62

It is not practical. I do not know why people do this, there is no benefit whatsoever. It amounts to misuse of something useful. Simply put, those videos are by people who don't know what they are doing and are misusing supercapacitors for a bizarre and senseless application they are neither well-suited to nor even practical. And they are offered on the ...


42

Direct answer to the question The direct answer to your question, assuming you intend to just connect the capacitor to the LED with a series resistor is no time at all. That is because a white LED takes more than 2.7 V to light. Check its datasheet. These things usually need a bit over 3 V. There are two options. The simplest is to use a LED with a ...


20

The voltage on a capacitor is proportional to the charge stored in it. That means that as the device draws current, the voltage will drop. You have to decide how low is acceptable. Do the math. For example, let's say the device can still run from 4.5 V. That means the capacitor voltage can drop 500 mV before the system doesn't work anymore. Let's use a ...


17

Since the product you purchased has no background information, you can't be certain. However, the convention for these stacked-disk type capacitors is polarity mark points to negative lead. This is the same as is the convention with conventional electrolytic capacitors. For example, the Eaton KR-5R5V474-R: Has its datasheet show: Similarly, for the ...


15

First to answer your question (Although that has been done in the comments already): No you do not need a resistor to limit the current. The Arduino will take just what it needs. Next: Your discharge over a small voltage range 5v down to 4.5V will be ~CV/I. Olin gave a figure for that. BUT! You can get a lot more time using a smaller and cheaper ...


14

Look at page 4, fig.12, graph of safe operating area. That is exactly what you need. You are talking about single pulse, right? You didn't mention any repetition or timing at all. If you open mosfet hard, say Rdson is 0.85mOhms. In case of 1000A the Vds will be less than 1V, so you have to look at the left side of graph. There is no line for 100ms pulse, ...


14

Low energy density: for a given amount of energy, they take up more space than almost all battery technologies High leakage: this document has some nice curves. It looks like they've lost half the voltage after a week. Bad discharge curves: normal batteries might give you 80% of the power before they've dropped 20% voltage. Capacitors give out power evenly ...


11

Your gate drive voltage is too low. That MOSFET needs 10V to turn on completely. 5V just barely clears the 4V threshold when the MOSFET just barely starts to conduct. DO NOT use the Vgsth if you intend to use your MOSFET at a switch. That is the voltage it just barely starts to conduct at. Use a Vgs at least as high as the one used to obtain the given ...


10

The problem is the fixed 15v output voltage of your DC-DC converter. Capacitors should be charged with a current output, not a voltage output. If the voltage on a capacitor is less than the output voltage of a power supply feeding it, it will take as much current as the supply cares to supply, in other words it behaves like a dead short. The DC-DC ...


10

From the images of the cell in test setup photo they appear to be similar to the Maxwell DuraBlue line of Ultracapacitors. See this datasheet for more information. The Maxwell application note 1007239, Test Procedures for Capacitance, ESR, Leakage Current and Self-Discharge Characterizations of Ultracapacitors, may be helpful. This line of "supercapacitors"...


10

I'd wager that the problem isn't an oscillation, it's just the initial inrush of current into the MOSFET that's killing it. When you connect your super-caps to the circuit, it will charge up the parasitic MOSFET capacitors \$\mathrm{C_{oss}}\$ and \$\mathrm{C_{rss}}\$. According to the datasheet, \$\mathrm{C_{oss}}\$ is only about 781pF and \$\mathrm{C_{rss}}...


10

The energy stored in a capacitor is given by : $$E= \frac{CV^2}{2}$$ Fill in the numbers for both 2.7 V and 900mV: $$E_{\text{full}} = \frac{100 \text{F} \cdot 2.7 \text{V}^2}{2}\approx365 \text{J}$$ $$E_{\text{end}} = \frac{100 \text{F} \cdot 0.9 \text{V}^2}{2}\approx41 \text{J}$$ In other words, we have \$41/365\approx 11\% \$ of the full capacity ...


9

What you have will work, although D12 is pointless. The problem is that when the cap is discharging onto the 5 V line, there will be a drop across D13. Using a Schottky as you show is a good idea, but the drop will still be around 1/4 volt. Another problem is that the voltage will go lower over time with the amount of charge drained from the cap. You ...


9

Assuming ideal conditions i.e. no leakage current in capacitor and other parts of circuit. Case 1: Your micro controller is running and drawing 20 mA. Lets assume your micro controller will work fine till voltage reaches 4V. However for atmega 328, you can make it run at even lower voltages if you choose to run it at a lower clock frequency. Assuming 20 mA ...


