Introduction: Super Capacitor UPS
For a project, I was asked to plan a backup power system that could keep the microcontroller running about 10 seconds after the power loss. The idea is that during these 10 seconds the controller has enough time to
- Stop whatever it is doing
- Save the current state to the memory
- Send out the power loss message (IoT)
- Turns itself to standby mode and waits for the power loss
The normal operation starts only after a restart. There are still some planning needed what could be the procedure if the power returns during this 10 seconds. However, my task was to focus on the power supply.
The simplest solution could be using an external UPS or something like that. Obviously, that is not the case and we needed something much cheaper and smaller. The remaining solutions are using a battery or a super capacitor. Exactly during the evaluation process, I saw a nice YouTube video about similar topic: Link.
After some considerations, the super capacitor circuit sounded as the best solution for us. It is slightly smaller than the battery (we want to use very widely used components, although I am personally not sure if the size reason is actually true), requires less components (meaning- it is cheaper) and most important- it sounds much better than a battery (consequences of working with non-engineers).
A test setup was built to test the theory and to control if the super capacitor charging systems works as should.
This Instructable shows more what has been done rather than explaining how to do it.
Step 1: The System Description
The system architecture can be seen in the figure. First, the 230VAC is converted to 24VDC that to 5VDC and in the end the microcontroller circuit is running at 3.3V. In the ideal case, one could detect the power failure already at the grid level (230VAC). Unfortunately, we are not able to do that. Therefore, we have to check if the power is still there at the 24VDC. Like this, one cannot use the AC/DC power supply storage capacitors. The microcontroller and all other important electronics are at the 3.3V. It has been decided that in our case the 5V rail is the best place to add the super capacitor. When the capacitor voltage is slowly decaying, the microcontroller can still work at 3.3V.
- Constant current – Iconst = 0.5 A (@ 5.0V)
- Minimum voltage (min. allowed voltage @ 5V rail) – Vend = 3.0V
- Minimum time that the capacitor has to cover – T = 10 sec
There are several special super capacitor charging IC-s available that can charge the capacitor very fast. In our case, the charging time is not critical. Thus, a simplest diode-resistor circuit is sufficient. This circuit is simple and cheap with some drawbacks. The charging time issue was already mentioned. However, the main drawback is that the capacitor is not charged to its full voltage (diode voltage drop). Nevertheless, the lower voltage can bring us some positive sides too.
In the Super capacitor expected lifetime curve from the AVX SCM Series datasheet (link) figure one can see the expected lifetime versus operating temperature and the applied voltage. If the capacitor has lower voltage value, the expected lifetime increases. That could be beneficial for as lower voltage capacitor could be used. That needs still to be clarified.
As will be shown in the measurements the capacitor operating voltage will be around 4.6V-4.7V – 80% Vrated.
Step 2: Test Circuit
After some evaluation, AVX super capacitors have been chosen for testing. The tested ones are rated for 6V. That is actually too close to the value we are planning to use. Nevertheless, for testing purpose it is sufficient. Three different capacitance values were tested: 1F, 2.5F and 5F (2x 2.5F in parallel). The rating of the capacitors is following
- Capacitance accuracy – 0% +100%
- Rated voltage – 6V
- Manufacturer part nr –
- 1F – SCMR18H105PRBB0
- 2.5F – SCMS22H255PRBB0
In order to match the output voltage with the capacitor voltage minimal forward voltage diodes are used. In the test VdiodeF2 = 0.22V diodes are implemented together with high current ones with VdiodeF1 = 0.5V.
Simple LM2596 DC-DC converter IC is used. That is very robust IC and allows flexibility. For testing different loads were planned: mainly different resistive load.
The two parallel 3.09kΩ resistors parallel to the super capacitor is needed for the voltage stability. In the test circuit the super capacitors are connected through switches and if none of the capacitors are connected the voltage can be too high. To protect the capacitors a 5.1V Zener diode is placed parallel to them.
For the load, the 8.1kΩ resistor and the LED are providing some load. It was noticed that the no load condition the voltage could go higher than wanted. The diodes can cause some unexpected behavior.
