A Brief Introduction To Electrical Electronics - Tech Projects/Documentations
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A Brief Introduction To Electrical Electronics

Introduction to DC circuits:

Case study: Electrical Electronics Circuits

Author: Eze-Odikwa Tochukwu Jed

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Introduction:

An electric circuit is a closed path consisting of active and passive elements all interconnected with the current flow confined to a closed path.

Active elements: Are elements that supply energy to the circuit. Common examples of active elements include: voltage sources, current sources, all different types of transistors (such as bipolar junction transistors, MOSFETS, FETs, and JFET), Diodes (such as Zener diodes, photodiodes, Schottky diodes, and LEDs).

Passive elements: Are elements that receive, consume, store energy i.e. resistors, capacitors, inductors etc.

Voltage Sources: A voltage source is an example of an active component in a circuit. When current leaves from the positive terminal of the voltage source, energy is being supplied to the circuit. As per the definition of an active element, a battery can also be considered as an active element, as it continuously delivers energy to the circuit during discharging.

Current Sources: A current source is also considered an active component. The current supplied to the circuit by an ideal current source is independent of circuit voltage. As a current source is controlling the flow of charge in a circuit, it is classified as an active element.

Transistors: Although not as obvious as a current or voltage source – transistors are also an active circuit component. This is because transistors are able to amplify the power of a signal.

Derived units:

UnitNameQuantity
symbol
Unit
Symbol
Expression
in other
units
Absorbed dosegrayGyJ/Kg
ActivitybecquerelBqs-1
Electric Capacitance faradCFC/V
Electric ConductancesiemensGSA/V
Electric InductancehenryLHWb/A
Electric Potential
difference
voltV,EVW/A
Electric ResistanceohmRΩV/A
EnergyjouleWJN.m
ForcenewtonFNKg.m/s2
FrequencyhertzfHzS-1
Illuminance luxElxlm/m2
Luminous fluxlumenlmcd.sr
Magnetic fluxweberWbV.s
Magnetic flux
density
teslaBTWb/m2
PowerwattPWJ/s
Pressure, stresspascalPPaN/m2
Quantity of
electricity
coulombQCA.s

Electric current: The concept of charge is based on atomic theory. An atom has positive charges (protons) in its nucleus and an equal number of electrons (negative charges) surround the nucleus making the atom neutral. Removal of an electron leaves the atom positive charged and addition of an electron makes the atom negatively charged. The basic unit of charge is the charge on an electron. The mks unit of charge is coulomb. An electron has a charge of 1.062 x l0-19 C. When a charge is transferred from one point in the circuit to another point it constitutes what is known as electric current. An electric current is defined as the time rate of flow of charge through a certain section. Its unit is ampere. A current is said to be of one ampere when a charge of 1 coulomb flows through a section per second.

i=\frac{dq}{dt}

If charge q is expressed in coulomb and time in second, 1 amp flow of current through a section is equivalent to approx. flow of 6.24 x 1018 per second through the section. Another method of defining 1 amp of electric current is as the constant electric current in two infinite parallel conductors separated from each other by 1 m, experience a force of 2 x 10-9 N/m.

Electrical potential: Law of conservation of energy states that energy can neither be created nor destroyed. However, it can be converted from one form to another e.g. Petrochemical to heat, Electro-chemical conversion, Photo-electric, etc. The movement of charges that contribute to current and the amount of work done per unit charge is the potential difference between the two points. The electronic charges flow from a lower potential to a higher potential and these contribute to electronic current, whereas the conventional current is considered to flow from higher potential to lower potential. If a differential charge dq is given a differential energy dw, the rise in potential of the charge

v=\frac{dw}{dq}

The units of potential or potential difference as derived from equation are joule/coulomb, and usually is termed in volts.

