Series coax termination for output really necessary? - Page 1
Forum section: electronics / projects

author: Martinn
author_rank: Frequent Contributor
timestamp: December 24, 2022, 09:11:00 am
content: Hi all,I am trying to build a low noise amplifier to measure PSU noise as described in LT AN118 (see my post herehttps://www.eevblog.com/forum/projects/measuring-switch-mode-supply-noise-hp461a-replacement-ada4895/)See my efforts so far below (straight out of AN118 and the ADA4895 datasheet).What I am wondering: When driving a 50 ohms coax cable, do I really need series termination at the output? The cable is going into the scope, which is terminated properly. So the only effect I could imagine is that reflections from scope input termination mismatch would at the (more or less shorted) amplifier output reflected again and give a comb-like amplitude error. Is that correct?BTW I checked the schematic of the HP461A - as far as I see, it does not have an output series termination, but requires it to be externally terminated with 50 ohms (also for proper biasing I assume).Series termination seems to be quite a waste in this concept becaucse the whole point of such a design is the net gain! Why waste 2x on output termination...?Thanks - Martin

author: ConKbot
author_rank: Super Contributor
timestamp: December 24, 2022, 09:38:56 am
content: In general, the series termination takes care isolating capacitive loads. and isolates the output cable from the feedback network. Notice on the opamp datasheet how little capacitive load it takes to start getting pretty significant gain peaking.If your output coax is a perfect 50 ohms, then yes the shunt termination will make it all good. If the coax isn't perfect and starts presenting a bit of capacitance, there goes your high frequency gain flatness.

author: jonpaul
author_rank: Super Contributor
timestamp: December 24, 2022, 10:05:04 am
content: What is the amplifier and what meter, scope, Analyzer?Signal level?A matched transmission line has Zsource = Zo line = Z load eg 50 Ohms.That achieves the best pulse, BW signaler fidélitéBut a 2:1 attenuation.Depending the amplifier, frequency of interest and calibration either no term or 50,75 Ohm maybe needed.See many spec sheet of the devices, app notes and papers.Jon

author: Martinn
author_rank: Frequent Contributor
timestamp: December 24, 2022, 10:22:18 am
content: Quote from: jonpaul on December 24, 2022, 10:05:04 amHi Jon,amplifier: See attached schematic, ADA4895 (20 and 40 dB gain).Frequency range: kHz to about 100 MHz (designed to capture switched mode PSU ripple)Scope... would currently be an elderly 300 MHz LeCroy with 50 input terminationSignal levels... up to some hundred millivolts I guess (as low as possible of course). Also depending on scope input noise.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 24, 2022, 10:29:10 am
content: It's indeed confusing how "series (source) termination" and "parallel (shunt/load) termination" are presented asalternatives, and the latter one is drawn as a single resistor. In reality, in a drawing which shows a signal source, 50 ohm line and 50 ohm shunt termination resistor, there isanother50 ohms hidden inside the signal source, only to confuse you!Source termination = single 50 ohm resistor at source -> no attenuationShunt termination = 50-ohm resistor at source PLUS 50 ohm resistor at load -> bog standard resistor divider which halves the voltage.

author: jonpaul
author_rank: Super Contributor
timestamp: December 24, 2022, 12:52:17 pm
content: Bonjour Martin,I am uninformed on the amplifier you show, my scopes and probes were sufficient over decades of SMPS design, mfg, compliance.Yes, matched système looses 1/2 Signal but retains wideband width and pulse fidelity.With no seris 50 ohm, the VSWR affects the recieved Signal at cable far end.For low freq eg audio, line hum, yes no série R is OK, above 100khz...1M use the 50 ohm.In general RF gen, EMI and SA are all 50 ohm matched systèmes.Siwastaja, The topics of SMPS EMI measurement are long conversations.Good réf on SMPS PSU PARD is the classic 1980s HP tech note on power supply measurements.For mains frequency and rough work, ripple just a 10x probe AC coup is OK.With wideband conducted EMI, a LISN on mains side, AC coupling and lots of protection. On load side AC coupling and perhaps a DC LISN.Besides a scope, an SA is required for real EMI testing or compliance.Most wideband scopes and SA have 50 Ohm input so the cables and coupling device should have 50 Ohm Zo.At low Signal level direct coax to a coupling cap and to scope, preamp.We often used current probes to distinguish CM from DM noise.Bon Fête de NoëlJon

author: Benta
author_rank: Super Contributor
timestamp: December 24, 2022, 10:38:15 pm
content: Yes, you need the amp outputs to be 50 ohms, which is done with series source resistors. You'll see why when you don't do it.A different comment: you realize that your "input protection diodes" aka power rectifiers have around 30 pF capacitance? Quite a lot at 100 MHz.

