We are designing a circuit that has two ICs with different supply range. it will use micro batteries as supply unit. What is the efficient way to feed two ICs together and waste minimum energy. Of course, with the lowest possible cost. any help would be highly appreciated
closed as too broad by Matt Young, Elliot Alderson, Chris Stratton, Nick Alexeev♦ Apr 29 at 0:02
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As Elliot Aldersen mentioned in his comment, it is not easy to help you with your design if you give us only "vague" design criterias. Nevertheless, I will try to give you my 50 cents:
The "most efficient" and "lowest cost" might be to connect the 3 L1154 batteries in series (nominal 4.5V) directly to your amplifier with 4V-5V input voltage range.
For the 1.8V chip, you might want to use an efficient buck type DC/DC converter to transform the 4.5V input voltage from the L1154 batteries to a 1.8V output voltage (Example for such a buck converter: TI TPS62243).
One problem you might encounter is that your L1154 batteries might not be able to deliver a total of 150mA into the load, because these type of batteries have a large output resistance. At least you will have a significant voltage drop from the batteries, so that it will not be possible to directly connect the amplifier to these batteries.
If this is the case, you would need to change your design: Connect the 3 batteries in parallel (nominal 1.5V output voltage) and use two boost type DC/DC converters, one for 1.8 output voltage and one for 5V output voltage.