# Calculating the right size for super capacitor

I have a circuit that draws a constant 0.15A and it is powered from a 9V power supply which is immediately stepped down to 3.3V through a linear regulator (LM1117T). My goal is to power the 3.3V portion of the circuit (through the 3.3V regulator if necessary) for 3 seconds after the 9V power is removed.

I used the calculator here and determined that a 0.22F capacitor would power my circuit for 8.25 seconds which is much longer than I need.

Unfortunately when I connect the capacitor it doesn't power the circuit for hardly any time at all. I added all the capacitors I had to the circuit (three parallel banks of two 5.5v 0.33F capacitors in series) for a total of 0.495F and it only powers my circuit from the 9v side for 1 second.

What am I missing here?

• Conservation of energy formulae in your skills yet? 0.5Wx3s= 1.5 Joules. What Ec=1/2CV^2 from 9V to Vcutout of Buck – Tony Stewart Sunnyskyguy EE75 Jul 2 '18 at 2:22
• The characteristics of your 3.3V regulator are very important. You said you were using a linear regulator, so you can't let the capacitor discharge to 3.3V, you can only let it discharge to the dropout voltage of the regulator (i.e. the minimum input voltage of the regulator that allows it to provide a regulated output) – Elliot Alderson Jul 2 '18 at 2:32
• @ElliotAlderson, in the calculator isn't the Vcapmin of 3.3v accounting for the regulators minimum input voltage? If I'm reading the datasheet correctly the dropout for my regulator is 1.2V. I realize that the regulator will not supply 3.3v if it isn't getting 3.3v, but it seems like the calculator is telling me that between 9v and 3.3v I will be able to draw 0.15A for 8.25 seconds. – ubiquibacon Jul 2 '18 at 2:48
• @TonyStewartolderthandirt, I sense your sarcasm, but I don't follow. Can you dumb it down for me? Where are you getting the 0.5W from and who is Buck? – ubiquibacon Jul 2 '18 at 2:50
• @ubiquibacon, if your dropout is 1.2V, then your minimum voltage is 3.3V (output) + 1.2V (dropout) = 4.5V (input). That accounts for only part of the discrepancy, though. The calculations are fairly easy remembering that 1 farad = 1 coulomb per volt -- this will give you about 6.6s of operation at 0.15A. I'd suggest measuring both the input and output currents operating from a fixed supply to see how much it's actually drawing and where it's going. Also, double check to see if you read the cap values correctly. – Cristobol Polychronopolis Jul 2 '18 at 12:36

There nothing wrong with using the calculator you have to provide an indication of the backup time you would achieve but several things wrong with your schema:

1. Using a linear regulator is not the best but can work for you. You have to consider the dropout voltage, which in the case of the LM1117 is about 1.5 V. (see the datasheet Note 4). Using a linear regulator will also increase the supercap value required over using a small SMPS.

2. The supercaps will be uncharged when you first turn the unit on, so will represent a very large charging current while they are charged. This may look like a short circuit to your 9 V power supply and it may shutdown. If there is no real short circuit protection it may potentially damage your power supply.
In addition, the regulator supply voltage (3.3 V) will rise very slowly which may upset an MCU starting, or peripheral resets and the like.
You should use a series resistor to lower the charge rate of the supercaps.

3. From your comments it seems you bought 0.33F 5V supercaps. Two of these in series will only give about 0.16F, not the 0.22F you show in your calculator.

4. Putting supercaps in series to get higher voltage is done all the time. However, you cannot do this simply, you have to ensure that the voltage across them is balanced and never exceeds their rating. Usually this is done by a circuit very like a battery BMS that ensure the terminal voltage is not exceeded. This may be too expensive for you, but I'll propose an alternative.

Now lets propose a circuit that may work for you:

simulate this circuit – Schematic created using CircuitLab

In this circuit, the power supply needs to be able to supply 9V @600mA peak, so you may need to revisit your power supply. The supercaps are charged via R1 and discharged via D3. The CUS10S30 has a Vf of about 0.25V @100mA and 0.35V @700mA.

The voltage delivered to the regulator is 8.65 - 8.75V

The Zener diodes allow a current shunt should either of the supercaps approach 5.1V

The supercaps will take approximately 5RC to charge, so about 15 seconds or so. During the charge time the power supply would provide initially 700mA tapering off to the 200mA load I allowed for.

The supercaps are 0.16F (for one pair of 0.33F in series)
The maximum voltage on the caps is about 8.75

The minimum voltage at which your output is regulated is (3.3 + 1.5 + 0.25) = 5.05V

At 150mA load, this should translate into approximately 3.5 seconds of backup capability and assuming decent caps with ESR less than 1 Ohm. As Tony pointed out in the other answer, the ESR will impact the result you obtain, but there are plenty of low ESR supercaps available.

I assume here that the problem you had with much smaller backup times was that the power supply was conducting extra current, which is why I put D4 in place.

With reference to the LM1117 datasheet, note that they recommend a 10uF capacitor on the input (read section 8.2.2.1.1). This would provide a low impedance path to control any instability ...they also recommend 10uF minimum on the output rail (read 8.2.2.1.3).

Update_1:

D1 and D2 are providing the protection against either supercap being charged to an excess voltage, they cannot be left out.
Consider that the values of the supercaps vary by 40% (one is 20% high the other 20% low) and then calculate the voltages across the caps as they charge in series. One will be high (the low C value) and the other low (the high C value). It is NOT simply half of the supply across each cap, so you CANNOT say that because the input voltage is 9V that the maximum charge voltage will be 9/2. Two conditions exist that will cause the voltage on each cap to vary.

