# How to regulate a 3.6 V battery to two power rails (3.3 V and 1.8 V)?

I have a 3.6 V battery from which I need two voltage rails, one that is 3.3 V and one that is 1.8 V. How should I design this circuit such that it is the most power efficient?

Here is what I was thinking:

• Using one voltage regulator to regulate the 3.6 V down to 3.3 V, and using voltage division after the 3.3 V to create a 1.8 V
• Using two voltage regulators to regulate the 3.6 V down to 3.3 V and 1.8 V

Are either of these good options? What else might you suggest?

Follow-up question about using two regulators: Should I put them in series or parallel?

i.e.

Series: Regulate 3.6 V down to 3.3 V, and then regulate this 3.3 V down to 1.8 V.

Parallel: Regulate 3.6 V down to 3.3 V, and separately regulate 3.6 V down to 1.8 V.

simulate this circuit – Schematic created using CircuitLab

• Yes. It depends on your current requirements... – Passerby Feb 10 '15 at 5:37
• It's probably wisest to regulate 3.6 down to 3.3 and then use another regulator to take the 3.3 down to 1.8. At the very basics, the less difference in voltage you're moving, the more efficient it is and the less strain you put on the components. Definitely don't use voltage division because then you're guaranteed to waste power. Also, my post assumes you're using Switch Mode Power Supply (SMPS) to do the regulating rather than linear regulators. Otherwise, if you're using linear regulators, you'll always waste the same amount of power. – horta Feb 10 '15 at 5:53
• I too suggest 2nd option – AKR Feb 10 '15 at 6:07
• My load current for the 3.3 V rail is 546.2 mA. My load current for the 1.8 V rail is 1.154 mA. – Christina Feb 10 '15 at 6:11
• Also, I believe my wording for the second option might be ambiguous. For clarification, do you mean to put the two regulators in series? Or parallel? I will add a diagram to my question for a visual. – Christina Feb 10 '15 at 6:14

My load current for the 3.3 V rail is 546.2 mA. My load current for the 1.8 V rail is 1.154 mA.

Your 1.8V rail is basically just a reference voltage. A voltage divider from a regulated source would work just fine. A 1.8V 1.154 mA load is like a 1500kΩ load, keep that in mind when you make your voltage divider. Efficiency isn't really an issue here. Even at 50% efficiency, that's 4mW.

As for the 3.3V rail, with a 3.6V battery, likely a lithium battery, you want a buck-boost switching regulator, as the voltage range of the battery will be above AND below 3.3V. You want it to regulate down, and then regulate up. Look for one based on your battery chemistry, as there are some that have built in battery protection, to prevent it from draining too low.

• What would you suggest for a NiMH battery? It's being charged by a solar panel, if that makes any difference. – Christina Feb 10 '15 at 6:52

The half-amp current drain at 3.3v is quite a bit for a 3.6v battery. I suggest you use a buck-boost regulator, such as the LTC3536, which can supply up to 1A and would allow the battery voltage to drop below 3.3v, say down to 3.0v or so and you would still get 3.3v out of the regulator.

Your spec for the 1.8v rail is much more modest, and there you can just a simple low-current buck regulator like the LTC3620 which can supply up to 15 mA.

• Thank you for the specific part suggestions. Should I take into account the ampere-hours as well? Or does only the amperage matter when choosing parts? I mention this because although the 3.3 V has a much higher current draw than the 1.8 V rail, it is only used in bursts (no more than 30 minutes a day in total, and no more 6 minutes at a time). – Christina Feb 10 '15 at 6:50
• @Christina Only the maximum current draw is important. It is always a good idea to leave some margin, and I chose a chip that can handle somewhat more than is needed (for example, I could have chosen a 600 mA part instead of the 1A part, but went with the higher value). – tcrosley Feb 10 '15 at 6:59
• I took a look at the datasheet for the regulator you suggested, and it seems like the efficiency is around 92%. From the LDO power dissipation equation (3.6 - 3.3 V) * .5462 A, I get that 163.86 mW of power is wasted, but the power efficiency is still considered 3.3/3.6 V = 91.67 %, is it not? How do I calculate the % efficiency for an LDO and the actual power dissipation for a buck-boost regulator? It would otherwise seem that the efficiency is the same percentage-wise.. which doesn't seem right. – Christina Feb 11 '15 at 8:06
• @Christina Interpolating the graph on page 1 of the LTC3536 datasheet, the efficiency for an input of 3.6v and 500 mA is around 90%. I believe your calculation for the LDO efficiency (92%) is correct. However the LDO is irrelevant because your battery will quickly drop to a voltage where the dropout voltage causes the output to go below 3.3v; this is why you need a buck-boost regulator. The datasheet I linked to does not have any formulae for calculating power dissipation. Here is a datasheet for another buck-boost regulator that does. – tcrosley Feb 12 '15 at 8:03