistorted power into a load near 300 ohms.QuoteHow are the transistors limiting current in that case? Your condition shows 12V into 60 ohms is 0.2A. The current is defined by the load resistance, and your model seems to be a constant-voltage amplifier into a physical resistance of the claimed value.QuoteSo BJTs don't exist? MOSFETs? Pentodes? Triodes with significant cathode degeneration? ...Photodiodes exposed to light?!All of these devices draw substantially constant current, either by themselves when under negative voltage bias (photodiode), or when wired in a common-(source/emitter/cathode) configuration and biased in the usual way.A 2N3904 might have a collector output impedance (h_oe) of 10kohms (at say Vce = 10V, Ic = 10mA), yet a voltage and current range of say 30V and +/-10mA, i.e. maximum power into a 3kohm load. Or maybe our supply is limited to 12V and an inductive load is not handy, so that it must be biased at 6V and the range is approx. +/-5V at 10mA, a 500 ohm load; perhaps it has emitter degeneration for a number of useful reasons, and its output impedance is even higher, 100s k, megs; in what way does this act to reduce its voltage or current range?(As it happens, (source/emitter/cathode) degeneration does reduce the voltage range, on account of its DC voltage drop; but if we include AC as well, that can be canceled out using an inductor. (Likewise, the output voltage range of this class A amplifier can be doubled.)Tim

author: TimFox
author_rank: Super Contributor
timestamp: March 10, 2021, 11:21:10 pm
content: Here is a good application note from NXP/Freescale (formerly Motorola) on RF power device impedances:https://www.nxp.com/docs/en/application-note/AN1526.pdfNote the paragraph on page 2, titled "Large-signal series equivalent impedances". They are discussing power amplifiers in the roughly 1 to 100 W power level. I quote:"Most often, the device output impedance is referred to as 'the complex conjugate of the optimum load impedance into which the device output operates at a given output power, voltage and frequency.' That is certainly a statement requiring some careful thought, especially since the term 'output impedance' is somewhat misleading. The designer new to high power devices should be aware that this so called 'output impedance' has no connection with the S22 small--signal measurement. Rather, as described in , it is the conjugate of the LOAD impedance at the fundamental operating frequency which allowed the transistor to 'function properly'." The note continues to discuss adjusting the load impedance for either maximum gain or maximum efficiency, as required.My main experience is with audio power amplifiers, where the source impedance (properly defined) never equals the load impedance, and that by a large factor, due to device characteristics (described in the previous post) and negative feedback. Even the single-ended triode folks, who often avoid feedback, typically operate with a load impedance two or three times the plate resistance, which is the actual output impedance of the tube circuit, not including the load resistance. Also, connecting such an amplifier to a loudspeaker load does not generally involve transmission line effects. However, audio signal generators often have an actual 600 ohm source impedance and are calibrated into a 600 ohm load impedance.

author: radiolistener
author_rank: Super Contributor
timestamp: March 10, 2021, 11:40:46 pm
content: Quote from: T3sl4co1l on March 10, 2021, 10:55:46 pm6 / 1000300 = 0.000006 A = 0.006 A max20 mA through 1 MΩ source+load will be impossible with ±6V op-amp. It needs for at least ±20000 V op-amp or impedance match circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 10, 2021, 11:55:20 pm
content: Quote from: radiolistener on March 10, 2021, 11:40:46 pmWhelp, pack it in boys, transistors don't exist. Guess I can always pivot to mechanical engineering...Tim

author: TimFox
author_rank: Super Contributor
timestamp: March 11, 2021, 12:04:22 am
content: radiolistener is still confusing output impedance with load impedance. It's unfortunate that Jacobi's Law, derived from Thévenin's Theorem, is so often mis-applied.Your examples of extreme output impedances obtained with appropriate negative feedback, 1 Mcorresponding to a good current source and 1 mcorresponding to a good voltage source, have the same constraints of maximum current and voltage into the load (of varying resistance) caused by the power supply to the op amp. You don't need 1 kV and 1 Mto create a 1 mA current source.Back at the beginning of this thread, I pointed out that when matching a receive antenna to a receiver input, that the AC version of Jacobi's Law will give the maximum signal into the input impedance of the receiver, but will not necessarily give the best signal/noise ratio. The equivalent audio result for selecting an input transformer for a microphone amplifier is to optimize the noise match (to the voltage and current noise generators of the op amp or input transistor), instead of the power match.

