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I am trying to design a split power supply from lithium-ion batteries to use with different portable audio projects.

In order to have a stable power source, since it comes from batteries, it has to be regulated.

These two LDOs can provide 300mA for the positive one and 200mA for the negative, so considering only the highest one I will have a PD of 0.4W

Do I have to calculate 4.2Vin as when the battery is full, even if nominal, and I guess under load, the voltage is 3.7V?

The junction temperature is 187.1°C/W

The increase of temperature is 74°C

Would I have problems considering that according to the datasheet the operating junction temperature is 125°C?

I have also read somewhere on that sheet that it should not be placed close to other chip etc., but how can that be accomplished on a small PCB while also considering that there is another LDO that needs to be placed?

Do you think I should change something or not?

enter image description here

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  • \$\begingroup\$ That's not the largest problem. The other lithium cell is shorted out, and the other one gets chared with double the voltage. \$\endgroup\$
    – Justme
    Sep 16, 2021 at 16:52
  • \$\begingroup\$ Hi, could you please be more specific? I don't see why is shorted \$\endgroup\$
    – Status Re
    Sep 16, 2021 at 17:00
  • \$\begingroup\$ The thermal resistance is 187°C/W for junction to ambient, 54.4°C/W for junction to PCB. You need to figure out actual junction temp for your construction which can be difficult. \$\endgroup\$
    – qrk
    Sep 16, 2021 at 17:14
  • \$\begingroup\$ @StatusRe Sorry, you used the same name GND but a different symbol so I guess they are not shorted after all. Still, bear in mind, that if you connect that device to a grounded power supply like your PC, the PC ground would not be the audio ground. Might be easier to just use single large supply than dual smaller supplies. \$\endgroup\$
    – Justme
    Sep 16, 2021 at 18:22
  • \$\begingroup\$ You will draw different currents from those cells. How do you plan to do balancing? \$\endgroup\$ Sep 16, 2021 at 18:36

2 Answers 2

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You could do this in a different way: use a single Li-ion cell with its charging system. This would reduce your system cost and size.

Your LDO to make +3V-ish is fine. Then use this one weird trick: a DCDC buck in 'inverter' connection to make -3V.

More about that weird trick here: https://www.maximintegrated.com/en/design/technical-documents/app-notes/3/3844.html

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  • \$\begingroup\$ That is interesting. My doubt is: would the negative supply be as "clean" as the positive one? \$\endgroup\$
    – Status Re
    Sep 30, 2021 at 5:29
  • \$\begingroup\$ If the DC-DC switching rate is high enough it can be out of range. You can also add an additional LDO after to further reduce noise. \$\endgroup\$ Sep 30, 2021 at 5:37
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Your calculations are correct with a few caveats:

Do you require the full 300mA output of this regulator? I do not recommend running a LDO (or really any component) at is maximum rated power for extended periods. My suggestion is to use a bigger regulator.

Do you plan on using the circuit while the batteries are charging? If so, then you will see a constant 4.2V at the input (assuming your batteries are balanced, possibly more, up to 8.4V).

The datasheet for that regulator specifies Rth(J-A) of 185.6 C/W but a Rth(J-B) of 54.5 C/W. Usually, the junction-to-ambient value will specify the conditions under which the value was obtained (e.g. on 1 square inch of 1 oz copper on 4-layer FR-4), but this datasheet does not and the application report referenced is heavy on theory but light on specifics. My guess (I just finished designing a board with this exact regulator) is that the Rth(J-A) is a minimum footprint value and as you add area to the input, output, and ground planes as shown in fig. 56, you'll approach the J-B value.

As for specific layout, the EN pin needing to be connected to IN kind of cramps your style because it would be a nice big single ground plane otherwise. As a data point, right now I have about 75mm^2 of 1 oz copper divided between the input, output, and ground planes on a top layer, a few (7) thermal vias underneath the package and by the ground pins that connect to the second layer ground plane and about 150mm^2 of copper on the bottom layer. It does fine running 70mA from a 5V supply.

Long story short: if you need 300mA, get something else; if you are going to be drawing less than that, give the regulator as much real estate as you can and you should be fine.

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  • \$\begingroup\$ Thanks for your answer. No I don't plan to use the circuit whilst on charge. In this case I can calculate a Vin of 3.7V? \$\endgroup\$
    – Status Re
    Sep 16, 2021 at 17:01
  • \$\begingroup\$ As for the EN pins i don't know yet where to connect them in order to be able to sink the current draw and save battery power when not in use. It will come later :) \$\endgroup\$
    – Status Re
    Sep 16, 2021 at 17:05

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