Super Capacitor LED

This is a simple charge and discharge circuit with over voltage protection. This device is useful for emergency lighting or renewable energy storage if you use bigger capacitors.

I used a super capacitor for this circuit that you can see in the photo. It all started when I purchased a super capacitor at Rocky Electronics on their website: https://rockby.com.au

I applied reverse voltage protection because the four diodes (D1, D2, D3 and D4) conduct in only one direction.

Initially I made the circuit without the diodes. I ignored the maximum voltage of my super capacitor that was only 5 V (because my other capacitors have a maximum voltage of 16 V) and thus burned my expensive super capacitors. A few weeks later I paid another $10 for two additional super capacitors and this time completed my circuit.

Warning: Do not connect this circuit to USB port.

Components: 0.1 Farad super capacitors - 2 (you might need spare), Bright LED - 2 (you might need spare), lead free solder, insulated wires, 1 mm metal wire, 1 Watt 100 ohm resistor - 3 (you might need spare), 0.25 Watt 100 ohm resistor - 3 (you might need spare), general purpose diodes - 10 (you might need to connect a few diodes in parallel), power source (AA/AAA batteries in series) - Do not connect to USB port.

Tools: Soldering iron, pliers, wire stripper, drill, drill bit.

Optional components: matrix board, encasement.

Optional tools: multi-meter.

The four diodes could have been replaced with a Zener diode. However, those are low cost high power and high current diodes that I used. The only look expensive (refer to make the circuit step). The maximum current is 1.5 Amps.

I drawn the circuit in free student edition software to save money. Thus I had to model the bright LED with three typical diodes.

Calculate maximum LED current:

Iled = (VcMax - Vled) / (Rd1||Rd2)

= (4*VdMax - Vled) / (Rd1||Rd2)

= (4 * 0.7 V - 2 V) / (100 ohms * 100 ohms / (100 ohms + 100 ohms))

= (2.8 V - 2 V) / (10000 ohms / 200 ohms)

= 0.8 V / 50 ohms

= 16 mA

Calculate the maximum power dissipation across the LED resistors:

Prdmax = (Iled / 2) * (Iled / 2) * Rd

= (8 mA)*(8 mA) * 100

= 0.0064 Watts < 0.25 Watts

= 6.4 mWatts < 250 mWatts

The other resistors will have to dissipate a lot more power:

Prsmax = Vrs * Vrs / Rs

(Vs - VcMax - Vd)*(Vs - VcMax - Vd) / Rs

= (5 V - 2.8 V - 0.7 V) * (5 V - 2.8 V - 0.7 V) / 100 ohms

= 1.5 V * 1.5 V / 100 ohms

= 0.0225 W

= 22.5 mW

Calculate charging time constant:

The charge time constant for capacitor voltage under 2 V is:

Time Constant = Rs1||Rs2 * (C1 + C2)

= 100 ohms * 100 ohms / (100ohms+ 100 ohms) * (0.1 F + 0.1 F)

= 50 ohms * 0.2 F

= 10 seconds

Calculate discharging time constant:

Usually we can calculate the discharging time constant if we know the equivalent resistance "seen" by the capacitors.

Assume the LED consumers 20 mA we can assume that the equivalent resistance is:

RledEq = Vd / Iled

= 2 V / 20 mA = 100 ohms

Thus the resistance seen by the capacitor is:

Req = Rd1||Rd2 + RledEq

To make the problem more complicated we assume that Ds diode is a short circuit (if Ds is not short circuit then Rs1 and Rs2 are infinite and can be ignored because they are not connected during discharge phase).

Req = Rs1||Rs2||(Rd1||Rd2 + RledEq)

= 100 ohms ||100 ohms||(100 ohms||100 ohms + 100 ohms)

= 100 ohms ||100 ohms||150 ohms

The the equivalent resistance equals to:

1 / Req = 1 / Rs1 + 1 / Rs2 + 1 / (Rd1||Rd2 + RledEq)

1 / Req = 1 / 100 ohms + 1 / 100 ohms + 1 / 150 ohms

Req = 1 / (1 / 100 ohms + 1 / 100 ohms + 1 / 150 ohms)

Req = 37.5 ohms

Thus you might think that the time constant is 0.2 F * 37.5 = 7.5 seconds. However, this assumption cannot be 100 % correct because when the capacitor voltage falls slightly below maximum LED voltage of 2 V, the LED almost turns OFF because the bright LED is a diode and possess non-linear characteristics. Thus the best way to predict the discharge time constant is via software simulations.

Simulations showed that the capacitor voltage discharges very quickly to 2 V but takes longer to fully discharge.

The maximum current is 14 mA, not 16 mA, probably because the maximum LED model voltage is slightly higher than predicted.

You will need high power resistors for Rs1 and Rs2 but low power resistors for Rd1 and Rd2. However, you can see that I used high power resistors (1 Watt) for both pairs because I had them in stock and it is easier to read the resistance value in the photo if you use bigger resistors.

I used a drill to make the hole for the bright LED.

My encasement can close without any screws. However, I have to apply force to open and close my box.

Testing showed that the LED is ON for many seconds after the power supply is disconnected.