Use n-channel MOSFET to ON/OFF my PIC circuit

[winson]

Hi friends,

I have try to use a n-channel MOSFET as a switch to ON or OFF my circuit(give power supply or cut the power). The MOSFET that i'm using now is STP36NF06L, i have facing problem with this MOSFET that is the circuit look like cant receive any power supply from my 9V battery(regulate by a 7805 IC) when i apply 5V to the gate of the MOSFET.

The connection is simple, the battery's positive terminal is connect to the input of my circuit(PIC microcontroller, LCD and other) which is the input pin of the 7805 IC and the common ground of the circuit is connect to the drain of the MOSFET. The MOSFET's source is connect to the negative terminal of the battery. Suppose when 5V is apply to the gate of the MOSFET then my circuit should be power-on and running but now what i observe is that the circuit does not running at all and the LCD also does not show anything.

Anybody can help?
Any suggestions will be appreciate.

Thanks.

[ArcMan]

... and the common ground of the circuit is connect to the drain of the MOSFET. The MOSFET's source is connect to the negative terminal of the battery.

There's something wrong with your circuit based on the above statement.  The negative terminal of the battery SHOULD BE the common ground of the circuit, and it should be connected to the source pin of the MOSFET.

Post a drawing of your circuit so we can look at it.

[Soeren]

Hi,

Your circuit IS correct as described (although not what I would choose).

When you apply +5V to the gate, where do you get the +5V from?  Hopefully not from the circuit which, at that moment, still isn't ground referenced.
You need a separate supply (+5V and ground) to turn the device on (and a gate-source resistor to turn it off when gate potential is removed).

A much better solution is to use a P-ch. device in between the +9V and the 7805s IN and then pull the gate low via a small signal NPN + resistors. It still needs to be a logic level MOSFET, the gate-source resistor should still be in use and a gate resistor of 10 Ohm to 100 Ohm wouldn't hurt either.
Tell me if you need a schematic of it.


Using an 7805 with a 9V PP3 means that you cannot use much of the power in the battery, since the 7805 needs around 8V to work and the end-of-life voltage for a PP3 is 5.4V - an LDO regulator is a better choice (or a DC-DC converter).

[winson]

Finally the thing is working, before that i have take the 5V from the output of the 7805 IC as the gate voltage and i thought when the MOSFET is OFF it still can provide 5V to the gate but before that i was not aware it cant, after that i was directly connect the gate to the 9V from battery and the MOSFET is succesfully turn on and the circuit also running.

Hi Soeren, since now n-ch is working so far, next time when i need the schematic for p-ch then request from you ok. Thanks

By the way, actually what is the special feature or difference of the LDO regulator if compare to the 7805? I have one LDO, call LE33CZ(3.3V regulator) and it's size is so small just like a transistor, i'm worry about it's power/heat dissipation ability since 7805 can dissipate a comparable more higher heat.

Thanks in advance.

[want2learn]

since 7805 can dissipate a comparable more higher heat.

Thats the thing with those type of regulators, they have to.

LDO means Low Dropout Regulator. Whereas the 7805 drops minimum 2v some LDO's only drop as little as 0.1v.

[Soeren]

Hi,

Have to clear up what seems to be a common confusion about LDO regs.
A 5V LDO and an 7805 will drop the exact same voltage when supplying the same current with the same input voltage above (U_out + U_drop) (The laws of thermodynamics still applies I believe ).

The only real feature of the LDO compared to a normal regulator is simply the minimum drop out voltage, which for an 7805 is minimum 2.5V (usually around 3V though), while the LDO can be anything down to 100 mV or less (depending on output current).

LDOs exists in different current and power capacities.
The higher current types usually have a smaller drop on a lower current (a 7.5A regulator supplying 1A can have a drop of eg. 1/3 of what a 2A regulator supplying 1A have).

An 7805 will go out of regulation when its input goes below around 8V, while a 5V LDO wil go down to at least 5.6 before loosing regulation - a big difference when using a 9V battery.

A 9V battery contains 6 cells. A single alkaline primary cell is up to around 1.65V for a short time when new and is considered flat at 0.9V.
That means that a new 9V battery is up to slightly below 10V while it's considered flat at 5.4V.
An 7805 will only use a minor part of the capacity of a 9V battery, while an LDO will use them up completely without loosing regulation.

[winson]

Hi,

Thanks for giving all the clear explaination! Yes, i agree with what you all said, now my circuit was powered by a 9V battery and when the voltage of the battery drop to certain level then it will be totally no output from the 7805 and it waste me a lot of money to buy the new battery. So after consider all the comment from you all them i think i should buy 5V LDO regulator for my circuit and i'll save a lot of money on battery.

Thanks a lot...