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From: rahulponna on 24 Jul 2010 22:29
rahulponna had written this in response to
http://www.electrondepot.com/electrodesign/Re-AC-to-DC-conversion-without-using-diodes-and-no-external-484462-.htm
:

markp wrote:

Hey mark, I have already looked at the LT3108 chip.yes it's used for
energy harvesting, but it needs a dc input voltage. I plan to use that
chip, but before that I have to rectify my AC voltage.



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From: Paul Keinanen on 24 Jul 2010 23:57
On Sun, 25 Jul 2010 02:25:32 +0000,
findprahul_at_yahoo_dot_co_dot_uk(a)foo.com (rahulponna) wrote:

>
>I am looking at a frequency of 950MHz. the input voltage is 200mV (peak).
>Also for precision rectifier I have to use external supply for the op amp.
>I need to make all my electronics work with the harvested energy.
>
Since this is a fixed frequency, just feed the antenna power
(typically at 30-60 ohms) into a 950 MHz resonator and make an
additional high impedance connection into the resonator.

A 1/4 wave resonator with a magnetic probe close to the grounded
feeding the low impedance antenna power and a capacitive probe close
to the high end of the resonator to extract the power might do it.
Rectify the increased voltage in the conventional way e.g. with a
simple RF probe (series capacitor, diode to ground).

It is not at all self evident that any active rectifier circuit would
work at such high frequencies.

From: mkaras on 25 Jul 2010 00:01
On Jul 24, 7:25 pm, findprahul_at_yahoo_dot_co_dot...(a)foo.com
(rahulponna) wrote:
> rahulponna had written this in response tohttp://www.electrondepot.com
>
> I am looking at a frequency of 950MHz. the input voltage is 200mV (peak).
> Also for precision rectifier I have to use external supply for the op amp..
> I need to make all my electronics work with the harvested energy.
>

Where is the 200mV peak 950MHz signal coming from? What kind of source
impedance is there?

Seems to me that it is essential to have answers to these before even
beginning to consider how to go about solving your problem. The source
impedance will have a direct impact on how much energy you can
transfer over into whatever circuit you can try to come up with.

mkaras
From: Paul Keinanen on 25 Jul 2010 00:32
On Sun, 25 Jul 2010 02:25:32 +0000,
findprahul_at_yahoo_dot_co_dot_uk(a)foo.com (rahulponna) wrote:

>I am looking at a frequency of 950MHz. the input voltage is 200mV (peak).

How is this measured ? Open circuit or loaded transmission line ?

200 mV peak is 140 mV rms and if that was unloaded voltage, the
voltage into a matched load would be 70 mV rms. Assuming 50 ohm
impedance levels, this is -10 dBm or 0.1 mW power available from the
source in matched condition.

If the original measurement was actually open circuit peak-to-peak,
the available power would be 25 uW.

From: Paul Keinanen on 25 Jul 2010 03:56
On Sun, 25 Jul 2010 07:32:00 +0300, Paul Keinanen <keinanen(a)sci.fi>
wrote:

>On Sun, 25 Jul 2010 02:25:32 +0000,
>findprahul_at_yahoo_dot_co_dot_uk(a)foo.com (rahulponna) wrote:
>
>>I am looking at a frequency of 950MHz. the input voltage is 200mV (peak).
>
>How is this measured ? Open circuit or loaded transmission line ?
>
>200 mV peak is 140 mV rms and if that was unloaded voltage, the
>voltage into a matched load would be 70 mV rms. Assuming 50 ohm
>impedance levels, this is -10 dBm or 0.1 mW power available from the
>source in matched condition.

If both the power source transmitter antenna as well as the energy
gathering antenna are simple omnidirectional Ground Plane antennas, at
950 MHz, the path loss of 1m distance (already in the far field for
that frequency) is 32 dB and at 10 m 52 dB and at 100 m 72 dB.

To get -10 dBm at the energy gathering antenna, the master transmitter
power would have to be +22 dBm (160 mW) at 1 m, +42 dBm (16 W) at 10 m
and +62 dBm (1600 W) at 100 m in free space. We are quickly talking
about power levels at which human exposure limits must be checked.

In order to transmit back any measurements to the master receiver,
co-located at the main transmitter site and assuming -120 dBm master
receiver sensitivity, the slave transmitter power would have to be -88
dBm @ 1 m, -68 dBm @ 10 m and -48 dBm @ 100 m.

At 100 m and assuming 100 % RF->DC->RF conversion efficiency, the
transmitter power would have to be -120 dBm + 72 dB + 72 dB = +24 dBm
(250 mW). Add to this the losses in the slave (perhaps 10-30 dB), so
clearly such distances (100 m) are not practical at these frequencies
and these kinds of antennas.

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