Diode with very low reverse leakage current?

[smpowell]

What's a good diode (preferably some jelly bean part) to use
when one needs an extremely low reverse leakage current?

The reverse voltage is normally between 1.33 and 1.50 VDC. The
diode is connected to a high resistance voltage divider that only
has 3 uA going through it. Its only function is to protect a
comparator (TS3V393C) from having a negative voltage applied
to it's inputs under non-operating conditions.

The diode (D1)is shown in the lower left hand corner of the first
schematic at:
http://octopus.freeyellow.com/slaveflash.html




Stephen Powell
Electronic Hobby Information
http://octopus.freeyellow.com/

 

[Jamie]

smpowell wrote:

What's a good diode (preferably some jelly bean part) to use
when one needs an extremely low reverse leakage current?

The reverse voltage is normally between 1.33 and 1.50 VDC. The
diode is connected to a high resistance voltage divider that only
has 3 uA going through it. Its only function is to protect a
comparator (TS3V393C) from having a negative voltage applied
to it's inputs under non-operating conditions.

The diode (D1)is shown in the lower left hand corner of the first
schematic at:
http://octopus.freeyellow.com/slaveflash.html




Stephen Powell
Electronic Hobby Information
http://octopus.freeyellow.com/

i tihnk what you mean is a low cut off voltage.

most diodes are good for reverse leakage for
basic things unless your talking about RF?
in any case i think what your looking for
with spec's like that is a germanium type
diode.

 

[Robert Monsen]

Larry Brasfield wrote:

"smpowell" <smpowell55@yahoo.com> wrote in message news:1105383574.235073.290180@f14g2000cwb.googlegroups.com...

What's a good diode (preferably some jelly bean part) to use
when one needs an extremely low reverse leakage current?


A small-junction NPN transistor, base-emitter junction can
be used to get leakage down where you appear to need it.
Tie the collector and base together.

According to Bob Pease in "Troubleshooting Analog Circuits" (page 66)
it's the base-collector diode you want to use. He suggests a 2N3707 or
2N930, although he also mentions 'some 2N3904', saying that there may be
a problem with some because of gold doping. His claim is 1pA leakage at 7V.

He also mentions JFET devices, in particular the 2N4117A and PN4117A,
which apparently have very small junctions, leading to even lower
leakage of typical 0.1pA, and guaranteed 1pA max.

Later in that chapter, he also mentions LEDs as having incredibly low
reverse leakages, but only if you keep them dark, and only reverse-bias
them up to 1V.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[John Popelish]

smpowell wrote:

What's a good diode (preferably some jelly bean part) to use
when one needs an extremely low reverse leakage current?

The reverse voltage is normally between 1.33 and 1.50 VDC. The
diode is connected to a high resistance voltage divider that only
has 3 uA going through it. Its only function is to protect a
comparator (TS3V393C) from having a negative voltage applied
to it's inputs under non-operating conditions.

The diode (D1)is shown in the lower left hand corner of the first
schematic at:
http://octopus.freeyellow.com/slaveflash.html

Stephen Powell
Electronic Hobby Information
http://octopus.freeyellow.com/

A small switching diode like a 1N4148 (can't get more jelly bean than
that) typically leaks about 25 na with 20 volts reverse bias.
http://rocky.digikey.com/WebLib/Diodes_Inc/Web%20Data/1N4148,1N4448%20Diodes.pdf
At a couple volts, I would expect the leakage to be quite a bit less.
That would drop something like 12.5 millivolts across your 500k pot.

If you can use a surface mount and are willing to go with less of a
bean, you might try the MA2J116:
http://rocky.digikey.com/WebLib/Panasonic/Web%20data/MA2J116%20(MA116).pdf
It has a typical reverse current of about 5 pA at 6 volts reverse and
20 degrees C.


--
John Popelish

 

[Larry Brasfield]

"Robert Monsen" <rcsurname@comcast.net> wrote in message news:HtednaqjO5pbkX7cRVn-rA@comcast.com...

Larry Brasfield wrote:
"smpowell" <smpowell55@yahoo.com> wrote in message news:1105383574.235073.290180@f14g2000cwb.googlegroups.com...

What's a good diode (preferably some jelly bean part) to use
when one needs an extremely low reverse leakage current?


A small-junction NPN transistor, base-emitter junction can
be used to get leakage down where you appear to need it.
Tie the collector and base together.


According to Bob Pease in "Troubleshooting Analog Circuits" (page 66) it's the base-collector diode you want to use. He suggests a
2N3707 or 2N930, although he also mentions 'some 2N3904', saying that there may be a problem with some because of gold doping. His
claim is 1pA leakage at 7V.

With all due respect for Mr. Pease, I'm going to stick my
neck out and disagree. Most of the leakage of a diode comes
from surface leakage and thermally generated carriers in the
depletion region. For a normally built BJT, the B-E junction
has less exposed surface than the B-C junction and also has
a smaller volume depletion region, being both narrower and
occupying somewhat less area. So, unless one requires a
reverse breakdown higher than the several volts available
from a B-E junction, it is the better choice. (I am willing to
be proven wrong on this, but it will take some evidence.)
I suspect Mr. Pease's advice was directed toward the case
where more than 3 to 6 Volts of reverse standoff is needed.

