On Sat, Jun 18, 2011 at 5:45 PM, David wrote: > All, > > I have been going round in circles today considering how to drive a > logic level FET. I have tried a few things on breadboard but don't > trust the results, many of the weird observations could be to do with > the breadboard. > > I wish to drive a IR LED at 40khz (essentially the Sony IR protocol), > somewhere between 50mA and 100mA max LED Ifwd. In testing I have been > using a 2N7000. > > I am confused around whether or not to use resistors, there is a lot of > conflicting advice online. The options seem roughly to be: > > A) Connect the gate directly to the PIC pin, no resistor. Lower parts > count, which is always useful. > If you're using a MOSFET, you don't (and I think shouldn't) use any resistors. MOSFETs have a very high input impedance and are voltage driven. Drive the gate directly. > B) Use a low (100Ohm or similar) resistor in series between PIC output > and the FET. Provides some protection for the PIC. > I may be wrong here, but protection from what? The gate on a MOSFET is insulated. Theoretically infinite impedance. > And for an alternative problem: > > Y) No resistor between gate and GND, once again lower parts count and no > worry of the PIC having to source more current when pulling the pin high. > If the PIC pin is configured as output, it's a driver - so it's low impedance to either logic high, or logic low, depending on what you "write" to the port. So you don't need a pullup/pulldown resistor since the gate will be up or down either way. > Z) A high (>=3D10k) resistor between gate and GND. Most useful if the PI= C > could ever be disconnected from the FET (which it cannot in this > instance). However, also provides some protection during ICSP, power on > (before TRIS is set) and will ensure the FET always turns off. > If the pin is ever configured as an input (high impedance), then the gate will be floating. What's the worst that could happen? Probably nothing. If you really want to feel cozy inside knowing the gate is off by default, you can put a 10k (100k?) resistor to ground. Very low current through the resistor. > I am not particularly tight on space nor constrained by a few extra > pennies, but it would be good to know which combination of (A|B)+(Y|Z) > is correct and for what reasons. > I vote A and Y because I don't think the resistors serve any useful purpose in this case. --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .