Help understanding Darlington CB leakage current and its ram

Just when I think I'm advancing as an engineer I come across something
that is simple in theory but completely stumps me. I am going through
a product design review. I am stuck on a comment a senior design
engineer made.

Sorry for the lack of ASCII artwork depicting my circuit but I don't
know how to do that, nor can I read circuits others have posted in
their threads. They look garbled. I'll try to describe this verbally.

I have a circuit that operates off a 13.5V supply. I am switching a
relay coil on and off with a FMMT614 Darlington NPN transistor acting
as a low side driver. One side of the relay coil connects to the 12V
supply, the other side connects to the transistors collector. The
emitter is grounded. The base is pulled to ground through a 100K ohm
resistor. Base current is supplied by a microcontroller output port
through a 10K ohm resistor. I think that's about it for the time
being. Simple I hope.

Oh yeah, I have a back to back zener/rectifier diode clamping network
across the relay coil to suppress transients to a safe level protecting
the transistor.

This circuit must operate from -40C to 85C. It must survive
environments as high as 125C without damage.

I thought this was a no-brainer. That's when my mentor suggested I
check out the collector to base leakage current of the transistor. He
did not elaborate any further.

I have two problems:

1.) Why did he say this?
2.) This data is not furnished in the data sheet nor in any
application note I found so far.

I assume he asked me to check this since the Darlington is a high
current gain device. A small amount of leakage C->B can turn into a
substantial current flow C->E. Could the device turn itself on and
activate the load!

I've read a few threads posted here through the years that argue where
C-B leakage current goes. Into the base load (pulldown), into the B->E
junction, both? That I'm not clear on. I don't mean to open that can
of worms again but feel I must. Can any amount of C->B leakage current
flow into the B->E junction? My thoughts are that enough leakage
current has to flow to develop a proper bias voltage across the base
pull down resistor. If this voltage approaches/exceeds the turn on
requirement of 1.5V for this transistor base current will flow. If the
amount of C->B leakage current does not create a high enough voltage
drop across the base pull down transistor (arbitrarily say 300mV) then
no current will flow B->E. Is this hypothesis flawed? Can C->B
leakage current sneak right through to the B->E junction regardless
without voltage drop developed across the base pull down resistor?

Another problem. How do I characterize the leakage current? It is not
published in the data sheet. I wrote the manufacture (ZETEX) regarding
this issue but their support team in India (being sarcastic here) has
not responded.

I hear leakage current increases with temperature and probably varies
depending on the potential across VCE when the device is off. But,
what are the hard numbers? I would set up an experiment to check this
out but do not have the necessary equipment to measure currents this
low in magnatude.

One might ask: Why not go to your mentor and ask him to elaborate.
Well, a little bit of this has to do with pride. He typically asks
such questions hoping you will go out, perform the research, and come
back to him with the answer. If you need to go to him again without
the answer it it like admitting defeat. Even if I set pride aside,
there is the fact that he is extremely busy. He probably has 20 design
reviews going at one time. I don't want to be a nuisance. I'd rather
try and iron things out here first.

That is if you don't mind helping.

Thank you

George W. Marutz II

 

[John Larkin]

On 21 Jul 2005 08:22:32 -0700, gerbermultit00l@yahoo.com wrote:

Just when I think I'm advancing as an engineer I come across something
that is simple in theory but completely stumps me. I am going through
a product design review. I am stuck on a comment a senior design
engineer made.

Sorry for the lack of ASCII artwork depicting my circuit but I don't
know how to do that, nor can I read circuits others have posted in
their threads. They look garbled. I'll try to describe this verbally.

I have a circuit that operates off a 13.5V supply. I am switching a
relay coil on and off with a FMMT614 Darlington NPN transistor acting
as a low side driver. One side of the relay coil connects to the 12V
supply, the other side connects to the transistors collector. The
emitter is grounded. The base is pulled to ground through a 100K ohm
resistor. Base current is supplied by a microcontroller output port
through a 10K ohm resistor. I think that's about it for the time
being. Simple I hope.

