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We are looking at using an AP2204K-3.3TRG1 to convert 6V from 4 AA batteries to 3.3V to power a microcontroller.

I found that the efficiency of a linear regulator is (ignoring quiescent current) roughly V(out)/V(in). If that's the case, it seems like we'll be losing 45% of our power to heating the pass element.

If we have 8,000 mAh of battery capacity, would our regulator effectively reduce the usable capacity to 4,400 mAh?

Would that mean that if we use 3 AA batteries instead and get a 73% efficiency converting 4.5V to 3.3V that we'd still have 4,400 mAh of useful battery life?

That is to say, if I have a 3.3V circuit and I'm using an LDO linear regulator, will I get more or less the same battery life from 3 AA batteries as I do from 4 AA batteries due to efficiency and heat dissipation?

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    \$\begingroup\$ How are you getting 8Ah? From 4 AA batteries of 2Ah each, in series? It doesn't work that way ;) \$\endgroup\$
    – marcelm
    Oct 25, 2018 at 20:29
  • \$\begingroup\$ Ohhhh!! I will do that. That’s probably where my misunderstanding began. \$\endgroup\$
    – D. Patrick
    Oct 25, 2018 at 20:31
  • \$\begingroup\$ Wait, mAh is directly convertible to wH as a function of voltage, right. Are wH also not additive in series? \$\endgroup\$
    – D. Patrick
    Oct 25, 2018 at 20:38
  • \$\begingroup\$ OK, batteries in parallel add capacity. Batteries in series don’t (presumably they average it?). I’ll update the question based on my understanding now. If I’m converting to wH, I can treat the system as one battery with multiple cells, right? \$\endgroup\$
    – D. Patrick
    Oct 25, 2018 at 20:49
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    \$\begingroup\$ Wh only depends on the number of cells. Each AA battery is around 1.3V * 2Ah = 2.6 Wh. Might be easier to round it to 2.5 Wh. With an LDO, your battery life will be almost the same whether you use 3AA or 4AA. However, you may not be able to fully discharge the 3AA batteries, since the regulator will drop out before they are fully discharged. \$\endgroup\$
    – mkeith
    Oct 26, 2018 at 2:30

3 Answers 3

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If we have 8,000 mAh of battery capacity, would our regulator effectively reduce the usable capacity to 4,400 mAh?

No, it is not a linear relationship.

Capacity is rated from full voltage (1.5V) down to the cutoff of 0.8V.
With time the battery discharges and capacity decreases. The battery voltage also decreases.

The cutoff voltage is 3.3V + 0.15 dropout voltage or 3.45V.
With four batteries the single battery cutoff voltage is 3.45V ÷ 4 = 0.8625V

The yellow area under the curve is unused capacity.

enter image description here


Would that mean that if we use 3 AA batteries instead and get a 73% efficiency converting 4.5V to 3.3V that we'd still have 4,400 mAh of useful battery life?

With 3 AA the single battery cutoff voltage is 3.45V ÷ 3 = 1.15V
Anything over 1.15V is unused.

The yellow area under the curve is unused capacity.

enter image description here

You would waste significantly less battery energy by using 3 batteries rather than 4.

The only gain you get by using 4 batteries is lowering the cutoff from 1.15 to 0.8625V giving you a slightly longer battery lifespan. But at the same time the energy between 1.15 and 0.8625V is waste.

Between the red lines is the extra run time gained with the fourth battery.
The yellow under the curve here is the capacity gained by using 4 instead of 3 batteries. enter image description here


A battery boost switcher for 3.3V is very simple.
An example is the Texas Instruments LV61225 Single-Cell High-Efficient Step-Up Converter

enter image description here

If the current is very low then look at LOW-POWER SYNCHRONOUS BOOST CONVERTER


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Battery voltage declines from the time the batteries are full until they are empty. When calculating efficiency for a linear regulator (LDO is a type of linear regulator), this must be taken into account somehow. The simplest way is to use the AVERAGE voltage of the batteries over the whole discharge curve. The average voltage of an AA alkaline battery is around 1.3V, although this depends on the load. So with 4 batteries in series, the average voltage will be around 1.3V * 4 = 5.2V. So the average efficiency of the LDO will be 3.3/5.2 which is around 63%.

If you use a buck converter, you can probably count on getting 85% with relative ease. Higher efficiencies are also possible, but you will have to do everything very carefully. So you will get roughly 30% longer battery life with a buck converter.

