enter image description here Hello, I have a question about battery charging and power supplying. I'll go with picture for better understanding

Starting from using 12V 1A wall adapter, Power will be supplied to My Raspberry PI through 5V rail. My plan is as follows : 1. Down convert 12V to 5V. 2. The converted voltage will be connected to both 5V rail and VIN for battery charger (LTC4067) 3. When wall adapter is connected, output of 12V to 5V regulator will supply power for RPI directly through 5V rail. 4. When wall adapter is disconnected, stored energy in 4.2 LI-PO battery will be supplied to RPI.

I know that recommended input voltage for RPI is 4.75~5.25 and expect that 4.2V LI-PO battery supplies quite insufficient voltage. My questions are here : a. Is boosting output voltage of LTC4067 to 5V necessary? b. If so, is it okay to just connect the boosted voltage to 5V rail?( in aspect of backward protection)

I'm really really unskilled novice and I don't know much about constructing circuit. I would be appreciated if you answer this question.

  • \$\begingroup\$ The LTC4067 already provides power-path management, the out pin will provide the battery or vin depending on source availability and current limits. \$\endgroup\$ – sstobbe Jul 24 '17 at 9:14
  • \$\begingroup\$ @sstobbe Yeah, but the 4.2V from the battery is not enough to power his Pi. Otherwise what you suggest would be the best solution, for sure. \$\endgroup\$ – nickagian Jul 24 '17 at 9:18
  • \$\begingroup\$ @nickagian Thats the point of using a charger with power path management, following the 4067 would be a buck-boost converter for 5V, though you could get away with just a boost converter with non-sync rectification output is Vin-1Vd for large inputs. \$\endgroup\$ – sstobbe Jul 24 '17 at 9:33
  • \$\begingroup\$ @sstobbe I totally agree that this is the point of using such an IC! IMO from this point of view, it is probably the wrong IC for the OP's application, since he cannot directly use the battery to power his circuit. \$\endgroup\$ – nickagian Jul 24 '17 at 10:34
  • \$\begingroup\$ I just found alternative that I use directly 12V to charge battery based on your advice. Thanks :) \$\endgroup\$ – 김현일 Jul 25 '17 at 6:21

enter image description here

Re-layout the circuit as above will give better efficiency and ease of design.

  • \$\begingroup\$ lol @ "your fruit". \$\endgroup\$ – Reinderien Jul 27 '17 at 3:44

a. Is boosting output voltage of LTC4067 to 5V necessary?

Well, obviously the Pi will not operate with voltages under 4.75V. That means you indeed have to boost the voltage of the battery when the main supply from the wall adapter is not present.

b. If so, is it okay to just connect the boosted voltage to 5V rail?( in aspect of backward protection)

Here I suppose you mean if it is ok to connect the boosted 5V from the battery to the 5V rail. But I don't understand the backward protection concern. Anyway, that should be OK to do, but somehow you have to make sure that the boost converter is disabled while the main voltage from the wall adapter is present and only enabled when the battery has to provide current.

But what do you need the OUT pin of the LTC4067 for? You should connect the boost converter directly on the battery, no need to use the OUT pin.

Alternatively, you could connect your main 5V power rail ONLY to the OUT pin of LTC4067 (and the output of the boost converter, as described previously) but NOT to the 12V to 5V converter.

  • \$\begingroup\$ Thank you for your advice! I just know that 12V to 5V regulating is not necessary If I use another battery charger IC. \$\endgroup\$ – 김현일 Jul 25 '17 at 6:24

Take a look at the Adafruit Powerboost 1000c. It does everything for you. It is open source and you can take the schematics and put it on your own PCB.

  • 1
    \$\begingroup\$ Will it run from a 12 volt input power? \$\endgroup\$ – Andy aka Jul 24 '17 at 8:28
  • \$\begingroup\$ No. Either you use your 12v->5v regulator you planned anyway, or you use another wall adapter like a USB adapter for your cellphone. \$\endgroup\$ – Felix Kunz Jul 24 '17 at 15:08
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    \$\begingroup\$ So please make that clear when you leave an answer so that lesser mortals don't presume you can. \$\endgroup\$ – Andy aka Jul 24 '17 at 16:14
  • \$\begingroup\$ Well I'll consider it if cost issue is solved ;). thanks anyway \$\endgroup\$ – 김현일 Jul 25 '17 at 6:22

The ancient oldschool method:

Use a 5VDC regulated wall transformer, the type with an internal switching supply. Then buy a DC power "barrel connector," the mate of the connectors commonly used on wall-transformer supplies. These have an internal switch and three terminals. The switch is used to connect the battery whenever the wall transformer is unplugged from your barrel connector. (Of course, this won't work if you unplug the wall transformer from the wall.)

Another common method is to pair a 5VDC regulated wall transformer with "diode AND logic," a pair of diodes, one each in series with the battery and with the external DC supply. In that case the higher voltage powers your device, while the diode to the lower voltage battery gets reverse-biased. But diodes waste power and also lower the input voltages by 0.7VDC. That's why the usual solution is the switch on the barrel-connector


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