Is there a better way to power this circuit from a battery?

Question

I am trying to learn how to nicely power a wifi chip by a rechargeable battery, and have the following arrangement so far (decoupling not shown):

The ESP8266 spends most of its time in deep sleep, but wakes periodically to take a sensor reading and transmit it over WiFi.

• When asleep (10 minutes): current = 20μA
• When active (1.2 seconds): current = 80mA (increasing to 180mA for short spikes during transmission)

The battery I am using is a single cell 1200mAh LiPo battery, which then feeds the MCP1700-330 3.3V LDO. Using the arrangement above, I measure the battery voltage dropping as follows during operation:

Currently, the battery life is around 10 months. I am wondering if there is a better way to power the circuit that I can use? (which doesn't add much extra complexity / cost). I am just learning, and this is my first attempt to power something with a battery.

I chose this arrangement because the MCP1700 has a very low quiescent current (1.6μA), and it can regulate the maximum battery voltage of 4.1V nicely down to 3.3V, as accepted by the load devices. I know that the dropout voltage is around 180mV, but am not sure what happens when the battery voltage drops below this (couldn't find it in the datasheet). Does it just track the input? Is it bad practice to do this?

I was planning to add undervoltage detection next, to disconnect the battery when its voltage reaches 3V, in order to protect it. But it seems like the battery is pretty much empty at 3.4V (or is this an effect of the LDO becoming unregulated?).

None of my devices actually need 3.3V, so does it make more sense to use a lower regulator like 2.6V (minimum specified by the ESP)? Or would this then result in too much lost energy (burnt as heat on the LDO) for the largest battery voltages? Should I use a different battery even?

I have read about buck-boost converters being used for this kind of thing (because they can ensure for example 3.3V even when the battery voltage drops below this), but it seems to me that they add more complexity for not much efficiency gain (because the currents are low).

Any feedback would be great, as I am confused as to what the best / standard way would be to go about this. Thanks!

---

EDIT - Some estimations of wastage

Assuming 1.2 seconds awake time, and 600 seconds sleep time.

Current flowing through LDO regulator when ESP is asleep = 20μA (from my circuit) + 1.6μA (quiescent current of LDO) = 21.6μA.

Current flowing through LDO regulator when ESP is awake= 80mA (dominated by my circuit, i.e. neglect LDO quiescent current).

When the LiPo battery is fully charged to 4.2V:

• Voltage dropped across the regulator is 4.2V - 3.3V = 0.9V

• Power dissipated by LDO when ESP is asleep: P = I*V = 21.6μA x 0.9V = 19.4μW.

• Power dissipated by LDO when ESP is awake: P = I*V = 80mA x 0.9V = 72mW.

• Energy wasted by LDO when ESP is asleep: E = P*t = 19.4μW * 600 sec = 11.6 mJ per cycle.

• Energy wasted by LDO when ESP is awake: E = P*t = 72mW * 1.2 sec = 86.4 mJ per cycle.

• Total energy wasted per cycle = 86.4 + 11.6 = 98 mJ per cycle.

When the LiPo battery is discharged to 3.4V (before nonlinear behaviour starts):

• Voltage dropped across the regulator is 3.4V - 3.3V = 0.1V

• Power dissipated by LDO when ESP is asleep: P = I*V = 21.6μA x 0.1V = 2.2μW.

• Power dissipated by LDO when ESP is awake: P = I*V = 80mA x 0.1V = 8mW.

• Energy wasted by LDO when ESP is asleep: E = P*t = 2.2μW * 600 sec = 1.3 mJ per cycle.

• Energy wasted by LDO when ESP is awake: E = P*t = 8mW * 1.2 sec = 9.6 mJ per cycle.

• Total energy wasted per cycle = 1.3 + 9.6 = 10.9 mJ per cycle.

Over the linear part of the battery's discharge from 4.2V down to 3.4V:

• Average energy wasted due to asleep = (11.6 + 1.3)/2 = 6.5 mJ per cycle
• Average energy wasted due to awake = (86 + 9.6)/2 = 47.8 mJ per cycle

It takes 9.5 months to go from 4.2V to 3.4V, which equates to 41000 cycles. So the total energy wasted

• due to being asleep is 6.5 mJ/cycle x 41000 cycles = 266 J
• due to being awake is 47.8 mJ/cycle x 41000 cycles = 1960 J

Total wasted energy is 2.23 kJ of which 12% comes from being asleep, and 88% from being awake.

Note, for this calculation, I have only accounted for when the battery voltage is dropping linearly (not including the non-linear behaviour after the "knee"). I have also assumed that the LDO regulator's quiescent current, and the ESP8266 sleep current are constant at 1.6μA and 20μA, respectively. This may not be the case - especially with the LDO, but I couldn't find info in the datasheet.

Energy required by my circuit:

• When asleep, energy required is 21.6μA x 3.3V x 600 seconds = 43 mJ per cycle. Total energy = 43 mJ x 41000 cycles = 1750 J.

• When awake, energy required is 80mA x 3.3V x 1.2 seconds = 317 mJ per cycle. Total energy = 317 mJ x 41000 cycles = 13000 J.

Therefore, over the linear discharge time of 9.5 months:

• When awake, we have 13 kJ used by my circuit and 2 kJ wasted by the LDO.
• When asleep, we have 1.75 kJ used by my circuit and 0.26 kJ wasted by the LDO.

This makes me think that even if the power supply (LDO or buck-boost) could perfectly convert the battery voltage into 3.3V with zero wastage, the gains in battery life wouldn't be huge (on the order of 10%), because most of the energy is actually consumed by my circuit. So might be better to use 2.6V instead of 3.3V...

---

EDIT #2 - Effect of different supply voltages

As requested by Neil_UK, I have removed the LDO from the circuit and varied the power supply voltage from 2.5V to 3.6V (as can be tolerated by the ESP8266). It can be seen that the deep sleep current varies from around 15.5μA up to 21μA. There is no meaningful change in the active current, which sits at around 77mA always:

I did not measure any change in time required to connect, but this ofcourse could be because I always have a strong enough WiFi connection. It would be better to measure RF output power, but I don't have the equipment, so this will have to do as a proxy for now.

Answer 1

This is not really an answer, but there's too much here for a comment.

Let's try to understand the consumption, before going about fixing it.

We have 600 s of sleep at around 22 uA (LDO+MCU), and then 1.2 s of activity at 80 mA. The latter has an average current of 1.2x0.08/600 = 160 uA during the 600 s cycle. The MCU/WiFi consumption therefore dominates the total, so efforts should first be directed to reduce this.

One observation is that increasing the cycle time will immediately bring big improvements. RemyHx has suggested in comments using a different MCU ESP32-C3 that can read the sensor in its sleep. An alternative would be to wake the existing MCU only long enough to read the sensor in most cycles, and then periodically wake for longer to dump the results over WiFi. I'll wager it doesn't take 1.2 s to read the sensor.

Does the consumption of the MCU during the active periods vary with power supply voltage? Measure the active current at voltages from 2.5 V to 3.6 V, and see if there's a difference worth having. This may inform your choice of the final regulated output voltage. You should check the sleep consumption as well, but this has a lower impact on the total.

Using a buck regulator gives you the possibility of providing more Ah to your load than you consume from your battery. Taking the average battery voltage to be 3.7 V, and assuming outputting 2.6 V, and using an 85% efficiency figure (guessed at from the data sheet of Andy's suggestion of the LT8608), then you can provide 3.7x0.85/2.6 = 1.21 times the battery current to the load. Is an extra 21% worth having via this route?

If you were to use just a buck regulator, you might find that its sleep performance wipes out that 20% gain. Although the LT860 claims a few uA quiescent with no load, that will rise when it switches. It claims an input current of typically 56 uA and max 96 uA when taking 6 V down to a 2.7 V load at 100 uA, so anywhere from 50% to 90% efficiency. You would need to get one and measure it carefully with your 20 uA sleep load.

Note that the LT8608 is good for 45 V and 1.5 A. Higher voltage means higher resistance components. Higher current means larger area components so higher switching charge and consumption. While it's worth having a part number to design from, there should be parts available better suited to your requirements. I would expect a properly designed part aimed at 5 V and 500 mA could be a bit better efficiency at 100 mA, and much better at the 20 uA output.

A possible way to improve the buck sleep performance would be to use it together with an LDO to handle sleep. It may be possible to simply put the LDO and buck outputs in parallel, with the LDO programmed for 2.5 V, and the buck to 2.6 V. I would expect that each would tolerate its output driven in this way, but you would need to test this. On waking, the MCU would first enable the buck, and then disable it last thing before sleeping.

A much harder to quantify difference is MCU supply voltage versus WiFi consumption. I would expect the rail voltage to affect RF power output. If you operate at low rail voltage, does this increase the time required to make a connection back to the AP, due to missed/retried transmissions?

Can you use an MCU that uses Bluetooth, preferreably LE, or even 6LoWPAN? I worked on a design a while ago that was powered by current transformer from domestic mains, and used a TI MCU chip that incorporated 6LoWPAN, to talk to its base every second. I don't recall what the total current consumption was, but it was commensurate with your power levels.

Answer 2

Any feedback would be great, as I am confused as to what the best / standard way would be to go about this.

I see no reason why you wouldn't choose a switching buck regulator that can work down to 3 volts and still produce an output of 2.6 volts (as you state you can live with). A buck regulator is slightly more complex of course but, if you want to extend battery life, then it's the way to go. But, you have to pick the right buck regulator that has a low quiescent current: -

Image from data sheet.

You'll have to do the math and work things out but, it looks viable to me.

Remember that a buck regulator is ideally 100% power efficient in that it ideally doesn't dissipate power when dropping voltage. A linear voltage regulator does lose power by its very nature.

There are probably decent alternatives to the AD8608 from the usual sources like TI or On-semi.

Answer 3

Abandon the regulator:

Use Primary Lithium cells:

• a CR123A/CR17345 3V Lithium battery @1500mAh. Low discharge, no loss in regulation, or,
• 2xAA Energizer L91 "1.5V" (1.8V when fresh) @3000mAh
• Other Primary Lithium cells 3V or 3.6V at the capacity/size that works for you.