I'm trying to understand the valid operating characteristics of BJTs but specifically the Fairchild FJPF5027. My circuit is like a resistive voltage divider with the high side resistor replaced by the BJT and being driven by another voltage divider to program the base current. The low side resistance, the "load", is from 100kOhms to
10MOhms or even higher.The FJPF has a collector current and emitter current cutoff of 10uA. This means that if the load is around 10MOhms with the voltage of 100V then the maximum current is 10uA and the transistor will not pass any current. It is in the "off state"? What this means is that even for lower loads such as 50MOhms will not be driven?
Are there solutions to this? I thought about adding a parallel resistance to the main load that will always drive the BJT out of cutoff. Say 100kOhm in parallel which will give a maximum collector current of 1mA. Unfortunately this ends up requiring the resistor to potentially dissipate 20W. I can increase the resistance by a factor of 10 to make this more reasonable yet then my "drive" current is only
100uA.I want the BJT to be "on" and not operate close to the cutoff so that the load ends up being cut off from any power. I also want the BJT to work somewhat in the linear region. Do I have the wrong BJT? What exactly is the operation of the BJT near the cutoff? Is it sharp or smooth? i.e., Will 15uA "bias" the BJT so that it will function in the linear region?
More generally if I am working with low power high voltage designs do I need to look for specific BJT's? e.g., Will high power high voltage bjt's create such problems as stated above?
Also, is it difficult to replace bjt's with mosfets excluding the gate driving differences? From what I can tell mosfets do not have a minimum drain current needed to operate. If I use a mosfet instead of a bjt will it work better in this situation. The goal here is to have a programmable voltage divider circuit but the load needs to be on the load side.