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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

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closed as too broad by Matt Young, Elliot Alderson, Chris Stratton, Nick Alexeev Apr 29 at 0:02

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ What is the nominal voltage of the battery? Are the batter(ies) connected in parallel or in serial? \$\endgroup\$ – Stefan Wyss Apr 28 at 17:11
  • \$\begingroup\$ @StefanWyss The circuit uses 3 * 1.5 micro batteries. they are connected in serial. \$\endgroup\$ – paradisal programmer Apr 28 at 17:21
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    \$\begingroup\$ How much current does each IC draw? How long do you expect the device to operate? Exactly what kind of battery are you planning to use..."micro batteries" doesn't help. \$\endgroup\$ – Elliot Alderson Apr 28 at 17:27
  • \$\begingroup\$ @ElliotAlderson one of them uses 30 mA and the other one is an amplifier which at least uses 120mA (it depends on speaker). we doesn't have limit in this area but we want to have battery life to be as long as possible. \$\endgroup\$ – paradisal programmer Apr 28 at 17:36
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    \$\begingroup\$ I'm sorry but you can't create a design if your requirements are "as long as possible" and "lowest possible cost". Those are contradictory requirements, and anything that we might suggest will fail to meet one of those design criteria...so why waste our breath? Please try to formulate realistic requirements. \$\endgroup\$ – Elliot Alderson Apr 28 at 18:08
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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).

Edit 1:

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.

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  • \$\begingroup\$ Thank you, could you please introduce an IC, buck type DC/DC converter? \$\endgroup\$ – paradisal programmer Apr 28 at 19:29
  • \$\begingroup\$ How can any solution for so many unknowns be possible? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 29 at 5:38
  • \$\begingroup\$ @paradisalprogrammer I added some further details to my anwer. \$\endgroup\$ – Stefan Wyss Apr 29 at 5:58

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