Those resistor values sound a little high to me, for no extremely
rational reason.

Oh yeah, I have a back to back zener/rectifier diode clamping network
across the relay coil to suppress transients to a safe level protecting
the transistor.

This circuit must operate from -40C to 85C. It must survive
environments as high as 125C without damage.

I thought this was a no-brainer. That's when my mentor suggested I
check out the collector to base leakage current of the transistor. He
did not elaborate any further.

I have two problems:

1.) Why did he say this?
2.) This data is not furnished in the data sheet nor in any
application note I found so far.

I assume he asked me to check this since the Darlington is a high
current gain device. A small amount of leakage C->B can turn into a
substantial current flow C->E. Could the device turn itself on and
activate the load!

I've read a few threads posted here through the years that argue where
C-B leakage current goes. Into the base load (pulldown), into the B->E
junction, both? That I'm not clear on. I don't mean to open that can
of worms again but feel I must. Can any amount of C->B leakage current
flow into the B->E junction? My thoughts are that enough leakage
current has to flow to develop a proper bias voltage across the base
pull down resistor. If this voltage approaches/exceeds the turn on
requirement of 1.5V for this transistor base current will flow. If the
amount of C->B leakage current does not create a high enough voltage
drop across the base pull down transistor (arbitrarily say 300mV) then
no current will flow B->E. Is this hypothesis flawed?

All sounds right to me.

Can C->B
leakage current sneak right through to the B->E junction regardless
without voltage drop developed across the base pull down resistor?

Well, if the leakage is there, it's real collector current, but it's
likely small. If your base-ground resistor diverts most of the leakage
to ground, the collector leakage will *not* be multiplied by beta,
which could make it a lot bigger.

Another problem. How do I characterize the leakage current? It is not
published in the data sheet. I wrote the manufacture (ZETEX) regarding
this issue but their support team in India (being sarcastic here) has
not responded.

Measure a few parts?

I hear leakage current increases with temperature

Assume it doubles every 10 degrees C.

and probably varies
depending on the potential across VCE when the device is off. But,
what are the hard numbers? I would set up an experiment to check this
out but do not have the necessary equipment to measure currents this
low in magnatude.

Most any handheld DVM can measure nanoamps... on its *voltage* range.
Just measure the voltage drop across a 1M resistor (1 mv per na) or
even across the meter's internal resistance, usually 10 megs. A Fluke
and a 9-volt battery and a couple of clip leads will quickly measure
Iceo, Icbo, and Icex.

One might ask: Why not go to your mentor and ask him to elaborate.
Well, a little bit of this has to do with pride. He typically asks
such questions hoping you will go out, perform the research, and come
back to him with the answer. If you need to go to him again without
the answer it it like admitting defeat. Even if I set pride aside,
there is the fact that he is extremely busy. He probably has 20 design
reviews going at one time. I don't want to be a nuisance. I'd rather
try and iron things out here first.

He sounds OK to me. And he's likely aware that some darlingtons,
especially the bigger ones, can be very leaky.

John

 

[PeteS]

Some quick thoughts, without going in to any real depth.

I pulled the data sheet, and it's a little sparse - I assume (but do
not know) that there is no internal resistor (pull from input device
base to lower emitter) ; Assuming that to be so, then it won't affect
other things.

The base in a darlington, is the base of the actual top collector
device. As (when off, anyway) a reverse biased PN junction, there will
be leakage current which increases linearly with temperature, and can
usually be assumed to be roughly equal to that of an ordinary bipolar
device (unless other guidance is given). (When on, it's more like a low
quality current sink towards the emitter).

For typical PN reverse leakage, I suggest you look at some PN diode
data sheets (they usually specify such things).

