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Super Cap Power Backup - Am I Overthinking this?

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JonSea

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I need to power power for a brief interval to a data logger circuit to save the last measurements when power is removed. This data logger is powered by the device under test and on an error condition (which we are trying to catch), the device under test shuts off the power. I need my data logger to run long enough to store the last measurements in EEPROM, so power for even a few seconds will be overkill.

The circuit shown below is what I have in mind, but it seems like it may be over-complicated. Supply voltage is 5 volts, and the micro (PIC18F25K22) should be happy do to 3 volts.

Power comes in from the left. A number of LEDs are connected directly to the 5v supply and switched by micro port pins. These do no need to function when power is removed. Power to the rest of the circuit is via D1, bypassing the super cap under normal operation. The super cap is charged via D2 and the resistor. There is no need to charge the cap quickly. Tens of minutes will suffice as the device under test runs long periods without failure. D2 is needed to keep the super cap from powering the LEDs in the event power is lost. D1 also serves to keep the super cap from supplying the LEDs if power is lost.

When power is lost, the circuit is powered from the super cap via D3. Diode D3 is required to keep the super cap from charging without the current limiting resistor in the path.

I am somewhat concerned about the voltage drops of D2 and D3. Will the voltage on the super cap via D2 eventually rise to the full supply voltage as the charge current approaches 0? I have to content with the voltage drop of D3 while the super cap is powering the circuit during power failure; I am less concerned about that if the voltage on the cap eventually reaches supply voltage.

Am I making this needlessly complicated? Suggestions on how to implement a super cap back up?

Super Cap Backup Circuit.jpg
 

audioguru

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A silicon diode has a forward voltage drop of 0.6V at low currents and 0.8V at high currents. Then your cap will charge to about 4.4V and the output to the Vcc will be 3.8V for only a moment because the voltage of a capacitor drops quickly, unlike a battery that keeps its voltage up until it is almost dead. I think the max allowed voltage for a super cap is 2.5V so you will need two or three in series and a circuit to balance their charging.

A Schottky diode (1N5817) has a forward voltage drop of only 0.2V to 0.3V.
 

MikeMl

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Are you powering down the micro, or letting it run as normal? For example, an Arduino mini that draws ~15mA while executing a program at full speed can be put in a sleep mode where it draws less than 2uA in standby. A supercap would do fine supplying 2uA, but might not supplying 15mA...
 

JonSea

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I just need the micro to work long enough to save about 8 bytes to EEPROM (I2C). After that, the program will stop executing and power allowed to drain to nothing.

I have planned on Schottky diodes....sorry for the incorrect symbol. Am I missing anything in needing 3 diodes?

Thanks for the replies.
 

audioguru

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A 1F capacitor charged by 5V though a 0.2V Schottky diode and delivering 4.6V through another Schottky diode has its voltage drop to 3V by a 15mA load in about 1/10th of a second.
 

ronsimpson

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Why D2? Is it to stop the LEDs when running from the SuperCap? If so, shut off the LEDs when no power is detected.
upload_2018-1-5_19-38-4.jpeg
 

Pommie

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As this appears to be temporary to fault find, why not just use batteries?

Mike.
 

JonSea

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As this appears to be temporary to fault find, why not just use batteries?

Mike.
It's actually intended for long term use in remote locations. The equipment under test may fail numerous times before anyone goes to the site to investigate.

The equipment being monitored was built in the 1970s. Eight identified fault conditions leading to the equipment being shut down are not logged locally, and those 8 fault conditions result in a single remote failure indication.

If a technician is sent to investigate, some (most) of the fault conditions don't leave a trace of what happened. For example, 2 of the fault conditions are over temperature and low flow of cooling air. If the equipment hasn't been restarted remotely and is stone cold when the technician gets there, there's going to be no indication that low air flow led to overheating.
 

MikeMl

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I think you are overthinking it:

61.png


You can use V(cc) < 2.5V to generate the interrupt to start saving and shutdown...

Do you have a better estimate of Load current, Pic current, how long it takes to save the variables, how much sag the Pic will tolerate?
 
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JonSea

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I need to figure out the load current of everything on the board and how long it takes to tidy things up. It shouldn't take long to save the data because that will already be in progress, as the supply voltage flag is a monitored value and the change causes a save.

Overthinking the circuit around the super cap and underthinking the rest with the assumption that I'd have far more time then was required.

Time to sharpen the pencil and get to work.

The D2 on the input of the super cap...it is required to prevent the device under test from sucking power from the cap when its supply is cut.

Random realization in my not-fully-awake state. ^^^
 

MikeMl

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...The D2 on the input of the super cap...it is required to prevent the device under test from sucking power from the cap when its supply is cut...
Yes, D1, the Schottky diode in my schematic, allows C1 to charge when power is up, and prevents C1 from discharging back into the "loads" when power disappears.

Notice that my "supercap" is only 1mF, not 1F!
 
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