# Basics of Electronics section 1. by TME Education

Hi, we are Raptors team from Lodz University

of Technology. For four years we have been building mars

rover and this is the latest, the third version. Our group consists of 12 members from different

faculties. It can be divided into four plus one extra

group. We have mechanics and designers, electronists,

IT developers and PR. We cooperate with the PhD students at the

University of Lodz in field of the science analysis. Due to the small number of team members and

interdisciplinary character of the project we perform majority of tasks together developing

and learning in different fields. Our rover is a professional, well developed

construction that can serve as a Martian Rover as well as a rescue robot. We have proven this on competitions. We took second prize at World Robot Summit

in Japan 2018 as well as fourth and fifth at University Rover Challenged in USA in 2018

and 2017. We are well prepared and we are pleased to

compete with other robotics teams all over the world. Hi I’m Marcin and this is Łukasz and we are

really pleased to welcome you to webinars of basics of electronics. These webinars were prepared in cooperation

with Students Scientific Robotics Association “SkAnER” and Raptors team. Raptors team have been building a modular

mobile, multi-purpose robot for over three years. We will start the series of workshops by getting

familiar with the basic physical quantities that will appear during the course. It is necessary to understand electronics

and move freely in the topic. The basic physical quantities, appearing in

electrical circuits are: current, voltage, resistance and power. Of course these quantities do not describe

all the physical properties of electrical circuits but it is necessary to understand

dependencies occurring in electrical systems. Electric potential Electric potential is the potential energy

of the charge related to its value or in other words the capacity of the load to do the work. Imagine a water tank filled with water to

a certain extent. Electric potential could be described as the

pressure generated by this water at the bottom of the tank. The force created by this pressure would make

the water flow through the outflow if we open the valve at the bottom. Electrical voltage

Electrical voltage is difference in electric potentials between two points imagine two

water tanks connected with a pipe and a valve between them. One of the tanks is filled to a greater extent

than the other one. The higher is the water level in the tank

the higher pressure is generated. Electrical voltage could be compared to the

difference in pressures in both tanks. Electric current

Electric current is the amount of charge that has flown at a given time or in other words

the rate of the charge flow. Continuing the situation from the previous

scene, if you open the valve between tanks the water will flow from the tank with higher

pressure to the tank with lower pressure. Electric current could be described as the

amount of water that has flown from one tank to the other tank. Electrical resistance

Electrical resistance is a measure of how difficult it is for current to flow for a

certain substance. It can be compared to a pipe connecting two

tanks. The longer and the narrower the pipe is, the

more difficult it is for the liquid to flow through it. In the actual conductor electrical resistance

is directly proportional to the wires length and inversely proportional to its cross-sectional

area. Resistance of the conductor may result from

the number of electrons in this conductor or from the density of the crystal lattice

of this conductor. This can be compared to a clogged pipe. The more dirt is in the pipe or the less water

can be found in its cross section, the slower the water flows through this pipe. Electric capacity is the ability to store charged. This quantity can be compared to a volume

of a tank and the charge stored in it could be compared to a water in the tank. Electric capacity of the element is the amount

of charge accumulated in this element divided by the voltage on this element. Power

Power is the amount of work per unit of time or in electricity a product of voltage and

current. In practice, power tells us how much energy

will be processed at a given time. In electronics the two most important physical

quantities are current,I and voltage, U. While designing, servicing or testing various

electronic devices we must measure these values across the whole system and that is why we

must necessarily learn the basic relations between those two quantities. In the 19th century German physicist, Gustav

Kirchhoff, formulated two basic laws related to voltage and current which we will constantly

use. Kirchhoff’s first law, or also called Kirchhoff’s

current law, explains how the electric current flows in and out of the nodes in the electrical

circuit system. A good analogy would be a flow of two river

branches which meet together in a certain point and create a bigger river. The overall mass of the water in the system

stays the same so the amount of water from the first branch plus the amount of water

from the second branch should give us the amount of water in the new branch, which is

described in terms of current. as the first current – I1 plus the second current –

I2 equals to the third current – I3. Kirchhoff first law states that at any node

in the electrical circuit the sum of the currents flowing into that node is equal to the sum

of the currents flowing out of that node. Note is simply a point which connects branches

and current flowing into the node has an arrow pointing towards it. Current flowing out of the node has the arrow

in the opposite direction. Let’s look at this example. Here we’ve got

six branches of which three flow current into the node and current in the other three flows

out of the node. I1 equals 2.5 amperes, I2 equals 2.2 amperes,

I3 equals 2.7 amperes, I4 equals 2.1 amperes, I5 is unknown and I6 equals 2.4 amperes. Let’s write down the equation. On the left

side we’ve got current which flows into the node and on the right side we’ve got current

