2-stage voltage multiplier question

[lbminh2015]

Dear all,

I'm doing an RF energy harvesting circuit in UHF band. My question is about the voltage multiplier. My 2-stage multiplier (using 2 HSMS-285C diodes) follows Karthaus-Fischer's voltage-doubler.

A 0.1F supercapacitor is put right at the output of the multiplier. This DC voltage will be provided to run an MCU and a sensor. The minimal voltage for operation of MCU and the sensor are 1.9V and 2.56V, respectively.

As in theory, the DC output voltage of the multiplier will be:
Vout = 4 * V_RFpeak - 4 * V_fwd (forward voltage of the diodes) - V_load . So say, if Vpeak is 300mV, the maximum voltage I will get is 1.2V, ignoring Vfwd and Vload.

Here is the problem. When I measured the input AC voltage at the receiver antenna (Powercast) terminal, I got a very small signal with the maximum amplitude of 200 - 350mV. BUT I could get 2.6 VDC with full load as told above (MCU & sensor on full operation). Not all the times I could get this result, but sometimes I could. Please refer my picture below.
Probe1: Vin, RF
Probe 2: Vout

Please help me explain this phenomena. It seems my measurements went wrong, or my theory got sth wrong. Thank you very much.

 

[lbminh2015]

I could not insert my image in the 1st post. Here is my result.

 

[jim hardy]

So say, if Vpeak is 300mV, the maximum voltage I will get is 1.2V, ignoring Vfwd and Vload.

Can you really ignore that ?
https://docs.broadcom.com/docs/AV02-1377EN

 

[Tom.G]

Looks like you are "getting something for nothing!"
What is the RF frequency?
What is the bandwidth of the device you used to measure "the input AC voltage at the receiver antenna"?

EDIT: Do you have a resonant circuit in there somewhere?
Were you measuring on something that could be a transmission line? At 1GHz the distance between a peak and a null is about 3 inches (7.5cm)

 

[tech99]

You will notice from Jim Hardy's post that the diode has high resistance for small voltages. For this reason, voltage multipliers do not give an advantage for small voltages. I think the lowest diode drop is from Germanium diodes.
To get the best energy from your antenna, you need to match the antenna to the high resistance of a single diode, which may be 1000 Ohms or more depending on signal voltage. It might be an advantage to choose an antenna with a high feedpoint impedance, such as a full wave dipole, or to use an impedance step-up device.
It is also possible to use diodes in parallel to reduce resistance.

 

[lbminh2015]

Dear all, thank you very much for your comments. I'm sorry not having mentioned the details of what I'm doing. I'm doing exactly an RFID transponder at 915 MHz. The reader I'm using has the output power of 650mW. At the distance of 15-20cm where I got the result as mentioned above, the received power, measured by a spectrum analyzer, is 3 dBm.

Please refer to my schematic is attached in this post.

@jim hardy : In practice obviously I could not ignore Vforward which is about 150 - 250mV (max) as noted in datasheet. What I wanted to say is that, even without any drop voltage at the diodes the maximum output voltage I can get must be less than 2.6V, which was the result I really got! So, WHY can it be like that?

@Tom.G : The carrier frequency is 915 MHz. The used oscilloscope has the bandwidth of 1 GHz and was set to full bandwidth for all 2 probes when measuring the signals.
- I do not have any resonant circuit; I have an impedance matching circuit (basic L-type) instead. However that circuit is not working. Currently I'm shorting the L simply by a small wire and leave the C open.

@tech99 : Yes I know the impedance matching is very important to harvest as most energy as possible. But I do not know how to implement that circuit in practice T_T The antenna I'm using is a microstrip one with S11 = -(15-20) dBm from Powercast. It's input impedance is around 50 Ohm.
For the parallel diodes, could you please tell me the multiplier's topology you were mentioning?

My question is still there. Even without any resonant circuit or impedance matching (or I really got it with just a wire?), the output voltage is much higher then it should be in theory. Thank you so much for your support and I hope I can explain my results with your help.

