How do you design a circuit in electrical engineering?

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How may a circuit be designed?

Any loop through which material is transported is a circuit. The charge that electronics carry serves as the material for an electronic circuit, and the positive terminal of a voltage source serves as the source of the electrons that make up this charge. The circuit is said to be complete when this charge travels from the positive terminal through the loop and reaches the negative terminal. However, this circuit is made up of a number of parts that have a variety of effects on the charge flow. Some may prevent the passage of charge, while others may just store or dissipate it. Some produce energy, while others need an outside source. For the students of the electrical engineering college in Jaipur, this procedure is simple.

Why We Need To Build A Circuit?

We occasionally might need to run a motor, a bulb, etc. These are all what we refer to as loads: lamps, motors, and LEDs. Each load must have a specific current or voltage to begin operating. This voltage could be either an AC or a steady DC voltage. A circuit cannot be created solely from a supply and a load, though. A few extra parts are required to ensure good charge flow and to process the charge given by the source so that the right amount of charge reaches the load.

Most of the circuits or electronic devices require a DC voltage for their operation. Students of best engineering colleges in Jaipur 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.

Process Of Converting AC Voltage To DC Voltage

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 including 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.

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 conducts and pulsating DC voltage is obtained across the bridge rectifiers by the experts of top engineering college in Jaipur.

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.

Steps To Build The Circuit

Step1: Circuit designing

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

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 by the experts of BTech colleges Jaipur 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 turn’s 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. 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 students of BTech electronics engineering colleges in Jaipur have the idea of the values for each component and the whole circuit diagram, drawing the circuit using circuit building software and simulate it becomes easy.

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