forward leakage of a diode ...

[free_electron]

I'm having trouble finding the forward leakage of a diode. nobody seems to specify that ...

for example : if on a 1n4148 ( standard silicon diode ) i put 0.1 volts : how much will it leak ( theoretically nothing since it is below the 0.7 volts , but in practice it will leak due to impurities )

here's my problem : i need to protect a high impedance 1 volt input. my idea was to put two or three silicon diodes in series ( 1.2 volts .. 2.1 volts )  ( the chip can handle 2.7 volts , but the signal to be measured is guaranteed never to be above 1 volt , besides the a/d reference is 1.024 volts ...) so i thought to clamp the signal , past the input series resistor ( there is a 1 meg resistor from the input to the sensing amplifier. sticking the two diodes there would solve the problem.
problem is the diodes have an impact on the signal to be measured ! they leak...

finding TVS diodes that go that low is not doable... zeners that low are simply stacked silicon diodes.

and no manufacturer seems to specify the leakage current in the 'non conductive' zone of a diode. trial and error... i don't feel like measuring this on a bunch of diodes and find what works best.

i may have to do the trick keithey uses in their electrometers : a pair of b-e junctions in antiparallel ( but stacked )

[duak]

I recall that either Bob Pease or Jim Williams had some info on the forward characteristics of various types of diodes.

The 1N4148/1N914 diodes are poor choices for this application because they are gold-doped to increase switching speed which has the side effect of increasing the forward current at low voltages.  Being in glass cases also makes them photosensitive.  It's better to use a bipolar transistor instead but the higher junction capacitance might be a problem.  There are low leakage non-gold doped diodes available too but no numbers come to mind.

Cheers,

[edavid]

I'm having trouble finding the forward leakage of a diode. nobody seems to specify that ...

for example : if on a 1n4148 ( standard silicon diode ) i put 0.1 volts : how much will it leak ( theoretically nothing since it is below the 0.7 volts , but in practice it will leak due to impurities )

No, it is not theoretically zero.  The junction equation is continuous from 0V up.

For some measurements, see The Art of Electronics 3rd ed figure 5.2

The winners are the same as for reverse leakage: 1N3595, 2N3904 BE junction, diode connected PN4117A

[T3sl4co1l]

No, it is not theoretically zero.  The junction equation is continuous from 0V up.

Indeed, continuous from some -V (not quite towards breakdown*) to +V.  The Shockley diode equation is:
\[i_f = i_s \left( e^{\dfrac{v_f}{\eta v_{th}}} - 1 \right) \]
Under forward bias, the -1 is negligible, but near zero, it is critical.  Simply put: smaller \$ \eta \$ and smaller \$ i_s \$ is all you need.

*IIRC, SPICE matches the avalanche curve at some point (piecewise continuous, or continuous derivative), then sets the exponent to cross through (BV, IBV).  Or you could use a TANH function to interpolate between avalanche and normal conduction (thus getting all continuous derivatives).  I don't know what's more physically realistic, but it probably doesn't matter enough to tell.

Tim

[fcb]

I use a pair of 1N3595 in one of my designs.  I seem to remember them being mentioned in the Keithley low level handbook, 1pA rings a bell.

[Leo Bodnar]

I'm having trouble finding the forward leakage of a diode. nobody seems to specify that ...

for example : if on a 1n4148 ( standard silicon diode ) i put 0.1 volts : how much will it leak ( theoretically nothing since it is below the 0.7 volts , but in practice it will leak due to impurities )

Are you taking the piss or just gently trolling?
I was just about to buy an electric car...
Leo

p.s. I got 33nA from https://www.diodes.com/assets/Datasheets/ds12019.pdf

[Tomorokoshi]

What is the bandwidth of the signal?

[nctnico]

I'm having trouble finding the forward leakage of a diode. nobody seems to specify that ...

for example : if on a 1n4148 ( standard silicon diode ) i put 0.1 volts : how much will it leak ( theoretically nothing since it is below the 0.7 volts , but in practice it will leak due to impurities )

Most diodes' datasheet has a forward current versus voltage graph so there is your answer.

[ejeffrey]

The low forward voltage behavior can be modeled  by the shockley diode equation, with the caveat that the ideality parameter eta is basically never specified in datasheets, although you may be able to estimate it from the provided parameters, or just guess that it is about 2.