9

It doesn't matter what voltage your power source is, as long as the capacitor is not subjected to a higher voltage than it is rated for. You could, for example, charge the capacitor from a 12 V source with a resistance in series with it. That will start the capacitor charging exponentially towards 12 V. You have to monitor the capacitor voltage and ...


9

This is Maxwell's process for measuring C from their test spec. \$C=C_{dcd}=\dfrac{I_5*(t_5-t_4)}{V_5-V_4}\$ The Capacitance is charged and discharged at rated current but measured by the rated from Urated to 50% Urated. Note that the voltage sags towards the previous voltage due to an additional RC time constant in parallel. (i.e. memory effect) Here it is ...


9

The meaning of the term supercapacitor has changed over time. In the 1980s, it referred to devices designed for very low current memory backup power and they had a relatively low output current capability (high ESR). I remember one that was 4F, 5V in a relatively small package but had a current capability of perhaps 10mA. Newer devices have low (and ...


9

Eaton HS model is not a regular supercapacitor, but it is a hybrid supercapacitor. It is a hybrid of regular supercapacitor and lithium-ion cell. So same restrictions apply due to the lithium-ion cell technogy - the lithium-ion part of the hybrid capacitor will get damaged when it is discharged below safe voltage.


8

For the connection of the battery to the MCU's VCC pin, you can use a simple dual Diode "OR" with low-forward drop diodes. This will mean as VCC is lost, the cap still stop charging and the diode input for VCC will drop, but the diode input for Cap-> MCU's VCC will continue on until the discharge curve shown by Whiskeyjack reaches a critical point where the ...


8

From Cooper Bussmann on supercapacitors: Voltage Supercapacitors are rated with a nominal recommended working or applied voltage. The values provided are set for long life at their maximum rated temperature. If the applied voltage exceeds this recommended voltage, the result will be reduced lifetime. If the voltage is excessive for a prolonged time period, ...


8

Another option could be to get a DC-DC step up converter which gives a 5V output. This can run on an input of 0.9V - 5V. Based on your 15F capacitor and a voltage drop of 4V (for a 5.5V supercapacitor charged to 5V) you could get 4000mV * 15F / 41mA (85% efficient) = 1,463s or about 24 minutes run time.


7

A supercapacitor or Electric double-layer capacitor (EDLC) is functionally no different from a polarized capacitor, at the schematic level-of-abstraction. Hence, there is no standard symbol for it (yet), as distinct from the polarized capacitor symbol. As with other schematic symbols, if there is a pressing need to indicate that a particular part on a ...


7

I recently read an article ... that seemed to indicate that "supercapacitors" are starting to approach the energy densities of batteries for some applications. It's possible that there may be niche applications where that is true (although none come to mind with a quick musing) but for even 'ordinary everyday' batteries they have a way to go yet as regards ...


7

This comparison is kilo-for-kilo. Disadvantages of supercapacitors Lower energy density than batteries. Higher self-discharge rates. Loses charge over time. Advantages of supercapacitors Can receive charge more quickly than batteries. Can output more current than batteries. Supercapacitors are better than batteries as a source of power, but they are ...


7

Yes, you can use switching power supply topologies to slowly and safely discharge a large capacitor bank. These can be controlled to create exactly the load current you need. Basically every switching power supply has some DC intermediate circuit with a capacitor bank, exactly as you described above. Still, the remark from the comments seems to be limiting ...


6

You can use them like that. The only drawback is the self-discharge of them internally. For a battery application this can be disastrous. For a mains powered application it's probably neglectable. If you are using them in series to get 5+ V rating, you might need to consider the difference in leakage current though them and balance them off with external ...


6

Putting capacitors in series to add up their voltage ratings isn't very wise. Their difference in leakage current will lead to an imbalance in the voltage they'll have across them (see https://electronics.stackexchange.com/a/80589/107479), and some of the capacitors could easily experience overvoltage (especially given the low voltage ratings supercapacitors ...


6

The 1N5223 zener is nominally 2.7 volts and can be as low as 2.565 volts or as high as 2.835 volts so if you can live with this variation then that's fine. The BAT54 being schottky will look like a small volt drop as the capacitor reaches full charge so the only problem here is what the residual current taken by the capacitor is - the higher the residual, ...


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