Wearable device Battery life calculation
Basic Explanation of my System:
The proposed design is a battery-powered wearable device that integrates advanced sensing and power management features. The system operates primarily from a single-cell Li-ion battery (3.7 V nominal) and includes a USB Type-C interface for charging and external power supply.
The device uses a battery charger with power-path management ( BQ24075) to ensure seamless operation:
When USB power is connected, the system is powered directly from the USB input while simultaneously charging the battery. When USB is disconnected, the system automatically switches to battery power without interruption. A fuel gauge IC monitors battery health parameters such as voltage, state of charge (SOC), and remaining capacity, reporting these to the microcontroller via I²C.
The system includes an electrochemical front end (AD5941) with a 16-bit ADC for high-precision sensor measurements.
Sensor signals are routed through a multiplexer (MUX) to the AD5941, which interfaces with an STM32 microcontroller development board for data acquisition and processing.
The power delivery circuit consists of a buck boost converter which will convert the battery voltage (3V to 3.7V) to 4V. The output of this buck-boost is connected to an LDO which will generate the 3.3V required for the board.
Below is my power tree and battery capacity calculation.May I know it is fine or not.
My power Delivery schematic is also given below. May I know your thoughts about this.Can I go ahead with this design. Mine is a 16bit ADC that is why I am going with this configuration.
1 answer
Working backwards from your devices to the input of the boost converter, your calculations seem correct. I am taking the current consumption of the individual blocks at face value as you show them.
Note that the input current you compute for the boost converter is based on the minimum battery voltage of 3 V. That's OK as long as you realize that's pessimistic. In reality, the battery voltage will vary from 3.7 to 3.0 V during that time. When at the higher battery voltage, the buck converter will draw less current.
If a 340 mAh battery is not a burden, then you're fine. If you want to trim that or extending run time would be useful, then there are some things you can do:
- Boosting to 4 V, then linearly dropping to 3.3 V is inefficient. Consider a buck/boost directly to 3.3 V.
- You could boost to a lower voltage. You should be able to find an LDO that works with only 200 to 300 mV headroom at 15 mA. That's not a big ask. Even boosting to the maximum battery voltage of 3.7 V gives you 400 mV headroom, but saves a continuous 4 mW.
- If you're going to stick with the LDO, then you don't really need a buck-boost converter, only a boost. You should be able to find something with better than 82% efficiency.
- Look at batteries carefully. 3.0 V for an "empty" lithium-ion battery seems high. There is a tradeoff between lowest voltage and lifetime, but 3.0 V still seems high. If that really is the spec for your battery, look around at other options.
I would seriously look into a buck/boost directly to 3.3 V.
I'm guessing that you are worried about noise to the supply of the A/D. That can be filtered with a few series chip inductors and capacitors to ground. Selecting a buck/boost with high switching frequency will make that easier. Pretty much any competent modern buck/boost will switch at 100 kHz or higher. That's not too hard to filter out.
The remaining digital chips can run directly off the 3.3 V switcher output. They won't care about a few 10s of mV ripple.
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