Note that you can also calculate the small signal impedance of a diode near DC by simply taking the derivative of the diode equation.  You get: dvf / dif = \eta v_{th} / i_s.  You may be shocked to find values < 1 megaohm even at zero DC voltage for common signal diodes.  That is at zero DC voltage!

If you need high impedance: use an ultra-low leakage diode such as the FJH1100, or use a transistor.  Alternately, you can use an LED -- they have orders of magnitude lower leakage than the best silicon diodes but have a correspondingly higher forward conducting voltage. Make sure you completely black it out to avoid photocurrent!  If you only care about low frequency you should probably bootstrap the diode by providing a buffered copy of the input to the other terminal of the diode.  That will keep the DC voltage across the diode very close to zero as long as the signal is in range.

[floobydust]

I see diode-connected transistors used when you need low leakage.
Temperature is very important, if your product will be in a hot environment. I believe leakage is log temperature.

[nctnico]

For the problem at hand I'm wondering what the chip's ESD diodes can handle and what the required bandwidth is. The solution may be as simple as sticking a series resistor at the input which limits the current into the ESD diodes which do exactly what you want: clamping without distorting the signals. As an extra measure you could put a zener diode or shunt regulator across the power rails to sink the current.

[TimFox]

In general, if you are trying to protect an input driven from a substantial source impedance (where the diode "leakage" current is important), you should establish a voltage with a suitable diode network to ground, biased from a resistor to the power supply (not much bias current required) and then connect a low-leakage diode from the input node to that bias voltage.  Depending on the voltage, this can be series-connected PN or Schottky diodes, or a suitable Zener.
I have used the base-collector diode of low-noise audio transistors (2N2484 NPN or 2N4250 PNP) and the gate-channel (source and drain connected together) of low-current JFETs to get low leakage current in the reverse-bias direction before the safety circuit conducts (with the low-leakage diode in the forward-bias direction).  When the input is very positive, the current will flow through the forward-biased diode to the diode network, limited by the source resistance to keep the diodes from smoking.

[free_electron]

I'm having trouble finding the forward leakage of a diode. nobody seems to specify that ...

for example : if on a 1n4148 ( standard silicon diode ) i put 0.1 volts : how much will it leak ( theoretically nothing since it is below the 0.7 volts , but in practice it will leak due to impurities )

Most diodes' datasheet has a forward current versus voltage graph so there is your answer.

except that curve is unreadable. i am interested in the 1millivolt,2 millivolt up to 0.4 volt range ... ( well below the knee of the diode )


What i'm trying to do is the following : i am sensing a small current through a sense resistor. i want to clamp the voltage across the sense resistor to maximum 1.2 tot 2.1 volts. i want the current to flow through the sense resistor, not the diodes
The sense resistor is in the order of  1 megohm. ( the source impedance is high and i can't change it .)

i want to protect the input from ESD events and nitwits connecting it directly to a voltage source. so i put a 100k series resistor and clamp it using a few diodes. ( it's a current source so series resistance doesn't matter )
i made a typo in the drawing. current is max 1 micro amp. so 0 1 microamp gives me nicely 0 to 1 volt which i can pick off using a nice unitiy gain instrumentation amplifier with a few pA input bias . (i could use gain in the amplifier but prefer not to ... )

signals are DC ...

i don't want any current going through the protection diodes (until the protection needs ot activate of course ) . any current loss there gives a measurement error. ideal diodes (0.6 volt forward voltage ) will work since they won;t trip until 3x0.6 is reached. clipping my input at 1.8 volts. but in the real world diodes even draw current at 0.1 volt ...

So i am trying to find a really good diode with very low leakage until it is supposed to conduct. And datasheets are very lacking in the 'non-conducting' area of the curves ...
Or maybe i'm going about this the wrong way.

I was thinking about this after watching the teardown of th electrometer . they use b-e junctions to do the protection. so i thought : that would solve it. problems is nobody specs that stuff...

[free_electron]

I'm having trouble finding the forward leakage of a diode. nobody seems to specify that ...

for example : if on a 1n4148 ( standard silicon diode ) i put 0.1 volts : how much will it leak ( theoretically nothing since it is below the 0.7 volts , but in practice it will leak due to impurities )

Are you taking the piss or just gently trolling?
I was just about to buy an electric car...
Leo

p.s. I got 33nA from https://www.diodes.com/assets/Datasheets/ds12019.pdf

how did you get to that number ? i don't see that anywhere in the datasheet.

[edavid]

If you use an op amp current to voltage converter, the voltage at the sensing node is zero, so there is less of a leakage issue.

