Capacitors Charging and Discharging Experiment for School Learning

PLEASE NOTE THAT THIS INSTRUCTABLE IS FOR BEGINNERS IN ELECTRONICS.

Capacitors are everywhere in our lives. But for some beginners in electronics, getting to grips with capacitors can be puzzling that they’re not as straightforward to understand as resistors, to know more about resistors, please kindly refer to Resistor Knowledge in a new tab. This instructable aims to walk you through how capacitors work under DC conditions using a circuit that’s easy to understand and build.

The diagram above shows a circuit that can demonstrate the process of charging and discharging capacitors. The charging circuit consists of S1, R1, a red LED, and electrolytic capacitors C1 and C2. The charging current is indicated by the red LED. On the right side of the circuit diagram, there’s a discharging circuit made up of S2, R2, a green LED, and capacitors C1 and C2. The discharging current is indicated by the green LED.

Material Requirements:

1 x 830 Points MB-102 Breadboard

1 x Red LED

1 x Green LED

2 x 470uF Capacitors

2 x 470Ω Resistors

1 x 2-Bit Slide Type Switch

10 x Jumper Wires (Random Colors Random Lengths)

1 x 4-slots AAA Battery Holder or Voltage Generator to produce 6V DC power source

Please note that both the LED and electrolytic capacitors have polarity requirements. Incorrect polarity may result in damage.

for the LED, the pin near the flat surface, or the shorter pin is the negative end.

for the electrolytic capacitor, the pin near the white band on its body, or the shorter pin is the negative end.

When you close switch S1 (while keeping S2 open), the power source starts charging capacitors C1 and C2 through resistor R1 and the red LED D1. At the moment the power is turned on, since C1 and C2 don’t have any charge yet, the voltage across them is zero. This means the current through the red LED is at its maximum, so the light is the brightest at the very beginning.

But as time goes on, the red LED starts to dim. That’s because the capacitors are building up more and more charge over time. This charge starts to oppose the charge from the power source. When the voltage across the capacitors gets close to the voltage of the power source, the capacitors are full of charge. There’s no more current flowing through the red LED, so it looks like it’s almost off.

Even when the capacitors are full, the red LED still has a very faint light. This is because capacitors have a small leakage current when they leave the factory, so they don’t completely block direct current in an ideal state.

The resistor R1 acts as a current limiter, slowing down the rate at which the power source charges the capacitor. The larger the resistance value of R1, the smaller the current flowing through the red LED when switch S1 is closed, which in turn dims the LED. However, this also extends the time the LED stays lit. If you want to experiment with this circuit, you can compare the effects of using different resistance values for R1. By observing how long the red LED stays lit, you can get a hands-on understanding of how the current-limiting resistor R1 affects the charging and discharging of the capacitor. This can help deepen your understanding of the factors that determine the charging time of a capacitor.

When capacitors C1 and C2 are fully charged, switch S1 is opened, disconnecting capacitors C1 and C2 from the power source. When switch S2 is closed, you will see the green LED D2 start to light up, indicating that capacitors C1 and C2 are discharging. This demonstrates the ability of capacitors to store electrical energy.

As the capacitor discharges, the stored charge decreases, causing a gradual drop in voltage across the capacitor and an exponential decrease in the discharge current. As a result, the brightness of the green LED quickly dims and eventually goes out. The larger the capacitance and the resistance of the current-limiting resistor R2, the longer the discharge time of the capacitor and the longer the green LED will stay lit.

It’s important to note that in this circuit experiment, due to the parasitic resistance produced by the PN junction of the green LED, the capacitor discharge is not complete. After the LED goes out, there will still be about a 2V voltage present at the terminals of the capacitor. This is because this 2V voltage is not enough to make the PN junction of the green LED conductive (the working voltage of the yellow LED is 3.0V to 3.6V), so there will still be a 2V voltage locked at the terminals of the capacitor. Touching the terminals of the capacitor with the end of a screwdriver can completely discharge the capacitor.

Brief Summary:

From the analysis of the charging and discharging process of the capacitors above, we can intuitively feel that the charging and discharging time of the capacitors is related to the limiting resistors R1, R2 and the capacitors C1, C2. The charging and discharging time constant Tau denoted by Greek letter ‘τ’, is the product of the resistance value ‘R’ and the capacitance ‘C’, that is,

τ=RC,

Where the unit of ‘R’ is ohms (Ω), the unit of capacitance ‘C’ is farads (F), and the unit of ‘τ’ is seconds (s).

This circuit uses two 470uF electrolytic capacitors in parallel to increase the capacitance. By connecting multiple capacitors in parallel, we can get an equivalent capacitor with a larger capacity. This is done to extend the time the LED light stays on, making it easier to observe the charging and discharging process of the capacitors.

Besides demonstrating the charging and discharging process of capacitors, this circuit also gives beginners in electronics a hands-on understanding of the characteristic of capacitors to prevent sudden voltage changes across their terminals.

From the waveform in the diagram above, you can see that when switch S1 is closed to connect the power source, the voltage across resistor R instantly increases to a value close to the power source. However, the real-time voltage across the capacitor follows a curve that gradually slows down and decreases in curvature. The voltage across the capacitor doesn’t instantly become equal to the power source like the resistor does. Instead, it gradually increases over time, effectively preventing a sudden voltage change across the capacitor.

This characteristic makes capacitors really useful in scenarios where you need to filter specific frequencies or compensate waveforms. They’re pretty versatile in their applications.

The characteristic of capacitors to prevent sudden voltage changes across their terminals is also evident when they’re discharging. The voltage at the positive terminal of the capacitor doesn’t just suddenly drop to the lowest value. Instead, it decreases over time, following a curve where the slope gets smaller and smaller. This is because the capacitor doesn’t let the voltage across its terminals drop sharply. Instead, it allows the voltage to decrease smoothly.

If you want to know more about the theoretical aspects of the capacitors, please refer to Capacitor Knowledge.

To get the experiment material for learning purpose please go to mondaykids.com.

If you found this instructable helpful, I would be honored if you could share it with your friends, family, and classmates etc. who are also learning about electronics. In the future, I will do my best to provide more interesting and easy-to-understand knowledge on electronic circuits to help you better grasp both analog and digital circuits. Thanks