Measuring Splitter/combiners

[MavamQ]

Here's what ChatGPT syas,
"
The question ask of ChatGPT: Show me two 50Ω transmitters feeding a two input lossless combiner, with 50Ω on the output. What is the power on the output.

Let’s do it rigorously with 50 Ω systems and actual voltages, so there’s no ambiguity.

---

System you asked for​

• Two independent 50 Ω transmitters, each delivering 1 W
• Feeding a lossless 2-way combiner
• Output terminated in 50 Ω
• Signals are same frequency and in phase

---

Step 1 — Voltage of each transmitter​

For a 50 Ω system:
P=Vrms2RP = \frac{V_{\text{rms}}^2}{R}P=RVrms2
So each 1 W source has:
Vrms=1×50≈7.07 VV_{\text{rms}} = \sqrt{1 \times 50} \approx 7.07 \text{ V}Vrms=1×50≈7.07 V

---

Step 2 — What a lossless combiner actually does​

A proper lossless combiner (e.g., an idealized Wilkinson combiner) ensures:

• Each source still “sees” 50 Ω
• Power from both sources is delivered to the output
• No reflections, no dissipation (ideal case)

The combiner does NOT directly short voltages together.
It transforms impedances so power combines correctly.

---

Step 3 — Total power at the output​

Total input power:
1 W+1 W=2 W1\,\text{W} + 1\,\text{W} = 2\,\text{W}1W+1W=2W
Since the combiner is lossless:
Pout=2 WP_{\text{out}} = 2\,\text{W}Pout=2W

---

Step 4 — Output voltage check (sanity check)​

Into 50 Ω:
Vout=2×50=10 V rmsV_{\text{out}} = \sqrt{2 \times 50} = 10 \text{ V rms}Vout=2×50=10 V rms
So:

• Each source contributes coherently
• The network transforms things so the combined result is 10 V rms into 50 Ω
• That corresponds to 2 W, not 4 W

---

Why not 4 W?​

The “4 W” result comes from naively adding voltages directly:
(7.07+7.07)2/50=4 W(7.07 + 7.07)^2 / 50 = 4\,\text{W}(7.07+7.07)2/50=4W
But that assumes:

• You physically forced both sources into the same node without impedance transformation

In reality:

• That condition would violate the 50 Ω match
• The sources would fight each other
• You’d get reflections or current redistribution—not extra power

---

The key constraint​

A passive, lossless RF network obeys:
Pout≤∑PinP_{\text{out}} \le \sum P_{\text{in}}Pout≤∑Pin
So:

• Best possible outcome → 2 W
• Never 4 W

 

[MavamQ]

Antenna Performance : co-phasing - IW5EDI Simone - Ham-Radio

Co-Phasing Co-phasing or “stacking” has long been a way to get high gain from antennas. Co-phasing involves placing two (or more!) identical antennas either side-by-side or one over top another (“stacking”) at a certain distance apart (usually a 1/2 Wavelength or more) and feeding the antennas...

www.iw5edi.com

Here is a ham combining two Antennas for 3db gain. Also has an example show 4 antennas for 6db gain. He's working with 12db gain antennas, 4 antennas add 12db +3db + 3db for 18db of gain when 4 antennas are phased together. Instead of a combiner he uses impedance conversion coax cables to make two 50Ω antennas in parallel (25Ω ) into 100Ω in parallel to match the 50Ω transmitter.

 

[prcguy]

So if I have $14 into a two-way splitter, do I have $28 dollars or $7? I'm so confused!!!

If you bought it in Kalifornia you paid to much due to taxes and your rights were divided in two. If you bought off eBay used and it shipped elsewhere you are probably good to go. I can go on but its better to stop here.

