from 6V to 3.3V, what's more efficient switching or LDO? - Page 1
Forum section: electronics / beginners

author: alexg
author_rank: Regular Contributor
timestamp: October 21, 2016, 02:48:37 pm
content: Hi All, I know that question is asked and reviewed often and I already did alot of homework and built few working regulators but my results are somewhat different so I wanted to ask someone with experience of dealing with both types of regulators in an actual projects.I want to squeeze the longest battery life to power microprocessor (and some peripherals) from 4AA batteries, average current consumption is about 5 - 6mA, what is better to use Switching Step Down, or Low Dropout Linear?Right now I have both and on fresh batteries they seem to consume the same current, but it will take somewhat longer time to see how batteries perform when close to depletion but I wanted to start building PCB right now and not sure what regulator will offer better battery life?

author: Kleinstein
author_rank: Super Contributor
timestamp: October 21, 2016, 03:01:08 pm
content: At 5 mA a reasonable quality switcher should get better efficiency than an LDO. It is only a very low currents like less than 0.1-0.01 mA where the no load current consumption of the regulator can shift it towards the LDO.Switching regulators can be quite different in consumption.With an LDO you could get a way with just 3 cells.A 5 mA consumption is quite a lot for battery operation. Chances are that one can reduce the consumption there.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: October 21, 2016, 03:08:59 pm
content: You need to very carefully select the switcher as many will have sufficient quiescent current draw that they won't be worth using for 5mA. Look for something with a max rated output of 100mA or less, as that's more likely to be designed for low quiescent current.Also consider off-mode operation - if your design has soft on/off using sleep, this requires additional care to avoid quiescent current losses with the switcher.How important is battery life for your application? Is it actually worth the addiitonal cost and complexity?

author: mariush
author_rank: Super Contributor
timestamp: October 21, 2016, 03:14:47 pm
content: It depends on your budget and the pcb area you can allocate to a voltage regulator.Linear regulators take very little space, with most linear regulators you only need a couple of ceramic capacitors at the input and output so they take very little space, but your efficiency is basically vout * 100 / Vin ... for 6v and 3.3v you're looking at 3.3 * 100 / 6 = 55%Note though that you may use regular non-rechargeable alkaline batteries which could be up to 1.65v when full and maybe 1.4v when close to depleted, or you may have rechargeable batteries which are close to 1.35v when full, about 1.15v when close to empty... so depending on battery type you can have 5.6..6.6v , or 4.8v.. 5.4vSo in some worst scenarios you could have an efficiency as low as 50%. Also note that Some micro LDOs (for maximum 50mA or 100mA that would work well for your project, with very low voltage drop) may only work with up to 5.5v so you have to check datasheets before picking one.Switching regulators can get you at over 90% efficiency but some have higher quiescent current (idle)... if you say the average is around 5mA in some cases the switching regulator may use more than 1mA just for its internals...There's all kind of switching regulators, with very low quiescent current, at higher frequency (which allows smaller parts but maybe less efficiency), there's even regulators with integrated inductor and mosfets to minimize external components.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: October 21, 2016, 03:28:08 pm
content: Some switchers might be worse, but if you pick the right one it will be significantly better than an LDO. It's not hard to do, and several manufacturers offer tools to design the switcher for you. For an example, enter your parameters in TI's Webench and you'll stumble across the TPS62745:http://www.ti.com/lit/ds/symlink/tps62745.pdfOver 80% efficiency including the quiescent draw for loads ranging from tens of microamps to hundreds of milliamps. At 5mA you should be solidly in the >90% range, compared to ~50% for an LDO. That's nearly double the battery life.

author: Gyro
author_rank: Super Contributor
timestamp: October 21, 2016, 04:56:48 pm
content: QuoteDischarge end-point is normally specified as 0.9V per cell for standard alkaline batteries. You probably don't want to go quite that far, but you could be looking as low as 3.6V for 4 AAs if you want full capacity out of them. With 3 cells you'd be sub 3v3.

author: alexg
author_rank: Regular Contributor
timestamp: October 21, 2016, 05:00:01 pm
content: Thanks everyone for reply, wow, I did not know that LDOs are so inefficient, I thought that's what Low Dropout means. I'm sticking to switching regulators then, I do get over 90% efficiency with one of them, thanks again all, I thought if LDO can get at least 80% then for the sake of less components and noise I could use it, but 50%, meh....

author: suicidaleggroll
author_rank: Super Contributor
timestamp: October 21, 2016, 05:18:37 pm
content: Quote from: alexg on October 21, 2016, 05:00:01 pmAll linear regs, LDO or otherwise, use a resistive element to drop the voltage while the current stays the same. So if your load draws 10mA from the reg, the reg will draw 10mA from its voltage source, and will burn off the voltage difference internally. As a result, the "efficiency" of a linear reg is simply the ratio of the output to the input voltage. For example, feeding 10V into a 5V linear regulator will ALWAYS give you an efficiency of 50%.If you want to improve the efficiency, you have to reduce the input voltage. With 9V in it's now 55%, with 8V it's 62.5%, and so on. A normal linear reg will stop you around 7V, if you try to go below that it won't be able to maintain a 5V output anymore. An LDO will let you drop the input all the way down to 5.5V or even lower, whichwouldimprove your efficiency, but unless you actually do that, nothing changes compared to a regular linear reg.

author: mariush
author_rank: Super Contributor
timestamp: October 21, 2016, 05:59:28 pm
content: Indeed, as suicidaleggroll says, if you plan on using LDO, best option is to get the input voltage as close as possible to the circuit voltage and to use a LDO with very low voltage drop (plenty of them with under 100mV voltage drop).In your particular case, best choices would be to go down to 3xAA batteries (3 x 1.2v..1.25v for rechargeable) or maybe switch to a 18650 lithium battery (or similar lithium battery) (between 3.5v and 4.2v when fully charged).With a lithium battery, maybe you could take advantage of the extra space available (an 18650 batter is smaller than 3-4 AA batteries) and add a small lithium battery charger inside your product. Chips like MCP73811 are super simple to use and would allow you to charge the battery from a usb port (they work with up to 6v) :http://ww1.microchip.com/downloads/en/DeviceDoc/22036b.pdfIf you were t need 5v, I would have also suggested a voltage doubler IC and to use 2 AA batteries or 4 batteries ( 2 in parallel + 2 in parallel , to get low voltage and more mAh). You'd have 2x1.25-1.3v = ~2.5v and a cheap voltage doubler IC would raise the voltage to around 4.8v-4.9v which is within the range of what a 5v regulator would work with.See LM2665 to see how simple such voltage doubling circuit could be:http://www.ti.com/lit/ds/symlink/lm2665.pdf