Hello, I am using a Seiko MS920SE lithium rechargeable coin cell in an embedded application. This is an 11mAH battery. I am trying to calculate the required charge time given my supply voltage and simple charge circuit. The charge circuit uses a fixed Vsupply of 3.3V. This supply is connected to an NSR0530HT1G Schottky diode with a forward drop of 370mV. The diode is then connected to a series current limiting resistor (50kOhms). Finally, this connects to the positive of the Seiko coin cell. As the cell voltage increases, this should reduce the charge current. Since this charge current is not constant, I am unsure what current to use when calculating the total charge time. I believe the cut off voltage for this coin cell is 2.0V. I appreciate any suggestions for how to calculate the total charge time given my charge circuit. Thank you
Hello @longboard,
This appears to be similar to the circuit described in the datasheet. For the 3.3 VDC supply, the resistor may be as low as 620 Ω.
It is difficult to give an exact answer as there is a variable voltage to current characteristic associated with the battery. We can think of this as the charge characteristic curves in reverse.
Recommendations:
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Lower the resistance into the 1 kΩ range and don’t worry about it.
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Perform imperial measurements using automated equipment to construct the charge curves. This could potentially be used to calculate the charge time from any starting position.
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Consider adding a dedicated battery protection IC. This TIBQ29729 Here is a representative example. This is an important consideration as we do not want the load to completely discharge the battery.
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Switch to a capacitor.
Please let us know if you would like to explore this idea further.
Sincerely,
Aaron
Hi Aaron,
Thank you for your reply and information. The charge circuit you posted is exactly what I am using. The design used a 49.9K for the series shunt resistance. However, I fully agree that reducing that value to ~1kOhms would be a good idea. I will also explore the battery protection IC that you mentioned to prevent loading the completely discharged battery.
I fully agree that the non-constant current nature of the charge current complicates the calculation of charge time. However, I am trying to make a rough estimate of the charge time for the coin cell. Below are the assumptions I am making…
- cut off voltage of 2.0V for the coin cell
- diode has a forward drop of 370mV
- R = series resistance = 49.9K for my existing charge circuit
- Vsupply = 3.3V
- battery capacity is 11mAh
Given those details, I think the initial charge current would be…
I(init) = (3.3V - 370mV - 2.0V) / R
I(init) = (3.3 - .370 - 2.0) / 49900 = 0.0186mA
Then taking 11mAh / I(init) = best case charge time
11mAh / 0.0186mA = ~591 hours = 24.6 days
Of course, as the battery voltage increases the charge current will decrease, so the equation above is not accurate but I think would be a best-case scenario. Does my logic seem reasonable for this?
During charging we have noticed that the battery voltage will rapidly increase from 0.7V to ~2.6V within ~12 hours. Given the 49.9kOhm series resistance I don’t think the coin cell has been charged much during only 12 hours, but regardless the cell voltage has quickly risen. I suspect that a quick rise in cell voltage during charging is typical for lithium chemistry, just as a quick drop in cell voltage during discharge. But I wanted to confirm if you don’t mind.
Thank you for your help and time.
Sean
Without analyzing everything you have posted, make note that your actual Vf voltage for the NSR0530HT1G is probably going to be significantly lower than your assumption of 370mV.
That spec is for a forward current of 100mA. You are down in the uA range, so Vf will be much lower.
Looking at the Vf vs. If graph:
At 25°C, a forward current of 0.02mA will only drop about 100mV, 0.1mA drops about 140mV, and 1mA would drop around 210mV. Also note that as the battery voltage rises, the current will drop, which reduces the Vf drop of the diode.
The datasheet for the battery recommends 620 Ohms or higher for charge current limiting. It would seem useful to choose a value closer to this than your 50k Ohm value in order to speed up the charge time. Even if you used a 5k Ohm value, you would speed up the charging nearly ten-fold. I guess it depends on your goals.
Hello @longboard,
Sure, it will do as a first approximation.
- The bulk charge occurs at lower voltages
- The final top off will take longer as the battery voltage increases.
Please help me understand why the high resistance is necessary. By my thinking, it is desirable to charge the battery as quickly as possible. This will then provide a measure of backup safety e.g., a power failure the same evening the product is placed in service.
Sincerely,
Aaron
P.S. Watch out for depth of discharge. This chemistry does not like to be fully discharged.
Hi David_1528,
That is a very good point you make regarding the reduced forward drop across the diode, given the current value. I agree that reducing the series resistor to something closer to 620 ohms (based on the datasheet), would reduce the charge time and seems like a good idea. This is an older design I have inherited. The original designer chose such a high value of series resistance to safeguard against over charging the coin cell. But moving forward, I think we should be safe with something closer to 1kOhms. Once we reduce the series resistance, I will monitor the current and voltage to measure how long it takes to fully charge the coin cell. I think this is the best way to go about determining how long the coin cell will take to charge, given the constantly changing current, due to the coin cell’s voltage rising as the charge progresses.
Thank you again,
longboard
Regarding overcharge risk, as long as you’re charging voltage does not exceed 3.3V, you can’t overcharge the battery, so that shouldn’t be a worry, with the 3.3V max supply volt caveat.
Hi Aaron,
Thank you for the information. What you said makes sense regarding the charge rates. The design for this circuit was done by an outside consultant and was a second rev. The first rev used a 1kOhm resistor, but the designer changed to a 49.9K for rev 2. We didn’t catch this change until recently after we started having some issues with rev 2 boards, specific to the real time clock, who’s voltage is maintained by this coin cell. The original designer was asked why the change was made to 49.9kOhms and the explanation was to prevent overcharging. However, given the 3.3V supply and series diode… I don’t think that overcharging is as concern. We are planning on having the boards re-worked to make the change back to 1kOhms for the reasons that you mentioned.
Thank you again for your help and time.
-longboard
Thank you David_1528, thank you for your comments. Our charging voltage should not exceed 3.3V. The 3.3V supply is regulated output (using a buck convertor), derived from a +12V battery. I agree that overcharging shouldn’t be an issue. I appreciate your help and comments.
-longboard