There is something else to consider. The power dissipation in the LDO will be equal to Iload * (Vin - Vout). Now, here, you have to use the max voltage, not average. So let's use 1.5V. That is Iloud * (6 - 3.3) = Iload * 2.7.

For 100mA that is only 270mW. Fairly manageable. But if your load is 500mA, then that is 1.35W, which will be difficult to manage. So higher currents should steer you away from an LDO and toward a buck converter.

Now let's consider quiescent current. If the regulator is going to operate for extended periods with light or no load, then quiescent current has to be considered. LDO's will have much lower quiescent current than buck converters, although some buck converters are low enough to run for many months (with no load) on AA batteries. Pay attention to quiescent current specification if that applies to you. Also, buck regulator efficiency at very light loads will not be much better than an LDO.

Now let's consider cost. If you are only building one, the price difference doesn't really matter. But if you are mass producing this, keep in mind that LDO's are WAY cheaper than buck converters (in volume). Although this may not be true if you 200mA or more of output current, because then you will need a big LDO or an LDO plus heatsink.

So those are all the tradeoffs. My guess is that you will be better off with a buck converter in your application unless the output current is under 50mA or so, in which case I would use an LDO.

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If we have 8,000 mAh of battery capacity, would our regulator effectively reduce the usable capacity to 4,400 mAh?

No. Charge from the battery, neglecting LDO current diverted to ground, appears in the load.

However, if you measure your primary battery in Wh, then yes, many fewer Wh will appear at the load, as you've 'wasted' the excess voltage of the battery. With that difference in voltage, a SMPS might be worth considering.

That is to say, if I have a 3.3V circuit and I'm using an LDO linear regulator, will I get more or less the same battery life from 3 AA batteries as I do from 4 AA batteries due to efficiency and heat dissipation?

You'll get more battery life from 4 AAs, as you will be able to drop to a lower voltage per cell before reaching the minimum voltage into the LDO. Assuming your LDO will work down to 300mV drop, then with 3 AAs you can only go down to 1.2v per cell, with 4 you can drop to 0.9v, a significant improvement in usable capacity.

You should use a different strategy for different cell chemistries if you want best efficiency. For cell types where the output energy is delivered with a large voltage swing like alkaline and rechargable lithium, an LDO will waste a lot of voltage at the start of life, you'll usually be better off with a SMPS. For cell types with a relatively flat voltage curve, like nickel, lead, primary lithium, and silver oxide, a careful choice of voltages will be reasonably efficient even with an LDO.

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  • \$\begingroup\$ Thanks!! I don’t think I understand the difference between mAh and Wh well enough. What do you mean charge from the battery appears in the load? I thought a AA battery was considered “dead” pretty close to 1.3V. Is it useful down to .825V? \$\endgroup\$
    – D. Patrick
    Oct 25, 2018 at 20:19
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    \$\begingroup\$ @D.Patrick Certainly not down to 0.825V, but I've always considered 1V per cell to be the cutoff point; there's very little usable capacity past that. 1.3V is definitely not dead, though. \$\endgroup\$
    – Hearth
    Oct 25, 2018 at 20:23
  • \$\begingroup\$ @Neil_UK, I converted the power capacities to Wh, but the proportions remain the same. Just to be clear, I may get more battery life out of 4 batteries provided the batteries effectively provide enough current below the dropout voltage? Thanks again. \$\endgroup\$
    – D. Patrick
    Oct 25, 2018 at 23:59
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    \$\begingroup\$ @D.Patrick To convert to Wh properly, you have to integrate over the varying battery voltage. With primary cells, you can use them down to whatever voltage they'll still deliver current, if the voltage is high enough. The difference between 3 and 4 cells is bigger if we allow a proper dropout on the LDO, let's say 300mV so 3.6v input. Now 3 cells give out at 1.2v, but 4 can run down to 0.9v each. You have to distinguish between run time and cost, 4 is less efficient than 3, but runs for longer. Without studying a charge/time graph carefully, it's not obvious which will be cheaper. \$\endgroup\$
    – Neil_UK
    Oct 26, 2018 at 4:23
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    \$\begingroup\$ @D.Patrick And of course I've just realised, the sum is correct for Wh, but for Ah it should be 100% (neglecting LDO current to ground) as all the currentr that flows in the load flows in the battery. \$\endgroup\$
    – Neil_UK
    Oct 26, 2018 at 4:25

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