thelightbrain
- 10th November 2004, 23:28
Trying to determine the BEST way for driving a "logic-level" N-Channel HEXFET from a PIC output pin (for example IRLD110 or IRLZ44N). I suppose I mean "most-reliable" method.
NOTE 1: The FET gate voltages are usually rated at 4.5V or 5V for "full on".
NOTE 2: The PIC HIGH output is rated as (Vdd - .7V)
NOTE 3: Assume we are driving an IR LED at 500mA to 2000mA using 40KHz to 56KHz PWM.
My problem is that I have seen this done three ways, and I know each of them WILL work. BUT...what are the advantages or disadvantages of each method? Here's MY understanding...
A: Connect the PIC output directly to the Gate -- Simplest approach, but this would appear to be problematic since the PIC's HIGH output will only be 4.3V, so the FET will never "fully" turn on, reducing drive current to the LED. Also causes more power dissipation in the FET due to higher "on" resistance.
B: Use a pull-up resistor between PIC output and the Gate -- This should bring the HIGH output back up to the full 5V. Should help FET operate at rated "on-resistance", but it adds another component.
C: Use a low impedance drive resistor -- I have seen this used for other high-current switching applications and so far I have been using this configuration for my designs. My understanding was that it helps prevent oscillations in the Gate drive. But there's alot of info about "miller effects", gate-charge, gate capacitance/inductance, etc that I don't quite understand.
I have attached a drawing of the three examples:
NOTE 1: The FET gate voltages are usually rated at 4.5V or 5V for "full on".
NOTE 2: The PIC HIGH output is rated as (Vdd - .7V)
NOTE 3: Assume we are driving an IR LED at 500mA to 2000mA using 40KHz to 56KHz PWM.
My problem is that I have seen this done three ways, and I know each of them WILL work. BUT...what are the advantages or disadvantages of each method? Here's MY understanding...
A: Connect the PIC output directly to the Gate -- Simplest approach, but this would appear to be problematic since the PIC's HIGH output will only be 4.3V, so the FET will never "fully" turn on, reducing drive current to the LED. Also causes more power dissipation in the FET due to higher "on" resistance.
B: Use a pull-up resistor between PIC output and the Gate -- This should bring the HIGH output back up to the full 5V. Should help FET operate at rated "on-resistance", but it adds another component.
C: Use a low impedance drive resistor -- I have seen this used for other high-current switching applications and so far I have been using this configuration for my designs. My understanding was that it helps prevent oscillations in the Gate drive. But there's alot of info about "miller effects", gate-charge, gate capacitance/inductance, etc that I don't quite understand.
I have attached a drawing of the three examples: