Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Antenna power output when mismatched

Status
Not open for further replies.

Roy_prime

Newbie level 4
Joined
Nov 8, 2013
Messages
7
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
69
Hi,

I've been studying about antennas and communication systems, when I came accross this statement:

"the power can actually be higher for a mismatched antenna!"

in https://www.antenna-theory.com/definitions/trp.php ...

Normally, I consider this site as good for basics.. But I'm not sure how the above statement can be true... If true, please explain how, if false, please tell why...
 

not really likely! If you have a well matched transmitter (i.e. match to a 50 ohm coax line), and there is some length of line between the transmitter and antenna....you will simply reflect back some of the power that could have been otherwise transmitted.

I could come up with a spurious situation where your statement is true...such as if the transmitter was badly mismatched, AND the antenna was badly mismatched, and there was a lenth of low loss cable in between. then are some ~single frequency the two mismatches and cable might form a bandpass structure...and transmit the energy despite the mismatches. BUT if you moved the frequency just a little, the structure will not longer resonate..and you will have very little transmitted power thru the air. i.e. IF you could get that to work, it would have to be a very narrow band signal
 

A PA sources maximum power when terminated with the complex conjugate of it's output impedance. There's no way to pull a higher power from it.
 

The statement is true for real world situations.

Generally a power amplifier is designed to deliver a certain amount of power into 50 Ohms (or other specified impedance). This does not mean that the output impedance of the amplifier itself is 50 Ohms (except for the conjugate matched situation as mentioned by FvM). Conjugate match situation is mostly met over a narrow range of frequency and output power range. When you change the frequency and/or input drive, you very likely deviate from conjugate match situation.

Many PAs are not operated under conjugate match situation. This is especially true for high efficiency switching topologies. Biff44 situation happens in real world. When you put a forward power coupler in the output of a PA, you may get more net output power by applying mismatch (load pulling). When there is no protection circuitry, you may destroy the amplifier as the efficiency may reduce significantly. Assuming that all sources have 50 Ohms output impedance, is a common cause of measuring error not noticed by many people.

The same situation happens when using an EMC measuring antenna (that can have weird output impedance). Gain for EMC antennas is given for a 50 Ohms load mostly (thereby deviating from the IEEE definition for power gain). When the input impedance of the measuring receiver isn't close enough to 50 Ohms, actual antenna gain can be more or less then the stated value. Of course all depending on frequency cable length, cable attenuation, etc. I usually mitigate these effects by checking with different cable lengths and/or adding attenuators to make sure that the measuring antenna sees 50 Ohms.
 

I try to imagine under which conditions you manage to increase the output power and simultaneously reduce efficiency? Don't say it's completely impossible, but it doesn't sound plausible at first sight.
 

@FvM: This example holds for a non-conjugate match.

If you have a somewhat voltage saturating/clipping class B or C amplifier and you reduce the loadresistance as seen from the active device (for example go from 500 Ohms to 400 Ohms), the first harmonic output current may increase more than the first harmonic output voltage drops (due to the voltage saturation issue and internal feedback). So the output power will increase in this example.

Due to the increaased output current, the average DC input current to the PA must increase also, hence the DC input power increases. Tthe output voltage swing (measured across the active device) reduces somewhat, hence the efficiency drops.

The phenomenon is very strong when playing with a class E output stage. There the device is switching, hence the gain is well below the maximum available gain for the device when operating under (for example) class B or C.

Another example is an audio amplfier. It is designed for (for example) 4 Ohms load, but its output impedance is well below that. So if there is no protection present, it will deliver more output power into 3 Ohms (if the supply can deliver the DC input power of course).
 

Another example is an audio amplfier.
Audio amplifiers are operated as voltage source rather than impedance matched. I assume that class E might be operated beyond impedance matching, too.
 

For a system with a PA delivering +20 dBm, whatever the impedance is, can max EIRP be +20 dBm, assuming 100% effective and perfect matched antenna.

"the power can actually be higher for a mismatched antenna!"

If 50 Ohm is considered as "matched" but PA impedance resulting in max power output not is 50 Ohm, can this occur. The word "mismatched" is a bit mismatched to use in a such case.
As mentioned in the link, +23 dBm radiation from a +20 dBm transmitter, noway.
Total system output power can never be higher then input power. Else it would solve a big energy problem.

Max possible output power at fundamental frequency is not always only depending on optimal conjugate matching load impedance.
As example, I recently impedance matched TI CC2520 (assume class E), No shield-can, no SAW-filter.
In this particular case was goal to optimize for max TX power efficiency and still keep harmonic radiation levels below -30 dBm. Every % of improved efficiency was important due to limited battery capacity.
Variations in load impedance at its harmonics in this chip, also affect fundamental output power level, even if fundamental impedance is kept constant.
Guess it can be explained as a function similar to a tank-circuit for class C. Manufacture information regarding this is in data sheet zero. In alternative sources have TI mentioned it, so it is not just me that have noticed this.
Have tried to Google a more clever explanation why harmonics impedances affect in this way, but no result.
 

@E. Kafeman: See the 20 dBm as mentioned in the datasheet of an amplifier (into 50 Ohms). Depending on design and topology, both the forward power and the net output power (Pfwd - Pref) can be above 20 dBm under certain non-50 Ohms load conditions. This is not the case when before doing the mismatch load test the amplifier was tuned for maximum output power into 50 Ohms.

Regarding your (assumed) class E device, it can reach that high efficiency only because of the harmonic current and voltage components present at the chip level. So the termination of the chip for harmonics is as important as it is for the fundamental component. The chip's output device is operating in its non-linear region, hence higher harmonics affect the fundamental component and vice versa. I noticed similar behavior in RF PA design. I can imagine your feelings regarding the design for a good trade-off between output power and efficiency.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top