or=\frac{Watt.sec}{Amp.sec}or=\frac{watt}{amp}

We have defined electric potential in terms of electronic charge flow. However we can also explain with reference to static charges, considering electric field due to such charge. Suppose we have a point charge Q in space located at some point. The very presence of this charge gives rise to an electric field emanating radially from this charge in all directions. The strength of the electric field decreases as we move away from the charge Q which practically reduces to zero at infinity. If we move a differential charge dq from = towards the charge Q, the differential charge experiences a force of repulsion or attraction depending upon whether the two charges Q and dq are of same polarity or opposite polarity respectively. Suppose Q and dq are positively charged. If we bring charge from infinity to point r1 from Q, the force experienced by the charge dq according to Coulomb’s law is given as:

df=\frac{Q.dq}{4兀£_{0}r_1^2}

Or in general if the distance is r from the charge Q the force experienced by the charge dq is:

df=\frac{Q.dq}{4兀£_{0}r^2}\overline{a{r}}

Where \overline{a_{r}} is a unit vector along the line joining Q and dq and indicates the direction of force. Now if this differential charge is moved through a distances dr, the workdone on dq against the electric field is:

dW=dF.dr\frac{Q.dq}{4兀£_{0}r_1^2}dr

or the total work done against the electric field by moving the charge from infinities to r is:

W=\int_{⍺}^{r} \frac{Qdq}{4兀£r^2}=\frac{Qdq}{4兀£r}

or the work done divided by dq is

\frac{W}{dq}=\frac{Q}{4兀£_r}=volt=potential

In fact by definition potential at any point r from the charge Q, due to charge Q is the work done against the electric field when a charge of one coulomb is moved from infinity to that point. However, in our explanation we have not taken the test charge as unit charge but a differential dq << Q so that when we bring test charge in the electric field set up by charge (Q) if the electric field is not disturbed and that is the basic requirement of any measurement i.e. any device that is used for measurement of some quantity in a circuit, should not disturb the quantity we are supposed to measure. However, we defined potential at a point due to a charge distribution is the work done per unit charge when it is moved from infinity to that point.

\frac{W}{dq}=\frac{W}{It}

the derived unit for power (P) is the watts such that

p=\frac{W}{t}

hence

\frac{W}{dq}=\frac{W}{It}=\frac{P}{I}=V

To find potential difference between two points say r1 and r2, we get

\frac{Q}{4兀£}\begin{bmatrix}\frac{1}{r2}- \frac{1}{r1}\end{bmatrix}

Where r2 and r1 are measured from the charge Q. It is to be noted that when a unit charge moves from a point of higher potential (closer to Q in this case) to a point of lower potential (relatively farther from Q) it gives up energy. However, when a unit charge moves from a point of lower potential to one of higher potential, it receives energy. In the former case the potential difference is known as voltage drop whereas in the latter case the voltage rise. If potential is multiplied by the current

\frac{dq}{dt}

v\times i\frac{dw}{dt}\times\frac{dq}{dt}=p

Which gives rate of change of energy with time and is equal to power.

thus power p=v\times i

since p=\frac{dw}{dt}

Lets briefly run a few calculations with some examples

Example 1: A person receives a severe shock when he is subjected to a current of 25mA for a time duration of 30ms. How many electrons passed through the person.

Solution:

Q = It = 25 x 10-3 x 30 x 10-3 = 75 x 10-5 C

Since 1C = 6.242 x 1018 electrons

= 75 x 6.242 x 1013

 = 4.68 x 1015

ANS= = 4.68 x 1015

Example 2: A device stores 500J of energy and releases this energy in the form of an electric current of 40A which has a duration of 15msec. Determine the average voltage across the terminals of the device.

Solution:

Since Energy W=VIt

500 = Vx 40 x 15 x 10-3

V=\frac{500}{40\times 15\times 10^{3}}

= 833 volts

ANS = 833 volts

In latter topics we will be discussing resistance, capacitance and a whole lot more along with calculations including lots of differential equation theorems.

References:

www.allaboutcircuits.com/textbook/direct-current

www.escomponents.com/blog/2019/7/31/active-amp-passive-components-what-is-the-difference-between-the-two

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