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My rant about transmitter affected by reflected power (per K3ZD)

pgib8

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This guy on youtube drives me crazy, he says that reflected power does not go back to the transmitter and that the transmitter doesn't see the reflection. I disagree with him so he deleted my comment at first until I tell him again with a different account. The worst part is he's pretending to educate people, dispelling myths apparently. He says "true: reflected power does not flow back into the transmitter and cause damage".
I ask him among other things: how come after accidentally transmitting a bunch without having an antenna connected, that my radio stopped producing power.
He answers:
"You blew up your rig because you were transmitting into an open circuit. The rig is designed to transmit into a 50 ohm load. An open circuit presents a huge impedance mismatch to the output stage so it overheated and failed. It had nothing to do with reflected power..."
He tells me "It's a complex subject and it will take a while to wrap your head around it. If you really want to understand this stuff it takes some time and effort."
and "You gotta do some reading."
Another thing he says: "When reflected power encounters a mismatch at the load, its phase is reversed and heads back to the transmitter, where it encounters a large impedance mismatch so is totally reflected back to the load where it is eventually radiated."
So what I think is that this guy doesn't understand impedance like at all, I'm not sure he's ever seen a smith chart let alone know what it means.
Here is a link to his channel but I recommend you don't go there because it will probably make you mad: youtube.com/@NoMoreRadioMyths (K3ZD, Mark the Ham Florida Man).
 
The math of VSWR relationships to all s-parms can make some renegades make wild claims. Power is a bidirectional conduction property. The idea of matched impedance with 50% of the power lost is to satisfy the MPT theorem.

So move on, don't sweat it.

https://www.amphenolrf.com/vswr-conversion-chart If your VSWR is better than 1.4 or RL > 15 dB it is no myth that improving 2:1 VSWR will significantly improve your output.

That said I once worked with a guy who designed all Bell's microwave repeaters had said his splitter design of 3.4 dB enabled a huge cost reduction in the spacing of repeater towers with the 0.1 dB reduction. Another tech I worked with in the 70's got a certificate for Moon bouncing his VHF transmitter to all 5 continents. He sweated over a 0.1 dB loss in his connectors until he overcame the cause of the problem. (Jack Askew).. Some RF guys are essentric but they do have a lot of experience. Jack once told me he had red eyes after a day of tweaking a 100W UHF transmitter, the next day I heard the designer redesigned the alum. lid with tuning holes to protect his eyes from mismatched radiation leakage while tuning filters.
 
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"When reflected power encounters a mismatch at the load, its phase is reversed and heads back to the transmitter, where it encounters a large impedance mismatch so is totally reflected back to the load where it is eventually radiated."

We have four steps:
1. Initially the power goes to a mismatched load.
2. Its phase is reversed and heads back to the transmitter.
3. At the transmitter, where it encounters a large impedance mismatch, the power is totally reflected back to the load.
4. From the load is eventually radiated.

So, in first step the power will not be radiated but will be (eventually) radiated in the fourth step.
In conclusion, is just a matter of time until the power will be radiated from a mismatched load, and don't have to do anything to happen this, just wait patiently.
 
3. At the transmitter, where it encounters a large impedance mismatch, the power is totally reflected back to the load.
This is at least one wrong assumption.
The power reflected by the load travels through the 50 Ohm line towards the 50 Ohm transmitter, and is absorbed by the transmitter's internal resistance.
 
If he is keen on his comments, he can try a 100 MHz, 200W RF Power Amplifier without Load. Let's see what'll happen..
He can use this amplifier hereafter as a French Frites Cooker.
 
How did you not know that your Tx would not get damaged from an open circuit?

Did you have a cable without an antenna?

