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BJT cutoff specs

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.
 
M

MooseFET

Jan 1, 1970
0
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.

All MOSFETs have a source to drain leakage current. This sets a
minimum current for them. At low amounts of gate bias, their
characteristics vary quite a bit from unit to unit.
 
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.

Hello James,

The Icbo is just the collector base leakage current. It does not mean
that the transistor doesn’t function below that current. Your
transistor will function in its linear region below 10uA.

I don't know the number of units that has to be made, but this leakage
current (at 100V Vcb) and low junction temperature will be extremely
low. Just try to measure it for some devices, you will probably see nA
range values.

You should design your driver in such away that you can shunt Icb to
the emitter (or below). When the cb leakage current goes into the
base, it will be amplified by HFE.

Best regards,

Wim
PA3DJS
www.tetech.nl
 
P

Phil Allison

Jan 1, 1970
0
<[email protected]

The Icbo is just the collector base leakage current. It does not mean
that the transistor doesn’t function below that current. Your
transistor will function in its linear region below 10uA.

** Really??

Not likely with any high voltage, fast switching BJT with Hfe of only 40 at
best.

http://www.fairchildsemi.com/ds/FJ/FJPF5027.pdf

Bet it will have less than unity Hfe at 10uA.


...... Phil
 
J

James Rollins

Jan 1, 1970
0
Hello James,

The Icbo is just the collector base leakage current. It does not mean
that the transistor doesn’t function below that current. Your
transistor will function in its linear region below 10uA.

But how well will it function? Is 10uA the "knee" in the graphs?
 
J

James Rollins

Jan 1, 1970
0
It would help if you posted a schematic.

As I said, it is simply a resistive voltage divider with the high side
resistor replaced with the bjt. I'm not sure how much clearer a
schematic would make that.
"Collector cutoff current" is OFF-state leakage, collector-to-base
leakage in this case. It's probably not as high as 10 uA in real
life... that's just a value they can easily test for. If you are
suggesting that there is some minimum "on" current, no, there isn't.

How well will the bjt behave near the cutoff?
This is maybe a big transistor to use to switch your apparently light
loads. Bigger transistors do have more off-state leakage.

I've looked for high voltage bjt's with lower cutoff but they all seem
to be 10uA. Maybe this is a somewhat generic number.
 
But how well will it function? Is 10uA the "knee" in the graphs?

Hello James,

What "knee" do you mean?

At low Ic you will have a low HFE (lower then published in the
datasheet), as Phil also mentioned.

How you will bias your voltage dividing transistor and why do you take
a 800V/3A device? Can you mention something more about the
application?

Best regards,

Wim
PA3DJS
www.tetech.nl
remove a,b and c when using PM
 
R

Rich Webb

Jan 1, 1970
0
As I said, it is simply a resistive voltage divider with the high side
resistor replaced with the bjt. I'm not sure how much clearer a
schematic would make that.

I'm surprised that John L. didn't mention it but the free LTSpice has
become a sort of lingua franca around here for exchanging schematics.
It's (not surprisingly) rather vendor specific but a lot better than
hand-waving. http://www.linear.com/designtools/software/
 
J

Jim Thompson

Jan 1, 1970
0
You can also hand-scribble it and post to any of the free
picture-hosting sites, or to a.b.s.e.

The OP sounds a bit confused about how transistors work, and about the
terminology,

Yep. He needs to go over to "BASICS".
so his schematic might have all sorts of problems. I've
seen some whoppers.

Agree, LTspice is about the only universal, free schematic format.
It's also a great simulator. It must be seriously hurting the
for-big-bucks simulator companies.

John

Not really. It lacks in sophisticated post processing of the data. If
it could match PSpice's Probe, it'd be a winner.

And PSpice is an order of magnitude cheaper than the "for-big-bucks"
simulators.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Stormy on the East Coast today... due to Bush's failed policies.
 
C

Charlie E.

Jan 1, 1970
0
Yep. He needs to go over to "BASICS".


Not really. It lacks in sophisticated post processing of the data. If
it could match PSpice's Probe, it'd be a winner.

And PSpice is an order of magnitude cheaper than the "for-big-bucks"
simulators.

...Jim Thompson

Hi Jim,
Don't know if you noticed, but Altium recently lowered the price on
their stuff to just $4K. That should drive EMA crazy... :cool:

Charlie
 
J

Jim Thompson

Jan 1, 1970
0
Hi Jim,
Don't know if you noticed, but Altium recently lowered the price on
their stuff to just $4K. That should drive EMA crazy... :cool:

Charlie

It sure is enjoyable to have the EMA guys calling, nearly weeping,
trying to sell me something ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine Sometimes I even put it in the food
 
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