reverse breakdown of BE junction

[Walter Harley]

In an oscillator simulation I was playing with, I noticed that the simulator
claimed that voltages on a 2n3055's B-E junction were reaching -40V.

To me, that seemed improbable. I somehow assumed that that the base-emitter
junction of a bipolar transistor would have a relatively weak reverse
breakdown voltage, even for a power transistor. I don't know why I assumed
that.

But searching datasheets, I can't even find anywhere where this
characteristic is described. Anybody got some info? I don't care so much
about the 2n3055 in particular, just wondering about this characteristic of
BJT's in general.

 

[Al]

In article <c4ptc7$461$0@216.39.172.65>,
"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

In an oscillator simulation I was playing with, I noticed that the simulator
claimed that voltages on a 2n3055's B-E junction were reaching -40V.

To me, that seemed improbable. I somehow assumed that that the base-emitter
junction of a bipolar transistor would have a relatively weak reverse
breakdown voltage, even for a power transistor. I don't know why I assumed
that.

But searching datasheets, I can't even find anywhere where this
characteristic is described. Anybody got some info? I don't care so much
about the 2n3055 in particular, just wondering about this characteristic of
BJT's in general.

Veb of a Motorola 2N3055 is 7 Vdc. For a 2N3904, it's 6 Vdc.

Al

--
There's never enough time to do it right the first time.......

 

[Walter Harley]

"Al" <no.spam@here.com> wrote in message
news:no.spam-123594.18465604042004@news.verizon.net...

Veb of a Motorola 2N3055 is 7 Vdc. For a 2N3904, it's 6 Vdc.

Ah, that explains why I couldn't find it on the datasheets. I didn't
realize that's what Veb referred to.

Okay, next question: Veb is listed as an "absolute maximum". Does that
imply that exceeding it causes permanent damage to the transistor? How
come? (After all, Zener diodes aren't damaged by being run at their reverse
breakdown voltage.)

Thanks,
-walter

 

[Al]

In article <c4q7dr$sdr$0@216.39.172.65>,
"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Al" <no.spam@here.com> wrote in message
news:no.spam-123594.18465604042004@news.verizon.net...
Veb of a Motorola 2N3055 is 7 Vdc. For a 2N3904, it's 6 Vdc.

Ah, that explains why I couldn't find it on the datasheets. I didn't
realize that's what Veb referred to.

Okay, next question: Veb is listed as an "absolute maximum". Does that
imply that exceeding it causes permanent damage to the transistor? How
come? (After all, Zener diodes aren't damaged by being run at their reverse
breakdown voltage.)

Thanks,
-walter

The reverse breakdown is often used to generate random noise. The
current is limited by a resistor. There are failure modes associated
with this. If the current is too large for that particular structure
and/or doping concentrations, the crystal lattice will be damaged. After
all, it was not designed to be a zener. You might not care if you are
building a random noise generator.

By listing it as a maximum, it does not mean that you should not exceed
that value. It means that the value could be lower and still be within
spec. If you found a transitor with a maximum Veb of 7, and you measured
it to be 10, it is defective! But if it is 6.5, that's OK. A minimum may
be specified as V(br)ebo, which is the reverse breakdown voltage,
emitter to base, open circuited. Whew! And it could be, say 6.5 Vdc

Al

--
There's never enough time to do it right the first time.......

 

[Hal Murray]

Okay, next question: Veb is listed as an "absolute maximum". Does that
imply that exceeding it causes permanent damage to the transistor? How
come? (After all, Zener diodes aren't damaged by being run at their reverse
breakdown voltage.)

It probably depends upon the rest of the circuit.

If you connect it across a big power supply and turn the knob
up until it breaks down, then it will suck a lot of current and
get real hot real fast.

If you connect it to the same power supply through a 10K resistor,
now when it breaks down the resistor will limit the current to
(roughly) 1 mA. That probably won't destroy anything.

There may be mechanisms in there that will shorten the lifetime.

I remember the first time I got to play with a curve tracer.
(many years ago) I turned the knob up until the transistor I
was testing broke down. Breakdown and recovery right there
on the screen, many times a second. There was another knob
that turned into how much energy got put into the transistor.
Turn it up and the curves shifted with temperature. Turn it
back down and they return to normal.

--
The suespammers.org mail server is located in California. So are all my
other mailboxes. Please do not send unsolicited bulk e-mail or unsolicited
commercial e-mail to my suespammers.org address or any of my other addresses.
These are my opinions, not necessarily my employer's. I hate spam.