[Leo Bodnar]

So i am trying to find a really good diode with very low leakage until it is supposed to conduct. And datasheets are very lacking in the 'non-conducting' area of the curves ...

Diode is not a relay or a fuse.
Datasheets are written with expectations that you roughly know what is happening outside the measured graphs.  They usually only mention things that you do not expect the part to do.
It sounds harsh but the best you can do is take a break from your design, honestly admit to yourself that you don't understand how semiconductor diode works and read a good electronics engineering college coursebook or TAOE. They have much better talent at teaching than a random internet forum.

But if you still demand the secret curves - here they are.
Pease says ~60nA for 1N4148 at 0.1V

[free_electron]

It sounds harsh but the best you can do is take a break from your design, honestly admit to yourself that you don't understand how semiconductor diode works

i'm not even going to go there ... i don't need a theory lecture. i wanted to see the Vf curves at low currents and find a diode i can buy at digikey that has the lowest If for the highest Vf

But if you still demand the secret curves - here they are.

That's what i wanted to see. i didn't want to do that experiment. It figured that Pease of all people would have done that.
So a yellow led will do the trick. up to 1 volt it draws less than 1 nanoamp.
Now to find a nice smd yellow LED that i can put in a dark enclosure

[agaelema]

Very interesting topic.

I'm reading some pages and is common the use of 2N3904 as low leakage diode using BE junction (if the voltage is between the limits) or BC junction, but generally reverse biased, not forward biased as the diodes in your circuit.
- https://www.electronicspoint.com/threads/diode-with-very-low-reverse-leakage-current.26296/
- https://electronics.stackexchange.com/questions/16966/transistor-as-low-leakage-diode

Maybe can you use the BE/BC junction conected between the input and an auxiliar supply (another regulador or zener) of 1.2V. In a normal condition the diode will reverse biased (like in the picture).

[David Hess]

LEDs painted black also make good low leakage diodes but of course they have a low peak reverse voltage so not really any better than a transistor base-emitter junction.

If the signal is low frequency, then the diode clamps can be bootstrapped to follow the input voltage reducing leakage further and then a second second of diode clamps used to clamp the bootstrap output.  I have seen this done in several HP designs and it is also used now on very low input current CMOS operational amplifiers.

Bob Pease discusses this subject in his book.

[free_electron]

Unfortunately those things (biased or bootstrapped clamps don't work in my case. i don't have aux supplies. My system runs off a coin cell (cr2032) and has to run for 2 years.

All i want is clamp diode that prevents the opamp input to get more than roughly 1.2 volts across it. ( the a/d power rail is 1.8 volts. and i want to protect the input from large incoming voltages such as esd spikes , metallic noise on the cable etc.
The main CPU shuts down the analog front end when not sampling to preserve power so the clamp needs to still work .... the opamp is the first active element. i need to protect that as it sees the two wires coming from the sensor (the big bad outside world). my idea is : stick a series resistor in the loop ( it is current anyway , so series resistance doesn't matter ) . then clamp it. but : the clamp is not allowed to drain anything until its trip point. neither from the loop (as that falsifies my reading ) , nor from the system power (i don't have the power budget )

Classic zeners don't work. A 1.8 volt zener leaks 1uA at 1 volt ... i only have 1 uA coming in ....
TVS diodes are even worse. a 1.8 volt TVS diode( trip voltage : 1.5 volts ) sucks 10uA at 1 volt ... well below her trip point.
Specialised clamp diodes begin at 2.5 volt but they pull 50nA at 2 volt ( before they conduct ). their behavior is not specified (no curves from 0 volt to their snap point ) , and even if such curve existed it will be non linear. i have a 10 bit A/D.
So i can digitize 1 nA. that diode would already throw my system off by 50nA ... not good.

The system is basically a coin cell operated LoRaWan node that samples the current coming from an electrochemical cell. That cell is a current output (0..1uA into 1Mohm giving 0..1volts) that is being digitized. I want to protect that 1Mohm input by placing a protection diode across it. Since we are dealing with a current mode output : anything i place in parallel with the 1Megohm falsifies my reading. Especially if its behavior is non-linear ... classic diodes leak too much in forward mode. (  iwas thinking of stacking two in series ... 2x0.7 volts .. but diodes also conduct at 0.1 volt and 0.2 volt ... i can't have that ( at least not if they conduct microamps ... )

I didn't feel like doing the legwork of spending a whole day playing with various diodes to find out which one leaks the smallest current.


