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I'm developing a very very low power system, and it has some components that need 3.3V. These components are actually spec'd to operate anywhere between ~1.8V and 3.7V, so theres some wiggle room there. Most of the time, these components will be taking in the range of uA's, maybe even under 1uA. The catch is, sometimes they will need up to 30mA. The power supply will be a 7.2V lithium something battery, which obviously needs to be regulated down. I imagine the voltage of this will swing between about 7.9V to say 6.9V (obviously will check).

Since its difficult/expensive to find a buck converter that a) has a low part count, b)has a quiescent current low enough to make it worthwhile, and c) can work down at the uA level but still supply those 30mA bursts, I'm thinking about just using diodes.

If I just chain a load of diodes (say 6 or 7) in series, that will bring the voltage down to what I need. Obviously the actual voltage will vary somewhat with current, although I imagine not that much when the max current is only 30mA. However, I am a bit nervous because I've never seen this being an accepted design. Can anyone see any issues with this? The main benefit is that there is no extra quiescent current draw, and its super simple and cheap. Obviously half the power is being wasted, but that's only the same as adding like another 2 or 3uA on average, which is great.

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  • \$\begingroup\$ how much time do the 30mA bursts last? \$\endgroup\$
    – berto
    Commented Mar 29, 2016 at 2:15
  • \$\begingroup\$ Your duty cycle must be really low if the linear losses equate to only 3uA aditional leakage. I had a similar project with a duty cycle of ca. 1s per minute, it was still worth it to use a low quiescent current switcher from TI. \$\endgroup\$
    – svens
    Commented Mar 29, 2016 at 2:25
  • \$\begingroup\$ This is a wireless alarm sensor, so the 30ma is a for a radio transmission when the sensor is activated. It lasts about 10ms to 100ms, and the sensor might only be activated once per day (or less or more). \$\endgroup\$
    – BeB00
    Commented Mar 29, 2016 at 2:37
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    \$\begingroup\$ LM317 plus 2 resistors has got to be cheaper than your chain of 6 or 7 diodes. Or look for a lower-Iq linear regulator if the 100 uA adjust-pin current of LM317 is a problem. \$\endgroup\$
    – The Photon
    Commented Mar 29, 2016 at 2:49
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    \$\begingroup\$ TLV704, maybe? Iq=3uA, output regulation spec'ed down to 1 mA (just set the nominal to 3.0 V instead of 3.3 and you should be good). The main issue with diodes is the forward voltage could actually vary quite a bit when current goes from 1 uA to 30 mA. \$\endgroup\$
    – The Photon
    Commented Mar 29, 2016 at 3:30

3 Answers 3

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The problem with a string of diodes as a voltage regulator for microamp current levels can be found in the IV curve of pretty much any diode - there is very little, if any, voltage drop when you are only drawing microamps. So you're effectively unregulated in that range, and then the voltage drops kick in when you actually need power.

Forward voltage at low current

I suppose you might be able to choose a resistor based on the actual quiescent current draw to hit 3.5 V and add a large enough capacitor on the circuit side that 100ms of 30ma will only drop 1.5V or so (about 2000 uF, if I didn't fudge the calculation). That would take a long time to start up, of course, while the capacitor charged. I guess you could have a voltage regulator tied to a "Push after changing battery" button to get it charged the first time. Either that, or figure some way to switch on a real regulator only when the radio needs to transmit.

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  • \$\begingroup\$ Yup that basically invalidates the whole idea :/. I'm not sure why I assumed diodes were more stable at lower currents, even though if you just think about the IV curve for 5 seconds, that can't be true. Having a button isn't really feasible here, and a resistor would waste too much power anyway, so I guess I'll probably have to go with a voltage regulator. \$\endgroup\$
    – BeB00
    Commented Mar 29, 2016 at 7:47
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Consider using a basic linear low quiescent current regulator. You probably know that linear current regulators basically "burn" the excess voltage which can be a problem when stepping down from a higher voltage like 7V, but remember that the power dissipated depends on the current (P=IV) which in your case is extremely low. At 5uA, you'll be burning (7V-3.3V)*5uA or 18.5 micro watts.

I've seen voltage regulators with extremely small quiescent currents, in the < 10uA range, and even those could supply 100mA continuously.

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    \$\begingroup\$ See, for instance, onsemi.com/pub_link/Collateral/MC78FC00-D.PDF \$\endgroup\$
    – WhatRoughBeast
    Commented Mar 29, 2016 at 3:42
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    \$\begingroup\$ The TPS783xx has some pretty astounding performance characteristics. ti.com/lit/gpn/TPS783 : 500nA Iq, 150mA rated output. \$\endgroup\$
    – Peter Smith
    Commented Mar 29, 2016 at 8:45
  • \$\begingroup\$ Suggesting a regulator with a 5uA quiescent current seems insanely wasteful in a device that has an average current draw of less than that. You'd basically be halving the potential battery life (if avg was 5uA). \$\endgroup\$
    – MattCochrane
    Commented Oct 16, 2016 at 7:00
  • \$\begingroup\$ Just to add, this answer was the most useful, but techically didnt answer the question, (but I did upvote it), which is why i accepted the other answer. \$\endgroup\$
    – BeB00
    Commented Apr 14, 2017 at 16:06
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Well, why use 2 cells in series? Your circuit can operate from only one cell with only a low-quiescent micropower CMOS LDO.

So, remove one cell.

If you need more battery life, put two cells in parallel.

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  • \$\begingroup\$ The minimum (end) voltage of a single cell is around 1.2V. This is below the required minimum 1.8V of the device. Furthermore, a regulator would only make things worse by 1) drawing extra current and 2) dropping voltage which is already too low. \$\endgroup\$
    – neonzeon
    Commented Apr 13, 2017 at 16:50
  • \$\begingroup\$ Oh, I mixed it up with LiIon (where the end of discharge voltage would still work) since he mentioned a "lithium-something battery"... More info about the battery would be useful btw... unless your 1.2V figure applies to his "lithium-something battery"? \$\endgroup\$
    – bobflux
    Commented Apr 13, 2017 at 17:04
  • \$\begingroup\$ Lithium-ion cells would not work well in this application because they have a self discharge rate of 2-3% per month. \$\endgroup\$
    – neonzeon
    Commented Apr 13, 2017 at 17:48
  • \$\begingroup\$ Yeah, plus it makes no sense to have rechargeables in something drawing µAmps. But I can't find any "lithium-something" cells that would make the OP's 7.2V pack. My other hypothesis was the most common (Li/MnO2) which is 3V nominal and discharges down to 2V, which makes a single cell suitable for the OP's circuit. So, another option would be Li/FeS2 1.5V AAs, which discharge down to 0.8V... so you'd need several in series. None of these make his 7V2 pack though...... BTW, which chemistry did you have in mind for your 1V2 discharge voltage?... \$\endgroup\$
    – bobflux
    Commented Apr 13, 2017 at 18:21
  • \$\begingroup\$ The voltage requirement was stated by the OP as 1.8V-3.7V. The single cell end voltage of Li/FeS2 is close to 1V. Which is one of the reasons I suggested using a pair of them. You see, Li/FeS2 is lithium iron disulfide. \$\endgroup\$
    – neonzeon
    Commented Apr 13, 2017 at 18:50

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