author: Mechatrommer
author_rank: Super Contributor
timestamp: December 25, 2022, 05:50:22 am
content: in order max power transfer... Zin must be = Zout... ie Z of coax cable + termination is 50 ohm... that is Zout... Zin is at opamp output, series resistor ensure Zin as nearly as 50 ohm throughout BW, see graph page 15 attached below... at higher freq, ADA4985 will increase its internal Z to 100 ohm... so external series resistor + capacitor network will try to flatten Zin as much as possible to get flatter respond at the terminated coax end. if you want gain by removing series resistor, yes you'll get bigger signal at low freq, but you will not get flat respond anywhere near 100MHz, and depending on you coax length, you may get reflection even at low frequency due to Zin <> Zout mismatch guaranteed. you need to watch that MIT video demo'ing effect of impedance mismatch (reflection / standing wave) on a string of mechanical sticks with varying length (impedance), someone may find it for you, i forgot the keyword. series output resistance may also isolate opamp output from capacitive load of coax cable, second attached is the effect of capacitive load on the opamp output respond. fwiw.

author: ejeffrey
author_rank: Super Contributor
timestamp: December 25, 2022, 05:52:11 pm
content: Quote from: Mechatrommer on December 25, 2022, 05:50:22 amNo. The maximum power transfer theorem states that for a given source, maximum power is delivered if the load is matched. It does not apply to selecting the source impedance. Adding a resistor on the source will reduce the power delivered to the load by 6 dB.The reason for source termination is to, as everyone else said, is for signal integrity not power transfer. It will further damp reflection from imperfect load termination, isolate the output amplifier from load reactance, and present a higher impedance (100 ohm) load to the amplifier.

author: Martinn
author_rank: Frequent Contributor
timestamp: December 25, 2022, 06:33:45 pm
content: Quote from: Benta on December 24, 2022, 10:38:15 pmI was also wondering about this. Normally you'd expect something low leakage, low capacitance...The input section is straight from AN118https://www.analog.com/media/en/technical-documentation/application-notes/AN118fb.pdfpage 18. Jim Williams suggested this as LNA input protection for measuring HV SMPS noise (AN118 subtitle: "A Kilovolt with 100 Microvolts of Noise").I replaced the suggested MUR110 (ultrafast rectifier, 100V, 1 A) with something similar available at LCSC (I'll have the whole thing assembled from JLCPCB).The purpose of these diodes is to protect the following LNA from the kV supply you connect at the input. As for the capacitance - the whole setup is intended to measure (already very low) noise (ripple + switch spikes, hence the 100 MHz bandwidth) at the PSU output. There will be already some hefty filtering with large capacitors in place so the additional 100 pF will essentially have no effect on the noise (if it would, you would have put them into the output filter in the first place).Any better alternative?

author: David Hess
author_rank: Super Contributor
timestamp: December 25, 2022, 06:41:02 pm
content: Quote from: Siwastaja on December 24, 2022, 10:29:10 amWhen the object is to prevent reflections and present a resistive load to the source, then any of the three termination methods work. Series termination, shunt termination, and both with double termination are all acceptable alternatives depending on the application.In old designs it was common to see shunt termination used, with the source being either high (transconductance amplifier) or low (shunt feedback amplifier) impedance, because a double termination halves the voltage for a loss of 6dB in gain-bandwidth product, which means adding more amplifier stages.

author: Martinn
author_rank: Frequent Contributor
timestamp: December 25, 2022, 06:44:59 pm
content: Quote from: Mechatrommer on December 25, 2022, 05:50:22 amBut would this not mean that this amplifier is not suited to drive a terminated 50 ohms cable?In the 100x schematic from the datasheet (Figure 52), they show its performance driving a 1 kOhms load (it seems). Should I use a separate cable driver to properly handle the low impedance output?