author: emece67
author_rank: Frequent Contributor
timestamp: March 11, 2021, 01:14:18 am
content: .

author: radiolistener
author_rank: Super Contributor
timestamp: March 11, 2021, 01:23:22 am
content: Quote from: TimFox on March 11, 2021, 12:04:22 amNo, I don't talking about load impedance RL.I'm talking about output impedance RS:Quote from: TimFox on March 11, 2021, 12:04:22 amYes, I'm talking about Jacobi's Law maximum power transfer theorem. But why do you think it is mis-applied here?Maximum power transfer is not equals to efficiency. But power amplifier needs to transfer maximum power into the load.In case RL> RS, power amplifier will transfer smaller power into the load.So, this is the reason why it needs RL= RS.Yes, it's just η=50% efficient, but it provides the maximum power into the load for available voltage and output impedance.And this is what is expected from transmitter power amplifier. Isn't it?And yes, you can improve efficiency by reduce output impedance, but it will leads to a worse power transfer into the load!This is acceptable for a low power amplifiers, but this is bad for a power amplifiers, where maximum power on the load is expected.PS: it looks that there is misunderstanding because I wrote "efficiency" instead of "max power transfer". So, I fixed that mistake in original message.Quote from: TimFox on March 11, 2021, 12:04:22 amI know that the best impedance match is not always the best for SNR due to a thermal noise and LNA noise figure limitations.But power amplifier works with a strong enough signals, so noise doesn't matter here.How this is related with a power amplifier?Quote from: emece67 on March 11, 2021, 01:14:18 amWhen we're talking about power amplifier with 1 MΩ output impedance, the current will be limited with internal source impedance RS= 1 MΩ.

author: TimFox
author_rank: Super Contributor
timestamp: March 11, 2021, 03:06:33 am
content: Historically, Jacobi’s Law was cited to contend that no more than 50% efficiency could be obtained in electrical systems. That’s why many systems, including electrical distribution and RF power amplifiers are designed for efficiency rather than maximum power output or maximum gain. Remember that low efficiency amplifiers, etc., stress their internal components—that’s where the lost power goes.The output impedance (high or low) in the op amp circuits discussed above is achieved by appropriate feedback, but did not affect the maximum power that can be delivered to the load.

author: radiolistener
author_rank: Super Contributor
timestamp: March 11, 2021, 04:10:56 am
content: Quote from: TimFox on March 11, 2021, 03:06:33 amIs there any example of power amplifier with output power more than 20 W and low distortions (not CW), which is designed for high efficiency rather than maximum power output?Quote from: TimFox on March 11, 2021, 03:06:33 amDoes it means that op-amp output impedance depends on the load?

