Go Down

Topic: High side MOSFET issue (Read 1 time) previous topic - next topic

jtw11

Hi there,

I'm used to using N channel MOSFETs in a normal low side driving application - however, I've need to switch a load on or off (not switching as in PWM, but just an on or off switch), but on the high side of the load.

I'm reading multiple app notes etc as we speak, but I'm somewhat lost - it seems I must use a P channel FET, but these need a negative gate voltage? - whereas I only have a single supply, +12V.

I've also seen arrangements with a BJT driving the FET? I suppose really one of my main questions, why can I not simply use an N channel FET in the high side position?

Any comments as to driving P channels?

astrofrostbyte

#1
Feb 27, 2013, 06:46 pm Last Edit: Feb 27, 2013, 06:50 pm by astrofrostbyte Reason: 1
The basic circuit is something like this:

Your control signal will prob. be 5V  maybe 3V3.

N-Ch fet:
Your load is ref. to GND, if all the supply voltage 12V is over the load , than the Fet's source will be at 12V,   to 'open' the fet you need to pull the gate higher(pos) than the source.  And that higher voltage is not available.

The P-Ch mostfet gate controlled with a neg voltage.  if the source is at 12Volt,  the gate can easely be pulled to a lower voltage and thereby creating a negative voltage.
Gear: Arduino- Uno,Due,Ethernet,  OLS, Buspirate, J-Link, TDS1002, Rigol DG1022

pito

You need a p-channel fet and a resistor (ie 10k). The resistor from gate to Vcc. The transistor will be off when the gate is log1 (or highZ), the transistor will be on (low resistance) when log0 at the gate.

jtw11

#3
Feb 27, 2013, 06:54 pm Last Edit: Feb 27, 2013, 07:58 pm by jtw11 Reason: 1
Quote
The P-Ch mostfet gate controlled with a neg voltage.  if the source is at 12Volt,  the gate can easely be pulled to a lower voltage and thereby creating a negative voltage.


Right - that explains it perfectly. I wondered why the BJT was there!

Without the BJT, and even with a 5V logic high, Vgs would still be effectively 5v-12v = -7V, and the FET would be on - take the gate to 12V, then 12v-12v = 0V and the FET is off.

EDIT - Saying that, this is all irrelevant to me, I'm driving the FET from the output of an op amp acting as a comparator, so I can simply have the power supply to the op amp the same as the source voltage of the P FETs, that way - when the comparator output is high, Vgs = 0v, when the output is low, Vgs = -Vsource

Correct?

MarkT

If you are driving the gate from an op-amp you may have problems turning the MOSFET fully off - Vgs should be close
to 0V to guarantee fully switched off, and many opamps can't drive to the rails.   Thus the opamp ought to one
that drives to the +ve rail fully, or else add a BJT to ensure this happens.
[ I won't respond to messages, use the forum please ]

astrofrostbyte

to jtw11:
You can not lose the BJT,  a Vgs of -7V is ok, but you can not make a arduino pin higher that 5V.
On most digital pins there are so called 'rail diodes' that will 'short' the In-/output pin to the 5V supply if it goes beyond say -0.6V , +5.6V.
The BJT makes an 'open collector' pin that can take say 40Volts.

Like MarkT notes,  not all opamps are Rail2Rail,  there are many common(lm311) comparators that have an 'open collector' output .
Gear: Arduino- Uno,Due,Ethernet,  OLS, Buspirate, J-Link, TDS1002, Rigol DG1022

Go Up