P/S: the transistor in my circuit is for data backscatter. I'm not using currently.

 

[Tom.G]

Not all the times I could get this result, but sometimes I could.

That may be the pertinent observation. Just a 1/4 wavelength position change in the test setup, or your body, could account for the "sometimes."

I suggest you obtain one of the evaluation boards or receivers from the transmitter supplier (http://www.powercastco.com/products/powerharvester-receivers/) to nail down your test procedures, that should eliminate many unknowns.

Unfortunately, the "Multiplier circuit.jpg" in post #6 doesn't say what the different traces are displaying.

A couple other points:
The scope is being used about at its 3db bandwidth, thus probably reading low.
Watch out for the capacitance of the transistor on the antenna feed, it could cause a serious mismatch.

This is at the edge of my electronics knowledge. Hopefully there are some RF gurus around here to help.

 

[lbminh2015]

@Tom.G : Thank you very much for your suggestion. Yes I think so, the ambient environment obviously accounted for the "sometimes". Most of the times I could get 1.8 VDC. Btw, the transistor was not there in my real set up. I have not tested it (for backscatting) yet.

I have tested P110B-EVB and it brought good results as it should be. The measured AC signals right at the antenna's terminal (connected to the EVB) showed small amplitude, just like my current result.

In my opinion, my RF measurement set up and condition were not correct. The scopes are normal ones: Tektronicx TPP100 - 1GHz - 3.9pF-10 Mohm and Agilent 10073C - 1 Mohm - 6-16pF. I soldered a small wire of 10cm long and 1.5mm-in-diameter at the antenna's terminal to connect to the probe tip. Please find my set up attached. It looks very funny, I know, but I do not know how to measure all these things in a good and correct way.

Let's calculate the AC voltage amplitude according to Ohm law. Assume no load at the antenna, I connected it directly to the Spectrum Analyzer. The input power is 3dBm or 2mW as measured, while antenna's input impedance is approx. 50 ohm. So, Vrms = sqrt(P * R) = 316mV ==> Vpeak = 447mV

==> Vout,DC of 2-stage multiplier would be: 447mV * 4 - 4 * (V forward of diode, 150mV) = 1.8 V (approx.)!

So, the observed Vpeak must be more than 447mV! Am I correct?

 

[tech99]

It sounds as if your received power measurement is about right. But as I mentioned, voltage multipliers do not work very well with small voltages.
Therefore I think you need to get the L-match working in order to obtain a voltage step up. Can you give the circuit? It must show resonance in order to work.
The antenna seems to be just one wire, as a probe, but you must have two wires to form a dipole. One wire on the end of a coax cable is not an antenna.

 

[lbminh2015]

@tech99 : Thank you very much for your comment. Yes the matching circuit has been designed with a shunt C of 5.6pF and a series L of 16nH. But in practice, any time I soldered these elements to my circuit, the 2-stage multiplier could not boost the input voltage to the desired voltage. It got worse than it was without them!

As in my knowledge, the calculation of impedance matching can be simply done with ADS software using smith chart, based on impedance conjugate, very simple theory. However, I don't know how to calculate the load impedance because it varies a lot when exposed to RF frequencies. Moreover it depends on the operating state of the circuit (active, low-power, deep-sleep states). For impedance matching, I really do not know how to make it work!

For antenna, it is the https://kr.mouser.com/new/powercast/powercastlifetimepower/. I understand the signal will be at the center and antenna ground is the connector's ground. So the red wire from the board is the signal at SMA's center. AC amplitude I got is that signal compared to SMA's ground. Sorry for my poor English.

 

[Tom.G]

==> Vout,DC of 2-stage multiplier would be: 447mV * 4 - 4 * (V forward of diode, 150mV) = 1.8 V (approx.)!

So, the observed Vpeak must be more than 447mV! Am I correct?

That is about what I got when I did that calculation earlier, so we at least agree! Correct? To Be Determined.