I would be wary of a 100k pull down on *any* bipolar device - you
wouldn't need much leakage to give you significant problems. I usually
reserve such values for FETs.

You may want to reduce that to more like 10k (or even less). That will
still be the vast majority of the current drawn from your processor
output pin (the minimum current transfer is 15,000 for the device).

Others here will no doubt weigh in

Cheers

PeteS

 

[John Popelish]

gerbermultit00l@yahoo.com wrote:
(snip)

Sorry for the lack of ASCII artwork depicting my circuit but I don't
know how to do that, nor can I read circuits others have posted in
their threads. They look garbled. I'll try to describe this verbally.

Reading ASCII drawings requires that you set your newsreader to a
fixed width (per character) font, like Courier. This puts the
characters on a Cartesian grid.

I have a circuit that operates off a 13.5V supply. I am switching a
relay coil on and off with a FMMT614 Darlington NPN transistor acting
as a low side driver. One side of the relay coil connects to the 12V
supply, the other side connects to the transistors collector. The
emitter is grounded. The base is pulled to ground through a 100K ohm
resistor. Base current is supplied by a microcontroller output port
through a 10K ohm resistor. I think that's about it for the time
being. Simple I hope.

Clear enough.

Oh yeah, I have a back to back zener/rectifier diode clamping network
across the relay coil to suppress transients to a safe level protecting
the transistor.

This circuit must operate from -40C to 85C. It must survive
environments as high as 125C without damage.

I thought this was a no-brainer. That's when my mentor suggested I
check out the collector to base leakage current of the transistor. He
did not elaborate any further.

I have two problems:

1.) Why did he say this?

He probably got bit on the ass for neglecting this effect at one time.

2.) This data is not furnished in the data sheet nor in any
application note I found so far.

If you have no data on the input transistor's collector to base
leakage versus temperature, how do you know that your 100k resistor
will drain the leakage without the voltage building up high enough to
keep the darlington from being turned on by it?

I assume he asked me to check this since the Darlington is a high
current gain device. A small amount of leakage C->B can turn into a
substantial current flow C->E. Could the device turn itself on and
activate the load!

There are two transistors in a darlington. The first one can be turned
on by its collector to base leakage, and then the second one is driven
by that leakage current times the beta of the first one. The second
device can also be driven directly by its own collector to base
leakage. Many darlingtons contain an internal resistor base to
emitter of the second transistor to handle the second case, since you
rarely have access to this base.

I've read a few threads posted here through the years that argue where
C-B leakage current goes. Into the base load (pulldown), into the B->E
junction, both? That I'm not clear on.

Almost all of the CB leakage will pass through the resistor, till the
base emitter voltage of that device gets high enough to start pushing
current through that junction. That voltage may be as low as .3 volts
at high temperature. Then that base current is amplified by the
transistor beta and is applied either to the second base emitter
junction or to the parallel combination of that junction and any
internal base emitter resistor.

I don't mean to open that can
of worms again but feel I must. Can any amount of C->B leakage current
flow into the B->E junction? My thoughts are that enough leakage
current has to flow to develop a proper bias voltage across the base
pull down resistor.

Yes, it is the voltage that determines how much current passes through
the junction.

If this voltage approaches/exceeds the turn on
requirement of 1.5V for this transistor base current will flow. If the
amount of C->B leakage current does not create a high enough voltage
drop across the base pull down transistor (arbitrarily say 300mV) then
no current will flow B->E. Is this hypothesis flawed?

I don't think so.

Can C->B
leakage current sneak right through to the B->E junction regardless
without voltage drop developed across the base pull down resistor?

No.

Another problem. How do I characterize the leakage current? It is not
published in the data sheet. I wrote the manufacture (ZETEX) regarding
this issue but their support team in India (being sarcastic here) has
not responded.
(snip)

I would perform and experiment on several devices from different
batches with no input resistor and a higher temperature than you
expect the circuit to ever see (up to the rated temperature for the
device) and the collector voltage at least equal to the supply voltage
plus the zener voltage, to get an idea how bad the problem might be.
If there is an internal base resistor on the second transistor, the
problem may be slight. Then find out by experiment how low a base
resistor is needed to bring the collector current down to an
essentially fixed value (no amplification of leakage). Then use one
1/10th of that value.

