# RF to DC convertor [closed]

I am working on RF energy harvester and I want to increase the current. Should I connect more rectifiers in parallel to one antenna or more antennas to one rectifier?

Here is my schematic.
My antenna is a printed Helical with 2.3 dBi gain and 66% efficiency. My transmitter is 1 W 868 MHz.

I want a current of a few milliamps.

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My range is about 10m from the 1W transmitter.

I am working on a prototype of low power device with my friend, I think my friend use this article for this design.

The device uses Microamps and low voltage but the problem with bluetooth TX current is about 6mA.

My PCB is 4-layers and My dual band antenna is printed Helical

I want to use one antenna and more rectifiers as I see this product use one antenna and more rectifiers chips.
TI RFD-ASSY-01 – Microwave Energy Harvesting System

They say: "The assembly can produce up to 60mA continuous current and ~40% RF-DC conversion efficiency into a 1kOhm load from 18dBm…29dBm input power at 2.45GHz."
[[RM: ie 60 - 800 mW input. ]]

• ... or more rectennas together? Without a schematic, and specifications of the individual components, and measured performance of the components individually and together, no one here is able to answer. What you are describing is a development question. 'I have got this so far, how do I improve it?' You make measurements, careful measurements, guess what's going on, make changes, make more measurements, see if your guesses were right, rinse and repeat, understand what's going on, then make better design. You probably need to optimise a rectenna, then connect those at DC. – Neil_UK Jul 2 '17 at 6:07
• i.stack.imgur.com/jQ3Ch.png here is my schematic – user31562 Jul 2 '17 at 6:33
• Can you describe what your motivations were when you designed that circuit, I.e. where does it architecture stem from? What were the calculations you did when choosing component values? If all this really works at 868 MHz, then those 100 uF are far too large unless I'm missing something. Isn't your rectifier design unsound, or is this a diode cascade I don't understand? – Marcus Müller Jul 2 '17 at 7:20
• You have a fundamental problem of power. If you want mA (perhaps 5?) at 2v, that's 10mW (+10dBm) at 100% efficineyc, or if you can get 25% efficiency then you need +16dBm receive power. Can you get only 14dB path loss at 10m range at 868MHz? You can totally ignore the mechanicsm of rectificiation etc until you have the brute received power from the antenna. – Neil_UK Jul 4 '17 at 5:32
• I'm yet to see a single [energy-harvesting] question that makes sense. – Dmitry Grigoryev Jul 4 '17 at 12:21

I am working on RF energy harvester and I want to increase the current. Should I connect more rectifiers in parallel to one antenna or more antennas to one rectifier?

Think about what happens when an antenna emits RF power into its surroundings (assume a vacuum and assume an isotropic antenna). An isotropic antenna is like a perfect light bulb that emits light-power in all directions. So let's say it emits 1 watt in total and then ask the question how much power can be received at some distance.

If that distance was 1 metre and a receiving antenna could be constructed that totally surrounded the antenna, then 1 watt would be received. If the distance was 100 metres and a perfect receiving antenna could be constructed that totally surrounded the antenna, then 1 watt would also be received. In other words, all the power is collected (ignoring inefficiences of course)

Now here's the problem, if you only have a small antenna at some distance, the amount of power it can physically receive is very limited. In other words, at 100 metres with a receiver aperture of (say) 1 sq metre, the total amount of power that can be received is a tiny fraction of that being pushed through the surface of a sphere of radius 100 metres.

The surface of a sphere has an area of $4\pi r^2$ so at 100 metres radius that is 126,000 sq metres hence, your antenna with an aperture of 1 sq metres receives only 8 uW from the 1 watt source.

The aperture size of any antenna is this: -

Your antenna antenna has a gain of 2.3 dBi (the "i" stands for isotropic and it means that you have 2.3 dB more gain that an isotropic antenna) hence, at 868 MHz (0.346 metres wavelength), your effective capture area is: -

$\dfrac{0.346^2}{4\pi}\times 1.303$ = 0.0124 sq metres.

This area totally governs how much power you can collect from one antenna so, when you ask a question like: -

Should I connect more rectifiers in parallel to one antenna or more antennas to one rectifier?

You are missing the big picture - address the big picture of how you are going to collect more power before you ask about current (bearing in mind that the receiving antenna has to deliver power into a 50 ohm load for maximum efficiency and that sets the big scene).

Quick calculation to see how much current can be efficiently liberated at 1 metre distance from the 1 watt transmitter given the gains involved. If both transmit and receive antennas are 2.3 dBi gain then the effective power transmitted is 2.3 dB higher at 1.7 watts.

