The discussion revolves around generating an RF magnetic field at variable frequencies (1 to 20 MHz) using a specific RF amplifier. Participants explore practical approaches to connecting a current loop to the amplifier, impedance matching, and considerations for the application of the magnetic field.
Participants express differing views on the appropriateness of direct connection versus using impedance matching techniques. There is no consensus on the constancy of the load impedance across the specified frequency range, and the discussion remains unresolved regarding the best approach for practical implementation.
Participants acknowledge the complexity of accurately determining the coil's loss resistance and the implications of varying load impedance across different frequencies. The discussion also highlights the need for further clarification on the specific requirements of the magnetic field application.
It sounds like you can, it says it's protected from opens and shorts and is unconditionally stable. But it will work better with an impedance transformer to better match the very low load impedance to the 50ohm output. It's not clear how much current you'll get out of it when it's output is shorted, like most RF stuff it's only specified at 50ohms. Why don't you ask the guys that make it?Malamala said:Can i directly connect the loop to the RF amplifier?
I actually checked it carefully and in my case ##X_L = 10 \Omega##. If I connect the loop directly, and I want to compute the current through the loop, should I then divide the applied voltage by this 10 Ohm? or does the amplifier has an extra 50 ohms internally so i need to divide by 50 + 10 = 60?DaveE said:It sounds like you can, it says it's protected from opens and shorts and is unconditionally stable. But it will work better with an impedance transformer to better match the very low load impedance to the 50ohm output. It's not clear how much current you'll get out of it when it's output is shorted, like most RF stuff it's only specified at 50ohms. Why don't you ask the guys that make it?
OTOH, if you get enough current with a 50ohm series resistor, then I'd do that, the amplifier will be happier and it's easy to understand.
Also, at 20MHz a 50ohm inductor is only 400nH (not really a short circuit), so a reactive matching network (Smith Charts etc.) would be the best because of the wide bandwidth. You'll need to understand your load impedance to do this right.
It's hard for me to believe your load is constant 10 ohms from 100KHz to 20MHz, but whatever. Yes the amplifier will have a 50 ohm output impedance.Malamala said:I actually checked it carefully and in my case ##X_L = 10 \Omega##. If I connect the loop directly, and I want to compute the current through the loop, should I then divide the applied voltage by this 10 Ohm? or does the amplifier has an extra 50 ohms internally so i need to divide by 50 + 10 = 60?
Sounds like a 4:1 balun would get you pretty close to a match then.Malamala said:I actually checked it carefully and in my case XL=10Ω.
Practice will be decided by your application's requirements.Malamala said:How should I proceed in practice?
Great suggestion!Baluncore said:an automatic HF antenna tuner
The LZY-22+ costs US$2219. That is the advanced level.DaveE said:Still, you've jumped to the advanced level. This guy isn't there yet.