Series coax termination for output really necessary?

[Martinn]

Hi all,

I am trying to build a low noise amplifier to measure PSU noise as described in LT AN118 (see my post here https://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

[ConKbot]

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. 

[jonpaul]

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 fidélité

But 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

[Martinn]

What is the amplifier and what meter, scope, Analyzer?

Signal level?

Hi 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 termination
Signal levels... up to some hundred millivolts I guess (as low as possible of course). Also depending on scope input noise.

[Siwastaja]

It's indeed confusing how "series (source) termination" and "parallel (shunt/load) termination" are presented as alternatives, 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 is another 50 ohms hidden inside the signal source, only to confuse you!

Source termination = single 50 ohm resistor at source -> no attenuation
Shunt termination = 50-ohm resistor at source PLUS 50 ohm resistor at load -> bog standard resistor divider which halves the voltage.

[jonpaul]

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 Noël

Jon

[Benta]

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.

[Mechatrommer]

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.

[ejeffrey]

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... .

No.  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.

[Martinn]

A different comment: you realize that your "input protection diodes" aka power rectifiers have around 30 pF capacitance? Quite a lot at 100 MHz.

I was also wondering about this. Normally you'd expect something low leakage, low capacitance...
The input section is straight from AN118 https://www.analog.com/media/en/technical-documentation/application-notes/AN118fb.pdf
page 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?

[David Hess]

It's indeed confusing how "series (source) termination" and "parallel (shunt/load) termination" are presented as alternatives, 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 is another 50 ohms hidden inside the signal source, only to confuse you!

When 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.

[Martinn]

...at higher freq, ADA4985 will increase its internal Z to 100 ohm...

But 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?

[David Hess]

...at higher freq, ADA4985 will increase its internal Z to 100 ohm...

But 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?

The 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.

[Siwastaja]

The maximum power transfer theorem

It'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!

[TimFox]

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.

[Someone]

The maximum power transfer theorem

It'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!

Performance art? A "correct" explanation was already provided:

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... .

No.  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.

           Rs          Rl               
   -----/\/\/\/------/\/\/\/------   
  |                               | 
  __                              | 
 /Vs\                             | 
 \__/                             | 
  |                               | 
  |                               | 
-----                           -----
 ---                             ---
  -                               - 
V = IR
I = V/R

P = V*I
P = V^2/R

Vs = 1

Resistors Equal
Ps = 0.5^2/1 = 0.25W
Pl = 0.5^2/1 = 0.25W

Source = 0
Pl = 1^2/1 = 1W

Load = 0
Ps = 1^2/1 = 1W

Maximum 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.

[Mechatrommer]

The maximum power transfer theorem

It'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!

Performance art? A "correct" explanation was already provided:

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... .

No.  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.

           Rs          Rl               
   -----/\/\/\/------/\/\/\/------   
  |                               | 
  __                              | 
 /Vs\                             | 
 \__/                             | 
  |                               | 
  |                               | 
-----                           -----
 ---                             ---
  -                               - 
V = IR
I = V/R

P = V*I
P = V^2/R

Vs = 1

Resistors Equal
Ps = 0.5^2/1 = 0.25W
Pl = 0.5^2/1 = 0.25W

Source = 0
Pl = 1^2/1 = 1W

Load = 0
Ps = 1^2/1 = 1W

Maximum 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.

i 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...

...at higher freq, ADA4985 will increase its internal Z to 100 ohm...

But 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?

if 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.

[jonpaul]

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

[bson]

           Rs          Rl               
   -----/\/\/\/------/\/\/\/------   
  |                               | 
  __                              | 
 /Vs\                             | 
 \__/                             | 
  |                               | 
  |                               | 
-----                           -----
 ---                             ---
  -                               - 
V = IR
I = V/R

P = V*I
P = V^2/R

Vs = 1

Resistors Equal
Ps = 0.5^2/1 = 0.25W
Pl = 0.5^2/1 = 0.25W

Source = 0
Pl = 1^2/1 = 1W

Load = 0
Ps = 1^2/1 = 1W

Rs 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.

[David Hess]

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.

[Someone]

BIG QUOTE

Rs is a given quantity, and is never zero.  For a given Rs, the maximum power is transferred when Rl = Rs.

So why take a HUGE QUOTE and cut out the line that already said exactly that?

Signal to noise going in the wrong direction here!

[bson]

BIG QUOTE

Rs is a given quantity, and is never zero.  For a given Rs, the maximum power is transferred when Rl = Rs.

So why take a HUGE QUOTE and cut out the line that already said exactly that?

Signal to noise going in the wrong direction here!

I see:

Maximum 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.

I 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?

[Mechatrommer]

theoritical maximum power will converge to singularity resulting the source has to collapse into black hole..

[Someone]

BIG QUOTE

Rs is a given quantity, and is never zero.  For a given Rs, the maximum power is transferred when Rl = Rs.

So why take a HUGE QUOTE and cut out the line that already said exactly that?

Signal to noise going in the wrong direction here!

I see:

Maximum 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.

I 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?

You 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:

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... .

No.  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.

I 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).

Impedance matching is/was about getting maximum power into a resistor connected to a resistive voltage source.

People 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....

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!

Performance art?
[... acknowledgement of previous good answer, and expanding on where the confusion is....]
Maximum 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.

Talking 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 is not delivered 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.

[Mechatrommer]

having a bad day? too much negative power transfer in you? i wasnt replying your post...