How Can We Efficiently Harvest RF Energy to Charge Batteries?

[Joel Shearon]

Hey Physics Forum!

So I will try to give as much detail as possible and needed without dragging on.
We have an RF source at 5.8 GHz producing -17dBm and we eventually amplify it close to 27dBm (so we get as close to 30dBm out of source antenna as possible).
We are using 5.8 GHz slot antennas on receiving and sending end.
We have a stub rectenna circuit to convert the RF into a DC which we plan to charge up an energy bank (super cap or battery) to then power a sensor bluetooth MCU.

Problem with super caps: they match the source voltage which without boosting is like 20-100mV (no good for super cap -> MCU). We have looked into a ultra low voltage booster by Linear Tech but we also were considering batteries which could charge up slowly to capacity.

Any recommendations on batteries or methods for banking this RF->DC? Current is on the order or 30uA so we are also worried that if we boost from mV to V that the current will just be awful.

Any ideas on any of this?

 

[anorlunda]

to then power a sensor bluetooth MCU

Can that sensor bluetooth function with just mv suppply? If no, then you're still faced with the need to boost voltage.

 

[anorlunda]

You may also get help from these PF threads.

https://www.physicsforums.com/threads/rf-energy-harvester.827739/
https://www.physicsforums.com/posts/5279094/
https://www.physicsforums.com/posts/5214582/
https://www.physicsforums.com/posts/3548833/

 

[Joel Shearon]

Can that sensor bluetooth function with just mv suppply? If no, then you're still faced with the need to boost voltage.

In short, it needs 253 mW with roughly 13-14 mA to do what we need. This is not appropriate for a direct powering method. Boosting also won't work standalone since the current would not suffice so we are looking at supercars/batteries/energy banking in general. I will look at your links and reply accordingly.

 

[berkeman]

so we get as close to 30dBm out of source antenna

Holy crap! What in the world are you doing? Do you not care that you will be jamming the 5.8GHz WiFi band? What country are you in?

http://www.afar.net/tutorials/fcc-rules

 

[Joel Shearon]

Holy crap! What in the world are you doing? Do you not care that you will be jamming the 5.8GHz WiFi band? What country are you in?

http://www.afar.net/tutorials/fcc-rules

We are under 30dB...

 

[Joel Shearon]

We are under 30dB...

• "In the 5.8 GHz band the rule is less restrictive. The maximum EIRP allowed is 53 dBm (30 dBm plus 23 dBi of antenna gain)" not to mention this

 

[berkeman]

We are under 30dB...

I understand. But the band is intended for communication, not super inefficient power transfer. Why don't you just use a solar cell or something to harvest energy for your circuit, instead of jamming everybody's WiFi signal?

 

[Joel Shearon]

I understand. But the band is intended for communication, not super inefficient power transfer. Why don't you just use a solar cell or something to harvest energy for your circuit, instead of jamming everybody's WiFi signal?

Project purpose is dual source energy-harvesting. We are using solar and RF and using source selection in order to sufficiently power an MCU with or without light. RF is a part of the project goal and 5.8 GHz is the range for the equipment we have and this project has been approved and monitored under Georgia Tech's Emag faculty and we have taken serious precautions to obey the FCC Rules.

With that out of the way, I was really looking on here for advice on how to efficiently bank the DC-converted RF energy.

 

[tech99]

Why not use an antenna which has a high output resistance, such as a half wave end fed, which will provide higher voltage, or step up the received RF voltage using a transmission line transformer or LC network? Why use a DC converter when you already have your power source as AC?

 

[sophiecentaur]

Project purpose is dual source energy-harvesting.