author: David Hess
author_rank: Super Contributor
timestamp: December 25, 2022, 06:54:15 pm
content: Quote from: Martinn on December 25, 2022, 06:44:59 pmThe excess feedback lowers the output impedance, so within the closed loop bandwidth, the output impedance is much lower than the output impedance of the amplifier itself. The same applies to any shunt or series feedback amplifier.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 25, 2022, 07:42:49 pm
content: Quote from: ejeffrey on December 25, 2022, 05:52:11 pmIt's a hilarious "theorem" - elementary school math, Ohm's law applied, total no-brainer.But it's a good lithmus test - people who parrot teachings always remind others about "maximum power transfer theorem", and in 99% of cases, get it totally wrong.But I'll do everybody a favor and introduce what this theorem actually is:Given a voltage source and two resistors in series, one of which is fixed at X ohms and you cannot change its value, you can maximize the power by choosing another resistor = X ohms, too. THAT'S IT!

author: TimFox
author_rank: Super Contributor
timestamp: December 25, 2022, 07:50:32 pm
content: At best, the maximum power transfer theorem relates to getting the maximum power gain from a linear system (meaning at low output power).It rarely gives a result relevant to generating the maximum output power from a practical amplifier.

author: Someone
author_rank: Super Contributor
timestamp: December 25, 2022, 09:57:11 pm
content: Quote from: Siwastaja on December 25, 2022, 07:42:49 pmPerformance art? A "correct" explanation was already provided:Quote from: ejeffrey on December 25, 2022, 05:52:11 pmRs Rl-----/\/\/\/------/\/\/\/------| |__ |/Vs\ |\__/ || || |----- -------- ---- -V = IRI = V/RP = V*IP = V^2/RVs = 1Resistors EqualPs = 0.5^2/1 = 0.25WPl = 0.5^2/1 = 0.25WSource = 0Pl = 1^2/1 = 1WLoad = 0Ps = 1^2/1 = 1WMaximum power into either the fixed resistor or the power into the sum of both resistors is maximised when the other resistor is 0 (short), not when they are equal (half power -3dB in sum, quarter power -6dB in individual).Given a voltage source the maximum power dissipation is with the lowest value resistive load. Impedance matching is/was about getting maximum power into a resistor connected to a resistive voltage source.

author: Mechatrommer
author_rank: Super Contributor
timestamp: December 26, 2022, 06:26:59 am
content: Quote from: Someone on December 25, 2022, 09:57:11 pmi didnt say to where maximum power transfer is... i meant maximum power transfer to the series resistor at the opamp output, not at the coax end (fixed 50 ohm). i'm still not wrong! who's parrot teaching? (attached)happy new year...Quote from: Martinn on December 25, 2022, 06:44:59 pmif the datasheet doesnt give you example or specification about driving 50 ohm load, thats the hint that the opamp is probably not suited for it, you can either use separate buffer, or fondle the network until it gives what you want. ymmv.

author: jonpaul
author_rank: Super Contributor
timestamp: December 26, 2022, 08:11:27 pm
content: Seems confusion here ré transmission of information eg monitoring Signals with accurate measurement eg Oscilloscope Probe.Vs Maximum power transfer eg the transmission of RF power in a broadcast station.These have different requirements for the source, cable and termination.The OP seems to ask for the first category, instrumentation and transmission of signalés and information, not power.Jon

author: bson
author_rank: Unknown
timestamp: December 26, 2022, 08:54:20 pm
content: Quote from: Someone on December 25, 2022, 09:57:11 pmRs is a given quantity, and is never zero. For a given Rs, the maximum power is transferred when Rl = Rs.In the case of a zero Rs above, maximum power would be transferred when Rl is also zero. The power would be infinite as you'd be dividing by zero.

author: David Hess
author_rank: Super Contributor
timestamp: December 26, 2022, 09:35:02 pm
content: A matched source and load provide maximum power transfer, but this is not the criteria used for large signal devices. The output impedance of an RF power amplifier driving a 50 ohm load is typically only a couple ohms. While in theory much more power is available with matching, this also reduces efficiency, and would normally destroy the source. The matching network at the output of an RF power amplifier deliberately does an impedance transformation to mismatch the impedance of the power amplifier to limit the power to an acceptable level.

author: Someone
author_rank: Super Contributor
timestamp: December 26, 2022, 09:45:55 pm
content: Quote from: bson on December 26, 2022, 08:54:20 pmSo why take a HUGE QUOTE and cut out the line that already said exactly that?Signal to noise going in the wrong direction here!