1711228648563.png
 
I knew it would be bad but it was an accident. This is a really long time ago. I think I had moved the radio or something like that and forgot to hook up the antenna cable. Then when I started using it I was wondering why I wasn't receiving anything, then realized it, plugged the antenna cable in but it was already too late.
Regarding what @vfone said, I have a few comments and someone correct me if I'm wrong on anything:
A) When the reflection hits the transmitter and based on its timing, it will either require an increased current, present an increased voltage, or a mix of both. I guess if the transmitter isn't maxed out and can handle the current then it's not a big deal, same for the voltage. But if it's operating at or near the limit that's what can destroy it. In my case I would always have the power dialed to the max.
B) Assuming there is a tuner after the transmitter or something that reflects the reflection back towards the antenna, when the reflection reaches the mismatched antenna, again it depends on the timing, aka. cable length. If the timing is such that it produces a higher voltage at this point and with the appropriate antenna [I think this would be considered a voltage-fed antenna but I could be wrong], then indeed the power would get radiated and I'm pretty sure the same with current for the current-fed antenna. However as soon as you change the frequency or if the timing is different at the desired frequency, it can also have the opposite effect and then the power is not radiated. If this is true, which I think it is, power could bounce back and forth between tuner and mismatch but you would think if the transmitter keeps pumping in the same amount of power you'd soon have lightning bolts come out of the cable. Instead, the transmitter can't put out as much power. I think of it as a water pipe with air in it or something else, that takes up space and you won't be able to push as much water through.
 
"You blew up your rig because you were transmitting into an open circuit. The rig is designed to transmit into a 50 ohm load. An open circuit presents a huge impedance mismatch to the output stage so it overheated and failed. It had nothing to do with reflected power..."
Any reflections in the steady-state can also be equally expressed as a change in input impedance [from the matched case]. Personally, I find the notion that "Reflections cause damage" is generally a hand-waving explanation, as it doesn't adequately describe the actual problems. As far as I know, there are only two impedance extremes which cause failure modes:

- Reflection in-phase to the transmitting, which is equivalent to an open-circuit load impedance. The failure mode here is the voltage at the output of the amplifier is double what it would be from the matched case, which can obviously cause some components to fail. IMO this only occurs for poorly-designed amplifiers -- most modern amplifiers are designed with components having the appropriate voltage rating so that this doesn't happen.

- Reflection 180 degrees out-of-phase to the transmitting, which is equivalent to a short-circuit load impedance. The failure mode here is what you would expect: the front-end transistor (and associated biasing circuitry) isn't designed to handle a shorted output, and so it burns out. Again, I think this only happens due to poor design; it's my understanding that the transistor can be designed with an appropriate biasing network such that the transistor supplies a finite and tolerable short-circuit current.
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If this is true, which I think it is, power could bounce back and forth between tuner and mismatch but you would think if the transmitter keeps pumping in the same amount of power you'd soon have lightning bolts come out of the cable. Instead, the transmitter can't put out as much power. I think of it as a water pipe with air in it or something else, that takes up space and you won't be able to push as much water through.
Yes, total power provided to a system is a combination of the emitted waves and the incident. I think in your situation, the presence of the reflected waves cause the amplifier to emit significantly less power.
 
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Any reflections in the steady-state can also be equally expressed as a change in input impedance [from the matched case]. Personally, I find the notion that "Reflections cause damage" is generally a hand-waving explanation, as it doesn't adequately describe the actual problems. As far as I know, there are only two impedance extremes which cause failure modes:

- Reflection in-phase to the transmitting, which is equivalent to an open-circuit load impedance. The failure mode here is the voltage at the output of the amplifier is double what it would be from the matched case, which can obviously cause some components to fail. IMO this only occurs for poorly-designed amplifiers -- most modern amplifiers are designed with components having the appropriate voltage rating so that this doesn't happen.

- Reflection 180 degrees out-of-phase to the transmitting, which is equivalent to a short-circuit load impedance. The failure mode here is what you would expect: the front-end transistor (and associated biasing circuitry) isn't designed to handle a shorted output, and so it burns out. Again, I think this only happens due to poor design; it's my understanding that the transistor can be designed with an appropriate biasing network such that the transistor supplies a finite and tolerable short-circuit current.
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Yes, total power provided to a system is a combination of the emitted waves and the incident. I think in your situation, the presence of the reflected waves cause the amplifier to emit significantly less power.
Great explanation. The radio I was talking about is an original President Lincoln, which is from a very long time ago so I will give it a pass on the "poorly designed" but otherwise agree with you. In my case the transistor must have not been able to handle the voltage.
I would say that in a shot-circuit, maybe we can also say 90% short-circuit, or 80% or some other number, that we can say that given any mismatch there is a chance of damage, although it starts out at basically zero chance and the chance of damage starts increasing as the mismatch gets worse, and at some point we're rolling the dice. I envision it like a loop on the smith chart, once the mismatch goes outside of it damage happens, and maybe it's 2 loops, one for short and one for open.
But one more question @PlanarMetamaterials, what would happen if there is 100% reflection but on the Smith chart it is at the very top or bottom, so right between open and closed?
 