At 1 metre distance, the sphere area is 12.6 sq metres. The receiver has an effective area of 0.0124 sq metres hence, the received power is 1.7 x 0.0124/12.6 = 1.7 mW. If that power were collected with 100% efficiency, the voltage produced across a 50 ohm load is 0.29 volts RMS with a current of 5.8 mA RMS.

You will probably achive no more than 20% of this given inefficiences and the conversion to DC so don't bank on more than 1 mA at 1 metre and with a pitifully low terminal voltage.

• Thanks for your answer , My device power consumption is 100micro watt at 1.8V but my only problem is the bluetooth consumes about 6mA TX current. – user31562 Jul 2 '17 at 10:18
• Address the power problem by using more antennas or increasing directivity (gain) of the antennas. Alternatively, think about it like having a solar cell at the distance you need and how you would get more light onto it from a 1 watt LED. There is no conceptual difference between the RF and optical scenarios. If you can't get the power at the distance you either need more transmit power or more surface area. Laws of physics etc.. – Andy aka Jul 2 '17 at 10:25
• Thanks very much , If I use multiple antennas So I must connect them to one rectifier circuit because I need more current not voltage , right? – user31562 Jul 2 '17 at 16:04
• Each antenna needs to be individually rectified and the dc power levels combined but don't be under any illusion, using multiple antennas is tricky. Each antenna must operate in a separate field to prevent one antenna stealing power from another. – Andy aka Jul 2 '17 at 18:42
• So it's hard to put printed antennas with a space between them. – user31562 Jul 2 '17 at 18:49

You also need to provide as much information as possible as factors that may not seem relevant may prove vital. Things like source to target range are pivotal. Transmit antenna gain /.directionality is crucial if any distance is involved. If you are close enough to be in the near-field and electromagnetic approach may be better.

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Your circuit diagram appears to come from this paper - intro page to
Enhanced Passive RF-DC Converter Circuit Efficiency for Low RF Energy Harvesting -
and the main point of the paper seems to be that there MAY be a better solution available than this one depending on power levels and other factors.

Full paper - html: http://www.mdpi.com/1424-8220/17/3/546/htm
Full paper - pdf: http://www.mdpi.com/1424-8220/17/3/546/pdf

The paper provides details of efficiency versus number of stages (fig 13), loading (fig 11), for their suggested converter with a range of power levels.

The paper provides such a range of discussion that it would SEEM to answer your question far better than anyone could here with the amount of information that you have provided. If you are already basing your work on this paper you should say so and state your requirements with reference to the paper, what you have tried, what works and what you still cannot understand.

Note in their fig 15 the nearly 3:1 ratio between their simulated and actual results. Read the text, note the reasons that they consider cause this result and consider whether you can optimise these. For example, mismatches due to component tolerances are easily addressed. Schottky-diode-practical versus ideal-diode performance differences MAY be able to be addressed by working at a higher input voltage - perhaps by using an RF 1:N transformer, or higher Q resonance or ...?

• My question if I connect more rectifiers in the same design in parallel the current will increase or I must connect more antennas ? – user31562 Jul 2 '17 at 7:37
• @user31562 I am aware of what you asked. My answer is as above. Tell us more about your actual circumstances. What range? (crucial) Did the circuit come from that paper? Have you considered more than 2 of their stages? Have you looked at optimising what they indicate needs optimising? Does changing the Q of the cct help? Are you very accurately on the resonant cct. ||| Adding multiple cctc with their own antennas should increase power if they are not too close to each other. – Russell McMahon Jul 2 '17 at 8:54
• @RussellMcMahon Having read that paper, it looks like their analysis of the L is hogwash, treating it as energy storage rather than antenna matching, like they're playing around with some components and getting a reasonable answer for the wrong reasons? – Neil_UK Jul 2 '17 at 9:20
• My range is about 10m from the 1W transmitter , I am working on a prototype of low power device with my friend , I think my friend use this article for this design. The device uses Microamps and low voltage but the problem with bluetooth TX current is about 6mA , My PCB is 4-layers and My dual band antenna is printed Helical Sourece:ti.com/lit/an/swra416/swra416.pdf I want to use one antenna and more rectifiers as I see this product use one antenna and more rectifiers chips rfdiagnostics.com/product/rfd-assy-01 – user31562 Jul 2 '17 at 10:10
• @Neil_UK I have not yet read the paper properly and may not get to do so any time soon as it seems to be peripheral to the main point now we know that the desired range is 10 metres. | I was not confident that the circuit as shown would work overly well at low Vin values but without simulation I'm not sure what series resonant voltages are achieved. Even using Schottky diodes the D1+D3 voltage drop requires about 0.5V to start to add any useful voltage increase to the L3-C3 path - hence my comment re perhaps using an input transformer. Maybe ... – Russell McMahon Jul 3 '17 at 19:12