Powering 3.3v from 5v booster

Using Arduino Motors, Mechanics, Power and CNC
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I have two rechargeable batteries that make around 2.5v. I use a boost converter to get this to 5v to power an mcu and other things that rely on 5v. Now, I have an ic that requires 3.3v and I'm curious what the best way to handle this would be. Here's my brain dump but I'm sure I'm missing a better more obvious way (which is why I'm asking you smart people).

  1. I could (naively) use a low dropout 3.3v voltage regulator to step the 5v down to 3.3v. This is very inefficent however since a lot of power will be lost to heat and undesirable since I'm on limited battery power.
  2. Use a diode to drop the 5v to around 3.3. Since the switching boost ic is keeping the supply constant I need not worry about major power fluctuations (this assumptuion is wrong of course until I test it with an o-scope). Same problem as above as far as efficiency.
  3. Same as above but with a voltage divider? I'm not sure what the + or - of this would be.
  4. Use a second switching regulator to boost a parallel 2.5v line from the battery to 3.3v. This adds a lot of parts and cost but seems like the best way (yes, please prove me wrong).
  5. Opposite of above -- use a buck converter to drop the 5v to 3.3v but regulated. I honestly have no idea what the efficiency curve for this would look like. Seems like black magic or crossing the streams.
  6. Run the new ic at 2.5v with a couple of caps to smooth and pray. Yeah, I'm not doing this.

Thoughts?

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SilentDirge:
I have two rechargeable batteries that make around 2.5v. I use a boost converter to get this to 5v to power an mcu and other things that rely on 5v. Now, I have an ic that requires 3.3v and I'm curious what the best way to handle this would be. Here's my brain dump but I'm sure I'm missing a better more obvious way (which is why I'm asking you smart people).

  1. I could (naively) use a low dropout 3.3v voltage regulator to step the 5v down to 3.3v. This is very inefficent however since a lot of power will be lost to heat and undesirable since I'm on limited battery power.

Would work at the expense of efficency. You don't state the current demand at 3.3vdc so amount of loss not known.
2) Use a diode to drop the 5v to around 3.3. Since the switching boost ic is keeping the supply constant I need not worry about major power fluctuations (this assumptuion is wrong of course until I test it with an o-scope). Same problem as above as far as efficiency.

Diode drop is nominally .6vdc so maybe 3 series silicon diodes would be needed. I could favor this method.

  1. Same as above but with a voltage divider? I'm not sure what the + or - of this would be.

Voltage divide almost always not the proper method, unless the 3.3v current draw is a constant, and if that was the case a simple single series resistor would work.

  1. Use a second switching regulator to boost a parallel 2.5v line from the battery to 3.3v. This adds a lot of parts and cost but seems like the best way (yes, please prove me wrong).

Best = subjective judgement and decision by the decision maker :wink: I would favor #2 method over this one.
5) Opposite of above -- use a buck converter to drop the 5v to 3.3v but regulated. I honestly have no idea what the efficiency curve for this would look like. Seems like black magic or crossing the streams.

Total effiency would just be a calculation of the two converters in series. However this is not really different then #4.

  1. Run the new ic at 2.5v with a couple of caps to smooth and pray. Yeah, I'm not doing this.

Depends on the allowable voltage input variation allowed by the IC's datasheet. If allowed OK, but keep in mind that your battery voltage is not a constant but rather decreases over time as charge is used up.
Thoughts?

Working out the design decision is part of the fun of a project. Remember the journey is part of the reward. :wink:

Good luck;

Lefty

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Which IC?

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IC is ATTiny24 @ 5v to 3.3v EM-406 GPS module.

After some additional research it seems that while it is the most expensive option, the buck converter would result in highest battery efficiency. The reason being as opposed to a linear regulator or diode that convert the extra energy to heat the buck converter switches the energy on and off and stores it in a magnetic field (hence the inductor). This fast switching results in the energy being stepped down with less waste.

I plan to verify this with some qualitative tests and reply back here if I remember.