Thursday, December 14, 2017

Basic Electronics on the Go - Semiconductor Basics

From http://www.electronics-tutorials.ws/diode/diode_1.html

 If Resistors are the most basic passive component in electrical or electronic circuits, then we have to consider the Signal Diode as being the most basic “Active” component.

However, unlike a resistor, a diode does not behave linearly with respect to the applied voltage as it has an exponential I-V relationship and therefore can not be described simply by using Ohm’s law as we do for resistors.

Diodes are basic unidirectional semiconductor devices that will only allow current to flow through them in one direction only, acting more like a one way electrical valve, (Forward Biased Condition). But, before we have a look at how signal or power diodes work we first need to understand the semiconductors basic construction and concept.

Diodes are made from a single piece of Semiconductor material which has a positive “P-region” at one end and a negative “N-region” at the other, and which has a resistivity value somewhere between that of a conductor and an insulator. But what is a “Semiconductor” material?, firstly let’s look at what makes something either a Conductor or an Insulator.


Resistivity

The electrical Resistance of an electrical or electronic component or device is generally defined as being the ratio of the voltage difference across it to the current flowing through it, basic Ohm´s Law principals. The problem with using resistance as a measurement is that it depends very much on the physical size of the material being measured as well as the material out of which it is made. For example, if we were to increase the length of the material (making it longer) its resistance would also increase proportionally.

 Likewise, if we increased its diameter (making it fatter) its resistance value would decrease. So we want to be able to define the material in such a way as to indicate its ability to either conduct or oppose the flow of electrical current through it no matter what its size or shape happens to be.

The quantity that is used to indicate this specific resistance is called Resistivity and is given the Greek symbol of ρ, (Rho). Resistivity is measured in Ohm-metres, ( Ω-m ). Resistivity is the inverse of conductivity.

If the resistivity of various materials is compared, they can be classified into three main groups, Conductors, Insulators and Semi-conductors as shown below.

Resistivity Chart

 

 



(to be updated)

Saturday, November 25, 2017

Basic Electronics on the Go - Star Delta Transformation

From http://www.electronics-tutorials.ws/dccircuits/dcp_10.html

Star Delta Transformation



 We can now solve simple series, parallel or bridge type resistive networks using Kirchhoff´s Circuit Laws, mesh current analysis or nodal voltage analysis techniques but in a balanced 3-phase circuit we can use different mathematical techniques to simplify the analysis of the circuit and thereby reduce the amount of math’s involved which in itself is a good thing.

 Standard 3-phase circuits or networks take on two major forms with names that represent the way in which the resistances are connected, a Star connected network which has the symbol of the letter, Υ (wye) and a Delta connected network which has the symbol of a triangle, Δ (delta).

 If a 3-phase, 3-wire supply or even a 3-phase load is connected in one type of configuration, it can be easily transformed or changed it into an equivalent configuration of the other type by using either the Star Delta Transformation or Delta Star Transformation process.
A resistive network consisting of three impedances can be connected together to form a T or “Tee” configuration but the network can also be redrawn to form a Star or Υ type network as shown below.

T-connected and Equivalent Star Network

 

As we have already seen, we can redraw the T resistor network above to produce an electrically equivalent Star or Υ type network. But we can also convert a Pi or π type resistor network into an electrically equivalent Delta or Δ type network as shown below.

Pi-connected and Equivalent Delta Network.

 

  Having now defined exactly what is a Star and Delta connected network it is possible to transform the Υ into an equivalent Δ circuit and also to convert a Δ into an equivalent Υ circuit using a the transformation process. This process allows us to produce a mathematical relationship between the various resistors giving us a Star Delta Transformation as well as a Delta Star Transformation.

 

Delta Star Transformation

To convert a delta network to an equivalent star network we need to derive a transformation formula for equating the various resistors to each other between the various terminals. Consider the circuit below.

Delta to Star Network.

 

 Compare the resistances between terminals 1 and 2.

 Resistance between the terminals 2 and 3.

 

 

 Resistance between the terminals 1 and 3.

