Hi everyone,

I originally used an n-channel MOSFET with an Atmel doing PWM and it works great, however, I need the load to be on the other side of the MOSFET. Based on the way they work, I figured a P-Channel MOSFET will be perfect.

This is how I have it now:
http://www.physics.udel.edu/~watson/scen103/mos4.gif

However, using the P-channel MOSFET I can move it down:
http://www.physics.udel.edu/~watson/scen103/mos5.gif

My question, with the N-channel MOSFET, I fed the gate a positive voltage and it worked just as I wanted. However, with the P-channel, do I have to send the gate a negative voltage? As in, will the P-channel MOSFET be working a straight-through diode when the gate is not powered?

 

PMOS must have a negative vgs. The PMOS is used as a high side switch, so you'd have the source at +5v. When the pin connected to the gate is at +5v versus the normal ground, vgs=0V so it's off. When the pin is low, vgs=-5V so it's on.

Be aware that 5V doesn't turn on most MOSFETs very hard, so they will not be ideal, cool-runnning switches at higher currents. This problem is somewhat more pronounced with PMOS, which typically has lower transconductance (gain). Enhancement-mode MOSFETs with lower gate threshold voltages can turn on much stronger with 5V.

 

Thanks for the reply.

However, would an MCU be able to send the gate a low voltage?

 

Choose a "logic gate" pFET it can fully open with 5V supply.

 

Interesting, what are some part numbers from IR? So these would work with a positive 5V supply to the gate? I am looking for something that can do atleast 4As of load on it, hence I am using a power MOSFET.

Our could I just just a MOSFET driver?

 

Yes, when the output +5V, the device OFF. Output zero volt: device is ON. The load between drain and GND.

Attached Files

irf7425.pdf (91.8 KB, 42 views)

 

So my MCU would have to send the output high at all times to keep the MOSFET from turning on? Hmm, so in the event that the MCU (Atmel in this case) fails the MOSFET goes straight through?

I'm sort of confused as with the N-channel MOSFET it would stay off when the gate was off, which seems kind of logical to me. The pfet is inverted even in this case?

Also, shouldn't it be relative to the source voltage (12V in my case)?

 

Vgs must actually be negetive to fully turn a P-Channel fet on , zero volts on Vgs will not turn it on..
for Vds = 20V
VGS = -2.5V --->-13A
VGS = -4.5V --->-15A

as Vgs goes more negetive Id gets larger in magnitude..

 

LiquidKernal,
What do you mean, "12V source voltage"? Your sketch showed 5V and the microcontroller could drive the P-channel Mosfet's gate to its source voltage to turn it off and to ground to turn it on. The microcontroller can't drive the gate to 12V so the Mosfet will never turn off.
This can be fixed by adding a simple transistor inverter, so that when the microcontroller's output goes to ground, the transistor's collector goes high (12V), turnig off the Mosfet. This might also shut things down if the microcontroller fails.

Willi,
Of corse a P-channel Mosfet turns on when its gate is grounded. Because its source is at a positive voltage (then its gate is negative to it).

 

But AGuru.., will it be fully " on " ..or just kind of " on"..?

i have worked mostly with N-Channel FETs..and with them , , with Vds = 5V ., Vgs needs to be "fairly close" to 5V to fully turn them on..!!

 

the nice thing about CMOS chips is they seem to work rail to rail..ie put 5V to the Plus of the chip , a logic one on the output will be 5V..or 4.99v...
where the 74LS series .. a logic one is approx...3.5V

 

Thanks for the all the informative replies everyone, I appreciate it

I have decided to go with an inverting MOSFET driver, which will pull the gate high when there is no voltage from the MCU. So in the event my MCU looses power, the MOSFET stays off.

 

The sketch was just an example I found online, just to illustrate my idea. My actual design uses a 12V source.

 

When I was playing with the nFET and an Atmel, the Atmel would feed something around 4.1V to the gate and that would actually turn the FET fully on.