Sensitivity of a closed loop negative feedback control system

The transfer function of the above closed loop negative feedback control is given by

Sensitivity is the differential change in O/P to the applied change in I/P.

The transfer function above will be used as "T" in all the equations below. 

So T= C(S)/R(S) 

Note: This is the important concept in control system.

This post is written by our Guest Author  Charan!!! 

HARNESSING THREE RENEWABLE ENERGY SOURCES FROM ONE GENERATING SYSTEM

We all know that the major source of electrical energy production is from fossil fuels and water stored in dams. Well that’s a silly point I wrote which we all know since our primary school. But what caused me to make a necessary invention regarding to the energy sector is that the fossil fuel will be depleted in few more decades and to the way population is growing we can not completely rely upon our hydro dams. There is a necessary of shifting our focus on the very unpredictable RENEWABLE ENERGY SOURCES and harness as maximum energy as possible from those. There is a saying among renewable energy enthusiasts that “ Studying the pattern of Renewable energy source is as tough as studying the psychology of women :P ”

MAJOR PROBLEM IN RENEWABLE ENERGY SECTOR FROM MY EXPERIENCE IS THAT WE CANT GET ONE PARTICULAR ENERGY SOURCE PROFITABLE THROUGHT A CALENDER YEAR

Ø  Sun emits his powerful rays only for few months in a year.

Ø  Wind energy is concentrated only to certain geographical part of the earth and doesn’t flow continuously.

Ø  Tidal energy (wave energy ) is highly un-predictable and its said to be non-profit source for a nation like INDIA though we have more than 7000 km of coastline because of low tides.

WELL I BELIEVE THAT HURDLES ARE ALWAYS THERE TO BE CROSSED. Here in this article I tried to bring out the niche way of producing electricity from above three sources of energy through-out the year.

You are seeing the images of the adverse effects of TIDES, SOLAR RADIATION AND WIND. How amazing it would be if all these three energy sources can be integrated at one point and harnessed and produced electricity?, excited? 

 Yes its what this project is all about. Having a single or two generating system to the max and extracting energy from three or more source at a single place.

Ø  The coastal line is well blessed with all these three energy sources

Ø  The amount of solar radiation required for power production will be naturally high in the coastal line

Ø  Wind velocity required for the wind mills to rotate and produce decent amount of electrical energy will be available too.

Ø  The tidal energy though the researchers predict that its impractical to implement it in major parts of Indian coastline, we can overcome it through my specific design for harnessing energy even from the low tides .

        HOW THESE ENERGY SOURCES WILL BE INTEGRATED TO A SINGLE GENERATING SYSTEM ??      Here I will be explaining you the different designs used for WIND MILL  and TIDAL DEVICE. The usual wind mill ( HORIZONTAL AXIS ) design has its many limitation and are suitable for the high altitude places. And in country like INDIA where environmentalists oppose the wind mill installation since they are threat to birds as birds are being killed in many parts cause of the wind turbines

This one is the top view of the wind mill which has vertical axis. This type of wind mill usually can generate electricity upto 0.9 MW with all bearing and air frictional errors. This design in particular has been applied for patent. The risk of birds getting struck in the blades is minimised and hence least chance of birds getting killed.

This is the design of tidal turbine in which a device harnesses the waves energy and pushes the high density liquid which further rotates the turbine and further the generator.THIS TWO TURBINES WILL BE COUPLED TO A HEAVY LOAD BEARING SPIRAL SPRING AND FURTHER TO THE GENERATOR.

THIS MEDIATOR SPRING DEIVCE DESIGN IS UNDER PATENT SEARCH AND IT WILL BE DISCLOSED SHORTLY .

This is written by our guest writer- Keshava.N 

Two capacitors Initially charged ,Popular question in Analog electronics

This is the Popular question in analog electronics. 
Lets see how this works

C1 is initially charged with 2volts and C5 is charged with 1 volts.
And there is a Resistor of 10k between the two capacitors.