Here are the possible inconsistencies I see:

• In your above calculation the Diode Drop was not used.
• It looks like there is an antenna standing vertically at the left of the photo and the "T" shaped board laying flat on the bench is another antenna. If that is the case you are transmitting vertical(horizontal) polarization and trying to receive horizontal(vertical) polarization. You will get much more power transfer if they are oriented the same.
• About the power calculation. The spectrum analyzer most likely has a 50Ω input impedance to match the antenna and transmission line. Your receive circuit however will be close to open-circuit when it is fully charged. This will double the measured received voltage with respect to the spectrum analyzer measurement. (Perhaps the source of the factor-of-two over-voltage you were measuring?)
• When the receiver is fully charged, you have a poorly matched transmission line from the antenna. The voltage at the end of that transmission line will depend on the SWR (Standing Wave Ratio) and on the phase of the standing wave at the end.
• For accurate field measurements, which is essentially what you are doing, you usually need several feet clearance around transmitter and receiver to avoid reflections. SInce you are working very near-field, this clearance can be relaxed. (I don't have a good feel for just how 'relaxed' you can get away with.)
• Many of the wire interconnects look to be a significant fraction of a wavelength long. They could well be interfering with the radiated field. Try to get them at least a few wavelengths away from the trans/rcvr. Oh, and the scope is close enough to be a reflector too.
• The work surface does not appear to be metal but what is underneath it? Anything metal stored underneath there?
• This one is probably not important, but just in case, what is in the wall around the workbench? From the photo it could be concrete, which is alright in itself. But around here (Southern California, USA), concrete walls have much steel reinforcement in them. This could have a small effect on your measurements.

(Isn't RF FUN? Almost nothing behaves as it does at the lower frequencies.)

[funny coincidence]
I had a doctors appointment today and he kept losing signal to his laptop computer. I mentioned that if the problem was in just a few locations, he should try moving his WiFi Router a few inches or feet. He went to look at the router and said "Come here. Look at this!" One of the employees had placed a metal tray to hold paperwork right next to the router. After moving the tray and saying "a few words" to the employee, his laptop was again communicating well.
[end coincidence]

EDIT: PowerCast has a calculator for received power and voltage at:
http://www.powercastco.com/power-calculator/

 

[lbminh2015]

@Tom.G : Thank you very much for your time and your story as well :) That would be very useful to me. Could you please help me make clear some things in your comments:

• In your above calculation the Diode Drop was not used.

I thought it is the forward voltage drop of the diodes which is 150mV, which was included in my calculation?

• It looks like there is an antenna standing vertically at the left of the photo and the "T" shaped board laying flat on the bench is another antenna. If that is the case you are transmitting vertical(horizontal) polarization and trying to receive horizontal(vertical) polarization. You will get much more power transfer if they are oriented the same.

Yes you are right! The vertical one is for energy harvesting only and the T-shaped one is for backscattering with an SL900A RFID chip (although my future design will not have this chip).

• About the power calculation. The spectrum analyzer most likely has a 50Ω input impedance to match the antenna and transmission line. Your receive circuit however will be close to open-circuit when it is fully charged. This will double the measured received voltage with respect to the spectrum analyzer measurement. (Perhaps the source of the factor-of-two over-voltage you were measuring?

Are you meaning that when fully charged, the harvester circuit will get a power 2 times larger than the measured one by spectrum analyzer? So how to explain the AC signals I got from antenna's terminal? I'm sorry for my poor English. I can't get what you mean in the last sentence "Perhaps the source of the factor-of-two over-voltage you were measuring?" TT_TT

When the receiver is fully charged, you have a poorly matched transmission line from the antenna. The voltage at the end of that transmission line will depend on the SWR (Standing Wave Ratio) and on the phase of the standing wave at the end.

I found on Wiki:

So if SWR is taken into account, and coincidentally if the red wire's end point is at the position of Vmin, what I measured would be Vmin while the true voltage is larger. That explains the result I got. Is it possible?