author: rfclown
author_rank: Frequent Contributor
timestamp: March 11, 2021, 06:24:14 am
content: Quote from: radiolistener on March 11, 2021, 04:10:56 amMost that I've ever done. A power amplifier is designed for certain requirements: output power, gain, DC supply voltage, efficiency, linearity, etc. The designer must decide upon a topology to pursue, then find a suitable device(s). You don't start with "I want the maximum output power". Maximum of what? Someone could take a certain device, and for academic reasons try to design something to get the most efficiency or the most power out of it, but that is not how RF power amplifiers are designed. You want an example, so I just went to Digikey and searched for RF MOSFET, >=20W, and randomly picked an NXP part (because I've used NXP and had an idea of what the datasheet would probably look like (I used to work for Moto)). I picked AFT09MS031NR1. So just Google for the datasheet.I typed "aft09ms031nr1 datasheet" in Google, and the pdf was the third link down. Read the datasheet.https://www.nxp.com/docs/en/data-sheet/AFT09MS031N.pdfYou'll find that the manufacturer gives you a Z load (and source) to design for (page 7). This is not the conjugate match. It is the load, that they determined with load pull, which will give optimal power/gain/efficiency performance. Not max power, not max gain, not max efficiency. You build a circuit which will present that load (and source) to the device, bias it with the voltages they specify, and you will get the performance shown on the datasheet. They even give you an example circuit. You'll see from the data curves that they get over 70% efficiency at about 2dB compression. That is excellent. This datasheet is only showing CW curves, but you'll find other devices where they show linearity metrics for applications where that is a requirement.... I just went back to Digikey to pick another part that might show linearity metrics also (so chose a cellular band frequency). This datasheet is unusual in that it shows load pull results for maximum output power, and maximum efficiency. I've never seen that on a datasheet before (and I'm old and have seen alot of datasheets). Usually you are shown one source and load impedance at several frequency points which will give the datasheet performance results. The part is NXP A2T18S165-12SR3.https://www.nxp.com/docs/en/data-sheet/A2T18S165-12S.pdf

author: emece67
author_rank: Frequent Contributor
timestamp: March 11, 2021, 11:10:28 am
content: .

author: emece67
author_rank: Frequent Contributor
timestamp: March 11, 2021, 11:35:10 am
content: .

author: radiolistener
author_rank: Super Contributor
timestamp: March 11, 2021, 01:30:44 pm
content: Quote from: emece67 on March 11, 2021, 11:10:28 amAs I know, in RF the default equivalent for impedance is a series equivalent, so unless otherwise stated explicitly, the output impedance means the value for a series equivalent model.When someone talking about 50 Ω input/output/characteristic impedance, I understand it as related to a series equivalent ZS. Which is unable to put 20 mA through 1 MΩ with 6 V.Quote from: emece67 on March 11, 2021, 11:10:28 amYes, they have about 30-40% efficiency. Attempt to get better efficiency leads to a very high distortions. You can put some powerful MOSFET and use it in CW mode at higher efficiency, but this is not acceptable for AM or SSB, because it will produce too much distortion and a lot of power emission outside signal bandwidth.Quote from: emece67 on March 11, 2021, 11:10:28 amYes, if you want to get 12.5 W from the amplifier with 1 Ω impedance output, it needs to handle at least 5 Amps current at 5 Vpk. You can do it with parallel transistors.But if you don't match low impedance transistor output with 50 Ω load on the amplifier output, your amplifier will not use all power potential of your transistors. Isn't it?For example, you're needs to design 1 kW power amplifier for 50 Ω load to use it for AM/SSB/FM transmitter. It needs to deliver 316 V amplitude into 50 Ω load. Are you sure that 1 Ω output impedance is a good idea for that? What kind of transistors you will use?

author: virtualparticles
author_rank: Regular Contributor
timestamp: March 11, 2021, 02:39:46 pm
content: To measure the output impedance of a PA, one uses a "Hot S22" measurement. The PA is driven by a signal generator at some frequency fairly far away from the frequency where the S22 measurement will be made. This is to insure that the PA is actually biased and operating. The output impedance will often look different with and without a drive signal. Then a directional coupler is attached to the output to pick up the reflected signal from the PA output. This is routed to the Direct Receiver Access *(DRA) port "B" of the Vector Network Analyzer (VNA) along with any padding needed to reduce the signal to non-damaging levels. A 20 dB coupling factor is a good start for that. The high power output of the coupler is then attached to a pad to reduce the signal level and often then attached to the output of a booster amplifier. The input is attached to port 2 of the VNA. The stimulus signal then exits port 2, is boosted by the amplifier, reduced by the pad and then enters the coupler and then the output of the PA. Reflections are picked up by the coupler and measured through the DRA port. The coupler/DRA function is needed because the pad and the booster prevent measuring reflections from the PA directly.Instead of measuring S22 directly, one could also perform "Load-Pull" measurements to completely characterize a PAIn answer to an earlier question, it is not always possible to provide a conjugate match to the output of the PA. The PA may be unstable under that condition. That is why Load Pull is so useful as it maps out all possible gain vs load conditions.Cheers!Brian

author: emece67
author_rank: Frequent Contributor
timestamp: March 11, 2021, 03:16:52 pm
content: .