Of course, if your logic signal pulls down to essentially zero volts
(and there is no way the logic can become unpowered while the
darlington is powered), the effective pull down resistance is the
parallel combination of the base resistor and the series resistor to
the logic signal. So by the time you get the series resistor low
enough for good saturation at lowest temperature and device gain, the
effective low state resistance may be well below that needed to keep
the leakage under control under all circumstances with no shunt
resistor across the darlington.

 

OK, took your advice and looked at diode leakage on a variety of
diodes. Leakage ranges from the low single digit nA range clear up
into the double digit uA range. I was dead in the water without
knowing how to characterize this specific transistor in comparison to
one of the numerous diode models.

So, I decided to run a experiment. I took a transistor and connected
its collector to 13.50VDC. I connected a 979K (at 25C) resistor from
the base to ground. I connected the emitter to ground. Measured
leakage current at room temperature was 0.2nA. I next moved the
experiment over to an environmental chamber that is running a test at
70C. I placed the test sample inside (wiring all connected). The
resistor dropped in resistance due to the rise in temperature so I had
to compensate for this. The end result was 4nA of leakage current.
This seems awefully low but makes sense if you look at the levels at
room temperature. If you keep halving leakage current for every 10C
drop in temperature from 70C it comes out to roughly 0.18nA at room
temperature. Close to what I measured. The un certainty of my
measurement equipment could have had a role to play in this small
discrepancy or perhaps a perfect doubling does not occur at exactly 10C
rise. Whatever...

Do the levels of leakage I mention make sense?

At 120C leakage should increase to approximately 128nA. Across my 100K
pulldown this results in a 12.8mV drop across the load resistor. IMO
this is nothing to be concerned about (assuming my findings are valid).

 

He sounds OK to me. And he's likely aware that some darlingtons,
especially the bigger ones, can be very leaky.

You are absolutely right. He is OK. As a matter of fact, I have
nothing but respect for the man. He challenges me to push myself to be
better.

I'm sure you have all been there at one point or another. Trying to
impress your mentors. You know, show them that you're approaching
their skill base and soon may surpass them,,, At least that's how I
feel.

George W. Marutz II

 

Of course, if your logic signal pulls down to essentially zero volts
(and there is no way the logic can become unpowered while the
darlington is powered), the effective pull down resistance is the
parallel combination of the base resistor and the series resistor to
the logic signal. So by the time you get the series resistor low
enough for good saturation at lowest temperature and device gain, >the effective low state resistance may be well below that needed >to keep the leakage under control under all circumstances with no >shunt resistor across the darlington.

Unfortunately the Darlington always has voltage across it. The micro
can turn on and off freely. So, I cannot count on the base resistor to
be a stiffer pull down at all times.

George W. Marutz II

 

[John - KD5YI]

gerbermultit00l@yahoo.com wrote:

Just when I think I'm advancing as an engineer I come across something
that is simple in theory but completely stumps me. I am going through
a product design review. I am stuck on a comment a senior design
engineer made.

Sorry for the lack of ASCII artwork depicting my circuit but I don't
know how to do that, nor can I read circuits others have posted in
their threads. They look garbled. I'll try to describe this verbally.

I have a circuit that operates off a 13.5V supply. I am switching a
relay coil on and off with a FMMT614 Darlington NPN transistor acting
as a low side driver. One side of the relay coil connects to the 12V
supply, the other side connects to the transistors collector. The
emitter is grounded. The base is pulled to ground through a 100K ohm
resistor. Base current is supplied by a microcontroller output port
through a 10K ohm resistor. I think that's about it for the time
being. Simple I hope.