author: bson
author_rank: Unknown
timestamp: December 26, 2022, 11:15:22 pm
content: Quote from: Someone on December 26, 2022, 09:45:55 pmI see:QuoteI see it saying the max power into either the fixed resistor or the sum of both resistors is maximized when the other resistor is 0. Giving slack for the horrible english grammar, I interpret this to mean power is maximized when one resistor is 0. No? Maybe you meant something different?

author: Mechatrommer
author_rank: Super Contributor
timestamp: December 26, 2022, 11:32:52 pm
content: theoritical maximum power will converge to singularity resulting the source has to collapse into black hole..

author: Someone
author_rank: Super Contributor
timestamp: December 27, 2022, 12:32:10 am
content: Quote from: bson on December 26, 2022, 11:15:22 pmYou want to discuss something technical that requires strict and terse language and the complain when the explanation is strict and terse? The long post was concluded with the simple explanation for when impedance matching is required and works and people juts keep ignoring it.... oh, I see it must be the season of pointless argument.ejeffrey stated the basis plainly and clearly, to correct the misleading generalisation from Mechatrommer:Quote from: ejeffrey on December 25, 2022, 05:52:11 pmI requoted that good explanation (in full with context) when yet another incorrect generalisation was posted by Siwastaja.Mechatrommer then posts up some negative reaction trying to be clever with mansplaining, that was only repeating the conclusion that I had agreed with and already posted, and explicitly repeated in my post. But by removing that (very intentional) sentence it makes it look like you (bson) are correcting something (which never existed as the bolded part above already said that explicitly, plainly, and clearly).Quote from: Someone on December 25, 2022, 09:57:11 pmPeople are jumping around and arguing about different things, I put up a larger post which clearly separates the two, then user (Mechatrommer) comes in and try to remove that delineation and continue arguing on the confusion. Its stupid, misleading, and only making the information here harder to follow.There is nothing incorrect in the post I put up, but now we have to mop up a half a dozen posts of nonsense. Seems disruptive and unhelpful.But since you insist....Quote from: Someone on December 25, 2022, 09:57:11 pmTalking about a source with a fixed output impedance is something completely different to talking about two resistors in parallel.Maximum power is produced/created from a source terminated signal into a short. That isn't very useful if you are trying to get power into some load, but all these stupid (and incorrect explanations) skip over that and make a nonsense generalisation. A better generalisation is two resistors in series generate the most power when their series resistance is minimised.Coming back to termination and what ejeffrey put so well, given a fixed load termination the maximum power isnotdelivered by matched source impedance. Mechatrommer had incorrectly flipped the clauses around and ejeffrey corrected it. A source with zero impedance delivers the most power to a given load..... but as the op was realising and asking, then why add source impedance?But we cant even approach answering that because there is so much bickering and noise about the underlying basics, filling up this forum with noise and rubbish.

author: Mechatrommer
author_rank: Super Contributor
timestamp: December 27, 2022, 12:48:43 am
content: having a bad day? too much negative power transfer in you? i wasnt replying your post...

author: Martinn
author_rank: Frequent Contributor
timestamp: December 27, 2022, 11:53:47 am
content: So, summarizing so far:- Output series termination would for an ideal terminated transmission line not be necessary, as it would be equal to a resistive load- as real world matching will be less than ideal, an isolation resistor will be necessary to avoid peaking or oscillation- "matched termination" probably does not apply here as the output impedance of an ideal op-amp is zero.However, Mechatrommer gave a good hint:Quote from: Mechatrommer on December 26, 2022, 06:26:59 amThe lack of 100 Ohms load specifications might mean that asking both 1 GHz GBW and 100 Ohms drive is asking too much. In fact, the rather similar LT6230-10https://www.analog.com/media/en/technical-documentation/data-sheets/623012fc.pdfdatasheet specifies (p. 15) GBW vs load that GBW drops to 500 MHz for a 100 Ohms combined (including feedback) load.So it looks like I need a separate buffer, maybe in 2x configuration to make up for the series termination loss. There's a number of video op-amps that might just work (barely) - any suggestion? 100 MHz BW, 2x gain, +- 4.5 V battery supply?While I'm at it: The HP461A amplifier (which I am replicating here) has a switchable 20 and 40 dB gain output. This fits the original ADA4895 nicely as the datasheet also suggests splitting the gain into two 10x stages.For switching between 10x and 100x, I'd simply add BNC ouput connectors (plus buffer) for both stages. Now, video op-amps often have a disable pin that is meant to be used for multiplexing, which I could use for the 10x/100x gain switching (on a single BNC output). I am aware that there are some pitfalls with this approach, any opinions?