I don't quite understand your use of the term "harvesting". That term would normally refer to picking up 'free' energy that would otherwise go to waste. In your case, you are producing very unneighbourly levels of potential interference for other people and recouping just a tiny fraction of it. You would do much better with a good halogen lamp, a couple of large reflectors and a PV cell. At the same time, you would be producing no interference (except for a tiny amount of light pollution if you did it at night.
As for the batteries, you would need to calculate just how much energy would be falling on your receiver - taking into account the gains of transmit and receive antennae etc etc. That would give you an idea of the mean power available from your power link and, depending on the duty cycle you want for your end use, you would get an idea of the power you could be using. The areas for your antennae will need to be pretty big for transmission over a significant distance.
In short, you need to decide on all the parameters of your proposal before deciding whether or not it could possibly work. Numbers really count here. Why don't people believe that simple fact? (Ref today's UK budget announcements haha)

 

[russ_watters]

Is the high power source just for testing purposes?

 

[rbelli1]

At what distance are you looking to use this?

BoB

 

[davenn]

Project purpose is dual source energy-harvesting. We are using solar and RF and using source selection in order to sufficiently power an MCU with or without light. RF is a part of the project goal and 5.8 GHz is the range for the equipment we have and this project has been approved and monitored under Georgia Tech's Emag faculty and we have taken serious precautions to obey the FCC Rules.

It really is a waste of time ... as per sophiecentaurs' comments
You like all others before you really don't understand how very low the radiated power level is. 27dBm ( 0.5W (500mW)) at even a few ft from the source isn't going to generate anything useful at the receive end

And also, sophies' other comments are as equally valid you are generating 0.5W transmit and probably wasting several watts achieving that to recover a few microwatts ... can you see how pointless that is ?

Dave

 

[tech99]

The calculation is easy apart from the rectifier, and this needs reference to the characteristic curves and is not easy to work out. For small power levels, the rectifier will lose a much of the available energy in its resistance, and is really the weak link. That is why I suggest trying to use an antenna which gives the highest voltage so that the rectifier is more efficient.

 

[rbelli1]

The antenna needs to receive about 1/4 of the energy emitted by the base station. I was asking about distance so someone could calculate the minimum size of the antenna that would be effective for that. I would guess that it would be anywhere from unreasonably to impossibly large.

BoB

 

[sophiecentaur]

The calculation is easy apart from the rectifier

OK. Do the calculation, assuming the rectifier is ideal. How much power is the system going to yield? How long would it take to charge a cellphone battery? Very likely, the charger could not keep up with the quiescent requirements of the phone on standby. Just divide your source of 0.5W by more than 1000 (optimistic for a path loss) and divide by your battery volts. That will give you the most charging current you could expect. Assume that phone battery will be about 1000mAh capacity.
The "weak link" is the reality of the situation.
I have asked this question in the past: Can you find such a system to buy on the Internet? Ask yourself why not.

 

[RFDiagnostics]

The RFD102A is a broadband RF-DC converter that can be used for your application. The module works from 60Hz...6GHz. I agree with the other posters that there is not a lot of wireless power to be harvested out there and that any wireless energy harvesting system needs to be designed with a wireless source in mind. The strongest source of energy is a microwave oven and to get useful power you need to be within inches of the microwave near the door seam or the vents on the side. WiFi is too weak unless you are next to the router or if you have some custom code to get the router to transmit on all channels. One option to try with your system is to get an RFD102A-DET that is optimized for 2.5GHz and trim the wire dipole in half to get to 5GHz and then do minitrims to center the frequency response at 5.8GHz.

 

[sophiecentaur]

Trimming a dipole to get optimum match will help you by a few dB only. 'Harvesting', to be any earthly use, would need a gain of at least 30dB. You are talking in terms of an array / dish of several metres diameter and not a dipole. That's unless your dipole is in the near field of the source. (Contactless Charging country.)

 

[sophiecentaur]

The strongest source of energy is a microwave oven and to get useful power you need to be within inches of the microwave near the door seam or the vents on the side.

My Microwave oven is probably running for an average of 10 minutes a day. How much total energy, leaked to the outside world does that represent? You really need to get real in discussions like this one (which, to my mind, border on the surreal).
If you can get that close to your microwave oven, then buy a mains splitter and use that other mains outlet for as much energy as you want. It won't be free but it is certainly easy to access and the necessary equipment would be cheap and readily available.