Where does the other 50% of energy go in wireless power transfer?

[Akmalidin]

Hello,

I am a student doing a research at university.I am working on wireless electricity transfer.I have two coils forming transmitter and Receiver. My coils are with diameter 17 cm. I have reached 53% efficiency in 15cm.And It is not increasing anyhow whatever I do. I want to ask where other 50 % of energy go? Plus, I have reached 50% with 15V input. And I expected that increasing the voltage ,would increase the efficiency.But Increasing the input voltage , causing drop in efficiency. Why is that? I can reach 50% efficiency in 20 cm too. But ,cannot get more then 53% in 15 cm.Why?

 

[mfb]

I want to ask where other 50 % of energy go?

Electromagnetic radiation and heating of your coils.

And I expected that increasing the voltage ,would increase the efficiency.

Why did you expect that?

How does your load look like? The efficiency will depend on it.

 

[Akmalidin]

Load is the resistor connected to bridge rectifier in the receiver. I am using optimal best load for that voltage and distance.

 

[Baluncore]

Your two coils make a transformer, but without a magnetic core. There is too much flux leakage = radiation.

You can model it as a loosely coupled transformer or as two loop antennas. The antenna model is probably more accurate here.

What is the frequency of the energy source ?
How many turns are there on each coil ?

 

[Akmalidin]

I am pumping 4 Mhz.
coils have 20 turns each.

 

[berkeman]

Are you doing this inside a shielded room or shielded enclosure? If not, you may be violating rules about illegal RF transmissions.

 

[Baluncore]

Have you measured or calculated the inductance of the loops ?
Have you modeled the transmission lines and impedance matching through the network to the load ?
If you are not careful the symmetrical non-linear diode bridge will act as an odd harmonic generator and produce a comb of radio interference. It will radiate at 4MHz, 12MHz, 20MHz, 28MHz ...

 

[sophiecentaur]

I endorse the above two posts. You could be causing havoc all around you with your experiment.

 

[Baluncore]

We do not know your power level. To turn on the diode bridge will take about 2 volts, then there is the voltage drop across the load, say another 2V. That makes an RF input of 4 volts for 50% efficiency since diodes represent a significant loss. If you remove the diode bridge and couple from the secondary loop directly into the load, does your efficiency improve ?

If the generator is 50 ohm the current will be at least 4V/50R = 80mA to be 50% efficient. Power will be 4V * 80mA = 0.32 watt minimum. I guess you are radiating less than 100 mW. Most power will be going into the diodes and load. But that is only a guess.

What is the resistance of your dummy load ?
What is the output impedance of your generator ?
How have you made your diode bridge? Part Number/s ?
What is the maximum output power of your generator ?

You have been cautioned sufficiently about causing interference with RF services. If you are not able to get assistance from your test site then I can give you advice here, but you will have to fully disclose your experimental equipment. It is a long time since I have needed to prosecute a generator of interference. I found that I could reduce interference more effectively by offering advice, assistance and education.

 

[Akmalidin]

NO I am not conducting the experiments inside the shielded room.

 

[Akmalidin]

Yes I have measured and Transmitter is 140 microH and the Receiver is 100microH. When I tried both making the same inductance, I had low efficiency.Now they may be pretty matched as they are producing 50 %.

 

[Akmalidin]

How can I model the transmission line? I have tried to match both coils ,using capacitors did not help to match them but i by wrapping the wire I somehow matched not fully though.And resistance is also matched to the input resistance.

 

[Akmalidin]

I think diodes do not cause problem unless they get hot and I have about 1.2 V drop in diodes.I tried to measure the output power with oscilloscope and it was also 54%.And with diode I have 53 %. I assume that some parasitic inductance and capacitance is stopping me from getting high efficiency.
I have output resistance 300 ohm in 15 cm.
Output resistance of my generator is 64ohm.
And I made bridge accordingly ,shown in the picture.
Output power of my generator is 90 W .

 

[berkeman]

NO I am not conducting the experiments inside the shielded room.

Do you understand the issues with this? What could be some issues with this?

What country are you in? As alluded to by Baluncore, when you interfere with your local radio frequency services (Police, Fire Department, Medical Services, etc.), you can get a knock at your door...

 

[Akmalidin]

I work in a lab and the frequency I am using is free to use as communication with governmental departments are done using higher frequency.

 

[berkeman]

I work in a lab and the frequency I am using is free to use as communication with governmental departments are done using higher frequency.

And you were asked about harmonics. Your response is...?

Who is your supervisor? Has he or she approved your RF test plan?

 

[berkeman]

BTW, in the US the AM broadcast band starts at 1MHz. If you interfere with it , the legal problems can be worse than interfering with the police/fire/ems bands. Go figure...

 

[Baluncore]

Akmalidin. You are new to RF impedance matching. Matching is essential to designing efficient RF circuits that get close to 100% power transfer.

Your generator has an output impedance of 68R, it is driving a load of 300R through your air-core transformer. If the coils are analysed as a transformer then you will need a turns ratio of √(300/68) = 2.1 If the transmit coil has say 10 turns, then the receive coil will need 21 turns to match the impedance. If you measure the inductance, the receive coil should have 2.1² = 4.41 times the inductance of the transmit coil.

You measured the transmitter coil as 140 uH and the receiver coil as 100 uH. That ratio is not going to match the load to the transmitter. It is the opposite of what you need. You should wind two more coils, this time with a turns ratio of 10 to 21.

If you transmit 90 watt from a 68R impedance, then the generator voltage will be 78.23 volt while the current will be 1.15 amp.
With a 10 : 21 turns ratio transformer the voltage rises from 78.23 to 164.28 volts.
With a 10 : 21 turns ratio transformer the current falls from 1.15 to 0.5476 amps.
V/I in the primary was (78.23 / 1.15) = 68.025 ohms. V/I in the secondary is now (164.28 / 0.5476) = 300.00 ohms.

165V of RF is more than enough to give you deep burns if you get too close. Please take care.

Latvia is in ITU Region 1, where 4.000 MHz is allocated to maritime voice communications. Maybe you should consider moving your frequency up to 6.780 MHz which is a dedicated ISM band. Your stray radiation will not be noticed there.
https://en.wikipedia.org/wiki/ISM_band#ISM_bands

 

[Baluncore]

I think diodes do not cause problem unless they get hot and I have about 1.2 V drop in diodes.

Is that 1.2 volt across one diode or across the two conducting diodes ?
What is the part number of the diodes you used ?

 

[Akmalidin]

1.2 V IN Two diodes. I am at work now but they are fast diodes with 4 ns. And When I measure with oscilloscope , I am getting the same efficiency as in diodes. So I would say, diodes are almost lossless.

 

[Baluncore]

I believe you have just confirmed that you are using Schottky fast recovery diodes. That is good.

The 50% circuit efficiency is probably explained by the transformer impedance mismatch due to the wrong turns ratio of the two coils.

 

[Akmalidin]

I see. Thank you and I am going to try right ratio of turns in both coils.

 

[sophiecentaur]

So I would say, diodes are almost lossless.

Or rather that you couldn't measure the loss. There is still a dissipation of VI for the current flowing through any component (say 1.2V X 1A = 1.2W, roughly). We just try to minimise this effect when we can.
It would be pretty hard to measure and compare the true powers, with and without the diodes, to that sort of accuracy - so you could say the loss is 'negligible' rather than almost zero.