felezjoo PI(the best pulse induction metal detector that I made until now)

[waltr]

if I use two 9v batteries it works fine but they discharge in 30 minutes. if I use 3x6v 4ah in series (19.5V) lead batteries the 7812 becomes hot instantly and goes into protection. I do not understand why...

Guess the 9V batteries are to small 'transistor' battery- These are for low current draw and will Voltage drop on higher current.
Whereas, 4A-Hr lead acid batteries will hold their Voltage with quite a current draw.
Therefore, The 9V batteries Drop Voltage = less Voltage drop on 7812 = less heat. but very short operation time.
Then the lead acid batteries do not Voltage droop = higher V drop on 7812 + more current to circuit (TX) = more heat.

[dbanner]

The regulator is going into thermal shutdown, this should not happen as this circuit shouldn't draw 1.5 Amps. Something is wrong, sinking high current.

[dbanner]

950ma measured by the battery... what can it be?

The circuit shouldn't draw 1amp when it is at idle, this is too high I think.

[dbanner]

The regulator is rated Vin up to 26volts, Vout [email protected]. proper heatsink and she can perform.

[dbanner]

Guess the 9V batteries are to small 'transistor' battery- These are for low current draw and will Voltage drop on higher current.
Whereas, 4A-Hr lead acid batteries will hold their Voltage with quite a current draw.
Therefore, The 9V batteries Drop Voltage = less Voltage drop on 7812 = less heat. but very short operation time.
Then the lead acid batteries do not Voltage droop = higher V drop on 7812 + more current to circuit (TX) = more heat.

The tx powered directly from battery.
18 volts to input of regulator to supply 12 volts for this circuit should be fine. With good heat sink, it shouldn't go into protection shutdown.
Maybe check C15,C16. Maybe one or both are leaking.

[dbanner]

I think you need minimum 14.6v to maintain 12 volts loaded. But maybe battery voltage is too much, and regulator has to dissipate too much power, in addition to demands of circuit. But I still think close to 1amp quiescent state seems high for this circuit.

[gandolfi]

hi,

i have 10.5v beetwen output of 7812 and ground. Do you think is the problem ?

[qute1]

Yes it should be 12v thats the point or a regulator it regulates the voltage to 12v (for the 7812 and 5v for the 7805) 1amp draw is too high also you MUST have a component that is missing its magic smoke and gone short or a misplaced diode in the mosfet/hexfet coil tx stage that is blocking instead of passing (maybe faulty or the wrong way round) i had a similar issue on a minipulse the darn things are so tiny you need mcgoos to ensure it is correctly orientated! usually two near each other one faces opposite the other not same direction.
The short etc may have let some magic smoke out of the 7812 also and damaged it.....
best of luck

[gandolfi]

the detector was working but I had to cause a short circuit by installing it in a box. I did not connect the coil or the LCD because it has also unsoldered during its insertion in the box. I buy connectors to avoid this in the future.

I will replace the 7812. The 7805 gives 5v output. Do you see other components to test? How?

I am going to help with the schema of a member of the forum.
http://www.geotech1.com/forums/attac...6&d=1530988965

[basellafe]

hi,

i have 10.5v beetwen output of 7812 and ground. Do you think is the problem ?

its look like your battery less than 12v maybe 11v or reglautor is broken

[gandolfi]

my battery deliver 20v and 500 ma. i'm waiting new regulator

[basellafe]

battery should be at least 2amber not 500ma this will be empty soon

[Kostasp]

Unlike bipolar transistors (which are current-driven), MOSFETs are voltage-driven. From the IRF840 datasheet you can readily see that Vgs threshold is 4V (max), which means that it needs to have at least 4V to turn on (and preferably much higher) up to the maximum rating of 20V. From the datasheet graphs, it is clear that Id plateaus at about Vgs=8V. This means that the resistor value of 2k2 is perfectly ok, as this will apply 12V to the gate. MOSFETs have a very high input impedance, and (from the datasheet) it is possible to calculate this value for the IRF840 as being 200M (from Vgs=20V and Igss=+/-100nA). So there is no need to be concerned about gate current. In fact, a value of 10k would work just as well as 2k2.

HI Qiaozhi I need your help I start felejzoo tell me about R5 is better to use 2K2 or 10K have any mater to performance thanks

[dbanner]

He said it would work just as well.

[Kostasp]

He said it would work just as well.

One is the best witch one have a better performance thanks