 This now gives us three equations and taking equation 3 from equation 2 gives:

 

Then, re-writing Equation 1 will give us:



Adding together equation 1 and the result above of equation 3 minus equation 2 gives:


 From which gives us the final equation for resistor P as:


Similarly, resistor Q and R  can be found  :-





When converting a delta network into a star network the denominators of all of the transformation formulas are the same: A + B + C, and which is the sum of ALL the delta resistances. 

Star Delta Transformation

Star Delta transformation is simply the reverse of above. We have seen that when converting from a delta network to an equivalent star network that the resistor connected to one terminal is the product of the two delta resistances connected to the same terminal, for example resistor P is the product of resistors A and B connected to terminal 1.



The value of the resistor on any one side of the delta, Δ network is the sum of all the two-product combinations of resistors in the star network divide by the star resistor located “directly opposite” the delta resistor being found. For example, resistor A is given as:



 Resistor B is given as:


Resistor C given as:





Thursday, November 16, 2017

Basic Electronics on the Go - Maximum Power Transfer

From http://www.electronics-tutorials.ws/dccircuits/dcp_9.html

We have seen in the previous tutorials that any complex circuit or network can be replaced by a single energy source in series with a single internal source resistance, RS.

 When we connect a load resistance, RL across the output terminals of the power source, the impedance of the load will vary resulting in the power being absorbed by the load becoming dependent on the impedance of the actual power source. Then for the load resistance to absorb the maximum power possible it has to be “Matched” to the impedance of the power source and this forms the basis of Maximum Power Transfer.

The Maximum Power Transfer Theorem is another useful circuit analysis method to ensure that the maximum amount of power will be dissipated in the load resistance when the value of the load resistance is exactly equal to the resistance of the power source. The relationship between the load impedance and the internal impedance of the energy source will give the power in the load. Consider the circuit below.
 

Thevenins Equivalent Circuit.



In our Thevenin equivalent circuit above, the maximum power transfer theorem states that “the maximum amount of power will be dissipated in the load resistance if it is equal in value to the Thevenin or Norton source resistance of the network supplying the power“.

In other words, the load resistance resulting in greatest power dissipation must be equal in value to the equivalent Thevenin source resistance, then RL = RS but if the load resistance is lower or higher in value than the Thevenin source resistance of the network, its dissipated power will be less than maximum.

For example, find the value of the load resistance, RL that will give the maximum power transfer in the following circuit.


Maximum Power Transfer Example No1.

 

  Where:
  RS = 25Ω
  RL is variable between 0 – 100Ω
  VS = 100v
Then by using the following Ohm’s Law equations:



We can now complete the following table to determine the current and power in the circuit for different values of load resistance.

Table of Current against Power

RL (Ω) I (amps) P (watts)
0 4.0 0
5 3.3 55
10 2.8 78
15 2.5 93
20 2.2 97
RL (Ω) I (amps) P (watts)
25 2.0 100
30 1.8 97
40 1.5 94
60 1.2 83
100 0.8 64

Using the data from the table above, we can plot a graph of load resistance, RL against power, P for different values of load resistance. Also notice that power is zero for an open-circuit (zero current condition) and also for a short-circuit (zero voltage condition).


Graph of Power against Load Resistance



 From the above table and graph we can see that the Maximum Power Transfer occurs in the load when the load resistance, RL is equal in value to the source resistance, RS that is: RS = RL = 25Ω. This is called a “matched condition”.

One good example of impedance matching is between an audio amplifier and a loudspeaker. The output impedance, ZOUT of the amplifier may be given as between and , while the nominal input impedance, ZIN of the loudspeaker may be given as only.

Then if the speaker is attached to the amplifiers output, the amplifier will see the speaker as an load. Connecting two speakers in parallel is equivalent to the amplifier driving one speaker and both configurations are within the output specifications of the amplifier.

Improper impedance matching can lead to excessive power loss and heat dissipation. But how could you impedance match an amplifier and loudspeaker which have very different impedances. Well, there are loudspeaker impedance matching transformers available that can change impedances from to , or to 16Ω’s to allow impedance matching of many loudspeakers connected together in various combinations such as in PA (public address) systems.