which flows out of the node. I5 is unknown so let’s leave it alone on this

side of the equation. So I5 equals to I3 plus I4 plus I6 minus I1

minus I2. With the values we’ve got I5 equals to 0.7

amperes plus 0.1 ampere, plus 0.4 amperes, minus 0.5 amperes, minus 0.2 which is equal

to 0.5 amperes. Let’s consider the same example but this time

current I5 flows outwards the node. So let’s write down the equation. Again, on

the left side we’ve got current flowing into the node and on the right side we’ve got current

flowing out of the node. So I1 + I2 equals to I3 + I4 + I5 + I6. Now let’s leave the I5 alone on the right

side, so I1 + I2 – I3 – I4 – I6 equals I5. Let’s write down the values, 0.5 amperes +

0.2 amperes – 0.7 amperes – 0.1 amperes – 0.4 amperes is equal to I5. Summing this up, I5 is equal to minus 0.5 amperes. This is negative value which actually means

that this arrow should be pointed in the other direction towards the node. Adding electrical elements to our circuits

shows that nodes are as well connections between these elements, so the overall rules of calculating

the currents in this system are the same as in the previous example. What is more, if we’ve got a series of elements,

one after another, the nodes are actually between them and this kind of connection is called

serial connection. Kirchhoff`s second law, also called Kirchoff’s

voltage law, describes how the voltage is distributed among a closed loop. Original statement says that sum of all the

voltages in the closed loop is equal to zero. Loops are closed elements of the whole system

in this particular example we’ve got two loops one here and one here, and these are the shortest

paths where you don’t have shortcuts between elements. As you remember electrical voltage appears

only between two points in the system. For instance, here we have a resistor, the

voltage on this resistor would be measured between this and this point. Another thing to remember is that on the sources

like voltage source or current source, the voltage has the same direction as the current

flowing through the element. And on the other hand when we have resistors or diodes, voltage

has the opposite direction then the current flowing through the element. When we consider sources as the voltage source

and current source the voltages on these elements add up to the overall sum of the voltages

in the loop and on the other hand, on other elements voltages are subtracted showing that

there are voltage drops on these elements. Here we have a simple loop with one voltage

source and two receivers. On the voltage source the voltage has a corresponding

direction to the flowing current and on the receivers the voltage is in the opposite direction

to the flowing current. We don’t know the value of the voltage source

but we would like to have the value of U1 as 5 volts and U2 as 3.3 volts. The question is: what should be the value of

the voltage source to provide enough voltage for these two receivers? Here we have a closed-loop. Let’s write down the equation. Sum of all the

voltages in this loop should be equal to zero, so E – U1 – U2 is equal to zero. Let’s leave E alone on the left side. E is

equal to U1 + U2. Now, with the values, E is equal to 5 volts

+ 3.3 volts which is equal to 8.3 volts. Here we have another, a little bit more complicated

example with two anti serial sources which means that they are in diffrent directions

and three receivers. The first source has a voltage of 20 volts,

voltage of the second source is unknown, the first receiver has 15 volts, the second 12

volts and the third one is also unknown. Here we have three loops, first one, second

one and third one as a whole and two nodes one here and one here. Okey, so let’s write down three equations

for each loop. Firstly, the direction in which we will sum

all the voltages in the loop. First equation will be E1 – U1 – E2 is equal

to 0. Second equation will be U1 – U2 – U3 equals

to 0 and the third one is going to be E1 – U2 – U3 – E2 equals to 0. Here we’ve got only two unknowns: E2 and

U3, so we need only two equations the third one is an identity and it is not needed to

solve it. From the first equation, when we start adding

values we’ve got 20 volts – 15 volts – E2 is equal to 0, which means that E2 is equal

to 5 volts. From the second equation we know U1 and U2

we don’t know U3. When we solve this U3 is equal to 3 volts. In previous part we talked about the relationships

that occur in circuits between currents and voltages. Now we will talk about the law that is as

often used as Kirchoff’s law. The German physicist George Simon Ohm based

his work on the conductivity of conductors. He discovered that current is proportional

to the voltage and the element that shows the proportion between current and voltage

is resistance. The Ohm’s law was formulated in the following

way, the current flowing in a conductor is proportional to the voltage applied and proportionally

coefficient with the inverse of resistance. Now it may seem a bit hard but we will learn

how to use it. Let’s imagine that we have two containers

full of water. The level of water in the first container

is higher than the level of water in the second container. They have different potentials so there is

a voltage between them. Containers are connected to each other with

a pipe with a valve, if the valve is closed there will be no water flow through the pipe

to the second container. If we open the valve there are a maximum flow

from the first container to the second and if we set the valve between the closed and

the opened one we can notice that the water flow is proportional. In our case the water flow is current and

the capacity of this valve is resistance. Resistors are used whenever you want to reduce

