I have a circuit which produces a Morse code output through 2N7000 FET. The max current of the 2N7000 is around 200ma and I am using 13.6VDC. I must make this circuit switch three (3) identical but separate loads, each 13.6VDC at 750 ma. I planned to use an integrated PNP Darlington package, but I cannot deal with the high VceSAT numbers (2 volts, etc.). Any ideas come to mind for how I can do this? My design skills are very rusty.
Bradley
Moshi, Tanzania

 

Instead of 2N7000's, use a FET in TO-220 that can handle the current. Look at something like an IRF510. A parametric search will turn up lower saturation voltage drops, so you can use smaller packages, perhaps surface mount.

 

Thanks for your kind reply. Perhaps the IRF 510 or equivelent would be a good replacement for the 2N7000 in my application. I am limited to 20ma base drive maximum. Is that enough to drive the IRF 510 into saturation for my 3 amp 13.6VDC loads?

My source logic chip will toggle on/off and feed into the IRF 510. When 5V logic level from my chip is high, it drives 20 ma max to IRF 510 and turns on to allow 13.6V to be passed to load. When logic level goes low, it will shut off 13.6 volts to load. Will this work okay?

Thanks kindly for your help my friend.

 

There is no base drive when you use a FET. The voltage on the gate lead turns the device on and off. Your logic level voltage means that the 2N7000 or the IRF510 would not be suitable.

As a suggestion, look at the IRLZ24NSTRLPBFCT from International Rectifier. It's in DPAK, withstands 55 volts, and will pass 118 amps. It's full on with +5 on the gate, and full off with 0 volts. All for $1.83 through Digi-Key.

Just put about 100 ohms in series with the gate, and use it with the source grounded. You might wish to get a data sheet for fuller understanding.