In light of all the instructables using supercaps as a power source, I present...

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With the first equation, you can find the percentage of charge (Q/Q_max) X (100%), by substituting the time elapsed, resistance of charging circuit and capacitance of capacitor. Assuming that your cap is at zero charge before charging.

The product RC is also known as the time constant.

For example, with a circuit resistance of 10â¦ and a 6F capacitor, the time constant is 60 seconds.

Lets assume our circuit is as follows:

- 10â¦ resistance

- 6F, 5V capacitor

- 5V DC source

After one time constant (60s) has passed, the charge ratio would be 0.632, hence 63.2% of maximum charge has been attained.

The thing about capacitors is that as the charge in the caps increase, the rate of charge decreases, this decrease becomes more significant as the cap approaches full charge.

Technically, the time taken for a full charge would be ∞, but in practice 5 time constants is the time taken to reach full charge.

Next, how to calculate energy stored in a capacitor.

Plug in the values into the equation given above, couldn't be simpler.

Example: 1.5F, 5.5V cap would have energy = 22.6875J.

One last useful equation would be to calculate for instantaneous voltage in capacitor when charging.

Using the first equation, substitute Q for capacitor voltage and Q_max for voltage of DC source.

*note that capacitor voltage will never rise above that of the source, and supply voltage used for charging should not be greater than the capacitor's voltage rating. DO NOT attach supercapacitors in reverse polarity to a source, it probably won't explode but it hurts to see an expensive component go up in flames.

Hope this has been of use.

- To learn more, visit hyperphysics.

- To get your hands on some supercaps, order samples from Cooper Bussmann.

**Q/Q_max =1-e^(-t/RC)**&

**E=1/2 CV^2**With the first equation, you can find the percentage of charge (Q/Q_max) X (100%), by substituting the time elapsed, resistance of charging circuit and capacitance of capacitor. Assuming that your cap is at zero charge before charging.

The product RC is also known as the time constant.

For example, with a circuit resistance of 10â¦ and a 6F capacitor, the time constant is 60 seconds.

Lets assume our circuit is as follows:

- 10â¦ resistance

- 6F, 5V capacitor

- 5V DC source

After one time constant (60s) has passed, the charge ratio would be 0.632, hence 63.2% of maximum charge has been attained.

The thing about capacitors is that as the charge in the caps increase, the rate of charge decreases, this decrease becomes more significant as the cap approaches full charge.

Technically, the time taken for a full charge would be ∞, but in practice 5 time constants is the time taken to reach full charge.

Next, how to calculate energy stored in a capacitor.

Plug in the values into the equation given above, couldn't be simpler.

Example: 1.5F, 5.5V cap would have energy = 22.6875J.

One last useful equation would be to calculate for instantaneous voltage in capacitor when charging.

Using the first equation, substitute Q for capacitor voltage and Q_max for voltage of DC source.

**V_cap = V_source *[1-e^(-t/RC)]***note that capacitor voltage will never rise above that of the source, and supply voltage used for charging should not be greater than the capacitor's voltage rating. DO NOT attach supercapacitors in reverse polarity to a source, it probably won't explode but it hurts to see an expensive component go up in flames.

Hope this has been of use.

- To learn more, visit hyperphysics.

- To get your hands on some supercaps, order samples from Cooper Bussmann.

<p>I collected experimental data and found Tau was not constant for my EDLCs. My nominal values were R=33ohms, C=1Farad making the nominal Tau = 33, but the experimental value of RC rose during discharge. Here is an example of how Tau (equal to RC) changes over time during a capacitor discharge:</p>

<p>This is not an accurate model of how a supercapacitor charges. For one, RC is not a constant, it is a function, meaning the charge and discharging <strong>rate</strong> changes! Specifically, the value of Tau (Tau=R*C) changes.</p>

<p>hello guys;</p><p>i have sun tracker <br>project, i wanted anyone can help me to know the specifications of my <br>supercapacitors ( my dc source voltage is 20 volt ) i need to have 12 volt and 1.2 amp from my supercapacitor</p><p>thanks. </p>

<p>So what is the value of 'e'???</p>

Why do you need 10 Ohm resistance?

Connecting a supercapacitor without a resistor across a source could lead to a short as the capacitor will draw a large current.