Return to Answer

added 20 characters in body

edited May 28, 2013 at 1:02

184.9k
17
249
435

Here's a better way to drive a high-side P-channel MOSFET:

schematic

^{simulate this circuitsimulate this circuit – Schematic created using CircuitLab}

Note that the BJT Q1 is a current sink; the current value is the control voltage divided by R2. If your control circuit puts out 12V, the current will be 12 mA.

This current will create a voltage drop across R1 as well, and if you happen to make the two resistors equal, this voltage drop will be essentially equal to the control voltage supply (minus the V_BE drop of Q1). Or you can make the resistors different in order to get a particular relationship.

You need to be aware of two things: The current level that you set will have some bearing on how fast the MOSFET M1 switches; higher current means faster switching. However, keep in mind that Q1 has a lot of voltage across it:. When off, it has the full high voltage supply across it. And even when switched on, it has the high voltage minus 2× the control voltage across it. This means that it needs to be able to handle this voltage to begin with, but it also means that it will be dissipating power equal to that voltage drop times the current when on.

Here's a better way to drive a high-side P-channel MOSFET:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Note that the BJT Q1 is a current sink; the current value is the control voltage divided by R2. If your control circuit puts out 12V, the current will be 12 mA.

You need to be aware of two things: The current level that you set will have some bearing on how fast the MOSFET M1 switches; higher current means faster switching. However, keep in mind that Q1 has a lot of voltage across it: the high voltage minus 2× the control voltage. This means that it needs to be able to handle this voltage to begin with, but it also means that it will be dissipating power equal to that voltage drop times the current when on.

Here's a better way to drive a high-side P-channel MOSFET:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Note that the BJT Q1 is a current sink; the current value is the control voltage divided by R2. If your control circuit puts out 12V, the current will be 12 mA.

You need to be aware of two things: The current level that you set will have some bearing on how fast the MOSFET M1 switches; higher current means faster switching. However, keep in mind that Q1 has a lot of voltage across it. When off, it has the full high voltage supply across it. And even when switched on, it has the high voltage minus 2× the control voltage across it. This means that it will be dissipating power equal to that voltage drop times the current when on.

Source Link

answered May 28, 2013 at 0:56

Dave Tweed

184.9k
17
249
435

Here's a better way to drive a high-side P-channel MOSFET:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Note that the BJT Q1 is a current sink; the current value is the control voltage divided by R2. If your control circuit puts out 12V, the current will be 12 mA.

You need to be aware of two things: The current level that you set will have some bearing on how fast the MOSFET M1 switches; higher current means faster switching. However, keep in mind that Q1 has a lot of voltage across it: the high voltage minus 2× the control voltage. This means that it needs to be able to handle this voltage to begin with, but it also means that it will be dissipating power equal to that voltage drop times the current when on.