 

[prcguy]

Antenna Performance : co-phasing - IW5EDI Simone - Ham-Radio

Co-Phasing Co-phasing or “stacking” has long been a way to get high gain from antennas. Co-phasing involves placing two (or more!) identical antennas either side-by-side or one over top another (“stacking”) at a certain distance apart (usually a 1/2 Wavelength or more) and feeding the antennas...

www.iw5edi.com

Here is a ham combining two Antennas for 3db gain. Also has an example show 4 antennas for 6db gain. He's working with 12db gain antennas, 4 antennas add 12db +3db + 3db for 18db of gain when 4 antennas are phased together. Instead of a combiner he uses impedance conversion coax cables to make two 50Ω antennas in parallel (25Ω ) into 100Ω in parallel to match the 50Ω transmitter.

That coax phasing harness is a form of Wilkinson power divider and to finish it off it would have a 100 ohm resistor across the two output ports. Its a power divider and a combiner. It has 3dB theoretical loss plus coax loss and mismatch loss in each direction so dividing a single signal into two 50 ohm ports it has 3dB loss plus the other small losses.

In the combiner mode inputting identical signals in phase into the two identical ports you will have about 3dB gain at the common port due to the signals being in phase. How is that possible if its got 3dB loss? If you disconnect one of the inputs and only have one signal feeding into one of the two identical ports you will have 3dB LOSS going through the device. But its working as advertised because it will have 3dB loss through it no matter which direction the RF is flowing through it. Its the FACT that it provides 3dB gain with two identical signals into the identical ports and in phase because the process actually produces 6dB of gain leaving you with 3dB net gain after hardware combiner loss.

You have to look back at my comments on multipath where a direct signal to your antenna from across town can become as much as 6dB higher due to a reflection arriving at the same level at your antenna and exactly in phase and since there is no lossy combiner hardware to eat up 3dB of the 6dB increase from combining two signals in phase you can get the full 6dB increase. Does that make sense? Why is it not up to a 3dB increase from two signals arriving in phase at my antenna? If it doesn't make sense then explain how you can receive an unobstructed signal at some level but due to reflection arriving under perfect circumstances it can become 6dB stronger.

 

[prcguy]

Here's what ChatGPT syas,
"
The question ask of ChatGPT: Show me two 50Ω transmitters feeding a two input lossless combiner, with 50Ω on the output. What is the power on the output.

Let’s do it rigorously with 50 Ω systems and actual voltages, so there’s no ambiguity.

---

System you asked for​

• Two independent 50 Ω transmitters, each delivering 1 W
• Feeding a lossless 2-way combiner
• Output terminated in 50 Ω
• Signals are same frequency and in phase

---

Step 1 — Voltage of each transmitter​

For a 50 Ω system:
P=Vrms2RP = \frac{V_{\text{rms}}^2}{R}P=RVrms2
So each 1 W source has:
Vrms=1×50≈7.07 VV_{\text{rms}} = \sqrt{1 \times 50} \approx 7.07 \text{ V}Vrms=1×50≈7.07 V

---

Step 2 — What a lossless combiner actually does​

A proper lossless combiner (e.g., an idealized Wilkinson combiner) ensures:

• Each source still “sees” 50 Ω
• Power from both sources is delivered to the output
• No reflections, no dissipation (ideal case)

The combiner does NOT directly short voltages together.
It transforms impedances so power combines correctly.

---

Step 3 — Total power at the output​

Total input power:
1 W+1 W=2 W1\,\text{W} + 1\,\text{W} = 2\,\text{W}1W+1W=2W
Since the combiner is lossless:
Pout=2 WP_{\text{out}} = 2\,\text{W}Pout=2W

---

Step 4 — Output voltage check (sanity check)​

Into 50 Ω:
Vout=2×50=10 V rmsV_{\text{out}} = \sqrt{2 \times 50} = 10 \text{ V rms}Vout=2×50=10 V rms
So:

• Each source contributes coherently
• The network transforms things so the combined result is 10 V rms into 50 Ω
• That corresponds to 2 W, not 4 W

---

Why not 4 W?​

The “4 W” result comes from naively adding voltages directly:
(7.07+7.07)2/50=4 W(7.07 + 7.07)^2 / 50 = 4\,\text{W}(7.07+7.07)2/50=4W
But that assumes:

• You physically forced both sources into the same node without impedance transformation

In reality:

• That condition would violate the 50 Ω match
• The sources would fight each other
• You’d get reflections or current redistribution—not extra power

