radio frequency source impedance

[kbrf]

The radio frequency source can be modeled as a constant voltage source with a series characteristic impedance of say 50 Ohms. That means it can drive a 50 Ohm load.
Q1.Does that mean the E/H ratio is 50 as in vacuum E/H is 120π.
Secondly, when source is perfectly matched to a reactive load say 1+j45 ohms by a matching network, by maximum power transfer theorem, half of the power is transferred. Q2.Then where the other half power is dissipated?
I have used an L section matching at frequency of 13.56 MHz.
My detector says 100% forward power and 0% reflected power for an input RF power of 1 kilowatt.

 

[MikeMl]

About question 1.

What does the E/H ratio have to do with the output impedance of an RF source? The output impedance of the RF source is what it is designed to be, in this case 50Ω. The RF source will deliver its rated power to a load only if its impedance is also 50Ω.

What is the E/H ratio? ( I have never heard of it)

About question 2.

If an RF source designed for a 50Ω load is connected to anything other than 50 ± j0Ω, then the source will not deliver its rated power. If you can diddle the matching device such that its input port looks like 50 ± j0Ω, then the full rated power is delivered (minus any losses in the matching network).

You do not think of an RF source as being say a 200W source, half of which is dissipated inside the source, and half of which is delivered to a load. Rather, you just think of it as being a 100W RF source which has a rated output load impedance and a certain (measurable) efficiency (Pout/Pin), where some of Pin might be coming from a power supply and some might be coming from an input signal.

 

[kbrf]

Q: What is E/H ratio?
Ans: Whenever you set a radio frequency oscillator to give power to a load, there is always a current and a voltage in the total circuit. The ratio of voltage to current is what I have stated as E/H ratio. If the source has a characteristic impedance of 50 Ohm and a load of 50 Ohm is connected to it then the V/I or E/H is maintained at 50 Ohm throughout the transmission line and maximum power is transferred.

I could not understand: "Rather, you just think of it as being a 100W RF source which has a certain (measurable) efficiency (Pout/Pin), where some of Pin might be coming from a power supply and some might be coming from an input signal. "

What is the power supply in a RF source? (Is it the generator itself?)
What is input signal?

 

[MikeMl]

Well, I have several RF sources around here (transmitters, signal generators, etc), and they all plug in to 12Vdc or 120vac. They effectively generate RF power from a power supply..., as such they have an efficiency.

I have several (linear) RF power amplifiers whose output power includes feed-through of an input RF signal. For example, I have an RF amp which requires ~100W of drive, and it produces about 1200W of output power (into a 50 to 70Ω load). In that case, most of the 100W of drive appears at the output, so effectively, the other 1100W is coming from the amplifier. Since the amplifier is ~50% efficient, the power input is about 2.2KW from the AC line ( and I have to get rid of about a 1kW of heat).

Now this amplifier is referred as a "1200W amplifier" (not a 2400W amplifier which delivers half of that to the load).

Sounds like you are trying to make up an new phrase ("E/H ratio") for something that already has a widely accepted definition. Its called "output impedance", which just happens to be a ratio of voltage and current.

 

[JimB]

The radio frequency source can be modeled as a constant voltage source with a series characteristic impedance of say 50 Ohms. That means it can drive a 50 Ohm load.

Depends on the radio frequency source.
A signal generator which has a maximum output of 2 or 3 volts can be modeled as you suggest, it can drive any impedance load it likes.

A power amplifier could be designed to work into a 50ohm load but that does not mean that the output impedance is 50ohm.

Q1.Does that mean the E/H ratio is 50 as in vacuum E/H is 120π.

I have never heard of E/H ratio.
A quick googling for E/H ratio gave references to photoelectrics and hadron calorimeters, you seem to be making some incorrect comparison with another branch of physics.

Secondly, when source is perfectly matched to a reactive load say 1+j45 ohms by a matching network, by maximum power transfer theorem, half of the power is transferred. Q2.Then where the other half power is dissipated?

If the generator CAN be accurately modeled as a 50 ohm resistor in series with a perfect voltage source, then the other half of the power is dissipated in the resistor.
Dont lose sight of the fact that this is just a theoretical generator, a practical generator will be more complex.

I have used an L section matching at frequency of 13.56 MHz.
My detector says 100% forward power and 0% reflected power for an input RF power of 1 kilowatt.

So, you are using a power amplifier (1kW) on one of the ISM frequencies, I would suggest that the amplifier requires a 50ohm load, rather than it has a 50ohm output impedance.
Your "detector", I assume is some kind of VSWR meter or a directional power meter and shows that your matching network is doing a good job of converting your complex load to 50ohm resistive, which is what the generator requires to work correctly.

The efficiency of the generator is another question entirely.
Comparing power input from the mains outlet on the wall to RF out from the output connector the efficiency will be quite a lot less than 50%.

JimB

 

[RCinFLA]

No, a coax does not have a uniform E/H density. It varies with distance from center conductor and is fraction of wavelength (within useable max freq of coax).

377 (120 pi) ohms is the characteristic impedance of free space which is the uniform relationship between E and H field.