Hi.

I've got a simple question... it's been a while since my last
electronics lab,
and I'd appreciate a quick help/answer.

If I put two Avalanche/Zener diodes in series, and apply reverse bias,
what would be the breakdown voltage? In other words, if it looked
like...

a ----->|------>|------ b

and I applied a reverse bias at a & b, would the total breakdown
voltage
be the sum of the two values for each diode, or be some other value?

I think it should be the sum... but, again, it's been a while... and
I'm not sure.

Thanks in advance....

Cheers,
Jay.

> I've got a simple question... it's been a while since my last
> electronics lab,
> and I'd appreciate a quick help/answer.
> If I put two Avalanche/Zener diodes in series, and apply reverse bias,
> what would be the breakdown voltage? In other words, if it looked
> like...

Twice as much!

jay.plum@gmail.com wrote:
> Hi.
>
> I've got a simple question... it's been a while since my last
> electronics lab,
> and I'd appreciate a quick help/answer.
>
> If I put two Avalanche/Zener diodes in series, and apply reverse bias,
> what would be the breakdown voltage? In other words, if it looked
> like...
>
> a ----->|------>|------ b
>
> and I applied a reverse bias at a & b, would the total breakdown
> voltage
> be the sum of the two values for each diode, or be some other value?

The short (and not very good) answer is, yes, the total
breakdown voltage is (roughly) the sum of the individual
diode breakdown voltages.

> I think it should be the sum... but, again, it's been a while... and
> I'm not sure.

The difficulties begin when the two diodes are not similar.
For instance, if you put a 3.3 and a 22 volt zener, each
with the same power rating, in series those two diodes have
very different test current ratings for their breakdown
voltage rating. The 3.3 volt zener will leak current in a
more gradually increasing ramp as voltage increases, than
the 22 volt zener will. So, as voltage rises, almost all of
it will appear across the 22 volt zener. Then as the
current approaches something close to the rated test current
for that zener, its voltage drop will begin to stabilize,
and most of any increase in the total voltage will begin to
appear across the 3.3 volt zener. Unfortunately, before the
3.3 volt zener reaches its rated test current (where its
drop is 3.3 volts), the 22 volt zener will overheat and
short out.

So if you put zeners with different voltage ratings in
series, you also need to have zeners that have power ratings
approximately proportional to their breakdown voltages, if
you want the total breakdown voltage to be equal to the sum
of the individual breakdown voltages.

On Aug 16, 10:56 am, John Popelish <jpopel...@rica.net> wrote:
> jay.p...@gmail.com wrote:
> > Hi.
>
> > I've got a simple question... it's been a while since my last
> > electronics lab,
> > and I'd appreciate a quick help/answer.
>
> > If I put two Avalanche/Zener diodes in series, and apply reverse bias,
> > what would be the breakdown voltage? In other words, if it looked
> > like...
>
> > a ----->|------>|------ b
>
> > and I applied a reverse bias at a & b, would the total breakdown
> > voltage
> > be the sum of the two values for each diode, or be some other value?
>
> The short (and not very good) answer is, yes, the total
> breakdown voltage is (roughly) the sum of the individual
> diode breakdown voltages.
>
> > I think it should be the sum... but, again, it's been a while... and
> > I'm not sure.
>
> The difficulties begin when the two diodes are not similar.
> For instance, if you put a 3.3 and a 22 volt zener, each
> with the same power rating, in series those two diodes have
> very different test current ratings for their breakdown
> voltage rating. The 3.3 volt zener will leak current in a
> more gradually increasing ramp as voltage increases, than
> the 22 volt zener will. So, as voltage rises, almost all of
> it will appear across the 22 volt zener. Then as the
> current approaches something close to the rated test current
> for that zener, its voltage drop will begin to stabilize,
> and most of any increase in the total voltage will begin to
> appear across the 3.3 volt zener. Unfortunately, before the
> 3.3 volt zener reaches its rated test current (where its
> drop is 3.3 volts), the 22 volt zener will overheat and
> short out.
>
> So if you put zeners with different voltage ratings in
> series, you also need to have zeners that have power ratings
> approximately proportional to their breakdown voltages, if
> you want the total breakdown voltage to be equal to the sum
> of the individual breakdown voltages.- Hide quoted text -
>
> - Show quoted text -

Hi, John.

Guess I'll have to give this a bit more care than I thought...

Thanks a lot for your detailed reply.
You've been a great help~!

Cheers,
Jay.

On Aug 16, 10:47 am, "A n g l e r" <p k o n i u s z @ h o t m a i l .
c o m> wrote:
> > I've got a simple question... it's been a while since my last
> > electronics lab,
> > and I'd appreciate a quick help/answer.
> > If I put two Avalanche/Zener diodes in series, and apply reverse bias,
> > what would be the breakdown voltage? In other words, if it looked
> > like...
>
> Twice as much!

Hi, A n g l e r.

Thanks a lot for your quick reply.

Cheers,
Jay.

On Thu, 16 Aug 2007 01:28:57 +0000, jay.plum wrote:

> I've got a simple question... it's been a while since my last
> electronics lab,
> and I'd appreciate a quick help/answer.
>
> If I put two Avalanche/Zener diodes in series, and apply reverse bias,
> what would be the breakdown voltage? In other words, if it looked
> like...
>
> a ----->|------>|------ b
>
> and I applied a reverse bias at a & b, would the total breakdown
> voltage
> be the sum of the two values for each diode, or be some other value?
>
> I think it should be the sum... but, again, it's been a while... and
> I'm not sure.
>
> Thanks in advance....

It will be the sum of their zener voltages.

Of course, the current will be the same, so if they're drastically
different models, the conduction could happen at different places
in the knees of the two, but that shouldn't be a problem if the
current is more than just a few mA.

Cheers!
Rich