Oh yeah, I have a back to back zener/rectifier diode clamping network
across the relay coil to suppress transients to a safe level protecting
the transistor.

This circuit must operate from -40C to 85C. It must survive
environments as high as 125C without damage.

I thought this was a no-brainer. That's when my mentor suggested I
check out the collector to base leakage current of the transistor. He
did not elaborate any further.

I have two problems:

1.) Why did he say this?
2.) This data is not furnished in the data sheet nor in any
application note I found so far.

I assume he asked me to check this since the Darlington is a high
current gain device. A small amount of leakage C->B can turn into a
substantial current flow C->E. Could the device turn itself on and
activate the load!

I've read a few threads posted here through the years that argue where
C-B leakage current goes. Into the base load (pulldown), into the B->E
junction, both? That I'm not clear on. I don't mean to open that can
of worms again but feel I must. Can any amount of C->B leakage current
flow into the B->E junction? My thoughts are that enough leakage
current has to flow to develop a proper bias voltage across the base
pull down resistor. If this voltage approaches/exceeds the turn on
requirement of 1.5V for this transistor base current will flow. If the
amount of C->B leakage current does not create a high enough voltage
drop across the base pull down transistor (arbitrarily say 300mV) then
no current will flow B->E. Is this hypothesis flawed? Can C->B
leakage current sneak right through to the B->E junction regardless
without voltage drop developed across the base pull down resistor?

Another problem. How do I characterize the leakage current? It is not
published in the data sheet. I wrote the manufacture (ZETEX) regarding
this issue but their support team in India (being sarcastic here) has
not responded.

I hear leakage current increases with temperature and probably varies
depending on the potential across VCE when the device is off. But,
what are the hard numbers? I would set up an experiment to check this
out but do not have the necessary equipment to measure currents this
low in magnatude.

One might ask: Why not go to your mentor and ask him to elaborate.
Well, a little bit of this has to do with pride. He typically asks
such questions hoping you will go out, perform the research, and come
back to him with the answer. If you need to go to him again without
the answer it it like admitting defeat. Even if I set pride aside,
there is the fact that he is extremely busy. He probably has 20 design
reviews going at one time. I don't want to be a nuisance. I'd rather
try and iron things out here first.

That is if you don't mind helping.

Thank you

George W. Marutz II

Hi, George -

The data sheet gives Icbo of 10 nA max at 25C. At 125C that would increase
by 2^10 or to 10 uA. (Probably won't be this high since the spec is at 100V)

The Vbe(on) is given as .8V typical at 50 mA Ic and about 125C (Vbe vs Ic at
temperature graph on page 2)

So, Icbo is about 10 uA and 8 uA goes through the 100k leaving 2 uA for Ib.
At high temperature, your gain is over 65,000. This could lead to a
collector current of 130 mA (if the relay coil permits). If it was my
design, I would go as low as I could with the b-e resistance as others have
recommended. It may help with noise as well.

I realize you effectively have a 10k shunt via the logic input signal, but
what is the microcontroller low output voltage at 125C? And, as John Poplish
pointed out, what if, somehow, your micro output tristates to floating? Go
for the robustness.

Just some thoughts. Good luck.

John

 

All three of you made mention that my base pull down resistor is too
high of a value. Funny, so did the senior design engineer.

Why did I do this? Well, for one thing, I did not realize that using a
high resistance value pull down could cause potential problems. I know
better now.

My train of thought was to keep current being sourced by the micro pin
as low as possible (around 500uA as an arbitrary target). Higher
source current can cause other problems in my system that I will not go
into now (A/D accuracy). I decided that the high current gain of the
darlington over the temperature extremes I will be working with would
aid in this effort. I first chose the base resistor from the parts
list already on my BOM. I chose another resistor that was already on
the BOM for the base pull down. I wanted this value to be greater than
the base resistor so I would ensure I would be turning the transistor
on. The 100K part fit the bill.

