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Battery Power vs. depth
- Thread starter homebre
- Start date
Rainyday101
New member
That's a good question and as I don't know for sure so I could only speculate. Power however is the result of the voltage times the current, so voltage alone is not the only factor in power output, current drawn is also a factor. There are many 9 volt detectors that go plenty deep. More batteries connected in parallel give more battery capacity. I don't know for sure if the FCC limits detector transmit power, but if they do it is a mute point because max allowable power would be max allowable power dispite the battery voltage.
MarkCZ
Well-known member
No!
Metal detector use or have within them a voltage regulator. The regulator regulates the voltage down to a point lower than the MAX output of the batteries.
(For Example)
A detector that uses 12volts will likely run at 8 volts. So it would have a ceiling or a reserve of 4 volts to use before a low battery problem would happen.
A detector that uses 9volts for its power supply will be regulated down to somewhere around 5volts, giving it a 4volt ceiling as well.
Now one running on two 9volts is most likely still actually running at 5volts, not 18volts, in this case the batteries are "paralleled" to double reserve current and to help with extended run time. Some of these units will in fact run with only one battery in them (some of them)
So, what would happen if you took a 9volt unit and supplied it with 12volts, would you gain any performance? "NO" the reason is because the built-in regulator is still going to cut it back to 5volts, if it can handle that much over voltage! it could burn the regulator up even trying it!
Now, without a regulator you would have one quirky detector, it would work more like a light bulb, starting out Bright and minute by minute getting a little dimer (weaker and audio loss)
Why do the designers use diffeentr voltage supplies? There is a lot of reason for this. I do know that as more and more micro electronics are being used its taking less voltage and current to drive them.
Mark
Metal detector use or have within them a voltage regulator. The regulator regulates the voltage down to a point lower than the MAX output of the batteries.
(For Example)
A detector that uses 12volts will likely run at 8 volts. So it would have a ceiling or a reserve of 4 volts to use before a low battery problem would happen.
A detector that uses 9volts for its power supply will be regulated down to somewhere around 5volts, giving it a 4volt ceiling as well.
Now one running on two 9volts is most likely still actually running at 5volts, not 18volts, in this case the batteries are "paralleled" to double reserve current and to help with extended run time. Some of these units will in fact run with only one battery in them (some of them)
So, what would happen if you took a 9volt unit and supplied it with 12volts, would you gain any performance? "NO" the reason is because the built-in regulator is still going to cut it back to 5volts, if it can handle that much over voltage! it could burn the regulator up even trying it!
Now, without a regulator you would have one quirky detector, it would work more like a light bulb, starting out Bright and minute by minute getting a little dimer (weaker and audio loss)
Why do the designers use diffeentr voltage supplies? There is a lot of reason for this. I do know that as more and more micro electronics are being used its taking less voltage and current to drive them.
Mark
Rainyday101
New member
Mark CZ is right about the regulators. A 5 volt regulator needs at least 6.5 volts to obtain stable regulation. A regulator needs a minimum of 1.5 volts more than the regulator voltage to be stable. Just because a detector has a 5 volt regulator in it doesn't mean that all the circuitry is running at 5 volts. Some will be, some might not. Today a lot of solid state devices run at 3.3 volts.
Burning up the regulator with a higher voltage will depend on how much power the regulator is rated for and it's power dissapation will depend on the current draw through it and if the regulator has heat sinking on it.
More than likely the larger battery configurations are to provide more battery capacity for a more current hungry circuit. Is this transmit power? Don't know for sure.
Burning up the regulator with a higher voltage will depend on how much power the regulator is rated for and it's power dissapation will depend on the current draw through it and if the regulator has heat sinking on it.
More than likely the larger battery configurations are to provide more battery capacity for a more current hungry circuit. Is this transmit power? Don't know for sure.