Is a 50 Ohm Load Equivalent to a Resistor in HF Transmitter Circuits?

[temujin]

Hi,

I have a HF transmitter rated with 200mW output power. An antenna is connected to the transmitter via a 50 ohm coax cable. (The antenna is also matched to 50 ohm.)

Does this mean that from the transmitters point of veiw, a 50 ohm resistor is connected as load?

Can I find the current in the antenna circuit simply by applying ohm´s law?
(I^2 = P/Z). I´m pretty sure I can´t, but I´m not sure why...
If not, what does actually the power rating of the transmitter imply?

regards
t.

 

[mmwave]

Hi,

I have a HF transmitter rated with 200mW output power. An antenna is connected to the transmitter via a 50 ohm coax cable. (The antenna is also matched to 50 ohm.)

Does this mean that from the transmitters point of veiw, a 50 ohm resistor is connected as load?

That the idea. If the cable is perfect and matched to the load you have the equivalent of a 50 ohm load at the output of the transmitter. In reality the antenna won't be perfectly 50 ohms with no reactance and the cable isn't perfect but if you bought them from a reliable source it should be good enough.

Can I find the current in the antenna circuit simply by applying ohm´s law?
(I^2 = P/Z). I´m pretty sure I can´t, but I´m not sure why...
If not, what does actually the power rating of the transmitter imply?

To the extent you truly have a 50 ohm antenna & a 50 ohm characteristic impedance cable this is exactly how you would calculate it. To be any more accurate would require a measurement of the cable + antenna or else a SWR meter connected between transmitter & cable. From that you can work out the actual magnitude of the impedance. To do better you need a network analyzer.

When the source and load aren't perfectly matched the concept of current loses a lot of its value. The current will vary sinusoidally with distance. The maximum possible current (for load of zero ohms) would be twice the current in your equation. The max possible voltage (for a open load) would be calculated similarly. Don't do these experiments to prove it - you will destroy your transmitter.

The power rating is always for a specified load impedance giving a perfect match to the transmitter output, usually 50 ohms except for cable tv systems.

 

[ravenprp]

Hi t,

To answer your first question, yes, from the transmitter's point of view, a 50 ohm resistor is connected as the load. This is because the antenna and coax cable are matched to 50 ohms, so the transmitter "sees" a 50 ohm load.

However, you cannot simply use Ohm's law to find the current in the antenna circuit. This is because the input impedance of the antenna is not purely resistive, it also has reactive components such as inductance and capacitance. In order to accurately calculate the current, you would need to use more advanced equations that take into account these reactive components.

The power rating of the transmitter implies the maximum amount of power that it can safely output without damaging itself. This does not necessarily mean that it will always output that amount of power, as it will adjust its output based on the load it is connected to. So in your case, the transmitter has a maximum output power of 200mW, but it will adjust its output based on the 50 ohm load to ensure that it does not exceed that power rating.

I hope this helps clarify things for you. Let me know if you have any further questions. Best regards.