author: emece67
author_rank: Frequent Contributor
timestamp: March 11, 2021, 04:34:15 pm
content: .

author: antenna
author_rank: Frequent Contributor
timestamp: March 11, 2021, 05:25:43 pm
content: The reason transceivers are 50ohm stems from the reason we choose to make our coax 50ohm.When it comes to coax transmission lines, there are an infinite amount of impedances one could obtain by varying the dimensions. Turns out, there are only two impedances that matter. 30ohm and 77ohm. At 30ohm, coax can handle the most wattage. At 77ohm, the coax exhibits the least loss. Consider television reception, it is RX only. If the consumer is not putting wattage down the line, why not go with the least loss? When it comes to radio communication like amateur radio, we care about both receiving weak signals AND the ability to carry power. Therefore, the coax cables intended for radio communication are typically designed to incorporate the best of both worlds. But instead of going with 77ohm for TV and 53ohm for communication, some smart person thought it made more sense to round these numbers to 75 and 50 respectfully. For that reason, most transceivers are designed for 50 ohm.

author: radiolistener
author_rank: Super Contributor
timestamp: March 11, 2021, 07:28:37 pm
content: Quote from: emece67 on March 11, 2021, 03:16:52 pmYes, the port can be modeled with any model - series RLC, or parallel RLC.Both model can represent the same impedance.You can represent parallel RLC circuit with a series RLC equivalent. And vice versa.But RLC values for series and parallel equivalent circuit will be different.The default model is a series RLC.

author: TimFox
author_rank: Super Contributor
timestamp: March 11, 2021, 07:47:17 pm
content: Impedance:Z=R+j Xis a series model,Admittance:Y= 1 /Z=G+j Bis a parallel model.Gis not 1 /R, andBis not 1 /XThese are definitions, not defaults.

author: radiolistener
author_rank: Super Contributor
timestamp: March 11, 2021, 08:02:40 pm
content: Quote from: TimFox on March 11, 2021, 07:47:17 pmFor series RLC, impedance: ZS= sqrt(R^2+(XL- XC)^2) = R+jXFor parallel RLC, impedance: ZP= 1 / (sqrt(1/R^2 + (1/XL- 1/XC)^2)By default series RLC is used. Isn't it?Quote from: emece67 on March 11, 2021, 03:16:52 pmNo, I'm talking about the following sequence:[Transistor with 1 Ω output impedance] => [impedance transformer 1:50] => [50 Ω load]If you want to get 12.5 W on 50 Ω load in that scheme, the transistor needs to provide 5 Apk at 5 Vpk into impedance transformer input.Because impedance transformer input is 1 Ω. And transistor is loaded with 1 Ω.The power amplifier needs to deliverhigh power into the load. Not voltage and not current.12.5 W is what is expected.0.125 W is not enough.The load is fixed to 50 Ω, so you cannot cheat physics by reduce output impedance.Since voltage and current are limited. It just leads to a low power output.