He also mentions JFET devices, in particular the 2N4117A and PN4117A, which apparently have very small junctions, leading to even
lower leakage of typical 0.1pA, and guaranteed 1pA max.

I suppose those might be considered the "jelly bean"
part the OP requested. I was trying to not overdo
the low leakage requirement because the circuit in
question can tolerate a few nA harmlessly.

Later in that chapter, he also mentions LEDs as having incredibly low reverse leakages, but only if you keep them dark, and only
reverse-bias them up to 1V.

I almost mentioned that, but the LED forward drop would
make it useless in the OP's circuit. There, the diode is
supposed to conduct so that a "protected" circuit's input
protection network does not have to source a few uA
when gently pulled below ground.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Larry Brasfield]

"smpowell" <smpowell55@yahoo.com> wrote in message news:1105390589.097340.257520@f14g2000cwb.googlegroups.com...

The reverse voltage is normally between 1.33 and 1.50 VDC. The
diode is connected to a high resistance voltage divider that only
has 3 uA going through it. Its only function is to protect a
comparator (TS3V393C) from having a negative voltage applied
to it's inputs under non-operating conditions.


In that circuit, I see no need for the diode. The current is strictly
limited, and the comparator surely has its own input protection.
Why do you believe more is necessary? Is there an exremely
high voltage on the other end of that 100M resistor?

The other end is connected to a capacitor at -300 VDC, the other end
is connected to Vcc which is normally at +4-6 VDC, except if the
battery is disconnected while the flash's capacitor is still charged,
in which case, the -300 VDC is going to force the 3 uA into something.

My guess is that the Vcc rail would go negative with respect to ground;

Agreed.

as to whether the IC would withstand having 3 uA forced through it the
wrong way I don't have clue.

You can find a few clues in the spec sheet:
http://www.st.com/stonline/books/pdf/docs/4070.pdf

Could some familiar with the inner workings of the TS393C care to
speculate about this?

No need to speculate. The manufacturer's schematic shows
protection diodes to each rail from each input. There is no
harm in asking those diodes to conduct a few uA. They are
rated to take 50 mA, so your few uA are not abusive.

Hmm, would putting a reverse diode between Vcc & ground suffice?

Not necessary.

(Keep in mind that I learned most of my what I know about
electronics from Forest Mims's books!)

Never heard of him.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

Ken Smith wrote:

In article <HtednaqjO5pbkX7cRVn-rA@comcast.com>,
Robert Monsen <rcsurname@comcast.net> wrote:
[...]
He also mentions JFET devices, in particular the 2N4117A and
PN4117A,
which apparently have very small junctions, leading to even lower
leakage of typical 0.1pA, and guaranteed 1pA max.

You can get "low leakage diodes" that are in fact JFET die in a two
lead
package.

Those would be the Vishay (Siliconix) PAD (TO-18b) and JPAD (TO-92)
parts. Farnell stocks the PAD5 (5pA) and JPAD50 (50pA)- the former
costs a couple of dollars in small quantities, and the latter around
$1. There is a PAD1 part.

http://www.vishay.com/docs/70339/70339.pdf
-------
Bill Sloman, Nijmegen

 

[Fred Bloggs]

smpowell wrote:

The reverse voltage is normally between 1.33 and 1.50 VDC. The
diode is connected to a high resistance voltage divider that only
has 3 uA going through it. Its only function is to protect a
comparator (TS3V393C) from having a negative voltage applied
to it's inputs under non-operating conditions.



In that circuit, I see no need for the diode. The current is strictly
limited, and the comparator surely has its own input protection.
Why do you believe more is necessary? Is there an exremely
high voltage on the other end of that 100M resistor?


The other end is connected to a capacitor at -300 VDC, the other end
is connected to Vcc which is normally at +4-6 VDC, except if the
battery is disconnected while the flash's capacitor is still charged,
in which case, the -300 VDC is going to force the 3 uA into something.

My guess is that the Vcc rail would go negative with respect to ground;
as to whether the IC would withstand having 3 uA forced through it the
wrong way I don't have clue.

Could some familiar with the inner workings of the TS393C care to
speculate about this?

The diode is not needed. As you say , the flash capacitor is referenced
to the BATT(+) rail. When the battery is removed, the capacitor will
discharge through the BATT(+) rail, R1 , R2, and the 100M. The
comparator terminals will be no more than 500K/100Mx300V= -1.5V maximum
relative to the GND terminal. Since the supply voltage across the
capacitor will be only a few tens of mV, this will not damage the IC,
any substrate conduction will be limited to 3uA. If you reduce R2 to 50K
, R5=1M ,R3=27K, R4=1M, then you retain the same adjustment range while
reducing the maximum reverse discharge voltage to -150mV.