Series coax termination for output really necessary?

[Martinn]

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:

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.

The 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 LT6230-10
https://www.analog.com/media/en/technical-documentation/data-sheets/623012fc.pdf
datasheet 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?

[David Hess]

- as real world matching will be less than ideal, an isolation resistor will be necessary to avoid peaking or oscillation

If 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.

"matched termination" probably does not apply here as the output impedance of an ideal op-amp is zero.

Feedback reduces the open loop output impedance which is what counts.

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.

The 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 LT6230-10

...

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?

The 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.

[Martinn]

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.

[Martinn]

I'm wondering what it will be: A low noise preamp? A 120 MHz oscillator? A wideband noise source?

As it turned out, it started as 300 MHz oscillator. See follow-up
https://www.eevblog.com/forum/projects/measuring-switch-mode-supply-noise-hp461a-replacement-ada4895/msg4652395/#msg4652395

- Martin

[EPAIII]

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 source
50 Ohms source resistance
50 Ohms load resistance
100 Ohms total resistance
Vs = SqRt(P x R) = 10 V
Vl = 5 V
I = 0.1 A
Load Power = Vl x I = 0.5 Watts (as expected)

HIGHER LOAD IMPEDANCE:
1 Watt source (the power is CONSTANT)
50 Ohms source resistance
100 Ohms load resistance
150 Ohms total resistance
Vs = SqRt(P x R) = 12.25 V
Vl = 8.16 V
I = 0.055 A
Load 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 resistance
25 Ohms load resistance
75 Ohms total resistance
Vs = SqRt(P x R) = 8.66 V
Vl = 2.89 V
I = 0.115 A
Load 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_theorem

But 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.

[TimFox]

Actually, the maximum power theorem starts with a "black box" containing either a constant 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.

[Mechatrommer]

I don't believe there is such a thing as a constant power source.

there 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

[TimFox]

I don't believe there is such a thing as a constant power source.

there 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

The 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) varies dramatically as the load resistance is increased from zero through the matched value to open-circuit.

[Mechatrommer]

...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...

infinite 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...

A practical voltage source has a maximum current capability.
A practical current source has a maximum voltage capability.

since we can accept that, practical constant power also has maximum voltage and current capability.

[TimFox]

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."