Step 3: Theoretical Calculations
- Constant current – Iconst = 0.5A
- Vout @ power failure – Vout = 5.0V
- Capacitor charging voltage before diodes – Vin55 = Vout + VdiodeF1 = 5.0 + 0.5 = 5.5V
- Start voltage (Vcap @ power failure) – Vcap = Vin55 - VdiodeF1 - VdiodeF2 = 5.5 - 0.5 - 0.22 = 4.7V
- Vout @ power failure – Vstart = Vcap - VdiodeF2 = 4.7 - 0.22 = 4.4V
- Minimum Vcap – Vcap_min = Vend VdiodeF2 = 3.0 + 0.22 = 3.3V
- Minimum time that the capacitor has to cover – T = 10 sec
Time to charge a capacitor (theoretical): Tcharging = 5*R*C
R = Rcharge + RcapacitorSeries + Rsw + Rdiodes + Rconnections
For 1F capacitor it is R1F = 25.5 + 0.72 + 0.2 + ? + ? = 27ohm
If C=1.0F, Tcharging = 135 sec = 2.5 minuntes
If C=2.5F, Tcharging = 337 sec = 5.7 minuntes
If C=5.0F, Tcharging = 675 sec = 11 minuntes
From the assumptions, we can assume that constant power rating is approx.: W = I * V = 2.5W
In a capacitor, one can store certain amount of energy: W = 0.5 * C * V^2
From this formula, the capacitance could be calculated:
- I want to draw x Watts for t Seconds, how much capacitance do I need (Link)?
C = 2*T*W/(Vstart^2 - Vend^2) = 5.9F
- I want to draw x Amps for t Seconds, how much capacitance do I need?
C = I*T/(Vstart-Vend) = 4.55F
If we choose the capacitor value to be 5F:
- How long will it take to charge/discharge this capacitor with a constant current (Link)?
Tdischarge = C*(Vstart-Vend)/I = 11.0 sec
- How long will it take to charge/discharge this capacitor with a constant power (W)?
Tdischarge = 0.5*C*(Vstart^2-Vend^2)/W = 8.47 sec
If using a Rcharge = 25ohm the charging current would be
And the charge time approximately: Tcharging = 625 sec = 10.5 minutes
Step 4: Practical Measurements
Different configurations and capacitance values were test. To simplify the testing an Arduino controlled test setup was built. The schematics are shown in the previous figures.
Three different voltages were measured and the results fit relatively well with the theory. Since the load currents are much lower than the diode rating the forward voltage drop is slightly lower. Nevertheless, as can be seen the measured super capacitor voltage matches exactly with the theoretical calculations.
In the following figure, one can see a typical measurement with 2.5F capacitor. The charging time fits well with the theoretical value of 340sec. After 100 additional seconds the capacitor voltage has been rising only additional 0.03V, meaning the difference is negligible and in the measurement error range.
On the otehr figure, one can see that after the power failure the output voltage Vout is VdiodeF2 smaller than the capacitor voltage Vcap. The difference is dV = 0.23V = VdiodeF2 = 0.22V.
A summary of the measured times can be seen in the attached table. As can be seen the results do not fit exactly with the theoretical calculations. The measured times are mostly better than the calculated ones, meaning that some resulting parasitics were not considered in the calculations. When looking the built circuit one can notice that there are several not well-defined connection points. Additionally, the calculations do not consider well the load behavior – when the voltage drops the current goes down. Nevertheless, the results are promising and are in the expected range.
Step 5: Some Improvement Possibilities
One could improve the operating time if one uses a boost converter instead of the diode after the super capacitor. We have considered that, nevertheless the price is higher than a simple diode has.
Charging the super capacitor through a diode (in my case two diodes) means voltage drop and that could be removed if a special capacitor charging IC is used. Again, the price is the main concern.
Alternatively, a high side switches could be used together with a PNP switch. A fast thought possible solution could be seen in the following. All the switches are controlled through a zener diode that is powered from 24V input. If the input voltage drops below the diode zener voltage the PNP switch turns ON and the other high side switches turn OFF. This circuit is not tested and most probably requires some additional (passive) components.
Step 6: Conclusion
The measurements fit quite well with the calculations. Showing that the theoretical calculations can be used – surprise-surprise. In our special case, little more than 2.5F capacitor is needed to provide sufficient amount of energy for the given time period.
Most important, the capacitor charging circuit works as expected. The circuit is simple, cheap and sufficient. There are some mentioned disadvantages, however, the low price and simplicity compensates that.
Hopefully this small summary can be useful for somebody.
3 years ago on Introduction
Eaton do also manufacturer Super capacitors, if there is any requirement please do let me know @ firstname.lastname@example.org