 

[Watson A.Name - \"Watt Su]

"Hal Murray" <hmurray@suespammers.org> wrote in message
news:1071egn2j1pql71@corp.supernews.com...

Okay, next question: Veb is listed as an "absolute maximum". Does
that
imply that exceeding it causes permanent damage to the transistor?
How
come? (After all, Zener diodes aren't damaged by being run at their
reverse
breakdown voltage.)

It probably depends upon the rest of the circuit.

If you connect it across a big power supply and turn the knob
up until it breaks down, then it will suck a lot of current and
get real hot real fast.

If you connect it to the same power supply through a 10K resistor,
now when it breaks down the resistor will limit the current to
(roughly) 1 mA. That probably won't destroy anything.

The beta of the transistor will be permanently and irreversibly reduced;
yes, it *will* damage the transistor.

If you don't belive this, try it with a 2N3904 or other cheap
transistor. I didn't believe it, and tried it. I'm now a believer.

There may be mechanisms in there that will shorten the lifetime.

I remember the first time I got to play with a curve tracer.
(many years ago) I turned the knob up until the transistor I
was testing broke down. Breakdown and recovery right there
on the screen, many times a second. There was another knob
that turned into how much energy got put into the transistor.
Turn it up and the curves shifted with temperature. Turn it
back down and they return to normal.

--
The suespammers.org mail server is located in California. So are all
my
other mailboxes. Please do not send unsolicited bulk e-mail or
unsolicited
commercial e-mail to my suespammers.org address or any of my other
addresses.
These are my opinions, not necessarily my employer's. I hate spam.

 

[Hal Murray]

The beta of the transistor will be permanently and irreversibly reduced;
yes, it *will* damage the transistor.

If you don't belive this, try it with a 2N3904 or other cheap
transistor. I didn't believe it, and tried it. I'm now a believer.

How much power/current did you put into your test case?

--
The suespammers.org mail server is located in California. So are all my
other mailboxes. Please do not send unsolicited bulk e-mail or unsolicited
commercial e-mail to my suespammers.org address or any of my other addresses.
These are my opinions, not necessarily my employer's. I hate spam.

 

[Dan Dunphy]

The key word here is simulator.
When I was using spice, there were no breakdown models. It's been 15
years, so I don't know hwat has been added to the models, lately.
Dan

On 4 Apr 2004 21:06:15 GMT, "Walter Harley"
<walterh@cafewalterNOSPAM.com> wrote:

In an oscillator simulation I was playing with, I noticed that the simulator
claimed that voltages on a 2n3055's B-E junction were reaching -40V.

To me, that seemed improbable. I somehow assumed that that the base-emitter
junction of a bipolar transistor would have a relatively weak reverse
breakdown voltage, even for a power transistor. I don't know why I assumed
that.

But searching datasheets, I can't even find anywhere where this
characteristic is described. Anybody got some info? I don't care so much
about the 2n3055 in particular, just wondering about this characteristic of
BJT's in general.

Colorado Springs, CO
My advice may be worth what you paid for it.

 

[Walter Harley]

"Al" <no.spam@here.com> wrote in message
news:no.spam-C00775.21332304042004@news.verizon.net...

By listing it as a maximum, it does not mean that you should not exceed
that value. It means that the value could be lower and still be within
spec. If you found a transitor with a maximum Veb of 7, and you measured
it to be 10, it is defective! But if it is 6.5, that's OK. A minimum may
be specified as V(br)ebo, which is the reverse breakdown voltage,
emitter to base, open circuited. Whew! And it could be, say 6.5 Vdc

I think you may not be correct about that. Veb seems to get listed as a
maximum rating, not as an electrical characteristic. That is, it's not a
specification, like Iceo(max) ("we promise there won't be more leakage than
this under rated conditions"); it's a safety limit, like Tj(max) ("if you
exceed this, all bets are off").

If the current is too large for that particular structure
and/or doping concentrations, the crystal lattice will be damaged.

That makes sense.

Thanks,
-w

 

[Oliver Betz]

"Watson A.Name - \"Watt Sun, the Dark
Remover\""<NOSPAM@dslextreme.com> wrote:

[...]

If you connect it to the same power supply through a 10K resistor,
now when it breaks down the resistor will limit the current to
(roughly) 1 mA. That probably won't destroy anything.