I think few readers here didn't catch the thing that what I mentioned in #4 was a joke.
From my experience dealing with high power transmitters I can say that when I used a well designed transmitter having a well designed VSWR protection, the worst enemy is not a shorted load or an opened load, because in those cases the output protection of the transmitter will take care the situation.
The worst case that I met for a high power transmitter, was a complex load combined with a poor VSWR somewhere between 3:1 and 6:1.
This will fool the transmitter protection and burn your power amplifier.
 
After reading the original Uniden schematic that was transferred to the President Lincoln Brand around 1988, I now agree with the old fart who said it has nothing to do with reflected waves and has everything to do with mismatched impedance. In the classical sense when the signal rise time is much slower or longer than the wave propagation delay from the final stage to the open antenna port, we can use lumped RLC elements and Kirchoff Laws to explain everything. In reality the waves move back and forth so fast that you cannot see the reflections with a slow risetime in the VHF band .
A conventional 50 ohm generator just outputs twice the voltage with no load.

The unlisted Uniden transistors in the last few stages of bottom left corner drive a high Q low pass filter which is damped by the antenna load. It is driven from a CE common emitter with step-up coils in between . The High output power is converted from a low impedance CE amp that drives step-up transformer coils to raise the impedance. This would fail from over current if the coils saturate or the voltage amplifies much more than 2x Vcc.

Given that Uniden knows how to design transistors with a loaded Q LPF, I expect they warned users in the manual to never operate without an antenna. .

1711401770247.png
 
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After reading the original Uniden schematic that was transferred to the President Lincoln Brand around 1988, I now agree with the old fart who said it has nothing to do with reflected waves and has everything to do with mismatched impedance. In the classical sense when the signal rise time is much slower or longer than the wave propagation delay from the final stage to the open antenna port, we can use lumped RLC elements and Kirchoff Laws to explain everything. In reality the waves move back and forth so fast that you cannot see the reflections with a slow risetime in the VHF band .
A conventional 50 ohm generator just outputs twice the voltage with no load.

The unlisted Uniden transistors in the last few stages of bottom left corner drive a high Q low pass filter which is damped by the antenna load. It is driven from a CE common emitter with step-up coils in between . The High output power is converted from a low impedance CE amp that drives step-up transformer coils to raise the impedance. This would fail from over current if the coils saturate or the voltage amplifies much more than 2x Vcc.

Given that Uniden knows how to design transistors with a loaded Q LPF, I expect they warned users in the manual to never operate without an antenna. .

View attachment 189614
That's very interesting, a little bit over my head, I'll have to ponder that :D
On a different note, I checked on ebay if anyone is selling any of these and the one I found which appeared to have been in great shape, the guy didn't know anything about it and wanted to show that it works, in doing that he probably smoked it because in 2 of the pictures he's transmitting and I bet there is no antenna hooked up -> https://www.ebay.com/itm/276396574902
 
But one more question @PlanarMetamaterials, what would happen if there is 100% reflection but on the Smith chart it is at the very top or bottom, so right between open and closed?
So, if the termination is an ideal inductive or capacitive load?
I don't think the impedance description is very helpful here; I'd revert to the incident and reflected wave viewpoint. The reflections will be equal magnitude and +/-90 degrees phase-shifted from the incident waves. This will cause a sqrt(2) rise in both voltage and current magnitudes (math). So, these loads could be problematic -- but realistically, should be within engineering safety factors!
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I now agree with the old fart who said it has nothing to do with reflected waves and has everything to do with mismatched impedance.
The two are obviously related, so to say that one has "nothing to do with" the other is a bit misleading. Both descriptions are valid and useful in different contexts.
 