---

The key constraint​

A passive, lossless RF network obeys:
Pout≤∑PinP_{\text{out}} \le \sum P_{\text{in}}Pout≤∑Pin
So:

• Best possible outcome → 2 W
• Never 4 W

Nothing here explains how its possible to get 3dB gain from combining when the combining device has 3dB insertion loss. Try again or pay attention to the fact its a 6dB process and 3dB is lost in hardware. I also asked you to find an RF combiner that has less than 3dB loss like the .3dB total loss you hinted at. A make and model is what I would like to see but I will debunk it so be prepared for that because it doesn't exist.

 

[MavamQ]

Here is a ChatGPT discription of why differ.

1) What the mathematician is doing → adds fields at constant amplitude​

They assume two identical waves:

V1=V,V2=VV_1 = V,\quad V_2 = VV1=V,V2=V
Add them:

Vtotal=2VV_{\text{total}} = 2VVtotal=2V
Power into the same impedance:

P∝V2⇒Ptotal=(2V)2=4V2P \propto V^2 \Rightarrow P_{\text{total}} = (2V)^2 = 4V^2P∝V2⇒Ptotal=(2V)2=4V2
That is:

+6 dB+6\ \text{dB}+6 dB
✔ Implicit assumption: each source keeps its full amplitude
✔ That means total power doubled first, then fields added

---

2) What you have with two antennas → fixed incident field​

Now start from physics, not algebra:

• A plane wave arrives with some field EEE
• One antenna extracts power PPP
• Two antennas each see the same field, so:

P1=P,P2=PP_1 = P,\quad P_2 = PP1=P,P2=P
Total available power:

Ptotal=2PP_{\text{total}} = 2PPtotal=2P
That is already:

+3 dB+3\ \text{dB}+3 dB

---

3) Now combine them (e.g., Wilkinson power divider)​

The combiner is (ideally) lossless. For in-phase signals:

• Voltages combine coherently
• No power is lost (ignoring small real insertion loss)

So output power = sum of inputs:

Pout=2PP_{\text{out}} = 2PPout=2P
That is:

+3 dB relative to one antenna+3\ \text{dB relative to one antenna}+3 dB relative to one antenna

---

4) Where the “6 dB” went​

It never applies to this situation because the starting point is different.

Case	What is fixed	Result
Math addition	Voltage amplitude per source	+6 dB
Two antennas	Incident field (energy density)	+3 dB

---

5) The key physical constraint​

With antennas you cannot do this:

• You cannot keep each antenna’s voltage the same and avoid doubling the captured energy
• The field determines how much power is available per antenna

So:

Two antennas do not give you two full-strength voltages “for free”—they give you two shares of energy extracted from the same field.

 

[MavamQ]

I have an email friend that wrote me the following.

"I tried it just now with a MCL 50 ohm splitter/combiner, Siglent dual channel funct gen, and Siglent Spectrum Analyzer. 1Mc 0 dBm each signal into combiner = +3 dBm output from combiner.
The function generator automatically makes both channels 0 degrees (in phase). I switched one channel to 180 deg--output dropped to -56 dBm--decent cancellation."
I quizzed him about the insertion loss, he responded,
" Yes, a tenth or so insertion loss, I just rounded it all off to show the combiner gives the expected and theoretical +3dB (power) gain."

 

[MaineGuy]

ChatGPT....

AI is no match for NS.

 

[MavamQ]

ChatGPT....

AI is no match for NS.

I certainly agre AI has many problems and you often have to know pretty much where you going to correct it and hopefully get usable answers. What is NS?
And since someone is at still looking at the thread I have come up with what I think is a good explanation and shows the signal add to +3db with only insertion loses which can be as low as 0.1db or 0.2db with a good quality splitter/combiner.

The input signals to a 50Ω splitter/Combiner: in phase signals of 1mW, 0.224V.
In a Wilkinson combiner or similar, the two 50Ω inputs must be transformed to 100Ω at the output so they can be in put parallel and equal 50Ω at the output. When you transform a 50Ω / 0.224V signal to 100Ω impedance, the0 voltage also transforms, to 0.316V. (conservation of energy)
Now, the two 0.316V signals combine in parallel at the output, they don't add, but the two 100Ω/0.316V are put in parallel to be a 50Ω/0.316V and we have 2mW signal. The output is double, or +3db.

