I have a small board powered by a button cell battery. The voltage range for the microcontroller on the board is 1.9V to 5.5V, while a single battery provides 3V. This leads to quick battery depletion, so I’m considering using two button cell batteries with a total voltage of 6V and adding a diode for regulation. The diode I’m using is the one below. Is this diode made of silicon or germanium? Also, what is its voltage drop, and I’m unsure about its reliability with this setup.
Sorry but this is an electronics design question and outside the scope of the forum which is for discussion related to the KiCad ECAD software. Please ask in another forum, maybe EEVblog. This topic will be autoclosed in 3 hours.
Hi, Topfast
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You will get some responses saying that your question does not have any obvious relation to KiCad so does not belong on this forum.
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As a new participant, you are unable to post much. But anyway I do not see any information specifying a diode.
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It is very difficult to get germanium diodes these days. I have been engineering for almost 50 years, and germanium was pretty much obsolete when I started.
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It is not a great way to do the task but I think you would be better off using two diodes, such as a BAV99 which is a dual series surface mounted type. Or if you prefer axial diodes, use two 1N4148 diodes in series.
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An “LDO” (low dropout linear regulator) with very low quiescent current would provide better voltage regulation but a bit more total current drain. I do not know what is your expected current drain of the load so this may or may not be significant.
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I want to say something what was not said.
Using two batteries this way you won’t gain anything. The microcontroller takes higher current if powered from higher voltage and batteries connected in serie have the same mAh capacity as one of them. It will work even shorter.
The key solution is to keep microcontroller most time in deep sleep mode.
CR2032 - about 200mAh
AA - about 2000mAh
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I’ve got offline suggestion that thanks to batteries in serie will work down to lover voltage per battery the device may be will work longer.
My answer:
Atmega (we used 10+ years ago) are up to 5V5. So first one I found:
https://www.mouser.pl/datasheet/2/268/ATmega48A_PA_88A_PA_168A_PA_328_P_DS_DS40002061B-3050139.pdf
See chart 31.5 at page 329.
Working with 8MHz clock it consumes 2mA at 3V and 4.5mA at 5.5V. So the battery mAh will be used 2.25 times faster. Battery is not capacitor. Voltage don’t drop linearly. Voltage keeps close to constant until mAh are exhausted and then drops down relatively fast.
So mAh will be used 2.25 times faster - you lose 55% of time. Thanks to ability to work with 2 times lover voltage for each battery you will get may be 10% of time, I think. But certainly not more than 20%.