Transformer Impedance Matching

 


 The maximum power transfer can be obtained even if the output impedance is not the same as the load impedance. This can be done using a suitable “turns ratio” on the transformer with the corresponding ratio of load impedance, ZLOAD to output impedance, ZOUT matches that of the ratio of the transformers primary turns to secondary turns as a resistance on one side of the transformer becomes a different value on the other.

 If the load impedance, ZLOAD is purely resistive and the source impedance is purely resistive, ZOUT then the equation for finding the maximum power transfer is given as:



 Where: NP is the number of primary turns and NS the number of secondary turns on the transformer. Then by varying the value of the transformers turns ratio the output impedance can be “matched” to the source impedance to achieve maximum power transfer.

Sunday, October 29, 2017

Basic Electronics on the Go - Norton’s Theorem

From http://www.electronics-tutorials.ws/dccircuits/dcp_8.html

Nortons Theorem states that “Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor“.

As far as the load resistance, RL is concerned this single resistance, RS is the value of the resistance looking back into the network with all the current sources open circuited and IS is the short circuit current at the output terminals as shown below.




The value of this “constant current” is one which would flow if the two output terminals where shorted together while the source resistance would be measured looking back into the terminals, (the same as Thevenin).
For example, consider our now familiar circuit from the previous section







To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.


When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as (done with Mesh Current Analysis, it seems):



If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.


Find the Equivalent Resistance (Rs)

 

Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following Nortons equivalent circuit.


Nortons equivalent circuit.

 


 We now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.


Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:



The voltage across the terminals A and B with the load resistor connected is given as:


Then the current flowing in the 40Ω load resistor can be found as:


Nortons Theorem Summary

The basic procedure for solving a circuit using Nortons Theorem is as follows:
  • 1. Remove the load resistor RL or component concerned.
  • 2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
  • 3. Find IS by placing a shorting link on the output terminals A and B.
  • 4. Find the current flowing through the load resistor RL.
In a circuit, power supplied to the load is at its maximum when the load resistance is equal to the source resistance. In the next tutorial we will look at Maximum Power Transfer. The application of the maximum power transfer theorem can be applied to either simple and complicated linear circuits having a variable load and is used to find the load resistance that leads to transfer of maximum power to the load.

Sunday, October 22, 2017

Basic Electronics on the Go - Thevenin’s Theorem

From http://www.electronics-tutorials.ws/dccircuits/dcp_7.html

In this tutorial we will look at one of the more common circuit analysis theorems (next to Kirchhoff´s) that has been developed, Thevenin’s Theorem.

Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance connected across the load“. In other words, it is possible to simplify any electrical circuit, no matter how complex, to an equivalent two-terminal circuit with just a single constant voltage source in series with a resistance (or impedance) connected to a load as shown below.


Thevenin’s equivalent circuit.

  

Consider the circuit from the previous section.

 

Firstly, to analyse the circuit we have to remove the centre 40Ω load resistor connected across the terminals A-B, and remove any internal resistance associated with the voltage source(s). This is done by shorting out all the voltage sources connected to the circuit, that is v = 0, or open circuit any connected current sources making i = 0. The reason for this is that we want to have an ideal voltage source or an ideal current source for the circuit analysis.

The value of the equivalent resistance, Rs is found by calculating the total resistance looking back from the terminals A and B with all the voltage sources shorted. We then get the following circuit.





Find the Equivalent Resistance (Rs)

 

 

The voltage Vs is defined as the total voltage across the terminals A and B when there is an open circuit between them. That is without the load resistor RL connected.




We now need to reconnect the two voltages back into the circuit, and as VS  =  VAB the current flowing around the loop is calculated as:




This current of 0.33 amperes (330mA) is common to both resistors so the voltage drop across the 20Ω resistor or the 10Ω resistor can be calculated as:
VAB  =  20  –  (20Ω x 0.33amps)  =   13.33 volts.
or
VAB  =  10  +  (10Ω x 0.33amps)  =   13.33 volts, the same.
Then the Thevenin’s Equivalent circuit would consist or a series resistance of 6.67Ω’s and a voltage source of 13.33v. With the 40Ω resistor connected back into the circuit we get:


and from this the current flowing around the circuit is given as:








which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous circuit analysis tutorial.

Thevenin’s theorem can be used as another type of circuit analysis method and is particularly useful in the analysis of complicated circuits consisting of one or more voltage or current source and resistors that are arranged in the usual parallel and series connections.