For accurate field measurements, ...

Yes next time I'll try to set up my measurement in a larger workspace. The work surface is the table and I believe there is metal in it. I believe I can get this result because of a lot of reflection. If the experiment was set up in a reflecting-free space, the result voltage would be much smaller.

@tech99 : could you please tell me what kind of voltage booster I can use instead of a multiplier? I intend to use BQ25570, a DC-DC buck-boost converter of TI; however I still need 1 or 2-stage multiplier to get the the minimum voltage for its operation. Any topology other than an commercial chip or multiplier for voltage boosting?

Thank you again for all of your precious time and hopefully my questions are not so funny.

 

[jim hardy]

(Isn't RF FUN? Almost nothing behaves as it does at the lower frequencies.)

+1

Good anecdote too.

 

[tech99]

• For accurate field measurements, which is essentially what you are doing, you usually need several feet clearance around transmitter and receiver to avoid reflections. SInce you are working very near-field, this clearance can be relaxed. (I don't have a good feel for just how 'relaxed' you can get away with.)

I do not agree with the statement.
The antennas can be as close as about a wavelength (or even less) and the field strength readings will be quite accurate. Close measurements reduce reflection trouble because the direct signal is so much stronger.

 

[tech99]

Sorry did I misinterpret your advice? Are you saying to make measurements very close and have a large spacing from reflecting surfaces?
If so I agree.
The required spacing from reflections needs to be a few times the antenna spacing.

 

[lbminh2015]

Are you saying to make measurements very close and have a large spacing from reflecting surfaces?
If so I agree.
The required spacing from reflections needs to be a few times the antenna spacing.

Thank you for your interpretation. Does this mean that the measurements should be made very close to the antenna (about a wavelength ~ 38cm or less), and the measurement set up should be far away from reflecting surfaces to get accurate field measurements?

 

[Tom.G]

I thought it is the forward voltage drop of the diodes which is 150mV, which was included in my calculation?

The stated formula is:

447mV * 4 - 4 * (V forward of diode, 150mV) = 1.8 V (approx.)!

447mV * 4 = 1788mV or 1.8V approx. but after the " - 4 * (V forward of diode, 150mV)" is accounted for, the result would be 1188mV or 1.2V approx.
(NOTE: In the real world both are approximations. At full charge the diode current will be close to zero so the voltage drop will be perhaps 40mV... aaaand then there are all those stray circuit elements that we normally 'assume' can be ignored, mostly because we can't predict them!)

Are you meaning that when fully charged, the harvester circuit will get a power Voltage 2 times larger than the measured one by spectrum analyzer?

I'm saying that the fully charged voltage multiplier circuit will not be drawing current, thereby not loading the 50Ω source that is the antenna. Whereas the spectrum analyzer, with its 50Ω input impedance, will act as a 2-to-1 voltage divider against the 50Ω antenna impedance.

So how to explain the AC signals I got from antenna's terminal?

I'm assuming you measured the antenna signal level with the scope:

Tektronicx TPP100 - 1GHz - 3.9pF-10 Mohm

That 3.9pF input capacitance has reactance of around 45Ω at 915MHz, again loading the antenna with a 2-to-1 divider and not showing the open-circuit antenna voltage.

So if SWR is taken into account, and coincidentally if the red wire's end point is at the position of Vmin, what I measured would be Vmin while the true voltage is larger. That explains the result I got. Is it possible?

It depends on what you define as "True Voltage." What you measured was the voltage At That Point when it was Loaded By The Measuring Device.
Possible? Yes.

The required spacing from reflections needs to be a few times the antenna spacing.

Yes. Thanks for making that clear. Especially for initial testing, that is very convenient. Final tests of course should be under the expected operating conditions.

 

[lbminh2015]

@Tom.G:
Thank you so much for your very detailed explanation, and your time too. Now I understand what you mean and how I can improve my measurements as well as my explanation of the results I got.