I'd put my write up into chat and ask for a simple easy to follow explanation, I think it did a very good job, but I won't post AI unless I am ask.

 

[KevinC]

I certainly agre AI has many problems and you often have to know pretty much where you going to correct it and hopefully get usable answers. What is NS?
And since someone is at still looking at the thread I have come up with what I think is a good explanation and shows the signal add to +3db with only insertion loses which can be as low as 0.1db or 0.2db with a good quality splitter/combiner.

The input signals to a 50Ω splitter/Combiner: in phase signals of 1mW, 0.224V.
In a Wilkinson combiner or similar, the two 50Ω inputs must be transformed to 100Ω at the output so they can be in put parallel and equal 50Ω at the output. When you transform a 50Ω / 0.224V signal to 100Ω impedance, the0 voltage also transforms, to 0.316V. (conservation of energy)
Now, the two 0.316V signals combine in parallel at the output, they don't add, but the two 100Ω/0.316V are put in parallel to be a 50Ω/0.316V and we have 2mW signal. The output is double, or +3db.

I'd put my write up into chat and ask for a simple easy to follow explanation, I think it did a very good job, but I won't post AI unless I am ask.



View attachment 200843

Please show us a "TV splitter/combiner" designed like the above (like you originally asked about in your first post).

 

[prcguy]

I have an email friend that wrote me the following.

"I tried it just now with a MCL 50 ohm splitter/combiner, Siglent dual channel funct gen, and Siglent Spectrum Analyzer. 1Mc 0 dBm each signal into combiner = +3 dBm output from combiner.
The function generator automatically makes both channels 0 degrees (in phase). I switched one channel to 180 deg--output dropped to -56 dBm--decent cancellation."
I quizzed him about the insertion loss, he responded,
" Yes, a tenth or so insertion loss, I just rounded it all off to show the combiner gives the expected and theoretical +3dB (power) gain."

That's the 6dB increase from combining in phase and incurring a 3dB loss through the splitter. Connect just one sig gen through the splitter and measure the loss with or without the unused port terminated. You will have a theoretical 3dB loss plus the tenth or few tenths of a dB loss from resistive or other losses making it more than 3dB loss through the splitter. It doesn't matter what direction you measure RF into the common port to an output or into an output to the common port. Try that your have your friend try that and then tell us how you can have 3dB loss through a component but get 3dB gain. I'll be waiting for the explanation.

This is one of the most basic things in RF and you don't get it and you are too, well I'll hold back saying it but too something to even understand. Sorry I can't help you but if you continue with this everyone who is familiar with RF components will think you are a fool.

 

[prcguy]

Take a look at this catalog of dividers/combiners. RF Power Splitters/Dividers, RF Power Combiners

In this case look at the row that starts with model number and go over to column that says
"Insertion Loss (dB) Above Theoretical, Typ". Read that line again. So for a two way splitter model that says .3dB loss you have to add the 3dB theoretical loss to the .3 to get 3.3dB loss through a two way splitter. Pick an 8 way, it might say .8dB loss in that column. You have to add that .8dB to the thoretical loss of 9dB to get 9.8dB loss through an 8-way and that would be in either direction.

Now click on any model number and go to its data sheet. In this case I clicked on the ZFSC-2-5+ two way splitter. You will then find both the S1 and S2 loss through the device will be what I am stating which is 3dB loss for this two way splitter plus an additional .27dB to over 1dB depending on frequency. Right there the mfr is telling you something opposite of what you are claiming that in this case a two way divider or combiner has minimum 3dB loss in either direction and when combining two signals in phase you still have that loss and its the fact that combing two signals in phase gets you a 6dB increase and you lose 3dB of that through the splitter leaving you with 3dB net gain. That's it. Any other explanation is is incorrect.

