MOSFET cut=off

[brentab]

Hello all!
From my understanding!  a P-channel Enhancement MOSFET, (The cut-off region) The Gate potential must be more positive with respect to the Source.
As a example we will take the RF9Z24NPBF FET.

From the data sheet.

1. Vgs - Gate-Source Breakdown Voltage: = 20 V
2. Vds - Drain-Source Breakdown Voltage: - 55 V

? From the above statement, If i wanted a Drain-Source Voltage of 40 volts, how can i cut the FET off when i can only supply 20 volts maximum to the gate.
 Thanks brentab

[gregallenwarner]

Not a problem. If you apply 40 volts to the source, in order to turn the transistor off, you need Vgs > Vth. So apply 40 volts to the gate. Vgs is now 0 volts. To turn it on, just lower the gate voltage until Vgs is < Vth. If Vth is -4 volts, for example, set the gate at 36 volts or lower.

[brentab]

So am I interpreting the data sheet wrong?
From what you are saying with a Drain to Source voltage 40 volts, I can have a maximum of 60 volt on the Gate.
Thank brentab

[gregallenwarner]

When the datasheet specifies Vgs, that's Gate to Source voltage. It's not absolute voltage relative to ground. If you've got 40 volts on the Source, and you couple that to the gate through a resistor, then 40 volts will also be at the Gate. Gate voltage (40 volts) minus Source voltage (40 volts) equals 0 volts for Vgs. That's well above the threshold voltage (Vth, usually something like -4 volts or thereabouts), and so the transistor turns off.

Vth tells you how much you need to lower the gate voltage, with respect to the source, in order to turn it on. So, for example, if you use the resistor I mentioned above, and create a voltage divider by connecting the gate through another resistor to ground, you can hold the gate voltage at a specific value. If Vth is -4 volts, then Vgs needs to be -4 or less. If you have a 1/10 voltage divider, it makes the math real easy here (for our example). 40 - 4 = 36. So pick your voltage dividers such that you get this voltage at the gate.

To turn it back off, you can insert a switch in series with the lower resistor of your divider, so when it's opened, the gate voltage rises back up to the source voltage, and Vgs returns to 0.

[mij59]

Hi,

Take a look at some mosfet tutorials e.g.
http://afrotechmods.com/tutorials/2011/11/29/transistor-mosfet-tutorial/

Do not use the absolute ratings to control  the mosfet.

[brentab]

I thanks you for your replies
Quote
"When the datasheet specifies Vgs, that's Gate to Source voltage. It's not absolute voltage relative to ground"

I think where i was wrong, i was interpreting the Data Sheet that Maximum Vgs  20 volts being relative  to ground.
Thanks brentab

[brentab]

mij59

I have watched and read several tutorials on MOSFET Trying to under stand But in every example

the Vcc was either 12 volts or 5 volts not the higher 40 volts witch exceeded the Gate-Source Breakdown Voltage:  20 V

That why I posted 40 volt for  Vcc

I think grefallenwarner shed new light on how Vgs is defined.

Thank brentab

[mij59]

mij59

I have watched and read several tutorials on MOSFET Trying to under stand But in every example

the Vcc was either 12 volts or 5 volts not the higher 40 volts witch exceeded the Gate-Source Breakdown Voltage:  20 V

That why I posted 40 volt for  Vcc

I think grefallenwarner shed new light on how Vgs is defined.

Thank brentab

Yes the answer was already in your question, you only had to realize it.

Take a look at this, it maybe somewhat confusing. 
http://afrotechmods.com/tutorials/2011/12/04/p-fet-reverse-voltage-polarity-protection-tutorial/  

[gregallenwarner]

That's a good video tutorial on power polarity protection, but keep in mind, in that configuration, the P-channel's Drain and Source are reversed from what is normally done when using the MOSFET as a high-side switch. Just a trap for young players, as Dave would say.