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I have a device that operates with 3.2 V MAX, and I wish to power it from a standard 3.7 Li-Ion battery. A fully charged battery=4.2V.
I used a simple LM317 circuit to drop voltage to 3.2V and it worked perfectly.

Now the problem is as the battery starts to drain, the output voltage starts to become lower, so subsequently the LM317 output voltage will be lower as well, for example

  • batt=3.7V, LM317=2.5V

In which this case the voltage is not enough to keep the device up and running.

So what I need is some sort of dynamic regulator to keep the output voltage fixed at 3.2V until the battery voltage drop to 3.2V as well. Advice on good ways to achieve this would be appreciated.

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    \$\begingroup\$ The LM317 has about 3V minimal drop voltage. It doesn't function correctly when you set your output voltage higher than Vin-3V. Use a low-dropout regulator (LDO). \$\endgroup\$ – Janka Feb 4 '17 at 13:27
  • \$\begingroup\$ I think the key words you are looking for are "buck switching power supply". Linear power supplies work because they burn up the extra voltage. Not what you want in a battery operated device. \$\endgroup\$ – st2000 Feb 4 '17 at 13:42
  • \$\begingroup\$ Dynamic regulator you are looking for is also called buck boost regulator. Search for it on Google. TI and Libera technologies have many of them. \$\endgroup\$ – Umar Feb 4 '17 at 13:43
  • \$\begingroup\$ Boost won't be necessary in this application, as the device's maximum voltage of 3.2V is not that much higher than the minimum voltage of Li-ion batteries, and there's not much juice left in the battery at this voltage. A buck converter is the best tool for this job IMO, as it is also more efficient than a buck-boost converter. \$\endgroup\$ – Dampmaskin Feb 4 '17 at 13:49
  • \$\begingroup\$ BTW, I'm not sure there is a "standard" LION battery. Different impurities in the positive electrode results in different nominal voltages. Might want to check on this before you settle on a design. \$\endgroup\$ – st2000 Feb 4 '17 at 13:50
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Maybe something like this is what you really need: -

enter image description here

It's an LT3127 buck-boost regulator that can deliver the required output voltage (3.3 volts in the example above) from input voltages ranging from 2.9 volts to 5.5 volts. It can supply 1 amp too. The 320k and 182k resistors set the output voltage and, for example, with a 330k and a 220k, Vout will regulate at 3.0 volts.

Power efficiency is pretty good too: -

enter image description here

Because it's a switching converter, it doesn't burn off the excess input-output voltage as heat but rather takes less current than the load when the input voltage is higher than the output voltage. For instance, when supplying a load of 300 mA, the power out is (say) 3 volts x 0.3 amp = 0.9 watts. The efficiency implies that the input power will be 0.95 watts when the battery voltage is 4 volts. This means the current taken from the battery is only 240 mA (and not 300 mA+ as you would get with a linear regulator). This means an extended battery life.

Also, because the regulator will work at lower input supply voltages than the output voltage, battery life will be further extended but, take care not to over discharge your battery.

Here's a link to the search engine results from Linear Technology. Similar in principle to searching on TI's site too.

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  • \$\begingroup\$ +1. This is the correct answer. Use a buck-boost regulator and tweak for maximum efficiency at the current level you need. I've had good success with TI's TPS63020 for higher current (4A) applications. Always looking for better efficiencies though. Just be careful to watch the battery voltage and force the regulator off when the input voltage gets too low; you can permanently damage most batteries by over-discharging them. \$\endgroup\$ – akohlsmith Feb 5 '17 at 5:13
  • \$\begingroup\$ @tamadver if you are happy with this answer, please consider formally accepting it. It's the only fee "demanded" on EE for (hopefully) useful answers. \$\endgroup\$ – Andy aka Feb 9 '17 at 15:57
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LDO meant low dropout of 2.5V in the 70's for linear regulators using transistors.

Today with MOSFET's it depends on cost and size of RdsOn of the internal FET and thus dropout can be < 0.1V for higher cost.

Your choices are: Max Dropout Vin-Vout, @ rated current, adjustable, fixed, Vin_range, Vout_range. There are about 20k different LDO's .

I suggest something like this http://www.digikey.com/product-detail/en/texas-instruments/TPS73632DBVT/296-17424-2-ND/696630

Or search for your own. with filters Ctrl+click and check box for Stock http://www.digikey.com/products/en/integrated-circuits-ics/pmic-voltage-regulators-linear/699

  • At 4.7Vin 3.2Vout efficiency = 3.2/4.7=68% @400mA Pd= (4.7-3.2)*I
  • At 3.5Vin 3.2Vout efficiency = 3.2/3.5=91% @400mA
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    \$\begingroup\$ I would add that since the device is battery powered a dcdc bubo regulator would increase battery time a lot. \$\endgroup\$ – Vladimir Cravero Feb 4 '17 at 13:34
  • \$\begingroup\$ I believe LDO is the closest to what i need so far, still need to try. just to confirm i understand that now: vin=4.7v so LDO=3.2v as it goes lower Vin=3.5v LDO output still =3.2v. is this correct? \$\endgroup\$ – tamadver Feb 4 '17 at 14:15
  • \$\begingroup\$ Regulator Topology Positive Fixed Voltage - Output 3.2V Current - Output 400mA Voltage - Dropout (Typical) 0.075V @ 400mA Number of Regulators 1 Voltage - Input Up to 5.5V Current - Limit (Min) 400mA so yes \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 '17 at 15:14

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