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How to drop the voltage by a low value (let's say <150mV) without wasting current to supply a microcontroller that spend most of its time in deep sleep mode?

I'm working on a board with a microcontroller that will accept up to 3.6V as supply voltage but LiFePO4 and Li/SOCl2 can go slightly higher when fully charged (or during charge for LiFePO4). The microcontroller uses ~10μA during deep sleep and can use up to 500mA (mostly spikes) when active and transmitting over WiFi (ESP32).

I know that I can use an LDO with a low quiescent current but I am curious if there are better/cheaper alternatives. For example, I thought about using the drop voltage of a Schottky diode.

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    \$\begingroup\$ This can't reasonably be answered without information about the minimum/maximum current draw of your circuit. You could also consider using a different microcontroller. Probably an LDO is your best bet. \$\endgroup\$ – Spehro Pefhany May 21 at 2:05
  • \$\begingroup\$ @SpehroPefhany I edited my question. \$\endgroup\$ – DurandA May 21 at 10:26
  • \$\begingroup\$ The MAX 600x family of shunt regulators has a 1uA typical 0.5 uA min operating (Cathode) current. datasheets.maximintegrated.com/en/ds/MAX6006A-MAX6009B.pdf Two of appropriate voltage in series may allow you low drain mode to work OK, switching to another mode on throttle-up. || A S1318 family regulator can have 55 nA drain, 0.054V dropout [!!!] and bearable cost. (About $US1) | There are "any number" of regulator rated at 100 mA output and well under 0.1V dropout and sub uA Iq. | \$\endgroup\$ – Russell McMahon May 22 at 7:54
  • \$\begingroup\$ 5 x 100 mA capable regulators in parallel would give you 500 mA capability at about 0.5 uA iQ and about $3 cost. \$\endgroup\$ – Russell McMahon May 22 at 7:54
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Another possibility would be to have two independent connections to the supply voltage, one being used for low power mode and another one for a high power mode according to the following:

Right before driving the MCU into deep sleep mode, the active low pin _ACTIVE can be disabled (high), thus turning the PFET \$M1\$ off. This means, that the MCU will be supplied by the highly ohmic path formed by \$R_2\$ and \$R_1\$. This voltage devider must be designed to simultaneously provide a voltage smaller or equal than \$3.6V\$, and a current of approximately \$20\mu A\$ for example (just enough to keep the MCU up and running). Once you want to wake up the MCU, the pin _ACTIVE can be enabled (low) and the PFET will be turned on. Since it has a very low \$R_{DS,ON}\$, the highly ohmic path will be bypassed. The additional voltage drop that you require can be achieved by selecting a schottky diode with an appropriate forward voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

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Both of those batteries (more the LiFePO4 than the Li/SOCl2) have useful charge below the 3.3V required for a microcontroller. If you were hoping to squeeze the absolute maximum amount of battery life possible, you could use a buck-boost converter. They exist in small form factors and high efficiencies. Otherwise, use an LDO. You probably could use a schottky diode but you probably shouldn't. The forward drop of a Schottky is dependent on temperature and current, and is not tightly controlled like an LDO. They're also about the same price as an LDO (@ qty 1 on digikey, the cheapest Schottky diode is $.13 and the cheapest LDO is $.11).

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    \$\begingroup\$ Assuming you have a microcontroller with a strict 3.3v input voltage. Most will work down to 2.x. Only issue is ensuring the clock you want to work at works at the lower voltage. \$\endgroup\$ – Passerby May 21 at 1:28
  • \$\begingroup\$ Low quiescent current LDO are typically more expensive and there is still a not negligible quiescent current. \$\endgroup\$ – DurandA May 21 at 1:32

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