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| port500 |
Posted: December 22, 2010 11:28 am
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Newbie ![]() Group: Members+ Posts: 4 Member No.: 34,114 Joined: December 22, 2010 |
Hello,
I am looking for an RF transceiver with very low current consumption. I am currently working on an RF based Automatic Meter Reading Project (AMR). We are looking for 433MHz Modules. The modules that i have found so far have around 15-20mA Current consumption while transmitting. I am looking for a RF module with transmission current consumption less than 10mA. The required range is 100m. My requirements are: Frequence: 433MHz I(transmitting) < 10mA range>100m I have no financial constraints in this regard. Any help will be greatly appreciated. Thank You This post has been edited by port500 on December 22, 2010 11:44 am |
| kellys_eye |
Posted: December 22, 2010 01:02 pm
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![]() Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 9,985 Member No.: 2,735 Joined: June 21, 2005 |
Power consumption will relate to radiated power and to attain a certain distance (with a given antenna) you will require a certain power - the efficiency of the transmitter will be high anyway so there's no particular way of improving it.
It therefore follows that you will require a more efficient antenna system rather than a low power transmitter. Improvements in front-end design (at the receiver) might help permit a lowering of power at the transmitter too. Anyway, you could always lower the transmission times (once per hour?) to conserve power. Is power conservation your main aim? -------------------- May contain nuts
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| CWB |
Posted: December 22, 2010 01:07 pm
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![]() Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 19,523 Member No.: 15,154 Joined: May 15, 2008 |
hmmm ...
if you need 100 meter range with 10mA of current draw you will need to find a module that uses a higher voltage than most of the others . basically : you need a given amount of rf power to "cover" the required range . output power of the unit can be measured in watts (or milliwatts) . (roughly , in this case) E X I = P (watts) . what you will be doing is exchanging a higher voltage for lowered current consumption . --- no free lunches --- i take it that you have calculated what the tx/rx field requirements will be ? without knowing this you will not be able to determine what the transmitter output power requirements would be . -------------------- "Know how to solve every problem that has been solved"
R. Feynman '88 |
| Ice-Tea |
Posted: December 22, 2010 01:52 pm
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 2,884 Member No.: 462 Joined: October 07, 2003 |
With lowest baud rate and lowest output power I guess you should get close to the minimum in this datasheet:
Linky. Pretty damn sure you'll need a directional antenna to get 100m range for it, though. |
| Ice-Tea |
Posted: December 22, 2010 01:57 pm
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 2,884 Member No.: 462 Joined: October 07, 2003 |
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| Ice-Tea |
Posted: December 22, 2010 02:30 pm
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 2,884 Member No.: 462 Joined: October 07, 2003 |
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| port500 |
Posted: December 23, 2010 05:26 am
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Newbie ![]() Group: Members+ Posts: 4 Member No.: 34,114 Joined: December 22, 2010 |
The Module will be integrated with Energy meters and used as an AMR system. The aim right now is to find an IC/Module that uses the least amount of Power/Current, and has a minimum range of 100m. The antenna design will come at a later stage. And the requirement from the power company is to transmit meter Data once every 20 minutes.
Tx needs to be below 10mA while transmitting at sufficient power to achieve a range of 100m in an urban area This post has been edited by port500 on December 23, 2010 05:26 am |
| port500 |
Posted: December 23, 2010 05:49 am
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Newbie ![]() Group: Members+ Posts: 4 Member No.: 34,114 Joined: December 22, 2010 |
I already found the semtech SX1240 module, and so far it is the best solution. I am looking for something even more low power than it
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| kellys_eye |
Posted: December 23, 2010 12:57 pm
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![]() Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 9,985 Member No.: 2,735 Joined: June 21, 2005 |
Given it's to be integrated with energy meters, why the reluctance to draw the power from the actual supply being measured? It's not like it would be 'chargeable' (and even if it were, the customer couldn't complain).
I can't imagine for a moment that the device will be BATTERY POWERED -------------------- May contain nuts
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| Ice-Tea |
Posted: December 23, 2010 02:31 pm
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 2,884 Member No.: 462 Joined: October 07, 2003 |
Suggestion: supercap. Average the draw out over the inactive period.
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| Village Idiot |
Posted: December 23, 2010 09:13 pm
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![]() Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 1,402 Member No.: 11,398 Joined: October 08, 2007 |
Another option that no one has mentioned so far, to reduce transmitter current consumption and maintain 100m range, is to build a more sensitive receiver. A good quality low noise amp on the receiver front end is required.
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| port500 |
Posted: December 24, 2010 05:26 am
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Newbie ![]() Group: Members+ Posts: 4 Member No.: 34,114 Joined: December 22, 2010 |
We have decided to use the SX1240 and work on a better receiver design. Thank You for the feedback. Much Appreciated.
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| sherlock ohms |
Posted: December 29, 2010 11:06 am
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![]() Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 2,764 Member No.: 26,125 Joined: September 10, 2009 |
-------------------- "Quotation marks make sentences appear more meaningful."
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| CWB |
Posted: December 29, 2010 03:50 pm
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![]() Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 19,523 Member No.: 15,154 Joined: May 15, 2008 |
some people have no shame at all ...
-------------------- "Know how to solve every problem that has been solved"
R. Feynman '88 |
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