I've been working on a GPS tracker which is intended to run/idle on a battery for an extended length of time (2+ years). The tracker uses an accelerometer (LIS3DH) and remains in a sleep state until it is awoken due to movement. While the tracker is asleep, current levels need to remain as low as possible to help preserve the battery. The accelerometer is capable of running on 2-3uA in a sleep state but can only tolerate a voltage input of 1.7 - 3.6V. As I will be powering this from a single cell LiPo (3.7V+) I need to ensure that the accelerometer VCC doesn't exceed 3.6V.

Using either a SMPS or a linear regulator would achieve this but would consume too much current to be practical. I would like to know if I can use a diode's forward voltage drop to ensure that the input voltage isn't exceeded.

In a more conventional setting where a reasonable amount of current is being passed through the diode, I assume this would work. However, I am not certain 2-3uA will suffice. Is the voltage drop of a diode tied to the level of current passing through it?

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    \$\begingroup\$ I think the forward voltage drop will be close to zero with only 2uA. If you have any on hand you should rig up a way to measure it (power supply and resistor+diode). If not, take a close look at a few datasheets to see if any of the graphs are sufficiently "zoomed in" to see that much detail. \$\endgroup\$ – mkeith Apr 19 '19 at 6:15
  • \$\begingroup\$ Are you sure this will work? How often will the accelerometer be moved? If it is moved often, won't the thing stay awake? If the GPS is de-energized, then it will have to do a cold fix every time, won't it? Are you going to store ehemeris data so you don't have to do a cold start? Etc. I know this isn't what you are asking about. I just want to make sure you have thought about this a bit. \$\endgroup\$ – mkeith Apr 19 '19 at 6:17
  • \$\begingroup\$ microchip.com/wwwproducts/en/MCP1810 \$\endgroup\$ – Bruce Abbott Apr 19 '19 at 7:02
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    \$\begingroup\$ @mkeith I managed to find a 1N4007 and coupled it with a 1M ohm resistor at 3-4V. It achieves a 0.4V drop even at 3-4uA although consumes 0.3uA itself. I suspect something like a 1N4001 will impose a greater drop. The tracker is only intended to be moved if the item it is placed within is stolen. It can be disarmed/placed back to sleep remotely. Cold lock times are within bounds. \$\endgroup\$ – Junkers Apr 19 '19 at 8:01
  • \$\begingroup\$ @Bruce Abbott Seems to have excellent Iq but Ignd is in the order of 200uA which makes it unusable in this context. \$\endgroup\$ – Junkers Apr 19 '19 at 8:04

While most silicon diodes need around 700mV to conduct 'sensible' levels of current, different diodes can have very different saturation currents, and so very different voltage drops at sub-μA levels. For instance the beefy rectifier 1N400x series can conduct a few μA at 100mV.

Some diodes are specified as 'low leakage', which means their saturation current is very low. As part of a study into low leakage protection of amplifier inputs here, I measured the low current voltage of a BAS116 diode (not particularly thoroughly). Several interesting data points are

  Vf     If
300mV   40pA
450mV   45nA
640mV   16uA    

Note that these are measurements at ambient, not specifications or over temperature, the data sheet specs are much higher. You can rest assured that 1N4148 and 1N400x conduct orders of magnitude more current at the same forward voltages.

At the time (2016), BAS116 was listed as 'until stocks exhausted', so it may have become unavailable since then. However, I expect you can find alternative diodes advertised as 'low leakage'. They are easy enough to test on a bench with a cheap meter and a bit of care. With a 10M input resistance and a 200mV input range, you should be able to resolve 10pA.

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    \$\begingroup\$ BAS116 datasheet says typical Vf at 10uA is ~0.45V. Fully charged Lipo cell is 4.20V, -0.45V diode voltage drop leaves 3.75V, above the accelerometer's maximum voltage rating. At higher temperature diode voltage drop decreases, so I think this is a poor way to reduce supply voltage. \$\endgroup\$ – Bruce Abbott Apr 19 '19 at 10:17

Typical diodes change 0.018 volts per factor-of-2 change in current. Or 0.058 volts per factor-of-10. And various silicon processes I worked with did use the standard 0.018volts/2:1.

Assume 0.6 volts Vforward at 1mA.

At 4uA, which is 1/10 * 1/10 * 1/10 * 2 * 2, we'd expect 0.6v - 3*0.058 + 2*0.018

or 0.6 - 0.174 + 0.036 = 0.6 - 0.138 = 0.462 volts across the diode at 4uA.

Lets not pretend this is precision math here. So expect about 0.4 volts.


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