The beta of the transistor will be permanently and irreversibly reduced;
yes, it *will* damage the transistor.

I also would like to know at which currents you measured beta.

Own experiment with a BC548B: 10uA over two minutes caused the
collector current (at 3uA base current) to drop from 1,0mA to 0,95mA.

'Trouble Shooting Analog Circuits' writes "low current beta" without
specifying "low current".

The OP might also read the thread "Reverse biasing base-emitter
junction" starting at Message-ID
<edz67.2477$LP2.269655@bgtnsc06-news.ops.worldnet.att.net>, Google
doesn't show one of Winfield Hill's postings (he seemingly corrected a
typo in the subject) in this thread but you can find it via the MSG-ID
<3B5C9E25.114CD8C0@rowland.org>.

Or Message-ID: <36F8A917.C714EEC6@lucent.com> ff.

Also no collector current mentioned...

Oliver
--
Oliver Betz, Muenchen (oliverbetz.de)

 

[Walter Harley]

"Oliver Betz" <OBetz@despammed.com> wrote in message
news:4073cbd5.30342886@z1.oliverbetz.de...

The OP might also read the thread "Reverse biasing base-emitter
junction" starting at Message-ID
edz67.2477$LP2.269655@bgtnsc06-news.ops.worldnet.att.net>, Google
doesn't show one of Winfield Hill's postings (he seemingly corrected a
typo in the subject) in this thread but you can find it via the MSG-ID
3B5C9E25.114CD8C0@rowland.org>.

Or Message-ID: <36F8A917.C714EEC6@lucent.com> ff.

Thanks for those references! Thorough discussion, very informative.

I think I'm going to go stick a scope on the cheapo strobe oscillator
circuit that triggered the question for me... maybe I can prevent an early
death

 

[Watson A.Name - \"Watt Su]

"Hal Murray" <hmurray@suespammers.org> wrote in message
news:1072t90120jchf6@corp.supernews.com...

The beta of the transistor will be permanently and irreversibly
reduced;
yes, it *will* damage the transistor.

If you don't belive this, try it with a 2N3904 or other cheap
transistor. I didn't believe it, and tried it. I'm now a believer.

How much power/current did you put into your test case?

I don't think it makes much difference. I used just a few mA, not
enough to make the transistor even warm. After less than a minute, the
gain was much less, maybe 1/3, like originally 150, afterwards, maybe
40.

I suggest you try it, you'll be a believer. It does damage the
transistor.

--

 

[Al]

In article <1074sscgt1u1qcf@corp.supernews.com>,
"Watson A.Name - \"Watt Sun, the Dark Remover\""
<NOSPAM@dslextreme.com> wrote:

"Hal Murray" <hmurray@suespammers.org> wrote in message
news:1072t90120jchf6@corp.supernews.com...
The beta of the transistor will be permanently and irreversibly
reduced;
yes, it *will* damage the transistor.

If you don't belive this, try it with a 2N3904 or other cheap
transistor. I didn't believe it, and tried it. I'm now a believer.

How much power/current did you put into your test case?

I don't think it makes much difference. I used just a few mA, not
enough to make the transistor even warm. After less than a minute, the
gain was much less, maybe 1/3, like originally 150, afterwards, maybe
40.

I suggest you try it, you'll be a believer. It does damage the
transistor.

When you use it as a noise source, you don't care about the beta. And
you should be limiting your current to about 10 microamps.

R(limiting) = (Vcc - Veb)/10E-6 or 200K for a 9V supply and a Veb of 7.
I would use a 220K resistor.

Al

--
There's never enough time to do it right the first time.......

 

[Watson A.Name - \"Watt Su]

"Al" <no.spam@here.com> wrote in message
news:no.spam-052269.08471306042004@news.verizon.net...

In article <1074sscgt1u1qcf@corp.supernews.com>,
"Watson A.Name - \"Watt Sun, the Dark Remover\""
NOSPAM@dslextreme.com> wrote:

[snip]

When you use it as a noise source, you don't care about the beta. And
you should be limiting your current to about 10 microamps.

R(limiting) = (Vcc - Veb)/10E-6 or 200K for a 9V supply and a Veb of
7.
I would use a 220K resistor.

In my experience, the common 2N3904 and other similar transistors are
rated for 6V, and actually have a breakdown voltage of 8.5 to 9V, so it
wouldn't work with a 9V supply. You need at least 12V or so.

Al

--
There's never enough time to do it right the first time.......