Cytron Technical Forum

Hello, recently I have a problem for my project that 9v battery drain the current faster than i thought(Battery death very fast).
My project connect from battery then go though LM7805 only link with MCU, RF module x2, Graphic LCD. May I know how to calculate my project required how much current or power to maintain working?? How much current will be discharge from a non-recharge battery actually??

I found (Battery DC Boost Converters Using MCP1640) at website but i dont know how to connect to MCU that might cause MCU damage.
Any power electronic expert here can help me??

No, I am not expert But believe me, ordinary 9V battery cannot last for long, even those with "Super Heavy Duty" .

Actually when MCU execute codes, it draw power continuously, LCD also draw power, RF also draw power..... and the keywords is "continuously". When you multiply the current draw over time, it is not possible for normal battery to support that application.

There are much to study for low power devices, you can try Microchip website as there do produce ultra low power MCU (claim to be world lowest power MCU ). Your circuit will need to be very efficient, MCU need to utilize sleep mode, etc.

That true like robosang said.
You can do nothing with the 9V battery spec, even you boost current. You boost current but voltage sure drop. Must think, the power and energy from 9V battery still remain same.
Laws of nature, we can't produce energy but the energy convert from another type to another type. This means, you still can't boost energy for the 9V battery.

Yea...9V battery won't last long. I tried it before. A much better and not that costly battery is the AA rechargeable with, for example, 2000mAh. It sure helps your circuit last longer compared to the 9V battery with normally only rated at around 500mAh.

It is said that you wouldn't want to overcharge a battery, over voltage charging, over current charging, and only use up to 40% of the power left. Read it somewhere online.