While Thevenin’s circuit theorem can be described mathematically in terms of current and voltage, it is not as powerful as Mesh or Nodal analysis in larger networks because the use of Mesh or Nodal analysis is usually necessary in any Thevenin exercise, so it might as well be used from the start. However, Thevenin’s equivalent circuits of Transistors, Voltage Sources such as batteries etc, are very useful in circuit design.


Thevenin’s Theorem Summary

The basic procedure for solving a circuit using Thevenin’s Theorem is as follows:
  • 1. Remove the load resistor RL or component concerned.
  • 2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
  • 3. Find VS by the usual circuit analysis methods.
  • 4. Find the current flowing through the load resistor RL.
In the next tutorial we will look at Nortons Theorem which allows a network consisting of linear resistors and sources to be represented by an equivalent circuit with a single current source in parallel with a single source resistance.

Saturday, October 21, 2017

Basic Electronics on the Go - Nodal Voltage Analysis

From http://www.electronics-tutorials.ws/dccircuits/dcp_6.html

Nodal Voltage Analysis complements the previous mesh analysis in that it is equally powerful and based on the same concepts of matrix analysis.  Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit.
So by adding together all these nodal voltages the net result will be equal to zero. Then, if there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchhoff’s first law equation, that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch. For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node.


Nodal Voltage Analysis Circuit

 


In the above circuit, node D is chosen as the reference node and the other three nodes are assumed to have voltages, Va, Vb and  Vc with respect to node D. For example;

 


 As Va = 10v and Vc = 20v , Vb can be easily found by:


In the next tutorial we will look at Thevenins Theorem which allows a network consisting of linear resistors and sources to be represented by an equivalent circuit with a single voltage source and a series resistance.


Friday, October 20, 2017

Basic Electronics on the Go - Mesh Current Analysis

From http://www.electronics-tutorials.ws/dccircuits/dcp_5.html

 While Kirchhoff´s Laws give us the basic method for analysing any complex electrical circuit, there are different ways of improving upon this method by using Mesh Current Analysis or Nodal Voltage Analysis that results in a lessening of the math’s involved and when large networks are involved this reduction in maths can be a big advantage.

 For example, consider the electrical circuit example from the previous section.

Mesh Current Analysis Circuit

An easier method of solving the above circuit is by using Mesh Current Analysis or Loop Analysis which is also sometimes called Maxwell´s Circulating Currents method. Instead of labelling the branch currents we need to label each “closed loop” with a circulating current.
As a general rule of thumb, only label inside loops in a clockwise direction with circulating currents as the aim is to cover all the elements of the circuit at least once. Any required branch current may be found from the appropriate loop or mesh currents as before using Kirchhoff´s method.
For example: :    i1 = I1 , i2 = -I2  and  I3 = I1 – I2
We now write Kirchhoff’s voltage law equation in the same way as before to solve them but the advantage of this method is that it ensures that the information obtained from the circuit equations is the minimum required to solve the circuit as the information is more general and can easily be put into a matrix form.

For example, consider the circuit from the previous section.



 First we need to understand that when dealing with matrices, for the division of two matrices it is the same as multiplying one matrix by the inverse of the other as shown.



having found the inverse of R, as V/R is the same as V x R-1, we can now use it to find the two circulating currents.




Mesh Current Analysis Summary.

The basic procedure for solving Mesh Current Analysis equations is as follows:
  • 1. Label all the internal loops with circulating currents. (I1, I2, …IL etc)
  • 2. Write the [ L x 1 ] column matrix [ V ] giving the sum of all voltage sources in each loop.
  • 3. Write the [ L x L ] matrix, [ R ] for all the resistances in the circuit as follows;
    •   R11 = the total resistance in the first loop.
    •   Rnn = the total resistance in the Nth loop.
    •   RJK = the resistance which directly joins loop J to Loop K.
  • 4. Write the matrix or vector equation [V]  =  [R] x [I] where [I] is the list of currents to be found.
As well as using Mesh Current Analysis, we can also use node analysis to calculate the voltages around the loops, again reducing the amount of mathematics required using just Kirchoff’s laws.