So, if the termination is an ideal inductive or capacitive load?
I don't think the impedance description is very helpful here; I'd revert to the incident and reflected wave viewpoint. The reflections will be equal magnitude and +/-90 degrees phase-shifted from the incident waves. This will cause a sqrt(2) rise in both voltage and current magnitudes (math). So, these loads could be problematic -- but realistically, should be within engineering safety factors!
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The two are obviously related, so to say that one has "nothing to do with" the other is a bit misleading. Both descriptions are valid and useful in different contexts.

The termination required is 50 ohms or free space impedance converted to 50 Ohms with a suitable antenna.

The passive output is essentially a tank circuit driven by step-up auto-transformers. Transmission line theory does not add any useful understanding of the failure.
 
From my understanding the fast rise time essentially means a square wave instead of a sine wave and that would saturate the inductor(s). It would still have to be due to the reflection even though that reflection, it's just one that occurs instantly. I could have had a coax with open (or short) attached of suitable length that would have created the same condition at the connector, it just wouldn't have been instant but would be due to the reflection, or I could have had the same coax but of infinite length and it would have never failed so that's the only reason why I would say it's because of reflection, just not in the traditional sense of thinking of reflections occurring from the antenna. I can see though as one gets closer and closer to the active component(s) at some point it doesn't sound right to call it a reflection anymore even though technically it is at least I think. If the last component in the circuit was a series capacitor but due to a PCB defect it was shorted to ground it might sound better to say the circuit is not working because the capacitor is shorted out than the circuit is not working because the defect is causing a reflection, but I can see both being valid and referring to the same thing in my opinion.
 
I can see though as one gets closer and closer to the active component(s) at some point it doesn't sound right to call it a reflection anymore even though technically it is at least I think.
Yes. As I wrote above, the description by incident/reflected wave and the description by impedance at the transmitter are both valid and equivalent, and lead to the exact same results.
 
I don't agree with the old fart who said it has nothing to do with T-line theory, rather I agree it does not add anything to the accuracy in estimating the gain in voltage or explain the saturation of the core or the over-current or over-voltage of the semiconductors. I hope that explains myself clearer.

more...


A fast rise does not need to be a square wave or an impulse. It can be any waveshape.

Yet it is where the rate of change in dV/dt that causes the the ringing and overshoot on logic signals that can be clearly demonstrated with T-line simulations and theory with (high impedance|| small pF) loads and finite delays.
 
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Same thing in opposite direction. A by antenna received signal is transferred ta an impedance matched receiver and almost all power is is absorbed by receiver. Non ideal losses in antenna and cable can not be avoided but in general is all power absorbed by receiver in a matched system.
If a less perfect matched receiver circuit will part of received power by reflected all the way back to free space but as antenna load now is mismatched relative receiver will a part of signal reflects once again.

Assume VSWR 6:1, that corresponds to 3dB mismatch loss. If VSWR is 6:1 will 3 dB of by antenna recieved signal never be absorbed by receiver. RSSI will show 3 db less signal level.
Same as for a TX circuit with a normal impedance, if loaded by a VSWR 6:1 mismatch. 3dB of total emitted power will then be absorbed by transmitter.
Typical impedance for smaller powerstages such as cellphones, WiFi, BT is typical around 30-60 Ohm complex impedance. Some chip have very low impedance such as CC2500 working in class E.
Tx impedance is just a few Ohm for that chip and to get decent efficiency and power is matching very critical as well as decoupling, which is described in its application note. Even its Vcc should be impedance-tuned using a short coil before decoupling to ground for best efficiency. according to app.note.

In an ideal case for a transmitter with zero impedance, not even reactive impedance, can not reflecting signal be absorbed as for absorption is an impedance needed.
For a such case is Tx just as effective radiator whatever antenna is used. It is good enough with any short loop antenna, it will be a 100% effective antenna-radiator into space av long as its resistive loss not causes it to melt and VSWR will not even be something measurable as it not exist any losses.
 

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