Fortunately Bob Pease has done this already ... yellow LED. 1nA at 1.2 volts. perfect. well within the noise of my A/D.
Now i have a starting point to see if it will work ( i'll build it on bench and measure it. i have keithley electrometers en picoampere meters ) so i can sense the exact voltage across my 1meg/led parallel circuit and measure the exact current through the combination.

[agaelema]

Unfortunately those things (biased or bootstrapped clamps don't work in my case. i don't have aux supplies. My system runs off a coin cell (cr2032) and has to run for 2 years.

All i want is clamp diode that prevents the opamp input to get more than roughly 1.2 volts across it. ( the a/d power rail is 1.8 volts. and i want to protect the input from large incoming voltages such as esd spikes , metallic noise on the cable etc.

Why not?

If a new source is a problem, connect the BE junction directly to positive pole of battery (or the 1.8V of A/D rail that you cited). The voltage at the input can exceed the rail by Vbe and the OpAmp output will saturate. To protect the A/D you can put a serie resistor to limit the current and another diode connected between the input of A/D and the 1.8V rail.

A 2n3904 BE junction (second the one of the links) has 1pA of reverse leakage, much less then 1nA of yellow led.

[free_electron]

Unfortunately those things (biased or bootstrapped clamps don't work in my case. i don't have aux supplies. My system runs off a coin cell (cr2032) and has to run for 2 years.

All i want is clamp diode that prevents the opamp input to get more than roughly 1.2 volts across it. ( the a/d power rail is 1.8 volts. and i want to protect the input from large incoming voltages such as esd spikes , metallic noise on the cable etc.

Why not?

If a new source is a problem, connect the BE junction directly to positive pole of battery (or the 1.8V of A/D rail that you cited). The voltage at the input can exceed the rail by Vbe and the OpAmp output will saturate. To protect the A/D you can put a serie resistor to limit the current and another diode connected between the input of A/D and the 1.8V rail.

A 2n3904 BE junction (second the one of the links) has 1pA of reverse leakage, much less then 1nA of yellow led.

problems : The power rails are off when not sampling ...  so protection would not work (the bias voltage goes to zero so the diode would turn on... which would pulls current from the sensor, a current which is larger than the 1uA. which is a nono... the more current i pull the faster the sensor ages. it's complicated. this thing is a based on an electrochemical reaction. draw more current and the reaction speeds up : eating away the electrodes in the sensor...

Unfortunately i'm not a liberty to give more info on the sensor. Suffice to say it is made for a max current of 1uA into a 1M sense resistor. if you lower the resistor value then the current delivered by the sensor will increase and that ages the sensor. It is also not allowed to send current into the sensor as that destroys one of the electrodes due to an electromigration process.

This bloody thing is made for a 1meg load and gives 1 volt at full reading ( actually 0.98 volt). you can lower the resistor but that comes at the price of a shorter lifespan. There is some passive circuitry inside this sensor in the form of a resistor bridge. i don't have acces to this as the whole thing is potted... otherwise i would stick my signal conditioning in there ... and clamp in a low impedance domain  but that doesnt work.


normally the sense amplifier is mounted directly on the sensor. in my case there is a few meters of wire ... so i want a esd clamp there ...
i already verified that cutting the power rail of the opamp does not change the input current draw of the opamp. it is safe to shut down the entire input path without causing additional load to the sensor.

anyway. this is dancing around the subject. all i want to find is a diode that does not leak in forward mode.

[David Hess]

Unfortunately those things (biased or bootstrapped clamps don't work in my case. i don't have aux supplies. My system runs off a coin cell (cr2032) and has to run for 2 years.

Bootstrapping the input protection diodes does not require a separate or higher voltage supply but the buffers do draw more power.

All i want is clamp diode that prevents the opamp input to get more than roughly 1.2 volts across it. ( the a/d power rail is 1.8 volts. and i want to protect the input from large incoming voltages such as esd spikes , metallic noise on the cable etc.

What kind of operational amplifier are we talking about?

Obviously it has low input bias current but bipolar input parts (1) include back-to-back diodes across their inputs to prevent base-emitter breakdown from high differential voltages if that is a problem.

