regarding n channel mosfet used as a switch

Hi all,
i need a circuit which will control the power by using n channel mosfet.Let me explain clearly , gate G connected to io pin of microcontroller(3.3V) ,Drain D (5V), sourceS as output .When G=3.3V, S=D=5V.Please help me in design the circuit.If u have any idea other than this , please do the needful.
Thankx

shashavali
shashaec@indiatimes.com

I don't think this will work. To operate the MOS as a switch, Vg has to be equal to Vd. Note that for the transistor to conduct Vg-Vs>Vth in general. So, there would be a one Vth drop between the gate and the source. I think you might use a CMOS inverter operating at 5V supply instead (I am not quite sure this will work fine though, try to simulate it or try it out).

Hope this is useful

This cannot work. For an N-ch MOSFET, the gate must be more positive than the source to turn it on. Since the source will be at +5V with the MOSFET on, supplying power to the load, the gate must be at +10V or so.

There are circuits that allow you to drive an n-ch MOSFET in such an application, but they will be complicated and the cost and complexity not justified, unless the load needs such a high current that you REALLY need an N-ch MOSFET.

Otherwise, use a P-ch MOSFET, with the S connected to +5V, the D to your load and the G to the PIC. If the PIC is powered at +3.3V, then I recommend you drive the MOSFET not directly with the PIC pin, since the MOSFET may still turn on, but use an open-collector/ drain transistor with a pullup to drive the P-ch MOSFET.

Your idea is not work. For the maximum source is gate voltage - Vt. Show you need PMOS instead of NMOS. Btw, if you have to use NMOS. Try to increase gate voltage above required source voltage at least 1V for fully turn-on

If your input to G is a clock signal, I suggest you can double the volatge of clock.

The circuit of double the voltage can be found in bootstraped clock signal.

Yibin.

Shashavali,
Connect the uC output to the gate of an additional N channel Mosfet (Q1). Source of Q1 goes to ground. Drain of Q1 goes to +5V through a pull-up resistor. Drain of Q1 also goes to gate of Q2 (The FET that you're trying to control). You must invert the logic sense of the uC output. (A"1" on the uC output turns Q2 OFF.
Regards,
Jon

if you require just a switch (meaning u dont necessarily need to use mosfets)you could also try other alternatives

for example you could use a simple npn transistor with the base connected to the input through a resistor and the circuit connected between the collector and emitter