I might have to break down and add a new component to the BOM. Perhaps
a 4.7K base resistor and 10K pull down?

Thank you for your valuable input.

Ge0rge

 

John - KD5YI (so many John's around here )

So what you are saying is that cut off current is the same as leakage
current?

Hmm... I guess I had a misunderstanding of what cut-off current meant.
THANKS. If I had known this I could have saved myself a lot of
anguish. I learned a few valuable leasoons here today.

Ge0rge

 

[John - KD5YI]

gerbermultit00l@yahoo.com wrote:

John - KD5YI (so many John's around here )

So what you are saying is that cut off current is the same as leakage
current?

Hmm... I guess I had a misunderstanding of what cut-off current meant.
THANKS. If I had known this I could have saved myself a lot of
anguish. I learned a few valuable leasoons here today.

Ge0rge

Actually, George, I ASSUMED it was the same as leakage. If I'm wrong, I need
to know it and I hope one of the regular gurus will come along and correct me.

John

 

Actually, George, I ASSUMED it was the same as leakage. >If I'm wrong, I need to know it and I hope one of the regular >gurus will come along and correct me.

Sounds right to me. I checked your hypothesis before I responded the
first time. Go to:

http://www.electronicdefinitions.com/definition.php?defid=362

Ge0rge

 

[John Larkin]

On 21 Jul 2005 12:13:15 -0700, gerbermultit00l@yahoo.com wrote:

All three of you made mention that my base pull down resistor is too
high of a value. Funny, so did the senior design engineer.

Why did I do this? Well, for one thing, I did not realize that using a
high resistance value pull down could cause potential problems. I know
better now.

My train of thought was to keep current being sourced by the micro pin
as low as possible (around 500uA as an arbitrary target). Higher
source current can cause other problems in my system that I will not go
into now (A/D accuracy). I decided that the high current gain of the
darlington over the temperature extremes I will be working with would
aid in this effort. I first chose the base resistor from the parts
list already on my BOM. I chose another resistor that was already on
the BOM for the base pull down. I wanted this value to be greater than
the base resistor so I would ensure I would be turning the transistor
on. The 100K part fit the bill.

I might have to break down and add a new component to the BOM. Perhaps
a 4.7K base resistor and 10K pull down?

Thank you for your valuable input.

Ge0rge

You might consider using a fet instead of a darlington, especially if
the micro is a 5-volt part. A 2N7002 costs a few cents, saves one
resistor, and will sink over half an amp with +5 on the gate.

John

 

[John - KD5YI]

gerbermultit00l@yahoo.com wrote:

Actually, George, I ASSUMED it was the same as leakage. >If I'm wrong, I need to know it and I hope one of the regular >gurus will come along and correct me.


Sounds right to me. I checked your hypothesis before I responded the
first time. Go to:

http://www.electronicdefinitions.com/definition.php?defid=362

Ge0rge

Even so, I would support John Larkin's suggestion that you use a FET. No
drain on the micro and none of these other problems. It is a match made in
heaven.

Cheers,
John

 

Even so, I would support John Larkin's suggestion that you use a >FET. No drain on the micro and none of these other problems. It is >a match made in heaven.

The original design used a FET. That's why the resistor values are the
way they are. The darlington was supposed to be a drop in replacement.
FETs for this purpose cost 5x as much as the Darlington in my circuit
now. Their are only a hand full of options for a N-Channel MOSFET with
a 100V breakdown rating in a SOT-23 package. The limited number of
competitive products on the market dictate higher prices. When you
manufacture 2 million+ products a year this sort of thing adds up.

Thanks

Ge0rge

 

You might consider using a fet instead of a darlington, >especially if the micro is a 5-volt part. A 2N7002 costs a few >cents, saves one resistor, and will sink over half an amp with >+5 on the gate.

