I'd like some guidance in designing and sourcing components for a 5.5 Mhz tuned tank circuit capable of going up to 5.7 and down to 5.3 Mhz. Any help would be greatly appreciated.

 

Will not work beyond the generation of picowatts of power unless you're orbiting the planet at the F layer. At the F layer, then we're talking about 10mW per m2 of ion density.

 

Suggestion #1: Compute the geometric mean for 5.3 MHz and 5.7 MHz and call that your center frequency. A tank circuit for that center frequency can be fabricated for any combination of an inductor and a capacitor that satisfies:

\( f_c\;=\;\cfrac{1}{2\pi\sqrt{LC}} \)

As you are no doubt aware that problem has an infinite number of solutions. To narrow the solution set down we pick a reactance for the two components in parallel.

Suggestion #2: Pick a reactance of 100Ω each for the inductor and the capacitor. The two in parallel will have an impedance of 50Ω

This tank circuit will then be a pretty good match for an efficient antenna with an impedance of 50Ω.

For example:

\( \sqrt{(5.3\times10^6)(5.7\times10^6)}\;=\;5.4964\times10^6 \)

\( L\;=\;\cfrac{100 \Omega}{2\pi f}\;\approx\;2.9\;\mu\text{H} \)

and

\( C\;=\;\cfrac{1}{2\pi(5.4964\times 10^6)(100 \Omega)}\;=\;290\text{ pF} \)

The following simulation shows the expected behavior

 

Suggestion #1: Compute the geometric mean for 5.3 MHz and 5.7 MHz and call that your center frequency. A tank circuit for that center frequency can be fabricated for any combination of an inductor and a capacitor that satisfies:

\( f_c\;=\;\cfrac{1}{2\pi\sqrt{LC}} \)

As you are no doubt aware that problem has an infinite number of solutions. To narrow the solution set down we pick a reactance for the two components in parallel.

Suggestion #2: Pick a reactance of 100Ω each for the inductor and the capacitor. The two in parallel will have an impedance of 50Ω

This tank circuit will then be a pretty good match for an efficient antenna with an impedance of 50Ω.

For example:

\( \sqrt{(5.3\times10^6)(5.7\times10^6)}\;=\;5.4964\times10^6 \)

\( L\;=\;\cfrac{100 \Omega}{2\pi f}\;\approx\;2.9\;\mu\text{H} \)

and

\( C\;=\;\cfrac{1}{2\pi(5.4964\times 10^6)(100 \Omega)}\;=\;290\text{ pF} \)

The following simulation shows the expected behavior
View attachment 306503

Thank you papa bravo

 

Thank you papa bravo

Is that 2.9 microhenries, or millihenries?

 

Will not work beyond the generation of picowatts of power unless you're orbiting the planet at the F layer. At the F layer, then we're talking about 10mW per m2 of ion density.

Unless he lied in the book there should be a lot more energy to harvest than that.

 

Unless he lied in the book there should be a lot more energy to harvest than that.

He probably did lie. The power levels of RF energy available for harvesting are not worth the effort.

 

Is that 2.9 microhenries, or millihenries?

Do the calculation yourself. I compute 2.9 microhenries and the simulation pretty much confirms that value.

 

Unless he lied in the book there should be a lot more energy to harvest than that.

It's all about energy density. If it was a readily available energy source, life would have evolved to use it like sunlight or even heated water on the bottom of the ocean.

There's energy in moonlight but the levels are tiny. https://forum.allaboutcircuits.com/threads/super-moon-shine.100322/post-752870

What's the book? As I've said before books can be garbage (double so when Tesla is mentioned) just like online information.

 

I'm an amateur radio operator and on my rig is a pan-adpter that is basically a spectrum analyzer so I can "see" RF energy across an adjustable frequency range. The typical noise floor for that band is approximately -120 dBm. That is a power level 120 dB down from 1 milliwatt. That amounts to 1 femtowatt. If you are not familiar with SI prefixes, that is,

\( 10^{-15}\text{ watts} \)

Try to explain to me why that level of power would be worth harvesting.

 

Papabravo: great idea - to spend a 1 megawatt to keep plane in the air to grab the 1 femtowatt.
As used to say mister Gorbačev: "Economics must be Economic".

I have been taught that sailboat is ever seen most expensivest method how to travel completely costlessly. Therefore may add that to sucking out the free energy from cellphone towers may harvest much more benefit. Thus, if my phone application shows all the time 1...3 GHz energy field around me in the city environment between -120 and -90 dB that may happen for some devices sufficient energy indeed.

 

I'm an amateur radio operator and on my rig is a pan-adpter that is basically a spectrum analyzer so I can "see" RF energy across an adjustable frequency range. The typical noise floor for that band is approximately -130 dBm. That is a power level 130 dB down from 1 milliwatt. That amounts to 1 femtowatt. If you are not familiar with SI prefixes, that is,

\( 10^{-15}\text{ watts} \)

Try to explain to me why that level of power would be worth harvesting.

It would be a moral victory, PB. In some way, it would vindicate Elon, err... I mean Nikola.