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I'm looking for a diode-like component with a "resistance" (when linearized about V=0) of at least 20 Mohms at 75 deg C. The behaviour for larger voltages doesn't really matter so long as the diode is able to conduct about 50 uA by 2.5V. The capacitance must be no more than a few pF.

A regular signal diode like a 1N4048 has a linearized resistance of about 5 Mohms at room temperature, and much lower resistance at 75 deg C, so this obviously won't work. I've found a few "ultra-low leakage diodes" that could work but are quite pricey and not in the form-factor I'd like:

https://www.digikey.ca/product-detail/en/on-semiconductor/FJH1100/FJH1100-ND/1058005

https://www.centralsemi.com/content/product/bare-die/select/diodes/Ultra_Low_LeakageDiodes.php

I also found this thread suggesting that using the base-emitter junction of a BJT may be preferable:

transistor as low-leakage diode

The ideal solution would cost less than $1 in QTY=10,000 and take up no more space than an 0603 or 0805 component (so putting 10 diodes in series is probably not feasible).

Any advise?

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    \$\begingroup\$ I'm a bit confused about what you're asking. Are you talking about reverse bias current leakage? Because thats what all your links are talking about. \$\endgroup\$ – BeB00 Jan 22 '18 at 3:33
  • \$\begingroup\$ In the application, the diode will be forward biased by 10 mV or so. I need the "resistance" at this operating point to be over 20 Mohm at 75 deg C (ideally over 100 Mohm). I'm pretty sure a diode with low reverse leakage (i_s) will also have high "resistance" at a few mV of forward bias. Via the Shockley diode equation: i = i_s [E^(V/V_thermal) - 1] which linearizes about V=0 to: i = i_s / V_thermal which means that R_effective = V_thermal / i_s So a diode with a small value of i_s should have a high value of R_effective. Does that make sense? \$\endgroup\$ – Peter R Jan 22 '18 at 3:42
  • \$\begingroup\$ Don't forget that \$I_{SAT}\$ itself is HIGHLY temperature dependent. So much so that it dominates and overwhelms \$V_T\$. \$\endgroup\$ – jonk Jan 22 '18 at 4:04
  • \$\begingroup\$ Yes, Isat increases very quickly with temperature. If I can find a diode (or something) that works at the highest temperature my device needs to function at (75 deg C) then it should be fine for all lower temperatures. BTW -- how do you make the nicely-formatted equations? (I'm new here). \$\endgroup\$ – Peter R Jan 22 '18 at 4:38
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    \$\begingroup\$ @PeterR Instead of asking for advice, ask a specific question, you'll get better answers. \$\endgroup\$ – Voltage Spike Jan 22 '18 at 6:18
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Try a green LED. Big bandgap, low Is.

Edit: I just tried one for the heck of it at room temperature. Forward current of a generic 0603 green LED is around 100pA +/-50% at 1V and around 2nA at +1.2V.

Blowing controlled +180 degrees F (82 degrees C) hot air across it (no environmental chamber here to do a proper test) +0.3V conducts in the <200pA region, so in G\$\Omega\$ territory.

At least that seems promising to me for a quick test.

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  • \$\begingroup\$ Thanks. I'll try that this week. It would be a simple solution, but I have a suspicion that Is won't be low enough. Easy to test though... \$\endgroup\$ – Peter R Jan 22 '18 at 4:44
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    \$\begingroup\$ But be careful about photocurrent! \$\endgroup\$ – Janka Jan 22 '18 at 6:55
  • \$\begingroup\$ @Janka True, also true about 1N4148s to some degree. \$\endgroup\$ – Spehro Pefhany Jan 22 '18 at 14:32
  • \$\begingroup\$ The original Raspberry Pi had an uncovered voltage regulator chip on the board which tripped as soon someone photographed it with a flash. Since that moment, I think even hobbyists realize photocurrents are harmful to all ICs. \$\endgroup\$ – Janka Jan 22 '18 at 14:41

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