Regulated 3.3V from a Lithium-ion (or LiPo) battery

Question

Background

I wish to power my circuit with a Lithium-ion or LiPo battery (likely a battery with around 1000 mAh capacity). These batteries have a voltage that goes from 4.2V to 2.7V typically during their discharge cycle.

My circuit (running at 3.3V) has a maximum current requirement of 400mA -- although I should state that this is only the peak draw occurring about 5% of the time; the circuit draws only about 5mA the remaining 95% of the time).

Question

What would be the best way to convert the (changing) output voltage of a Lithium-ion battery into the required 3.3V to power my circuit with up to the peak current draw of 400 mA? By "best way", I mean most efficient voltage conversion so as to make the best use of battery capacity.

The tricky part for me has been the fact that the Li-ion battery voltage will be both sometimes ABOVE and sometimes BELOW my required final voltage! If it were only one of those two, I would probably have just used either a LDO regulator or a boost IC like TPS61200, respectively.

Answer 1

You should try with BUCK-BOOST DC/DC converter. There are available with efficiency above 90% Check out TI and Linear websites; there are "calculators" that would help you:

• http://www.ti.com/lsds/ti/analog/powermanagement/power_portal.page
• http://parametric.linear.com/buck-boost_regulator

Options:

• TPS63031
• TPS63001

Answer 2

• A linear regulator will do about as well as any alternative.

• Options of regulator parts that are suitable (inexpensive and with low dropout voltage of under 200mV at around 400-500 mA current) include the following: TPS73633, TPS73733, TPS79533, TPS79633, LD39080DT33, LD39150PT33, MIC5353-3.3, ADP124ARHZ-3.3

• Efficiency will be close to or over 90% for most of battery voltage range.

• Probably 80%+ of battery capacity will be available and leaving some capacity in the battery will add usefully to battery cycle life as LiPo and LiIon batteries "wear out less" if Vbattery does not drop too low.

• A buck regulator could get better efficiencies if extremely carefully designed but in many cases will not.

TPS72633 datasheet - fixed 3.3V out, <= 5.5V in. Well under 100 mV dropout at 400 mA across temperature range. About $US2.55/1 at Digikey, falls with volume.

TPS737xx datasheet up to 1A with 130 mV dropout typical at 1A.

LD39080... datasheet 800 mA, dropout OK.

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You say load is 400 mA peak over short periods but <= 5 mA for 95% of the the time. You do not say what battery capacity you wish to use, but let's assume 1000 mAh capacity - not a very large battery physically and common in cellphones etc.

If 3.3V is wanted then a regulator with Vin >= 3.4V is easily achieved and 3.5V in even more so.

So what % of battery capacity do we get at 0.4 C at room temperatures? Based on the graphs below - probably over 75% at 400 mA and close to 100% at 5 mA for a 1000 mAh battery. See below.

For Vout = 3.3V and 90% efficiency, Vin = 3.3 x 100%/90% = 3.666 = 3.7V. So up to 3.7V a linear reguator gives >= 90% - which it is possible to exceed with a buck converter, but only with great care. Even at Vin = 4.0V, efficiency = 3.3/4 = 82.5%, and it does not take long for Vin to fall below this, so in most cases efficiency of a linear regulator will be close to or above 90%, while using the majority of the battery capacity.

While I feel D Pollit's figure of 3.7V for Vbattery_min is too high in this case, using a figure of 3.5V or 3.4V will provide the large majority of battery capacity and will usefully prolong battery cycle life.

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Capacity as a factor of temperature and load: 400 mA = 0.4C.

The left hand graph below from a Sanyo LiPo datasheet which was originally quoted. At 0.5C discharge the voltage drops below 3.5V at about 2400 mAh or 2400/2700 = 88% of nominal capacity of 2700 mAh.

The right hand graph shows discharge at a current of C/1 (~= 2700 mA) at various temperatures. At a temperature of 0 C (0 degrees Celsius) the voltage drops below 3.5V at about 1400 mAh, but at 25 C it's about 2400 mAh (as per left hand graph) so as temperature drops we can expect a substantial drop in capacity, but down to say 10 C you'd expect 2000 mAh or more. That's at C/1 discharge, the 400 mA = 0.4C in this example, and the 95% discharge rate of 5 mA will probably give close to full nominal capacity.

Answer 3

I would try one of the following methods:

• boost the voltage until it doesn't drop below 3.3V and then regulate down to this value
• use two batteries in series
• try to redesign the circut; some ICs with nominal voltage of 3.3V will work even at 2.5V

Answer 4

Get an LFP (Lithium ferrophosphate) battery. Nominal Voltage is about 3.2V and the working voltage ranges 3.0 to 3.3V. Draining your lithium Ion battery from 4.7V down below 3.7V is just detrimental to its life as it is inversely proportional to the depth of discharge

Answer 5

To be honest, a LDO regulator is probably good enough. When a Li-Po cell gets down to 3.3V, it has delivered most of its power (see lipo discharge curve). Many devices (esp8266, nrf24l01, etc.) that state nominal 3.3V supply will operate well below 3.3V.

As a practical example, I built a speedometer with wireless transmitter and receiver/display modules using NRF24L01 modules for the wireless and BA33BC0T linear regulators. Both transmitter and receiver cell voltage are shown on the receiver's display and in practice they cut out around 3.1-3.0V. I ride in (these devices operate in) temperatures from 5 to 30 degrees C.

Keeping in mind that this LDO regulator's datasheet quoted a 0.3V-0.5V I/O difference (I think?) and the NRF24L01 quotes a supply range 3.0V-3.6V, this is really good for a Li-Po project.