author: David Hess
author_rank: Super Contributor
timestamp: December 27, 2022, 01:19:42 pm
content: Quote from: Martinn on December 27, 2022, 11:53:47 amIf the transmission line is properly terminated at the end, then the input looks resistive so no isolation resistor is required to isolate a capacitive load.QuoteFeedback reduces the open loop output impedance which is what counts.QuoteThe heavy load combines with the open loop output impedance to lower the open loop gain. Otherwise the operational amplifier can drive a 50 ohm load but with reduced output swing to account for current limiting.

author: Martinn
author_rank: Frequent Contributor
timestamp: December 29, 2022, 07:36:42 pm
content: I added two LT1818 as output buffers in a 2x configuration. I also reduced the input voltage of the coupling film capacitor to 250 V for a more manageable size. Input protection now has a much lower capacitance (and hopefully still protects the amplifiers).Files just went out to JLCPCB for PCB/assembly. I'm wondering what it will be: A low noise preamp? A 120 MHz oscillator? A wideband noise source? We'll see. JLCPCB is really cheap so there's a fatal bug, I can easily order a fixed version.Together with a 9 V battery and a switch it should fit in a hammond 1590B case.

author: Martinn
author_rank: Frequent Contributor
timestamp: January 21, 2023, 01:50:11 pm
content: Quote from: Martinn on December 29, 2022, 07:36:42 pmAs it turned out, it started as 300 MHz oscillator. See follow-uphttps://www.eevblog.com/forum/projects/measuring-switch-mode-supply-noise-hp461a-replacement-ada4895/msg4652395/#msg4652395- Martin

author: EPAIII
author_rank: Super Contributor
timestamp: January 23, 2023, 11:16:27 am
content: I remember doing the proof of the power transfer theorem back in my college days (1960s). But I still become confused by it when discussions like this arise. And I have to review things and remember what it actually talks about.The confusion in the power transfer theorem is in the fact that it is talking about POWER; not Voltage, nor current, but POWER. And that POWER must be held constant at the source. So the source can not be modeled with a constant Voltage source nor with a constant current source. It must be modeled with a constant POWER source and either the Voltage or current or both will vary when the total resistance, Rs + Rl, is considered.This means it applies to things like electric generators, which can generate only a certain amount of power. Or the TV transmitters that I operated where the power generated was approximately constant. This has nothing to do with the actual level of that power, high or low. It applies to micro-Watts and terra-Watts equally.So quick examples using 1 Watt power level.MATCHED IMPEDANCES:1 Watt source50 Ohms source resistance50 Ohms load resistance100 Ohms total resistanceVs = SqRt(P x R) = 10 VVl = 5 VI = 0.1 ALoad Power = Vl x I = 0.5 Watts (as expected)HIGHER LOAD IMPEDANCE:1 Watt source (the power is CONSTANT)50 Ohms source resistance100 Ohms load resistance150 Ohms total resistanceVs = SqRt(P x R) = 12.25 VVl = 8.16 VI = 0.055 ALoad Power = Vl x I = 0.45 Watts (with constant power it went down)LOWER LOAD IMPEDANCE:1 Watt source (the power is CONSTANT)50 Ohms source resistance25 Ohms load resistance75 Ohms total resistanceVs = SqRt(P x R) = 8.66 VVl = 2.89 VI = 0.115 ALoad Power = Vl x I = 0.33 Watts (again, with constant power it went down)This is not a proof: that would require calculus. But it does show that when the power is held constant, the theorem does work. And, as others have said, it does not apply to varying the source impedance.https://en.wikipedia.org/wiki/Maximum_power_transfer_theoremBut all of the above does not say it is a good idea to omit the source resistor when an AC transmission line is being driven. There are other reasons for that (reflections) and I can tell you from practical experience that there will be bad consequences at higher frequencies. With DC there are no problems that I know of.

