How to make an electric circuit step by step

If you are interested in understanding electronic circuits, one of the best ways to learn about electronics is to build a simple circuit. This simple circuit consists of just three components: a 9 V battery, a light-emitting diode (LED), and a resistor. Not only will you learn something about building circuits, but you can also you this completed circuit to practice using your multimeter.

Here is the schematic for this circuit:

You can build this circuit on a solderless breadboard. You'll need the following parts:

• Small solderless breadboard

• 470 Ω, 1/4 W resistor

• Red LED, 5 mm

• 9 V battery snap connector

• 9 V battery

• Short length of jumper wire (1″ or less)

Here are the steps for building this circuit:

1. Connect the battery snap connector.

   Insert the red lead in the top bus strip and the black lead in the bottom bus strip. Any hole will do, but it makes sense to connect the battery at the very end of the breadboard.

2. Connect the resistor.

   Insert one end of the resistor into any hole in the bottom bus strip. Then, pick a row in the nearby terminal strip and insert the other end into a hole in that terminal strip.

3. Connect the LED.

   Notice that the leads of the LED aren't the same length; one lead is shorter than the other. Insert the short lead into a hole in the top bus strip, and then insert the longer lead into a hole in a nearby terminal strip.

   Insert the LED into the same row as the resistor. Both the LED and the resistor are in row 26.

4. Use the short jumper wire to connect the terminal strips into which you inserted the LED and the resistor.

   The jumper wire will hop over the gap that runs down the middle of the breadboard.

5. Connect the battery to the snap connector.

   The LED will light up. If it doesn't, double-check your connections to make sure the circuit is assembled correctly. If it still doesn't light up, try reversing the leads of the LED (you may have inserted it backwards). If that doesn’t work, try a different battery.

Do not connect the LED directly to the battery without a resistor. If you do, the LED will flash brightly, and then it will be dead forever.

Before I go into details about how a circuit is designed, let us first know what a circuit is and why do we need to build a circuit.

A circuit is any loop through which matter is carried. For an electronic circuit, the matter carried is the charge by electronics and the source of these electrons is the positive terminal of the voltage source. When this charge flows from the positive terminal, through the loop, and reaches the negative terminal, the circuit is said to be completed. However this circuit consists of several components that affect the flow of charge in many ways. Some may provide a hindrance to the flow of charge, some simple store, or dissipate the charge. Some require an external source of energy, some supply energy.

There can be many reasons why we need to build a circuit. At times we may need to glow a lamp, run a motor, etc. All these devices-a lamps, a motor, LED are what we call as loads. Each load requires a certain current or voltage to start its operation. This voltage may be a constant DC voltage or an AC voltage. However, it is not possible to build a circuit just with a source and a load. We need a few more components that help in the proper flow of charge and process the charge supplied by the source such that an appropriate amount of charge flows to the load.

A basic Example – Regulated DC Power Supply to run an LED

Let us a basic example and the step by step rules in building the circuit.

Problem Statement: Design a regulated DC power supply of 5V which can be used to run a LED, using AC voltage as the input.

Solution: You all must be aware of the regulated DC power supply. If not, let me give a brief idea. Most of the circuits or electronic devices require a DC voltage for their operation. We can use simple batteries to provide the voltage, but the major problem with batteries is their limited lifetime. For this reason, the only way we have is to convert the AC voltage supply at our homes to the required DC voltage.

All we need is to convert this AC voltage into DC voltage. But it is not as simple as it seems. So let us have a brief theoretical idea about how AC voltage is converted into regulated DC voltage.