author: TimFox
author_rank: Super Contributor
timestamp: March 11, 2021, 08:35:19 pm
content: Earlier, I referred you to an NXP application note for high-frequency operation of semiconductors, which carefully distinguished between output impedance and load impedance in RF amplifiers.Here is a simpler exercise based, not on models or on theorems that may or may not apply to the situation, but on a data sheet and manufacturer’s typical operating conditions. This is the simplest possible power amplifier circuit: a vacuum triode connected to the load through an ideal transformer. At low frequencies, there are negligible reactances and no transmission lines to complicate the circuit, and there is no feedback applied.First, download the data sheet (RCA HB-3, 1947) fromhttps://tube-data.com/sheets/049/2/2A3.pdfwhich contains the characteristic curves (plate current vs. plate voltage at different grid voltages) from 1933 on page 3. For your convenience, RCA put a 2500 ohm load line on this graph.Second, look at the “Amplifier – Class A1” listing on the bottom of page 1. This circuit is simply1. The cathode operated from a filament transformer with grounded center tap.2. A negative grid bias voltage of -45 V, with the AC input superimposed on it.3. The plate connected through the primary of the output transformer to a DC supply of +250 V.4. The load and transformer ratio to present an AC impedance of 2500 ohms in series with the plate.5. The quiescent point (with zero AC input) for the plate is +250 V and 60 mA.6. RCA states that the full output power is 3.5 W (mean power). Therefore, the AC voltage across the transformer primary is 93.54 V rms, or 265 V peak-to-peak. The plate circuit swings between +117.5 V at 113 mA and +382.5 V at 7 mA.7. The corresponding load line on the characteristic curves must go through the quiescent point (+250 V and 60 mA) with a slope of -(1/2500) A/V. The RCA-plotted load line shows this quiescent point on an interpolated curve at approximately -43.5 V grid voltage, and draws the line from the 0 V grid line at the upper left to -87 V grid voltage at the lower right (the numerical values in the 1947 table are slightly different from the 1933 graph).8. Now, what is the output impedance of the “naked” triode? The relevant parameter on page 1 is the “plate resistance” = 800 ohms.9. This is approximately 1/3 the load resistance seen by the plate. The 2A3 (along with larger tubes such as the 300B and the 304TL, or the voltage regulator pass tube 6AS7G), with a low amplification factor µ were used in order to achieve a low output impedance without feedback.As an exercise for the reader, I suggest you repeat this process with a pentode or beam-power tube, such as the 6V6. These tubes have a higher plate resistance than triodes (and much higher µ) , and usually the source impedance is greater than the load impedance. From the RCA RC-24 tube manual, a single-ended 6V6 at +250 V and 47 mA quiescent point, uses 5000 ohms load resistance but has plate resistance of 50,000 ohms. We see that source or output impedance is not directly related to the load impedance,QED.

author: radiolistener
author_rank: Super Contributor
timestamp: March 11, 2021, 09:28:53 pm
content: Quote from: TimFox on March 11, 2021, 08:35:19 pmhm... what happens if it will be loaded with 800 Ω load?

author: TimFox
author_rank: Super Contributor
timestamp: March 11, 2021, 10:16:40 pm
content: Good question.An 800 ohm load would represent a load line with a slope of 1/800 instead of 1/2500 A/V.If you draw that higher-slope line through the same quiescent point, you will get the same swing in plate current in the negative-ward direction (from quiescent to near zero current), so for reasonable distortion you need to limit the current swing in the positive-ward direction to get a symmetric waveform. With roughly the same current into a smaller load resistance, the power output will fall by roughly the ratio of resistances, or 1/3 in this case.The quiescent point in the RCA datasheet is already at maximum power dissipation: 250 V x 60 mA = 15 W. From that point, the 2500 ohm load line extends symmetrically from near the zero grid voltage curve to the zero current axis.The design process for these amplifiers tries to fit the load line and quiescent point into roughly the center of a region defined by the zero-grid-voltage curve, the zero-current axis, and the power dissipation curve (hyperbola) that limits safe operation of plate voltage and current. If you tilt the load line you need to shift the quiescent point to allow symmetrical swing in current from 0 to quiescent to twice quiescent without exceeding the power dissipation. The limits to operation of the active device are not a rectangular boundary (as in the op amp example limited by the supply voltage and current capacity).In standard textbooks (see chapter 13.2 of "Radiotron Designer's Handbook, 4th Ed", 1952, RCA reprint 1960; and chapter 3.7 of "Radio Engineers' Handbook", F E Terman, McGraw-Hill, 1943 for class-A single triode power amplifiers) the theoretical treatment for an idealized triode suggests that the load resistance for maximum power is approximately twice the plate resistance (as opposed to the 2A3 suggested operation at three times the plate resistance).

author: emece67
author_rank: Frequent Contributor
timestamp: March 12, 2021, 12:45:24 am
content: .