1. The value of the caps is +/-20%
2. The leakage current will be different.

The Supercap manufacturers do supply balancing boards (like a BMS) for the leakage current, but you have to design your own overvoltage protection.

• Thanks. I put 0.22F in the calculator because it was one of the configs I tested. I put a 3.3v 0.22f cap on the output side of the VR expecting that it would power the 3.3v circuit without wasting energy through the VR. It seems the energy was back feeding through the VR though. When doing my tests I pull the plug from my board so the 9v PSU is completely removed. This results in the 3.3v circuit dying almost instantly. When I instead pulled the VR out of the board the 3.3v circuit did stay on for a short time. This is what makes me think it was back feeding when the VR was plugged in. – ubiquibacon Jul 2 '18 at 18:20
• I realize the image of the calculator is not correct for what I just explained though since it has 9v and 3.3v as opposed to 3.3v and 2.8v which is the minimum voltage required to run my 3.3v circuit. – ubiquibacon Jul 2 '18 at 18:33
• Also, what is that 10uf cap on the right side of the circuit for? – ubiquibacon Jul 3 '18 at 5:08
• @ubiquibacon ...added to answer. You have to read the datasheet for components you use. – Jack Creasey Jul 3 '18 at 5:38
• You said "The Zener diodes allow a current shunt should either of the supercaps approach 5.1V". With a 9V DC source, how could either caps ever go over 4.5V? What would happen if I eliminated D1 and D2? – ubiquibacon Jul 12 '18 at 5:39

EE101 Physics and Conservation of charge energy.

Update:

Your problem is not Specifying/Verifying C voltage and 9V source current limit which can demand 100A without due to cap. ESR.

A 9V battery cannot supply more than 1A without getting warm and are current limited by internal ESR.

Therefore your result is insufficient charge time with a battery but ok with a PS. However high ESR from clone supercaps with ESR>1 Ohm will cause significant Ohm's Law voltage drop even with 0.15A current reducing your storage time significantly.

These are common oversights and the reason why Maxwell Ultracap test specs are so detailed. Since these are obviously NOT MAXWELL Ultracaps you need the specs.

See these how high some are in ESR ( 150 Ohms ) which will fail from ESR*C=T (53%) in your application.

Note that this is a very inefficient way to store and transfer energy due to the losses involved. The rise in voltage was not specified due to current limit of supply.
So Ic=CdV/dt with current limit I for charge to reach initial voltage in 1 second and discharge to 3.3+1.1V (cutout).

simulate this circuit – Schematic created using CircuitLab

Understand the exact equivalent circuit is the key to solving this problem not using online calculators and make incorrect assumptions about your circuit.

Your Energy load is defined by power and time, if constant $E=P*t=V*I*t=3.3\ V* 0.15\ A*3s$ = 0.5 watt * 3 sec = 1.5 joules

A battery or capacitor will transfer energy by changes in energy stored, $Ec$ in $Ec=\frac{1}{2}C(V_1^2-V_2^2)$ = 1.5 joules from 9 V to $V_2=V_{cutout}$ of "buck type" SMPS converter, which I suggested from OKI.

For a linear regulator the cutout is given as 1.1 V above the output or 4.4 unless using efficient FET type LDO, thus the bipolar cheap LDO loses more energy than the load. So V2 = 4.4 V

This assumes C is constant over the time interval which is not strictly true for Maxwell Supercaps (see fine print) or batteries if not in the nominal midrange.

So expect something less than ideal. Perhaps 50% for times not >> T due to memory effects, depending on ESR*C=T

Since $V_1$ is 9 V and you can figure out $V_2$ from specs and C from above as you minimum size solution plus say 25% for efficiency losses.

Since you know math, I trust you can solve for C and hopefully lessons learnt for choosing a mismatched battery to cap regulator poor choices.

Next time define overall goal with output spec values and more efficient energy storage means such as a L-ion cell and FET LDO.

• Thanks for the detailed answer. I'm not a professional EE so I misunderstood some of the specs on the data sheet. I thought the dropout voltage of my LM1117 (1.2v) was actually 1.2v, not 1.2v + 3.3v. I had avoided using a buck converter since it is more expensive. I'll do new math tonight since I now know where I went wrong. Thanks again! – ubiquibacon Jul 2 '18 at 14:54
• A super cap can chargeup >>1000x faster than a battery of same size but also is >\10000 smaller capacitance. Decide what you really want to do and let us know, – Tony Stewart Sunnyskyguy EE75 Jul 2 '18 at 16:21
• A battery is a no go. I need something that will not need to be replaced and optimality will be nearly fully discharged after the 3 seconds has past. – ubiquibacon Jul 2 '18 at 18:05
• You source and load specs are weakly defined. – Tony Stewart Sunnyskyguy EE75 Jul 2 '18 at 18:17
• If after charging V=9V and then drops abruptly when removed, at 0.15A CC load then the voltage drop overall due to 6 caps ESR will drop voltage immediately V1-V2=I*2/3*ESR of each cap average. 2/3 since 2S3P config on 6 caps. So if each cap is 15 Ohms, drop V=1.5V , But if 45 Ohms each, then drop V= 4.5V and alsmost immediate low voltage from regulator . Your (ESR) mileage may vary. – Tony Stewart Sunnyskyguy EE75 Jul 2 '18 at 18:55