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I am designing a system that contains an RTC to for displaying the date and time on a user interface, as well as log events with time stamps. In this particular case, the system gets powered-on at intervals ranging from once a day to once every two weeks. When used, it runs on a DC power supply, which is derived from the mains AC power supply. The system can remain powered-on/used for any time duration between 30 minutes to 4 hours at a time. When powered-off, the mains power (and thus the DC power supply to the circuitry) gets disconnected. This is when the RTC needs to make use of a backup battery (or similar backup) solution in order to keep the RTC running. The RTC that I intend to use is the Microchip MCP79510.

Of course the most common solution for such an application would be to simply use a standard coin cell battery (such as the CR1220 or CR2032). However, the system needs to be designed in such a manner that the normal user does not have any physical access to the internals of the system, where the backup battery is located. Only authorized personnel should have such access. So in order to prevent visits to the system by such personnel just for the sake of replacing a battery every time it has been depleted (even though this could only be once every 5-10 years, and there could be a number of these systems implemented in the field), I am looking at an alternative solution to provide this backup power.

My first thought is to just use a rechargeable coin cell battery, with a small charging circuit. This should theoretically never require a battery replacement, unless the battery itself is malfunctioning. I have done some reading regarding charging these rechargeable coin cell batteries. I am well aware that non-rechargeable coin cell batteries are... non-rechargeable...

With that said, I have read that trickle charging should never be used for these types of batteries, since the constant current charging mechanism could lead to a charging voltage higher than the battery is rated for. Therefore, my understanding is that I cannot use a "charging" circuit that always keeps the battery at its fully charged voltage, as used with other types of battery applications. Am I understanding this correctly, or am I mistaken?

My first preference for this type of battery is the something like the Panasonic VL1220. For this particular battery, there is a document that outlines a few charging circuits, of which the below circuit seems to be the "standard" one to use:

VL1220 charge circuit

Doesn't this circuit constantly charge the battery, as is typically recommended against?

During further reading, I came across the Microchip MCP73831 charge management controller. While it seems like an attractive alternative to the discrete circuit solution, it also mentions a constant current charging functionality, which is (according to my understanding) not suitable for charging these type of batteries. Could this device be configured properly for charging rechargeable coin cell batteries, such as the case in my application?

An alternative solution to the battery setup is to use a supercapacitor. Although it is a much simpler solution, this would heavily rely on the system being powered-on before the supercapacitor is depleted, in order to recharge the supercapacitor. The operating voltages for these types of capacitors put a heavy restriction on the total backup time, especially if there exists the possibility of the system not being powered-up early enough to recharge the supercapacitor before it has already been depleted. Furthermore, the physical size of very high capacitance supercapacitors also becomes an issue. Therefore, the supercapacitor solution is probably not ideal.

UPDATE: From the comments thus far and Neil_UK's answer, it seems that the MCP73831 is not necessary, and that the above circuit would be sufficient for the application.

UPDATE 2:

In the case of using a rechargeable coin cell battery, is the expected lifetime of the battery itself in the same order of the non-rechargeable coin cell batteries? If this is the case, then perhaps it is inevitable that the battery will have to be replaced every 5-10 years, irrespective of whether it is the rechargeable or non-rechargeable type (which means I might just as well go with the non-rechargeable battery type).

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    \$\begingroup\$ That circuit charges using constant voltage, not constant current. It can't charge the battery at a voltage higher that 5*R2/(R1+R2). \$\endgroup\$ – Finbarr Jan 4 '18 at 12:24
  • \$\begingroup\$ So the voltage divider basically limits the maximum possible voltage applied to the battery by the charging circuit? \$\endgroup\$ – wave.jaco Jan 4 '18 at 13:04
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    \$\begingroup\$ The voltage divider, together with D1, IS the charging circuit. \$\endgroup\$ – Finbarr Jan 4 '18 at 13:50
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    \$\begingroup\$ D2 prevents 5V from being applied to the battery directly and bypassing the charging circuit. \$\endgroup\$ – Finbarr Jan 4 '18 at 19:22
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    \$\begingroup\$ The bypassing isn't achieved. D2 blocks it. Without it, D3 would bypass R1 and D1 and apply 5V (minus the diode forward drop) directly to the battery. \$\endgroup\$ – Finbarr Jan 5 '18 at 10:16
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Panasonic warn against constantly supplying a current, the sort of thing you can do with nickel chemistry batteries, if the current source can supply a voltage higher than 3.6v. With a 5v supply, there would be a great temptation to simply put a resistor in series to limit the current to the battery. That would work for NiMH, but would damage the Vanadium/Lithium battery.

The illustrated circuit does not supply 5v to the battery. R1 and R2 reduce the voltage, and D1 reduces it further. Warning, it does rely on the input voltage being a good 5v. If you supply (say) 9v to that charging circuit, it will damage the battery. Unless you have a good 5v, you should instead use one of the regulator circuits suggested by Panasonic, for instance #3 or #4.

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  • \$\begingroup\$ Thanks for the warning on the requirement of a stable 5 V supply. I assume that the values of the resistors for the voltage divider can just be recalculated if a different supply voltage is used, in order to supply the correct voltage to the battery? As a matter of interest, why would simply putting a resistor in series be damaging the battery, as mentioned in your first paragraph? \$\endgroup\$ – wave.jaco Jan 4 '18 at 18:31
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    \$\begingroup\$ At low current, a resistor drops very little voltage. If you have a single resistor from a 5v source, then the battery can charge to nearly 5v, as the charge current drops low. NiMH can take continuous current safely, VPL cannot. The resistive divider always pulls a current through both resistors, so always has a voltage drop. One of the standard results in elementary electronics is that a resistive divider using resistors of Rtop and Rbottom driven from a voltage of V volts looks like a voltage source of VRb/(Rt+Rb), in series with a resistor of (RtRb)/(Rt+Rb). \$\endgroup\$ – Neil_UK Jan 4 '18 at 18:52
  • \$\begingroup\$ Thanks for the clarification on the single resistor. I understand the derivation of the voltage at the voltage (resistive) divider. Could you just elaborate on why the effective series resistance in such a case is the parallel combination of Rt and Rb? From which point/which perspective is this effective resistance observed? \$\endgroup\$ – wave.jaco Jan 4 '18 at 19:28
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    \$\begingroup\$ Glad you asked, always better to come at these things from 1st principles, rather than maths. We measure the O/C voltage, easy, just the divider formula, and the S/C current, easy, just Vin/Rt (there's no voltage across Rb). We know voltage and current, so the ratio is effective series resistance. For the generalisation of this, see Thevenin's Theorem \$\endgroup\$ – Neil_UK Jan 5 '18 at 5:38
  • \$\begingroup\$ It all makes sense now. Accepting your post as the answer. \$\endgroup\$ – wave.jaco Jan 7 '18 at 18:55

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