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Posted by on January 31, 2008, 9:33 am
> For cordless tools, you could perhaps setup the system so the tag is
> powered both from an internal rechargable batter, and the tool's
> battery (which also recharges the tag's battery).
I thought of something like that, but the electronics would not be
simple. Just adding a resistor to step the voltage down would not do
the trick on such a low-current device. Still, if two button batteries
last for 9mos, two AAA's should last for years.
Regarding the number of tags. I just realized that you could just
purchase more readers. The system is designed so that one owner's tag
will not be picked up by another's reader. So, you could use one
tracker for remotes 1 to 8, and another for numbers 9-16.
Joe Dunfee
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Posted by PeterD on January 31, 2008, 2:13 pm
On Thu, 31 Jan 2008 06:33:18 -0800 (PST), cadcoke4@yahoo.com wrote:
>> For cordless tools, you could perhaps setup the system so the tag is
>> powered both from an internal rechargable batter, and the tool's
>> battery (which also recharges the tag's battery).
>
>I thought of something like that, but the electronics would not be
>simple. Just adding a resistor to step the voltage down would not do
>the trick on such a low-current device.
I'm an EE, and certainly it would work just that way. Say a 18.0 volt
tool, and the RFID tag takes 1.5 volts (single non-lithium cell) then
two resistors, in series between the 18.0 volts and ground on the
battery. Choose resistor values of, say, 9.1K connected to the +18V
connected to a 820 ohm then to ground. From the junction of the 9.1K
and the 820 ohm resistors, to ground yuo will have about 1.5 volts. If
you are going to charge a rechargable battery, you'd need more than
1.5 volts ideally, but I think you get the idea. Battery drain on the
main battery would be nominal. If you use both the rechargable tool's
battery, and the battery in the device, an isolation diode would be
needed to prevent the RFID's battery from draining through the 820 ohm
resistor. I low forward voltage diode would work, as would adjusting
the value of the 820 ohm resistor up slightly to comphensate.
Lithium cells, and double buttons would be about 3 volts, so the
resistor values would again need slight adjusting. The adjustment is
linear, reduce the 9.1K by about 820 ohms, and increase the 820 by the
same amount would double the voltage.
>Still, if two button batteries
>last for 9mos, two AAA's should last for years.
Been there, done that... (Used AA's not AAAs, but the same difference)
It's been about six or seven years now. I think they are starting to
get weak, but they still work! <bg> If you have room for the batteries
this is a viable way, but don't forget you'll need a battery holder,
mounting, etc, while the resistors won't need nearly as much to hold
them in.
>
>Regarding the number of tags. I just realized that you could just
>purchase more readers. The system is designed so that one owner's tag
>will not be picked up by another's reader. So, you could use one
>tracker for remotes 1 to 8, and another for numbers 9-16.
>
I'm not sure I"d want to be carrying around a bunch of readers
however... But that depends on how many tags you need to read.
Also you should institute a death penalty on those who lift tools...
That will cut down on losses too. (before I get flamed too much, that
death penalty is: "I see that tool in your trunk is not yours, please
put it back, go to the manager's office and get your pay, and don't
come back." (The job is dead!)
>Joe Dunfee
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Posted by on January 31, 2008, 10:45 pm
> I'm an EE, and certainly it would work just that way. Say a 18.0 volt
> tool, and the RFID tag takes 1.5 volts (single non-lithium cell) then
> two resistors, in series between the 18.0 volts and ground on the
> battery. Choose resistor values of, say, 9.1K connected to the +18V
> connected to a 820 ohm then to ground. From the junction of the 9.1K
> and the 820 ohm resistors, to ground yuo will have about 1.5 volts.
Wouldn't these two resistors continuously drain the battery? Perhaps
you mean that the power is only applied to these resistors when the
switch is pulled. Then the series resistors charge the transponder
battery? Is this right?
Joe Dunfee
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Posted by PeterD on February 1, 2008, 10:50 am
On Thu, 31 Jan 2008 19:45:31 -0800 (PST), cadcoke4@yahoo.com wrote:
>> I'm an EE, and certainly it would work just that way. Say a 18.0 volt
>> tool, and the RFID tag takes 1.5 volts (single non-lithium cell) then
>> two resistors, in series between the 18.0 volts and ground on the
>> battery. Choose resistor values of, say, 9.1K connected to the +18V
>> connected to a 820 ohm then to ground. From the junction of the 9.1K
>> and the 820 ohm resistors, to ground yuo will have about 1.5 volts.
>
>Wouldn't these two resistors continuously drain the battery?
The drain is very small, much less than the self-discharge rate of a
NI-CAD battery for exmaple. The battery would go dead on its own long
before the resistors drained it.
> Perhaps
>you mean that the power is only applied to these resistors when the
>switch is pulled. Then the series resistors charge the transponder
>battery? Is this right?
No, though you could do that, you are probably makeing it more complex
than it needs to be.
>
>Joe Dunfee
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Posted by dpb on February 1, 2008, 12:37 pm
PeterD wrote:
> On Thu, 31 Jan 2008 19:45:31 -0800 (PST), cadcoke4@yahoo.com wrote:
...
>> Wouldn't these two resistors continuously drain the battery?
>
> The drain is very small, much less than the self-discharge rate of a
> NI-CAD battery for exmaple. The battery would go dead on its own long
> before the resistors drained it.
Self-discharge is <<2 mA??? (Just asking, seems high ottomh....)
--
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