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.

One comment on “Basics of Electronics section 1. by TME Education”

  1. Kangethe Kelvin says:

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

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