The theory behind the circuit

1. AC voltage from the supply at 230V is first stepped down to low voltage AC using a step-down transformer. A transformer is a device with two windings –primary and secondary, wherein the voltage applied across the primary winding, appears across the secondary winding by the virtue of inductive coupling. Since the secondary coil has a lesser number of turns, the voltage across the secondary is less than the voltage across the primary for a step-down transformer.
2. This low AC voltage is converted to pulsating DC voltage using a bridge rectifier. A bridge rectifier is an arrangement of 4 diodes placed in the bridged form, such that anode of one diode and cathode of another diode is connected to the positive terminal of the voltage source and in the same way the anode and cathode of another two diodes are connected to the negative terminal of the voltage source. Also, the cathodes of two diodes are connected to the positive polarity of the voltage and the anode of two diodes is connected to the negative polarity of the output voltage. For each half-cycle, the opposite pair of diodes conduct and pulsating DC voltage is obtained across the bridge rectifiers.
3. The pulsating DC voltage thus obtained contains ripples in the form of AC voltage. To remove these ripples a filter is needed which filters out the ripples from the DC voltage. A capacitor is placed in parallel to the output such that the capacitor (because of its impedance) allows high-frequency AC signals to pass through get bypassed to the ground and low frequency or DC signal is blocked. Thus the capacitor acts as a low pass filter.
4. The output produced from a capacitor filter is the unregulated DC voltage. To produce a regulated DC voltage a regulator is used which develops a constant DC voltage.

So let us now get into designing a simple AC-DC regulated power supply circuit to drive a LED.

Steps in building the circuit

Step 1: Circuit Designing

To design a circuit, we need to have an idea about the values of each component required in the circuit. Let us now see how we are designing a regulated DC power supply circuit.

1. Decide the regulator to be used and its input voltage.

Here we require to have a constant voltage of 5V at 20mA with the positive polarity of the output voltage. For this reason, we need a regulator that would provide a 5V output. An ideal and efficient choice would be the regulator IC LM7805. Our next requirement is to calculate the input voltage requirement for the regulator. For a regulator, the minimum input voltage should be the output voltage added by a value of three. In that case, here to have a voltage of 5V, we need a minimum input voltage of 8V. Let us settle down for input of 12V.

2. Decide the transformer to be used

Now the unregulated voltage produced is a voltage of 12V. This is the RMS value of the secondary voltage required for a transformer. Since the primary voltage is 230V RMS, on calculating the turns ratio, we get a value of 19. Hence we have to get a transformer with 230V/12V, i.e. a 12V, 20mA transformer.

3. Decide the value of the filter capacitor

The value of the filter capacitor depends on the amount of current drawn by the load, the quiescent current (ideal current) of the regulator, the amount of allowable ripple in the DC output, and the period.

For the peak voltage across the transformer primary to be 17V(12*sqrt2) and the total drop across the diodes to be (2*0.7V) 1.4V, the peak voltage across the capacitor is about 15V approx. We can calculate the amount of allowable ripple by the formula below:

∆V = VpeakCap- Vmin

As calculated, Vpeakcap = 15V and Vmin is the minimum voltage input for the regulator. Thus ∆V is (15-7)= 8V.

Now, Capacitance, C =( I*∆t)/ ∆V,

Now, I am the sum of the load current plus the quiescent current of the regulator and I = 24mA (Quiescent current is about 4mA and load current is 20mA). Also ∆t = 1/100Hz = 10ms. The value of ∆t depends upon the frequency of the input signal and here the input frequency is 50Hz.

Thus substituting all the values, the value of C comes to be around 30microFarad. So, let us select a value of 20microFarad.

4. Decide the PIV (peak inverse voltage) of the diodes be used.

Since the peak voltage across the transformer secondary is 17V, the total PIV of the diode bridge is about (4*17) i.e. 68V. So we have to settle down for diodes with a PIV rating of 100V each. Remember PIV is the maximum voltage that can be applied to the diode in its reverse biased condition, without causing breakdown.

Step2. Circuit Drawing and Simulation

Now that you have the idea of the values for each component and the whole circuit diagram, let us get into drawing the circuit using circuit building software and simulate it.

How do you make a simple circuit for kids?