the voltage or limit the current flowing in a circuit. This can be shown on LED which dims when we

limit its current. The presented system consists of voltage supply

and the resistor. The circuit is closed so there is a current

flowing in a circuit. From the Kirochhoff`s second law we know that

the whole current will be deposited on this resistor. Thanks to Ohm`s law we know how big is current. The voltage supply gives us 10 volts and the

resistance is 1 kilo ohm so the result is 10 milliamps. If the resistance decreases the current will

increase and if the resistance increases the current will decrease. The next example is very similar to the previous

one, but instead of voltage source here we have current source. In short, current source

maintains the constant value, 20 milliamps, of current flowing in the circuit, attached

resistance is 400 ohms. Using the Ohm’s law we can calculate the voltage

on a resistor which is 8 volt. Let Us try to analyze a more complex circuit.

Here we have two voltage sources each 27 volts R1 – 40 ohms, R2 – 20 ohms, R3 – 20 ohms R4

– 45 ohms and R5 90 ohms. Our aim is to calculate free current flowing

in the circuit I1, I2 and I3. The first step will be drawing directions

of currents and voltages. As we remember, voltage on sources are in the

same direction as currents and on receivers are in opposite directions. Now we have to draw the circles inside the

loops. In this circuit we have two nodes, so you have to write only one equation from the

first rule of Kirchhoff and we have to independence loops so we have to write two equations from

the second law of Kirochhoff. Now we have to write the equations: current

one is current two and current tree, and now equations from the second law of Kirchhoff,

voltage on a first source minus voltage on a R3 minus voltage on the R4 minus voltage

on a R2 plus voltage on E2 minus voltage on the R1 is zero. And minus voltage on a R5 plus a voltage on

a R4 is zero. Now I will try to present voltage as Ohm’s

law so voltage first is first current multiplied by third resistance; voltage fourth is second

current multiplied by fourth resistance, voltage second is first current multiplied

by second resistance, voltage first is first current multiplied by first resistance and

voltage fifth is third current multiplied by third resistance. After all mathematical transformations we

get the result: current one is 0.6 amps I2 is 0.4 amps and I3 is 0.2 amps. Developing mathematical formulas may seem

a little bit complicated but if you use the rules wisely we can simplify many elements. Let’s recall what combinations of the elements

do we know so far. Serial connection is called when all

resistors are put in a line and end of the first resistor is connected to the beginning

conduct of the second. Parallel connection is called when all resistors

are connected with both ends. The resistance of the whole set can be calculated

in this way. So let’s make more complex example now, resistors

second and third are connected in parallel with serial resistor first. First of all we have to calculate the resistance

of the R2 and R3, and then we have to add it to the resistance of a R1, so the whole

resistance is like that. Now we can try to look at the connections

between resistors and sources between all elements in the circuit and try to find other

way to calculate this circuit. To calculate the currents in the circuit. First of all, we should try to connect those

sources and try to calculate the whole resistance of the whole circuit. Here are step by step solution how it should

be done. Having total resistance and having one voltage

source we can calculate first current which is 0.6 amps. Knowing the first current and the voltage

on third, second and first resistor we can go back to those equations and calculate the

second and the third current. If one resistor is significantly bigger than

the other one in serial connection we can omit the other one in the total resistance

and if one resistor is also significantly bigger than the other one in a parallel connection

we can omit this bigger one. So the total resistance is only the resistance

of the resistor second. The last method is not very popular but useful.

If we have to connect the unknown resistor with a known one and the voltage source first

of all we have to check the voltages on a source and on a resistor, then after all calculations

including Ohm`s low and Kirchhoff we know the value of unknown resistance. In practice, sometimes it is necessary to reduce

the voltage in a circuit. However, reducing voltage supply isn’t possible and adding a

serial resistor isn’t the optimal solution. This diagram shows the basic voltage divider. We want to calculate the voltage on the output

of the second resistor. When we put voltage to the input of the divisor the current starts

to flow through both two resistors. The current is equal to voltage divided by the sum of

those two resistors. The voltage we want to find is equal to the

voltage drop of the second resistor. This is the general formula for voltage divider. It is easy to calculate the voltage by

multiplying the voltage input by the resistance on which we want to calculate the voltage

and divide it by the sum of all resistances in the circuit. However, sometimes the calculations with the

voltage divider aren’t so simple, sometimes we have to look at the connection of many

serial resistors. Sometimes we may be interested in a voltage

such as this marked on the picture. The procedure is the same. Multiply the input

voltage by the resistance on which we want to calculate the voltage and divided by the

whole resistance in a circuit. Now we can sum up all the rules which have

already been presented on Ohm`s law and electric power formula. With all those formulas we can form an electric

formula wheel.

Glued to my laptop screen, to watch it till the end. #tmeeducation