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Is my supercap circuit OK ?

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Jan 10, 2009
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hello dear forum,

I want power supply the RTC of STM32 with supercapacitor instead of battery

I will for the first time use supercap

is this below circuit OK ?

thank you


So for Ic=330mA max and T=63%3.3V=RC and 99%Vin=5xRC **

How long will it take to power up your Cap to say drive a 1W white LED which as a threshold of 2.85V at say 10% of Imax or 10%*330mA and an ESR of ~1/Pmax =1

Then your Cap has an ESR of ??

How long would expect to run the LED with an added Rs=1Ω

If you start Vcap from 0V.

What is the ratio of charge time to run time from Imax to 10% of Imax.

What is this ratio? When Vinit=2.85V? Vfinal=3.3V Rs added to LED using Ohm's Law to drop 3.3 to 3V or rated Vf of your part =0.3V/330mA~1Ω

What is the Cap leakage in terms of seconds RpC? How long can you leave it unused for a change in 0.1V?

What heatsink Rca thermal resistance do you need to limit junction rise to 60'Cmax? 80'C?
Which is better? Why?

When you can answer these, you can master Ultracap use.
Supercap is a rather general term, without specific data we can't actually rate the circuit. I think it's basically o.k. for RTC supply. In case of doubt I would refer to the datasheet of supercap manufacturers.

My problem with supercap RTC supply is that the backup time is too short for applications where an instrument is only used occasionally. But they are o.k. for digital watches or radios and similar home appliances.
hello SunnySky,

thank you for your message

I need to backup the RTC of my STM32 uC for at least 12 hours in case of an electricity cut

I found this webpage for calculation

I think 0.3 F, 5 V cap will be OK

I need an approximate solution
all I need is to supply the clock of uC

can you please comment on my above circuit
maybe I need change the charging resistor to lower charging current

It would probably be better to connect the 3.3V directly to VBat. Then wire the diode and resistor in parallel, and together in series with the capacitor, cathode end to VBat. That gives you full 3.3V at VBat when the power is applied and only drops it by Vf when the power fails. At the moment you lose Vf even when the power is present so VBat probably never goes higher than about 2.8V.

I would increase the resistor to at least 100 Ohms to limit the charging current. You might find a 10 Ohm resistor passes more current than necessary to the capacitor and gets hot. In similar designs I use 1K.

Hello Brian

do you mean this ?


but if the power goes off there will be current flowing not only to Vbat pin but also towards the rest of the circuit

am I wrong ?

A 1W diode ESR is around 1 Ω
A 10W diode ESR is around 0.1 Ω
A supercap ESR varies from <0.1 Ω($!) to >>100Ω

This part is very good value, 0.5F, 3.5V, 25Ω
This means you don't need a current limiting R. While the heat is dissipated inside the cap, it is thermally insulating so dont let it get hot from constant max current charge discharge.

Series power Schottky diodes are for safety and switching. Never reverse voltage or go over-voltage. 10% less is safe margin.
anotherbrick - you are quite right but you didn't mention other things drawing current from the supply!

You can add another diode in series with the input but as with your original circuit, it will give a constant voltage drop at VBat. If you can live with that there isn't a problem. Please be aware of the charging current to the capacitor. As Tony points out, it's own internal ESR will limit the charging current to a level which will not damage the capacitor itself but you may still be drawing ~130mA current from you supply line. Make sure that wont cause you other problems. Using the method I described you get the same VBat voltage but you can use a series resistor to reduce the current to a much lower level and the VBat pin voltage also stays up while the capacitor is charging.

If you need discharge at higher current rates 10 mA with charge rates of up to 150mA from a higher source

Considering this one as a better option.
**broken link removed** which can handle a 1.7A charge rate with a 15'C rise and has an ESR of 55 mΩ 2.5F $5

Ideally a constant current source is best for charging the Cap from a voltage higher than the target Voltage, so it has a linear ramp from 0 to 100% in 1 Time constant instead of 500% of the RC time constant.
A 5V LDO with Rs from adj to Vo will act as a suitable current source.

But when you use a CC you must design a voltage limiter to prevent over charge. But since there is a drop on the LM317 of ~2V use better versions with 1V drop.

In this case you asked for 3.3V and you could also consider the energy of E=1/2CV² is 2.5x at 5V (25) vs 3.3V (10), which means possibly up to 2.5 times longer run time using a 5V to 3.3V DC-DC Buck boost regulator
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I fear most considerations in this thread are discussion questions beyond the problem stated in the original post.

It's about Vbat supply of STM32 RTC, voltage range is 1.65 to 3.6 V, about 1 µA at lower voltages. So the circuit in post #1 is exactly what you need, for only 12 h backup time with a rather small goldcap
Historically, mOBO's use Lithium coin cells for RTC backup because;
They are ;
- 3.0V over the useful lifetime
- then rise in ESR from 1k to 10k depending on size to at least 10x this value when the voltage dies at 2/3 original.
- however when double-layer ultracaps became popular , and cheap , some started to use these instead, in appliances with RTC because;
- power outages are short and 100 Ohm ESR was OK.
-leakage increases with C value in uA so leakage RC product is a figure of merit.

- now ultracaps have much lower ESR , higher Amp ratings for wireless irons etc with recharge docking bays

- and many other uses

The best way to understand the properties is to read the datasheet from a major supplier like Panasonic for either supercaps or Lithium cells.
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