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I know this question has been covered multiple times already; but as far as my reading goes I couldn't get any clear answers yet.

I am building a circuit based on a LiPo input source (4.2V to 3.3V) and I plan to shut-off any power withdraw at 3.3V in order to save the LiPo lifetime. (using voltage comparator on EN pin). The maximum current draw is around 500-600 mA, but for 80-90% of the time, the system will be on lower power mode (around 100mA). The operating voltage is 3.3V for as long as possible; until the LiPo voltage reached 3.3V (bottom discharge limit)

Therefore the dilemma and if I got my reading correct, so far this is what I understood. Under high load it is not favorable to use an LDO because of the increase of drop-out voltage; however it seems that a DCDC step-down would do the job perfectly. But according to my case I would like to have a second opinion; is there any reason to use one or another ?

LDO : http://www.ti.com/lit/ds/symlink/tps737.pdf

DCDC Step-down : http://www.ti.com/lit/ds/symlink/lm3671.pdf

edit : In addition; what do you think about the benefit of using an LDO that would not track bellow 3.3V and therefore get rid of the voltage reference? As opposed to the DCDC that will track voltage well bellow 3.3, hence the need to shutdown system.

Thank you.

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  • \$\begingroup\$ Is cost/complexity an issue? Why don't you calculate the efficience differences? Data for the LM3671 can be found at datasheet page 9, but its mostly (for the 100~600mA range) on the high ~80%. \$\endgroup\$ – Wesley Lee Nov 16 '16 at 5:25
  • \$\begingroup\$ cost is an issue and I want to aim for the cheapest best option; however complexity isn't as I have plenty of place on the PCB to route extra passives. So if I understood correctly the only discriminating factor would be efficiency ? \$\endgroup\$ – Waz Nov 16 '16 at 5:26
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    \$\begingroup\$ Its not clear by your question the output voltage of the system. Is it 3.3V? p.s.: 0.6A*3.3V@85% = ~2.3W for the buck converter, while 0.6A*4.2V = ~2.5W and 0.6A*3.5V= 2.1W for the LDO \$\endgroup\$ – Wesley Lee Nov 16 '16 at 5:38
  • \$\begingroup\$ oh sorry; forget to mention ! Yes indeed the system operating voltage is 3.3V \$\endgroup\$ – Waz Nov 16 '16 at 5:52
  • \$\begingroup\$ A pseudo schematics would help to understand what you are trying to achieve. \$\endgroup\$ – Marko Buršič Nov 16 '16 at 6:17
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Reducing a 4.2V source to 3.3V with the linear regulator represents an efficiency of about 3.3V / 4.2V ≈ 78.57% regardless of current. The efficiency increases as the source voltage decreases, up to 3.3V / 3.4V ≈ 97.06% (assuming a 100mV dropout worst case), with a small drop to 3.2V / 3.3V ≈ 96.97% before the cutoff is reached.

The switching regulator has an efficiency of about... we'll call it 90% at the currents you are considering. The linear regulator will only reach 90% efficiency with a source voltage of 3.3V / 0.9 ≈ 3.67V, which the battery will reach with about 35-45% of its energy left.

It is also possible that the final switching regulator design will have less efficiency than given in the datasheet, in which case the linear regulator is at even more of an advantage. And of course below about 80% it would be silly to use the switching regulator at all.

So the switching regulator will probably last longer on the same amount of battery assuming a high enough efficiency, but requires additional passives compared to a linear regulator and is noisier due to the pass element turning on and off. The linear regulator has a simpler design with less noise, but will likely burn through the battery a bit faster.

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I'd suggest that you would use a switching (Buck) convertor for this task. There are plenty available at very low cost, this would seem to be ideal for you...: https://www.adafruit.com/products/2745

Using a linear LDO regulator is always less efficient. The efficiency is the ratio of power in to power out. It you want to use a single LiPo cell then you cannot achieve a low voltage cutoff of 3.3V on your LiPo cell since even the best LDO must have a voltage drop across it. Consider the following excellent LDO, the Linear LT1763: http://cds.linear.com/docs/en/datasheet/1763fh.pdf The dropout voltage for the LT1763-3.3 is a minimum of 350mV and this increases with chip temperature. This means that at 25DegC the lowest voltage on the LiPo cell would be about 3.7V. A fully charged LiPo at 4.3V would see this regulator therefore drop 1V at the maximum output current for this device of 500mA....so it would dissipate 0.5W.

The fully charged total cell power is 4.3V @ 500mA (2.15W) and the output power is 3.3V @ 500mA (1.65W), therefore the power conversion efficiency is 1.65/2.15 or 77%.

At the low voltage cutoff (3.7V) the input power is 1.85W and the efficiency 1.65/1.85 giving 89%.

If you now look at the datasheet for the TI LM3671, notice on page 9 Figure 16 that the efficiency for output currents between 100-500mA is never below about 93% (and exceeds 95% for some of the voltage range) for all cell voltages from 3.6-4.3V

The Adafruit convertor I suggested is based on the LM3671, but most of the cheap small switching regulators like this one will return efficiencies in excess of 85% at mid range working currents.

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  • \$\begingroup\$ Cutting off a 4.3V LiPo battery at 3.7V leaves almost half the energy unused. And the asker has given a 3.3V cutoff. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 16 '16 at 7:07
  • \$\begingroup\$ But as I've shown above he cannot achieve this with either a liner LDO or a Buck convertor. If he wants to achieve a low voltage cutoff down at 3.3V cell voltage he would need a Buck/Boost convertor. These are available cheaply on Ebay, but their efficiency is much lower: tinyurl.com/zbuk2td \$\endgroup\$ – Jack Creasey Nov 16 '16 at 7:16
  • \$\begingroup\$ That's not what figure 3 of the LDO datasheet the OP linked says. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 16 '16 at 7:18
  • \$\begingroup\$ You have to read both Figure 3 and Figure 4......it's not possible to have no chip temperature rise at the currents he wants, so the dropout voltage will be above .15V giving 3.45V ...so he cannot achieve a 3.3V cutoff anyway. \$\endgroup\$ – Jack Creasey Nov 16 '16 at 7:29
  • \$\begingroup\$ If you correlate figure 4 to figure 3 you'll see that figure 4 assumes the device is operating at 1A, which is well above stated requirements. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 16 '16 at 7:32

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