author: TimFox
author_rank: Super Contributor
timestamp: January 23, 2023, 03:26:54 pm
content: Actually, the maximum power theorem starts with a "black box" containing either aconstant voltage source with fixed series resistance, or the equivalent constant current source with fixed parallel resistance equal to the series resistance for the voltage source.That box's circuit will deliver maximum power into the usual matched load resistance.I don't believe there is such a thing as a constant power source.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 23, 2023, 04:25:36 pm
content: Quote from: TimFox on January 23, 2023, 03:26:54 pmthere is..https://www.testandmeasurementtips.com/bench-power-supply-from-saelig-provides-up-to-80-v-or-50-a-at-750-w/and we can make one if that really matters...https://www.analog.com/en/design-notes/constantpower-source.html

author: TimFox
author_rank: Super Contributor
timestamp: January 23, 2023, 05:21:51 pm
content: Quote from: Mechatrommer on January 23, 2023, 04:25:36 pmThe constant-voltage source contained within a Thevenin-equivalent model (before the series resistance) is defined as one that produces a constant voltage, independent of load resistance.A practical voltage source has a maximum current capability.Similarly, the constant-current source contained within a Norton-equivalent model (without parallel resistance) produces a constant current, independent of load resistance.A practical current source has a maximum voltage capability.For the full equivalent models (contained in the black box with two output terminals), there is a finite open-circuit output voltage and finite short-circuit output current.A hypothetical constant power source would deliver a constant power independent of load resistance, which is absurd since there can be no power into either a short-circuit or open-circuit load.Obviously, with appropriate feedback, a piece of test equipment can be programmed to deliver a reasonably constant power into a reasonable range of load resistances, but that is not relevant to the maximum-power theorem.In the maximum-power theorem, the power delivered by the internal voltage or current source into the total resistance (finite output resistance in series or parallel with the external load resistance) variesdramaticallyas the load resistance is increased from zero through the matched value to open-circuit.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 23, 2023, 08:20:03 pm
content: Quote from: TimFox on January 23, 2023, 05:21:51 pminfinite voltage will be required to maintain constant current at open circuit, which will be equally absurd... similarly its impossible voltage potential to exist at close circuit...Quote from: TimFox on January 23, 2023, 05:21:51 pmsince we can accept that, practical constant power also has maximum voltage and current capability.

author: TimFox
author_rank: Super Contributor
timestamp: January 23, 2023, 09:22:32 pm
content: There is a lot of confusion here about the "maximum power theorem".That theorem follows from Thevenin's and Norton's theorems: Thevenin states that any collection of generators and resistances terminating in two terminals can be represented as a constant-voltage source in series with a series resistance. The value of the voltage source is the open-circuit voltage at those two terminals and the value of the resistance is that voltage divided by the short-circuit current.Norton states that that two-terminal box can be represented as a constant-current generator in parallel with a parallel resistance. That current is given by the short-circuit current and the resistance by the ratio of the open-circuit voltage to the short-circuit current.It follows immediately that for a given network, the value of the series resistance equals that of the parallel resistance.The next result to follow is the maximum power theorem, where the maximum power delivered to a load from those two terminals is into the matched load resistance, and is referred to as the "available power".Ideal voltage and current generators have zero series resistance and infinite parallel resistance, respectively, and are idealizations: practical implementations have finite resistance.Note that if one has a laboratory voltage supply or audio power amplifier with voltage feedback, the output resistance in normal operation is very low, but the circuit will not function properly into a short circuit, or into a load given by that low output resistance, and the maximum power theorem does not apply.For amplifiers, the theorem does predict the load impedance that gives maximum power gain at output power far below maximum, but not necessarily the maximum power output. For example, the audio amplifier might be designed for maximum output power into 4 ohms, but have a source impedance (due to voltage feedback) of, say, 0.1 ohms.Considering the commercial power supply operated in constant-power mode: there exists a reasonable range of load resistance where the power delivered is reasonably constant. Therefore, the maximum-power theorem does not apply, since the power delivered into the load is independent of load resistance.If you degrade that circuit by adding an external resistor in series with the supply output, then the maximum power will be delivered into a high load resistance, much higher than the external resistor, within the limitations of the supply. Similarly, for an external parallel resistor, the maximum power will be delivered into a low load resistance, much lower than the external resistor, within the limitations of the supply.My extended reply here is to refute EPAIII's contention: "And that POWER must be held constant at the source. So the source can not be modeled with a constant Voltage source nor with a constant current source. It must be modeled with a constant POWER source and either the Voltage or current or both will vary when the total resistance, Rs + Rl, is considered."