Since C1 is at higher potential than C5 , current flows through 10k and voltage of C1 tends to drop and that of C5 will increase . The current will  stop flowing after both capacitor reach common voltage.
That common voltage can be found using the equation

V=Q/C

V=(C1v1+c5V2)/ (c1+c5) {Refer figure}
Substituting the values

V=(1u*2++0.5u*1)/ (1u+0.5u)
V=2.5u/1.5u

V=1.66v. which is the voltage across each capacitors

Pulse width Modulation

Pulse width Modulation is the important concept in the field of analog electronics. It has various applications and it is not limited to one particular field. Let it be Communication systems or Power electronics or Microcontrollers.
Pulse width modulation is used  everywhere.So for a electronics/electrical engineer  it is very  essential to understand the concept. 

 Pulse width modulation , as the name suggests, is a technique in which the output pulse width is varied depending on a input signal. 
let's go ahead and study simple pulse width modulation

Circuit 

The circuit consists of triangular wave generator and a comparator.The inputs to the comparator are 

1)triangular wave

2) constant Dc voltage

Refer this tutorial for generation of triangular wave : TRIANGUAR WAVE

The comparator  is nothing about Opamp without feedback. So without feedback ,Opamp output will be Positive or negative saturation depending on the value of voltages at Inverting and Non inverting terminals. 

Two cases 

1) IF noninverting terminal voltage is greater than inverting terminal, output will be in positive saturation

2)IF noninverting terminal voltage is less than inverting terminal, output will be in negative saturation

So depending on the value of DC voltage, the Pulse width of the  square wave is varied. Hence the name Pulse width modulation.

Voltage Doubler.

This is the simple circuit which is evolved from two basic circuits.

Voltage doubler =Positive clamper +Peak detector. 

Suppose you have a sine wave of peak voltage 5v. i.e Peak to peak voltage of 10v. How to get output which is double the input voltage.

Output=2*Peak voltage.

Here is the circuit

The two circuits are already discussed 

POsitive clamper

Peak detector

In this circuit we just make use of these concepts to get voltage which is twice the input peak voltage.Hence the name voltage doubler

If you check after 5ms output will be double.Here you can see output is not 10v but approximately 9.8v. This is because diodes used are not ideal. Add 0.19 voltage to the positive clamper as shown below  to get exact 10v.(Those who are fond of exact results ;) ) 

The simulator used is Pspice. 

Peak detector using uA741 opamp

Hello all lets see the circuit of Simple Peak detector using Opamp

This is the Peak detector circuit

Initially 

case 1) IF vin>vc (1u capacitor)output(6)  i.e Point before the anode of the diode gets positive saturation voltage.So diode gets forward biased. So There will be feedback to negative terminal of opamp(i.enegative feedback)

Capacitor gets charged to maximum Input voltage.

If the input is less than capacitor voltage,Capacitor retains the previous maximum voltage itself(As there is no discharge path)

case 2) If vin<vc(capacitor) Point 6 is at negative saturation so the diode is reverse biased. The opamp is in open loop configuration.. The output is still the original "maximum input voltage which charged the capcitor"

The capacitor gets charged if vin>vc.

The first part R and C is a circuit which generates a random input voltage denoted by green probe. Red probe denotes the output.

Look at Red probe(output) which detects the  Peak of the input voltage and hence the name Peak detector.

Double Clipper/ Limiter circuit.

suppose you have a sine wave of 5volts. peak voltage. i.e (10v p-p)
like shown below. 

The popular question is..  Design a circuit which clips the sine wave at 2v(Say) in the positive direction and at -3v in the negative direction. I.e cut the wave form at 2volts and at -3v.

Solution : 

Yes you need diode and voltage supplies to do that. 

1)During positive half cycle of the input  

• when vin<2v,Both D7 and D8 are reverse biased.So output follows input  
• when vin>2v. D7 is forward biased and D8 is reverse biased.

D7 forward voltage is 0.7v.

So output is 0.7+1.3=2v.

2) During negative half cycle of the input.

• when Vin>-3v(i.e -1,-2) Both D7 and D8 are reverse biased and output follows input

• when vin< -3v,  D8 is forward biased. 

So finally we get a wave which is clipped at 2v and at -3v

This is the popular double limiter circuit. which will be asked in many competitive exams.Check the circuit in your analog labs. CHEERS!