 

[MavamQ]

prcguy, walk me through where my explanation is wrong in posy #49.
at this point I think I have figured out we are talking about two different things. It is correct that if you put two voltages in series the voltage doubles and you get 4 times the power or +6db gain. (this totally ignores impedances) But this not what a splitter/combiner does. It has to take two 50Ω inputs transform then up to 100Ω put them in parallel to get the impedance back down to 50Ω. The voltages don't add when in parallel, But the Voltage increase from the 50Ω to 100Ω transformation and then the paralleling of the two 100Ω impedances to get 50Ω, results in +3dbb of gain. Again, I think we are talking about to different things and our only quibble at this point is about the 3db loss in a combiner. I contend a good combiner only gas 0.1 to 0.2 db of loss. Please walk through post #49.and tell me where you disagree. Please all others chime in.
I'll add the chatGPT rewrite for a more clear explanation of my write up. It is very clear.

Start with two identical RF signals:
Each signal is 1 mW into 50Ω. That corresponds to 0.224 V RMS
The two signals are in phase
Step 1 — What the combiner must do.
In a Wilkinson (or any proper 2-way combiner), the goal is:
Combine two 50Ω sources, Maintain a 50Ω output match
To make that work, each input is impedance transformed from 50Ω to 100Ω before they are joined.
Step 2 — What happens during impedance transformation
Each signal still carries 1 mW of power (power is conserved in a lossless network).
When you transform: 50Ω → 100Ω, Power stays the same (1 mW)
Voltage must increase because: P=V^2 /R , So increasing resistance forces voltage up:
0.224 V (at 50Ω) → 0.316 V (at 100Ω)
Step 3 — Combining the signals
Now you have two signals:
Each is 0.316 V into 100Ω. They are in phase
These are connected in parallel: 100Ω ∥ 100Ω = 50Ω, The voltage at the node remains 0.316 V
Important point:
The voltages do not add to 0.632 V because they are not stacked in series
They are two sources driving the same node, not summed voltages
Step 4 — Final result
At the output: Voltage = 0.316 V, Impedance = 50Ω
Now calculate power: P=V^2 / R = (0.316)^2 / 50 = 2mW
Final takeaway
Each input contributes 1 mW, the combiner delivers 2 mW total
That is a +3 dB increase in power
The key insight:
The power doubles because two sources deliver energy into the same load, not because voltages directly add.

 

[MavamQ]

Take a look at this catalog of dividers/combiners. RF Power Splitters/Dividers, RF Power Combiners

In this case look at the row that starts with model number and go over to column that says
"Insertion Loss (dB) Above Theoretical, Typ". Read that line again. So for a two way splitter model that says .3dB loss you have to add the 3dB theoretical loss to the .3 to get 3.3dB loss through a two way splitter. Pick an 8 way, it might say .8dB loss in that column. You have to add that .8dB to the thoretical loss of 9dB to get 9.8dB loss through an 8-way and that would be in either direction.

Now click on any model number and go to its data sheet. In this case I clicked on the ZFSC-2-5+ two way splitter. You will then find both the S1 and S2 loss through the device will be what I am stating which is 3dB loss for this two way splitter plus an additional .27dB to over 1dB depending on frequency. Right there the mfr is telling you something opposite of what you are claiming that in this case a two way divider or combiner has minimum 3dB loss in either direction and when combining two signals in phase you still have that loss and its the fact that combing two signals in phase gets you a 6dB increase and you lose 3dB of that through the splitter leaving you with 3dB net gain. That's it. Any other explanation is is incorrect.

I have looked and looked over the Minicircuits site, Please show my any line that says the -3db applies to combiners like it does to splitters.
It only applies to splitters because you send 1/2 the signal to each output so each has to be -3db.
Please critic the latest explanations, mine or the chatGPT rewrite, It is better to debunk.

 

[prcguy]

Yes in the act of splitting or combining and maintaining the same input/output impedance you have to transform the 50 ohm input to some higher impedance so that you have 50 ohms at the common connection when combined. For a two way splitter the transformation would be to 100 ohms to get 50 ohms at the common point. Yes the voltage and current measurements will be different at different points inside the splitter/combiner but the power will be essentially the same at any point inside the splitter/divider except for very small resistive and other losses.

