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Author Topic: Designing coin cell + capacitor for periodic high current draw  (Read 935 times)
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Hi everyone,

I have a small radio circuit that sends a message every 4 seconds, then sleeps.
The circuit draws 30mA for ~10ms, then 0.004mA for 4000ms.

I'd like to power it from a coin cell such as a CR3032 (500mAh, 3V), but the coin cell datasheet recommends a constant discharge of 0.20mA, otherwise the capacity suffers considerably.

It most likely won't even do the 30mA I require, I hear that the maximum draw on these button cells is 15mA, and even that kills the cell much faster.

I'd like help designing and dimensioning a circuit where a capacitor would slowly charge (at 0.20mA or less) during those 4000ms of sleep, and then source those 30mA for 10ms in place of the coin cell.

Thanks in advance!
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I could be as simple as installing a series resistor from the battery to a 'super cap' wired to the other side of the resistor and ground, and powering the load from the positive side of the cap. Not sure if the best method to proceed is with a rigorous mathematical analysis or just guess at some component values and test.

Lefty

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You might want to spend some time with these charge / discharge calculators along with retrolefty's idea of just using a resistor.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html#c2
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No need for anything fancy, lets start with the fundamental equation for a capacitor:  C dV = I dt.  (This is just another way to say capacitance is charge divided by voltage-change)

Lets say the 30mA current is needed for 20ms (conservative assumption) and the voltage on the capacitor should only fall 0.3V during this discharge period.  We'll assume the cell can provide negligable current compared to 30mA.   Thus C times 0.3 equals 0.03 times 0.02 : this gives 2mF (but noone says milli-farads so we'll call it 2000uF - the next standard value 2200uF)

So a 2200uF electrolytic is fine.  However we must check the leakage current rating of the capacitor is good enough for our desired battery life - go to the datasheet for this.  Leakage at 3.6V is likely to be less if the capacitor has a higher rated voltage (such as 10V or 16V rather than 6V3).

Also note that when the cell is first connected it will take some time to charge up the capacitor.  If the cell can provide about 5mA, say, then it will take about 1.5 seconds to charge up to 3.6V (this might be too slow to ensure a clean power-on reset, but otherwise isn't problematic)
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Thanks everyone for the replies, it was exactly what I needed.

For future reference, here is a design I simulated for:
- (duty cycle) 10ms @ 12mA ::: 4000ms @ 0.004mA
- For a max 0.3v drop: C=0.010*0.012/0.3=400uF

Picture:
- 750kohm load is the 0.004mA part, 250ohm load is the 12mA part
- Switch on the right lets me turn on the 12mA part for 10ms

I then added the 2k resistor to limit current draw from the cell.

Coin cell load from zero: 1.5mA (goes down to 0.2mA in 1.6 seconds)
Not a problem since this only happens the first time the cell is inserted.

Coin cell load after 10ms 12mA pulse: 0.15mA (less than the 0.20mA recommended by manufacturer)
Circuit voltage after 10ms pulse: 2.71v.

Great! Thanks!


* CoinCellPlusCap.png (38.88 KB, 860x640 - viewed 28 times.)
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Why are the 750k and the 250 ohm resistors in parallel? That will just waste your energy.
You should use the calculated load of the circuit when on instead of the "250 ohm" resistor.
You should use the top resistor to limit current draw while charging.
The 750k resistor seems totally useless -- current will be drawn by the capacitor when charging, not by anything else.
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Why are the 750k and the 250 ohm resistors in parallel? That will just waste your energy.
You should use the calculated load of the circuit when on instead of the "250 ohm" resistor.
You should use the top resistor to limit current draw while charging.
The 750k resistor seems totally useless -- current will be drawn by the capacitor when charging, not by anything else.

The 750kohm and 250ohm are just emulating my real circuit parts, specifically an ATmega328P in power-down mode (4uA with watchdog enabled) and a radio module that's doing 12mA in TX mode.

Sorry for the confusion. I couldn't draw them in the simulator, so I just drew their equivalents (so-so).
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gotcha -- makes much more sense :-)
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