# Capacitor Power Bank

#### ambrose457

Joined May 5, 2020
1
I am working on a home project that uses an actuator to open and close a sliding door.
The problem that I'm trying to solve is how to open the door when the power goes out.

To open the door the actuator requires 12VDC at 5Amps for 15 seconds.

I want to build a circuit using capacitors or super capacitors.
I do not want to use a battery.

Any suggestions on the design will be appreciated.

#### shortbus

Joined Sep 30, 2009
8,043
Before spending much time, effort or money on the project. look for links on something called *RC time constants*. If you're honest with yourself, you'll see a battery is a better thing.

#### Teljkon

Joined Jan 24, 2019
173
I would also be interested in how to do this i have a few super capacitors my self. That I want to hook up to air pump for inflating tires. mostly as a exercise to learn about super capacitors.

#### Papabravo

Joined Feb 24, 2006
14,664
The equation you are working with starts with a known voltage on the capacitor. Then you have a load that the capacitor discharges through.

$V_C\;=\;V_0e^{\frac{-t}{RC}}$

One time constant is equal to the product of the capacitance and the load resistance.
At t= 0, the voltage on the capacitor is equal to $$V_0$$
At t = RC, one time constant later the voltage on the capacitor is $$0.368V_0$$
At t = 2RC two time constants later the voltage on the capacitor is $$0.135V_0$$
At t = 3RC three time constants later the voltage is $$0.0498V_0$$

Point is, a capacitor will discharge so quickly through a low impedance load that it is hardly worth the effort to charge it up. Worse the voltage decays exponentially. So:
1. You can't win
2. You can't even break even
3. You can't get out of the game

#### Hymie

Joined Mar 30, 2018
877
If you consider that the capacitor charge or discharge is due to a constant current (rather than into a resistive load), then the mathematics become very simple.
The change in voltage can be expressed by the following equation.
V = (I x t)/C
Where:-
I is the current in amps
t is the time in seconds
and C is the capacitance in Farads
Taking your situation, if we accept an allowable voltage drop to 9V (3V drop) after 15 seconds, then transposing the above equation we get
C = (5 x 15)/3, which equals 25 farads.
A 16V, 25 farad capacitor is going to be quite expensive.

#### Teljkon

Joined Jan 24, 2019
173
If you consider that the capacitor charge or discharge is due to a constant current (rather than into a resistive load), then the mathematics become very simple.
The change in voltage can be expressed by the following equation.
V = (I x t)/C
Where:-
I is the current in amps
t is the time in seconds
and C is the capacitance in Farads
Taking your situation, if we accept an allowable voltage drop to 9V (3V drop) after 15 seconds, then transposing the above equation we get
C = (5 x 15)/3, which equals 25 farads.
A 16V, 25 farad capacitor is going to be quite expensive.

Yeah but 11 X 2.7v 100f capacitors isn't. You will have to excuse my ignorance but why wont a bank of super capacitors like this work?

#### shortbus

Joined Sep 30, 2009
8,043
Physics?

"EDLCs share the same equivalent circuit as conventional capacitors. The first order model is represented by the circuit below. It is comprised of four ideal components. The series resistance Rs which is also referred to as the equivalent series resistance (ESR). This is the main contributor to power loss during charging and discharging of the capacitor. It is also comprised of a parallel resistance Rp which affects the self-discharge, a capacitance C and a series inductor L that is normally very small as a result of the cell construction. "
From; https://www.tecategroup.com/products/ultracapacitors/ultracapacitor-FAQ.php

#### Teljkon

Joined Jan 24, 2019
173
Thank you for the info.

#### Teljkon

Joined Jan 24, 2019
173
Did you guys just assume that Chinese persons gender. Gasp For shame.

#### Teljkon

Joined Jan 24, 2019
173

#### Audioguru again

Joined Oct 21, 2019
2,054
The voltage in a capacitor drops very quickly when beginning a discharge.
But the voltage of a rechargeable battery stays up at the beginning and during a discharge.
So of course a battery is better than a capacitor to do some work.

#### Hymie

Joined Mar 30, 2018
877
OK, slapped wrists for non-pc gender terminology.

The capacitor module appears to be made up of six 100F, 2.7V capacitors in series, to give a 16V, 16.67F (claimed to be 20F) capacitor. Quite what the semiconductors and other components on the board I doing, I don’t know for sure – perhaps they are trying to equalise the voltage across each capacitor (or to ensure it cannot exceed 2.7V).

That said, the product is claimed to assist in engine cranking – given that a starter motor can draw up to 200A, if only the capacitor was providing this energy – then the capacitor would be completely discharged within 1 second; I doubt you would notice any difference in engine starting/cranking performance with this module fitted to your car.

#### Teljkon

Joined Jan 24, 2019
173
OK, slapped wrists for non-pc gender terminology.

The capacitor module appears to be made up of six 100F, 2.7V capacitors in series, to give a 16V, 16.67F (claimed to be 20F) capacitor. Quite what the semiconductors and other components on the board I doing, I don’t know for sure – perhaps they are trying to equalise the voltage across each capacitor (or to ensure it cannot exceed 2.7V).

That said, the product is claimed to assist in engine cranking – given that a starter motor can draw up to 200A, if only the capacitor was providing this energy – then the capacitor would be completely discharged within 1 second; I doubt you would notice any difference in engine starting/cranking performance with this module fitted to your car.

A YouTuber by the name of Laser hacker I think does this.