J
James Rollins
- Jan 1, 1970
- 0
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.
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.