The bottom line continues to be, for a two way splitter/divider you will have at least 3dB loss through it in either direction and preferably with the unused port terminated otherwise you will have lots of amplitude ripple and it will appear at some frequencies that the 3dB loss is a bit more or less at some frequencies, so lets assume you did the measurement with the unused port terminated. So with a device that has 3dB loss through it, and that has two equal level signals applied to the two output ports and lets use 0dBm as an example, if you only get 3dB gain at the input port from combining you would loose that 3dB through the device giving you 0dBm at the common output port due to the 3dB loss per side of the combiner. But you really do get a 3dB gain at the output which would be pure magic and not possible. Except for the fact the act of combining in phase actually gives you a 6dB increase and not a 3dB increase and with the internal 3dB loss of the combiner you finally get the 3dB increase which we measure and know.

This must be the 15th time I've stated the same thing but in every case its true.

 

[prcguy]

I have looked and looked over the Minicircuits site, Please show my any line that says the -3db applies to combiners like it does to splitters.
It only applies to splitters because you send 1/2 the signal to each output so each has to be -3db.
Please critic the latest explanations, mine or the chatGPT rewrite, It is better to debunk.

If you looked in the catalog and went to a data sheet for any splitter then looked at the loss, its bi-directional. Here is data from one particular splitter/combiner. Look at the total loss column then look at note 1 at the bottom.

If you give me some time I will connect a two way splitter/combiner on my bench and show you the loss using it as a splitter and a combiner with pictures of each. One setup should make the point which would be a single signal of say 0dBm into one output port of the splitter with the other input port terminated. The common port will measure 3dB loss plus a few tenths of a dB more for about -3dBm. Then I will remove the termination and feed that port 0dBm in phase with the first port and the common port will now have about 3dBm minus a few tenths of a dB for internal losses. If I show you that in pictures of a measurement I just did can you then explain how the splitter shows 3dB loss with one signal applied but 6dB gain or 3dBm measured output when two signals are applied? Will that change your mind?

 

[KevinC]

Will that change your mind?

Survey says....no.

 

[MavamQ]

Yes in the act of splitting or combining and maintaining the same input/output impedance you have to transform the 50 ohm input to some higher impedance so that you have 50 ohms at the common connection when combined. For a two way splitter the transformation would be to 100 ohms to get 50 ohms at the common point. Yes the voltage and current measurements will be different at different points inside the splitter/combiner but the power will be essentially the same at any point inside the splitter/divider except for very small resistive and other losses.

The bottom line continues to be, for a two way splitter/divider you will have at least 3dB loss through it in either direction and preferably with the unused port terminated otherwise you will have lots of amplitude ripple and it will appear at some frequencies that the 3dB loss is a bit more or less at some frequencies, so lets assume you did the measurement with the unused port terminated

If you measure a combiner with one input port terminated,

• At the junction, the signal splits equally between the two branches
• One branch goes to the output
• The other branch goes into the terminated port

So:

• Half the power goes to the output
• Half is absorbed in the termination
• That is why you measure -3db loss with you method.

This must be the 15th time I've stated the same thing but in every case its true.

Or not.

 

[prcguy]

If you measure a combiner with one input port terminated,

• At the junction, the signal splits equally between the two branches
• One branch goes to the output
• The other branch goes into the terminated port

So:

• Half the power goes to the output
• Half is absorbed in the termination
• That is why you measure -3db loss with you method.

Or not.

Leave the unused port unterminated. It will still show about 3dB loss but with some amplitude ripple across the frequency range. At some points it will be a little more than 3dB and other points a little less than 3dB but an average of 3dB loss unterminated.

 

[MavamQ]

Survey says....no.

I have no problem changing my mind, I jsut see more evidence to my way of seeing. I think I debunked his latest contention of measure insertion loss with one port terminated. The combiner shows -3db loss, but only because the termination absorbs half the power and the other half goes to the output. Please read #58 and debunk it, I'm here to learn.