With a 3 volt supply voltage, protecting each input with a pair of low leakage diodes like the mentioned 2N4117 JFET or a 2N3904 base-collector junction which go to the supply voltages is usually sufficient if the operational amplifier's internal protection diodes are not.  Add resistors at the inputs and also between the protection diodes and the operational amplifier inputs.  The later resistors also isolate the capacitance of the diodes from the inverting input which would otherwise reduce phase margin.  Do not make the resistors any larger than necessary.

problems : The power rails are off when not sampling ...  so protection would not work (the bias voltage goes to zero so the diode would turn on... which would pulls current from the sensor, a current which is larger than the 1uA. which is a nono... the more current i pull the faster the sensor ages. it's complicated. this thing is a based on an electrochemical reaction. draw more current and the reaction speeds up : eating away the electrodes in the sensor...

Unfortunately i'm not a liberty to give more info on the sensor. Suffice to say it is made for a max current of 1uA into a 1M sense resistor. if you lower the resistor value then the current delivered by the sensor will increase and that ages the sensor. It is also not allowed to send current into the sensor as that destroys one of the electrodes due to an electromigration process.

This bloody thing is made for a 1meg load and gives 1 volt at full reading ( actually 0.98 volt). you can lower the resistor but that comes at the price of a shorter lifespan. There is some passive circuitry inside this sensor in the form of a resistor bridge. i don't have acces to this as the whole thing is potted... otherwise i would stick my signal conditioning in there ... and clamp in a low impedance domain  but that doesnt work.

That sounds a lot like a standard cell.  They have similar requirements.  I am not recommending you do this but the Linear Technology LT1012, LT1008, and LT1097 datasheets show an example where a depletion mode MOSFET with its gate tied to the positive supply is used to disconnect a standard cell from the operational amplifier input so removing power does not damage the cell.  This would be difficult to do in a 3 volt circuit.

When shutting off the power to the operational amplifier, why not leave the input protection diodes connected to power?  They will be reverse biased anyway.

i already verified that cutting the power rail of the opamp does not change the input current draw of the opamp. it is safe to shut down the entire input path without causing additional load to the sensor.

That is a little odd.  Most CMOS operational amplifiers have protection diodes between the inputs and the supply pins so if the supply voltage drops, the inputs end up shorted to the supply pins.  The high series input resistance solves this problem but it also solves the problem for the external clamp diodes anyway.

(1) Super beta input bipolar operational amplifiers are suitable for megohm input resistance circuits and have the advantage over many low input current JFET and MOSFET parts of not having their input bias current double every 10C which is important if high temperature operation is a consideration.  If you are running on a 3 volt coin cell however, they are probably not suitable do to higher supply current unless the amplifier is being shutoff.  They do have the advantage of higher precision and lower noise however.

[ejeffrey]

Most diodes' datasheet has a forward current versus voltage graph so there is your answer.

except that curve is unreadable. i am interested in the 1millivolt,2 millivolt up to 0.4 volt range ... ( well below the knee of the diode )

Doesn't matter.  There is actually no such thing as a "diode knee".   The current is exponential following the shockley diode law that multiple people have posted about, there is no threshold.  0.6 V is just a convenient reference point.  The IV curve flattens out at high current due to self heating and the bulk resistance in series with the junction, but you can extrapolate from the data sheet values down to much lower currents.  For instance, the 4148: https://www.diodes.com/assets/Datasheets/ds12019.pdf you can see that at low currents, the IV curve is a straight line on the log log plot.

Another example, the ultra low leakage FJH1100: https://www.fairchildsemi.com/datasheets/FJ/FJH1100.pdf

You can read off from the table that the voltage drop is about 80 mV / decade at low currents(corresponds to an ideality parameter around 1.3).  So you will get 1 nanoamp at ~300 mV forward bias, and 1 picoamp at around ~50 mV forward voltage.

So you could put 4 of these in series if you wanted to get about 1 nanoamp @ 1200 mV.  But the LED solution is much better (and several orders of magnitude cheaper!)

The curves/ratings are usually at 25 C.  The scale current is exponential in temperature, so if your device has to operate at 35 C, the forward current will be much higher at the same voltage.

[T3sl4co1l]

i'm not even going to go there ... i don't need a theory lecture. i wanted to see the Vf curves at low currents and find a diode i can buy at digikey that has the lowest If for the highest Vf

Then give up now. They don't exist.  I don't know of any manufacturer that makes promises about current flow at low voltages.

You must solve this problem either by:
1. Give up. Assume that most diodes don't leak much down there, and just go with it.  Accept the consequences.
2. Measure all incoming diodes (as a spot or sample check, or up to 100% test, it's up to you) to confirm their properties in this application.  Accept the monstrous increase in unit cost.

Tim