help---RF to DC convertors

Hello to all,
I am engaged in a project wherein I propose to use RF power to operate a wireless sensor network.. The problem is that I am not getting the desired voltage levels after conversion of RF to DC ...
I am thinking of using a switch or something like that to charge up a capacitor and then use the large voltages generated by accumalated charge on the capacitor... Could anyone of you suggest something for the same?? Also could charge pump dc-dc convertors serve the purpose?? (in sensor nodes low power is essential , so i cant have the dc-dc convertor if it takes lots of power..)..
Thanking you in advance..
Shantanu

use an voltage amplifier instead...

any suggestion i'm newbie too

This is an example of DC amplifier

You could use voltage multiplier if your sensor does not need too much current.
Give us data: RF frequency, RF power level, required DC voltage and load current.

First youwill have to connect a RF receiver then detect it with Very High frequency special purpose diodes then a filter capacitor

The problem is that I am not getting the desired voltage levels after conversion of RF to DC ...

Click to expand...

What antenna and conversion circuit have you tried?

Have you got a good impedance match?
eg if your antenna is 50ohm then 100uW of receive energy gives you 0.071V
If you need 3.3V then you need to step up the volage.

when you said "operate" you mean feed the sensor from RF signal?
in other words, obtain a DC voltage and use it for 3.3V? as you said.
if that is your intention just remmeber that the rf signal attenuates as
fast as 20 dB per decade at least. The RF signal propagate in every
direction and you have to transmit energy you should use a very
large antenna gain (almost non-dispersive as the laser) or make the
channel lest dipersive(use a cable) so the energy does not scatter all
over. even if you have a big antenna gain AND a short distance we
could be talking of a signal at least 40 dB below the power at the
transmitter. in order to have a 3.3VDC after rectification you would need
(3.3)^2/377= 2.8 mW at the receiving antenna. 2.8 mW is 4.47 dBm
at the receiving antenna. Asuming a 40 dB of loss due to propagation
you wil need 44.7dBm of power or 28W. You will need a very powerful
transmitter. for that short distance, that assuming the regulations in your
country allows that much power to be trnasmitted.

Hi to all,
I want to thank you all for the replies...
First of all, I am proposing to use RF energy to power sensor nodes.. I am doing simulation using ADS2004 A... I plan to use RF freq. of 2.4 GHz, then use a Rf to DC convertor (its a modified rectifier and regulator - which works on this high a freq and my work on it is pretty much over. .) .. The problem is that I am getting tooo low voltage levels at the output..
Sensor nodes typically require aboout 50 microwatt power... So even if low Rf power is available, it could be just fine....
I am thinking of using a charge pump (it would raise the voltage level to about 3.3 V0 , but problem is if tis feasible..
ANother doubt is that ppl use formula--- Path loss=4o dB + 20 x (distance in meters)... Is it correct one for 2.4 GHz?? If not ,could you suggest a better one for calculating power loss...
Thanking you in advance..
Shantanu...

Added after 2 hours 11 minutes:

other important information--

Required dc voltage= 3.3 V
current in order of micro-amperes as the sensor node requires only a few mirowatts for operation... Frequency of RF to be used=2.4 GHz... Also I do not propose to use antenna of more than 2db at node (as it will consume power too!!!)

First if you are going to use 2.4 GHz i am assuming you
are going to use the ISM band. The ISM band is rectricted
in US to 30 dBm EIRP maximum. Second the link budget
is going to depend on the type of wireless link you have:
a) if the link is NOS the formula is aproximated:
Pr=Po-10nlog(d[Km]) whera A is the power at 1KM and
n is between 3.0 and 4.0
b) For LOS the formula is:
Pr=PoGtGr(lambda)^2/(4(pi)d)^2 which then became
Pr=Po+Gt+Gr-32.6-10log(d[Km))
If you need 50 microwatt= -13 dBm and asuming a G=6dB
the max distance would be (LOS)
10log(d[Km])= Po-Pr+Gt+Gr-32.6=30 dBm+6+13dBm -32.6=
16.7 dB= 43 Km

If the system is unidirectional you can use a bigger receiving
antenna to increase the distance but if the system is bidirectional
you can't put more than 30 dBm EIRP. However, you will need
some of the power for information processing. Lets say you need
about 2dBm of signal to process, lets say you need another
2 dBm for the upconverter, and the upconverter has an efficiency
of 75 %. You loss about 5 dB, then the distance will be
16.7-5= 11.7= 14 Km. The antenna is a passive device, does not
consume energy, the use of a 2 dB gain antenna does not make
sense to me.

I hope this help.

Added after 8 minutes:

I forgot. At 14 Km you will receive the average of the
signal. You will need to increase the margin in order to
have a reliable link. Assuming LOS and Rice distribution
you could expect a variability of up to 6 dB easily. If
your sensor network is not going to be miision critical
and it does not need high reliablity you can forget about it
and just tolerate retransmission. However, good engineering
justment will tell you to add at least 3 dB more for margin.
This will put the maximum distance at 8.7 dB = 7.41 Km

Added after 1 hours 31 minutes:

I made a mistake: the second formula must
include a term dependent of the frequency so
Pr=Po+Gt+Gr-32.6-10log(d[Km)) will
be Pr=Po+Gt+Gr-32.6-10log(d[Km))-10log(f(MHz))
so for 2.4 GHz 10log(2400) = 33.8 dB
If you add 33.8 dB you will have less than 1 Km
of distance. So, the project is not viable.

jallem said:

in order to have a 3.3VDC after rectification you would need
(3.3)^2/377= 2.8 mW at the receiving antenna.

Click to expand...

I'm not convinced by this maths. Where do you get the 277 figure from?

I am thinking of using a charge pump (it would raise the voltage level to about 3.3 V0 , but problem is if tis feasible..

Click to expand...

I think this is probably the wrong approach. If you have enough RF power available you can transform it to more volts and less current before the rectifier.

its a modified rectifier and regulator - which works on this high a freq and my work on it is pretty much over. .

Click to expand...

As I mentioned before you have to get a good impedance match.

you will have less than 1 Km
of distance. So, the project is not viable.

Click to expand...

500M would be incredible. I'd be impressed if it could get more than 10meters.
I think it is safe to say that it will only work when there is clear line
of sight from transmitter to sensor. Walls, people, trees vegitation etc
in the path at 2.4GHz attenuate the signal a lot.
If want want to do more than tens of meters you need to consider the fresnel effect.