Thanks for the input John. However, read my comment to
John - KD5YI. I need a part with a higher VDSS rating. The 2N7002,
and many other comparable devices, just would not cut it. I desire as
small of package as is possible. The SOT-23 fits the bill for now.
Unfortunately you don't find many SOT-23 N-Channel FET offerings with a
VDSS of 100V or greater. The price takes a considerable jump over the
60V counterparts.

Oh, one last question. What resistor would you eliminate? The Gate
pull down? Why?

Ge0rge

 

[John - KD5YI]

gerbermultit00l@yahoo.com wrote:

Even so, I would support John Larkin's suggestion that you use a >FET. No drain on the micro and none of these other problems. It is >a match made in heaven.


The original design used a FET. That's why the resistor values are the
way they are. The darlington was supposed to be a drop in replacement.
FETs for this purpose cost 5x as much as the Darlington in my circuit
now. Their are only a hand full of options for a N-Channel MOSFET with
a 100V breakdown rating in a SOT-23 package. The limited number of
competitive products on the market dictate higher prices. When you
manufacture 2 million+ products a year this sort of thing adds up.

Thanks

Ge0rge

From the Mouser catalog:

Fairchild BSS123, 100V, 6 Ohms, SOT-23, $.08/250.
Supertex VN2110K1, 100V, 4 Ohms, SOT-23-3, $.24/1000.
Siliconix SI2328DS-T1, 100V, 0.25 Ohms, SOT-23, $.55/500

I didn't look any further. I don't understand why you would need a 100V
device to operate a 12V relay, but I'm sure you do. If you find that you
could use a 60V device, you'll have a much wider selection. But, you knew
that already.

Good luck.

John

 

[John Larkin]

On 22 Jul 2005 05:27:06 -0700, gerbermultit00l@yahoo.com wrote:

You might consider using a fet instead of a darlington, >especially if the micro is a 5-volt part. A 2N7002 costs a few >cents, saves one resistor, and will sink over half an amp with >+5 on the gate.

Thanks for the input John. However, read my comment to
John - KD5YI. I need a part with a higher VDSS rating. The 2N7002,
and many other comparable devices, just would not cut it. I desire as
small of package as is possible. The SOT-23 fits the bill for now.
Unfortunately you don't find many SOT-23 N-Channel FET offerings with a
VDSS of 100V or greater. The price takes a considerable jump over the
60V counterparts.

Oh, one last question. What resistor would you eliminate? The Gate
pull down? Why?

Ge0rge

Assuming your uP starts up with its output port pin tri-stated, you
would still need a pulldown to ensure the fet is off. But you wouldn't
need a series resistor.

You can, incidentally, leave the gate lead of a 2N7000/7002
disconnected, and turn it on/off by briefly charging or discharging
the gate by touching V+ or ground with one finger and then tapping the
gate lead with the other. Once on or off, it will stay that way for a
while... about a week is the longest I've verified.

John

 

Snip:

Fairchild BSS123, 100V, 6 Ohms, SOT-23, $.08/250.

Excellent. Never came across this one. I was using the $0.55
Siliconix part.

I didn't look any further. I don't understand why you would >need a 100V device to operate a 12V relay, but I'm sure you >do.

Unsuppressed battery feed / automotive load dump.

Ge0rge

 

[John - KD5YI]

gerbermultit00l@yahoo.com wrote:

Snip:


Fairchild BSS123, 100V, 6 Ohms, SOT-23, $.08/250.


Excellent. Never came across this one. I was using the $0.55
Siliconix part.

In that case, the Rds of the Fairchild part may be too high for you.

I didn't look any further. I don't understand why you would >need a 100V device to operate a 12V relay, but I'm sure you >do.


Unsuppressed battery feed / automotive load dump.

Ah! As I recall from SAE J1211, you can have transients well over 100V at
times. But, I don't know if that is applicable to your particular case.

Ge0rge

Cheers,
John