Lithium coin cells are rated for fairly low standard current draws, on the order of 1 to 5 mA. Also, while they allow greater pulsed current draws (i.e., periodic bursts), this appears to be detrimental to cell capacity (and also can cause a drop in the voltage during the pulse).
I am bringing up this topic out of interest in applicability of coin cells for general use-cases (such as LEDs or more recently low-power wireless transmission), so I don't have a specific circuit in mind.
But imagine two scenarios, one a low-duty cycle and one a more demanding case:
- Case A: Load draws 25 mA for 25 milliseconds once every 2.5 seconds.
- Case B: Load draws 50 mA for 100 milliseconds once every 1 seconds.
I'm interested in an analysis of whether a capacitor-based reservoir can be applied to (and thus, whether it is wise to) run either of the pulse-draw cases above off a coin cell.
Note 1: In both cases, I'm considering a generic situation with Coin cell --> 3.3V Boost regulator --> LOAD [microcontroller + LEDs with series resistors + Wireless module + etc]. And the Cap/Supercap parallel to the load supply.
Note 2: I'm aware that one could use Li-ion/LiPo batteries but they have higher self-discharge (whether due to their chemistry or due to their protection circuitry), so they may not be ideal for, say, a wireless temperature logger that transmits once every hour.
Relevant documents: The following datasheets show various pieces of information, including pulse discharge characteristics, operating voltage vs. load, etc.:
In addition, the following documents discuss some empirical assessment / qualitative discussions about running somewhat large loads (with peak current draw on the order of tens of milliamps) using a coin cell:
TI App note: Coin cells and peak current draw
Nordic Semiconductor App note: High pulse drain impact on CR2032 coin cell battery capacity
Freescale App note: Low Power Considerations for ZigBee Applications Operated by Coin Cell Batteries
Jennic App note: Using Coin Cells in Wireless PANs
Answer
The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.
ΔV = I × Δt / C
Solving for C gives:
C = I × Δt / ΔV
Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:
C = 25 mA × 25 ms / 0.1 V = 6.25 mF
The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.
For the second example, the numbers work out to:
C = 50 mA × 100 ms / 0.1 V = 50 mF
Average current = 50 mA * 100 ms / 1.0 s = 5 mA.
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