# Steps to Build an Electronic Circuits

##### What is a circuit and why do we need to build a circuit?

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.

Here our choice of the software is Multisim.

Below are the given steps to draw a circuit using Multisim and simulate it.

1. On your windows panel, click on the following link: Start >>> Programs –> National –> Instruments –> Circuit design suite 11.0 –> multisim 11.0.
2. A multisim software window appears with a menubar and blank space resembling a breadboard, to draw the circuit.
3. On the menu bar, select place –> components
4. A window appears with the title-‘select the components’
5. Under the heading ‘Database’ – select ‘Master Database’ from the drop-down menu.
6. Under the heading ‘group’- select the required group. If you want to go for a voltage or current source or ground. If you want to go for any basic component like a resistor, a capacitor, etc. Here first we have to place the input AC supply source, hence select Source –>Power Sources –> AC_power. After the component is placed (by clicking the ‘ok’ button), set the value of RMS voltage to 230 V and frequency to 50Hz.
7. Now again under the components window, select basic, then transformer, then select TS_ideal. For an ideal transformer, the inductance of both coils is the same, to achieve the output we have the change the secondary coil inductance. Now we know the ratio of the inductance of the transformer coils is equal to the square of the ratio of turns. Since the turns ratio required in this case is 19, therefore we have to set the secondary coil inductance to 0.27mH. (Primary coil inductance is at 100mH).
8. Under the components window, select basic, then diodes, and then select the diode IN4003. Select 4 such diodes and place them in a bridge rectifier arrangement.
9. Under the components windows, select basic, then Cap _Electrolytic and select the value of the capacitor to be 20microFarad.
10. Under the components window, select power, then Voltage_ Regulator, and then select ‘LM7805’ from the drop-down menu.
11. Under the components window, select diodes, then select LED and from the drop-down menu, select LED_green.
12. Using the same procedure, select a resistor with the value of 100 Ohms.
13. Now that we have all the components and have an idea about the circuit diagram, let us get into drawing the circuit diagram on the multi sim platform.
14. To draw the circuit, we have to make proper connections between the components using wires. To select wires, go to Place, then wire. Remember to connect the components only when a junction point appears. In multisim, the connecting wires are indicated by red color.
15. To get an indication of the voltage across the output, follow the given steps. Go to Place, then ‘Components’, then ‘indicator’, then ‘Voltmeter’, then select the first component.
17. Now click on ‘Simulate’ then select ‘Run’.
18. Now you can see the LED at the output blinks, which is indicated by the arrows going green in color.
19. You can verify whether you are getting the correct value of voltage across each component by placing a Voltmeter in parallel.

Now you have an idea about designing a regulated power supply for loads that require a constant DC voltage, but what about loads that require variable DC voltage. I leave you with this task. Furthermore, any queries regarding this concept or electrical and electronics projects Please provide your ideas in the comments section below.

1. CRS007 says:

Very good sir
I learnt many practical things here
Thank u so much

1. Tarun Agarwal says:

Hi
Thank you for giving us a perfect rating

2. Subburaj says:

Sir i searched lot of websites but i can’t get detailed solution. You open my eyes to learn electronis…my heart says this is the best article at all…..

1. Tarun Agarwal says:

Hi Subburaj
Thank you for taking the time to review your stay with us

i searched a lot. but now i find you…… you explained very well; circuit components value calculation…. kindly make more articles or videoes on it. because this topic related work is not in adequate quantity on google……. i shall be thankfull to you. regards!

1. ankit Lohiya says:

Hi Tarun,

4. Mhfooth says:

Thanks for a very helpful information.
Could you help me about the inverters I mean how can I get AC from DC power supply step by step like this way of explaining.

5. Moisés a Guimarães says:

Hello Friend! Very good content presented! How to make 3.7vdc for 5vdc?

6. febin mohamed says:

Hello sir/madom,I want to know the circuit design of a automobile brake indicator circuit.

As i am Beginner i could not understand the full circuit so I want to Know How we are Selecting the Components value ,Also I want to know all formula and i want to know about like how we are converting the Schematic to wiring diagram for this all please brief me with one one Example so that it will be helpful for me to learn .

Thanks And Regards
Mahesh

8. Alok Sharma says:

Hi there,
I am looking to get few reliable circuits build to test some controllers operations, I cannot understand the whole thing. I was wondering what kind of help i can get from you guys
thanks

9. chandrasekar says:

i want to know how to select transformer in detail ,i cannot understand with this brief explanation..

10. mujeeb says: