1
\$\begingroup\$

I am working on a project in which I require three separate DC voltages from a single Li-ion battery (cell) of 2600 mAh capacity. I want my Project to be portable, and I only have the single battery to work with. How can I get voltages of 3.3 V, 5 V, and 12 V from my battery?

I measured the battery voltage and it is 4.2V but I understand that the battery voltage will range from about 3V when fully discharged to 4.2V when fully charged, so the power supply needs to accommodate this battery voltage range.

I am using an Arduino Pro Mini 5 V 16 MHz for my project.

\$\endgroup\$
6
  • \$\begingroup\$ How much current do you need for each voltage? How tightly do they need to be regulated? How important is energy efficiency? \$\endgroup\$ – David Schwartz Sep 30 '16 at 22:53
  • \$\begingroup\$ I deleted the Arduino tag because the Arduino has nothing to do with the question. \$\endgroup\$ – Matt Young Sep 30 '16 at 23:46
  • \$\begingroup\$ 5v Li-ion battery? This question seems %&^*%&^*%; Li-ion chemistry implies a battery voltage between around 3.2 and 4.2 volts per cell, depending on charge state, power draw, chemistry particulars, etc. \$\endgroup\$ – Matthew Elvey Oct 1 '16 at 2:38
  • \$\begingroup\$ @MatthewElvey You are right, I rechecked the voltage of the battery, it is 4.2v dc. \$\endgroup\$ – Aniansh Oct 1 '16 at 3:02
  • \$\begingroup\$ A single LiIon cell will be 4.2V when fully charged and about 3V when discharged. Under voltage discharge protection MUST be used. \$\endgroup\$ – Russell McMahon Oct 1 '16 at 12:11
4
\$\begingroup\$

I would use SEPIC or buck-boost converters adjusted individually for the voltages you need.

However, you should reconsider if you really need all 3 voltages. I did it once on one of my design and regreted it, as debugging the circuit was very exasperating.

Also, if one of the power buses is very low power (say, you only need a few miliamps on the 3.3V) you can use a cheap voltage regulator to obtain said voltage from one of the others if power efficiency is not key.

Also, if you could provide more details we could come up with several proposals ;)

\$\endgroup\$
2
  • \$\begingroup\$ I want the 3.3v for a bmp 180 sensor but I guess if i use an Arduino nano for the job, I can directly use the 3.3v from the board(Earlier I was planning to use the pro mini which doesn't have onboard 3.3v supply). \$\endgroup\$ – Aniansh Oct 1 '16 at 3:00
  • \$\begingroup\$ @aniansh That should work. But remember to consult the arduino's specification and check how mucho amperage it can deliver on the 3V3 rail. Then check that the sensor's peak current is below that limit (which it probably is). Worst case, the regulator breaks and you have to replace it. \$\endgroup\$ – andresgongora Oct 1 '16 at 9:00
4
\$\begingroup\$

For 3.3 volts a standard linear or buck regulator is the simplest approach. For 12 volts use a boost regulator and for 5 volts tee-off a linear or buck regulator from the 12 volt output. Depending on current taken on each power rail this may change.

\$\endgroup\$
3
  • \$\begingroup\$ What's the advantage of connecting the 12V output to the 5V regulator's input rather than connecting the battery directly to the 5V regulator's input instead? \$\endgroup\$ – Jordan Melo Oct 1 '16 at 2:46
  • 1
    \$\begingroup\$ To create a regulated 5 volt output from 5 volts input requires a sepic or buck-boost or flyback design and the circuit is more complex so, if you are using a boost circuit to create the needed 12 volt supply it makes sence to tee-off its output and use a simple buck regulator for 5 volts. Simplicity and PCB space are usually important. \$\endgroup\$ – Andy aka Oct 1 '16 at 9:10
  • \$\begingroup\$ @jordan Efficiency. Say your appliance consumes 1A at 5V. If you use a switched buck converter with an efficiency of 85%, your circuit will deliver 1A@5V (5W) and dissipate additional ~17% (1/0.85) of those 5W as heat (0.85W wasted). On the other hand, a voltage regulator works by dissipating the excess voltage on a internal resistor. So if your voltage regulator is powered at 12V, and outputs 5V, it has to get "rid" of 7V. Because the input and output current are the same for such a device, it will dissipate as heat 7W (7V * 1A) in our example. \$\endgroup\$ – andresgongora Oct 1 '16 at 9:14
4
\$\begingroup\$

Before even thinking about implementation I want to ask you question: How much current will be consumed through 3.3 V, 5 V and 12 V? If we suppose that efficiency of power level converters is 100%, what is the maximal time your appliance will work properly using the battery?

You also should take into account that chips require A specific level of voltage to operate, e.g. datasheets should state range, for example 5 V ± 10%, meaning if your main power level will drop below 4.5 V, the whole appliance may malfunction, because its main brains will fail to operate.

\$\endgroup\$
5
  • \$\begingroup\$ Will using a 12V battery make things easier? \$\endgroup\$ – Aniansh Sep 30 '16 at 19:30
  • 2
    \$\begingroup\$ Aniansh, the voltage level and capacity of the battery will of course change the equation. The art of your task is to find optimal match. There're a number of circuits available to drop 5v to 3v3 and increase 5v to 12v; finally you can see mobiles which work for weeks. If I would be you I would not go for shortcut like just taking car battery for application, but try to predict the current values, and see how many A*h you may need to run the device, and then see how you can effectively convert these A*h into battery voltage, its capacity and select best power conversion circuits. \$\endgroup\$ – Anonymous Sep 30 '16 at 19:36
  • 1
    \$\begingroup\$ Wow, thank you for giving me a direction. I think I will have to do a bit more research about this. \$\endgroup\$ – Aniansh Sep 30 '16 at 19:39
  • \$\begingroup\$ This is just general guidelines for design, yet you do not answer any part of the question at all, except asking a few of your own. While it might be useful 'food for thought' it is not really an answer. \$\endgroup\$ – jbord39 Oct 1 '16 at 2:40
  • \$\begingroup\$ @jbord39 there's no right answer to the question because it lacks vital information. Currently marked answer also asks questions, but provides generic answer which in reality may not fit the application. \$\endgroup\$ – Anonymous Oct 1 '16 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.