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Post Reopened by Russell McMahon♦
Post Closed as "Not suitable for this site" by Andy aka, ocrdu, evildemonic, Russell McMahon♦
Changed several incorrect and important assumptions re battery voltages. LiIon = 3v - 4.2 V range approx.
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Russell McMahon ♦
Russell McMahon ♦
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My circuit needs 3.3 volt, and I have 2 options with a Li-ion battery pack.

Option 1: I can put 2 cells in series and get approximately 74.4 to 6V (7.4 volt mean) and then using buck converter reduce it 3.3 V. In that case my battery capacity will be 2000 mAh.

Option 2: In another case, I can put the cells in parallel and get 34.2 to 3V (3.7 volt mean). In this case, I am going to get battery capacity of 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh but cannot use the range from 3.3V to 3V without using a buck boost converter.

Which option will give me more battery life for my circuit?

My circuit needs 3.3 volt, and I have 2 options with a Li-ion battery pack.

Option 1: I can put 2 cells in series and get approximately 7 volt and then using buck converter reduce it 3.3 V. In that case my battery capacity will be 2000 mAh.

Option 2: In another case, I can put the cells in parallel and get 3.7 volt. In this case, I am going to get battery capacity of 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh.

Which option will give me more battery life for my circuit?

My circuit needs 3.3 volt, and I have 2 options with a Li-ion battery pack.

Option 1: I can put 2 cells in series and get approximately 4.4 to 6V (7.4 volt mean) and then using buck converter reduce it 3.3 V. In that case my battery capacity will be 2000 mAh.

Option 2: In another case, I can put the cells in parallel and get 4.2 to 3V (3.7 volt mean). In this case, I am going to get battery capacity of 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh but cannot use the range from 3.3V to 3V without using a buck boost converter.

Which option will give me more battery life for my circuit?

"cell", not "battery", Li-ion not "Li-iron"
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Davide Andrea
Davide Andrea
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Li-ironion battery 2 in parallel or 2 in series - Capacity vs voltage, which one will give more battery life?

My circuit needs 3.3 volt, and I have 2 options with a Li-ironion battery pack.

Option 1: I can put 2 batteriescells in series and get aproxapproximately 7 volt and then using buck converter reduce it 3.3 V. In that case my battery capacity will be 2000 mAHmAh.

Option 2: In another case, I can put my batterythe cells in parallel and get 3.7 volt, in. In this case, I am going to get battery capacity equals toof 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh.

Which option will give me more battery life for my circuit?

Li-iron battery 2 in parallel or 2 in series - Capacity vs voltage, which one will give more battery life?

My circuit needs 3.3 volt, and I have 2 options with Li-iron battery pack.

Option 1: I can put 2 batteries in series and get aprox 7 volt and then using buck converter reduce it 3.3. In that case my battery capacity will be 2000 mAH.

Option 2: In another case, I can put my battery in parallel and get 3.7 volt, in this case, I am going to get battery capacity equals to 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh.

Which option will give me more battery life for my circuit?

Li-ion battery 2 in parallel or 2 in series - Capacity vs voltage, which one will give more battery life?

My circuit needs 3.3 volt, and I have 2 options with a Li-ion battery pack.

Option 1: I can put 2 cells in series and get approximately 7 volt and then using buck converter reduce it 3.3 V. In that case my battery capacity will be 2000 mAh.

Option 2: In another case, I can put the cells in parallel and get 3.7 volt. In this case, I am going to get battery capacity of 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh.

Which option will give me more battery life for my circuit?

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Akshay
Akshay
  • 29
  • 3

Li-iron battery 2 in parallel or 2 in series - Capacity vs voltage, which one will give more battery life?

My circuit needs 3.3 volt, and I have 2 options with Li-iron battery pack.

Option 1: I can put 2 batteries in series and get aprox 7 volt and then using buck converter reduce it 3.3. In that case my battery capacity will be 2000 mAH.

Option 2: In another case, I can put my battery in parallel and get 3.7 volt, in this case, I am going to get battery capacity equals to 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh.

Which option will give me more battery life for my circuit?