2.5V to 10V Boost Converter simulation

Calculation

Vo=10V

Vi=2.5V

For Boost Converter, the duty cycle of the pulse required is D=1-(Vi/Vo)=0.75

It is normal to operate coils at a frequency which is not perceived by the human ears. so let's keep the frequency at 20kHz . i.e the switch is operated at this frequency.

The above equation of duty cycle is valid only if the boost converter is operated in continuous mode. i.e Inductor current never falls to zero.

To make sure of that, the inductance of inductor should be greater than the inductance L

L=D*(1-D)*(1-D)*R/(2*f)

=0.75*0.25*0.25*100/(2*20000)

L=117uH. We have chosen a value of 200uH to make sure the below converter operates in continuous mode. 

The output capacitor C required to limit the output ripple voltage to 1 percent is determined

C>=D/(R*0.01*f)=37.5uf. I have chosen a standard 47uf.

Note: The Duty cycle has been calculated assuming that voltage drop across the diode is 0V. But the diode used in the above circuit has practical voltage across it. So we have made duty cycle as 77 here to get the required output voltage.

The Simulation can be found out here -->http://tinyurl.com/qgao6jy

Difference between diffusion capacitance and depletion/transition capacitance

The reactance of a capacitor is given by Xc=1/2*pi*f*C. At lower frequency, Xc is very very large and we can treat it as open circuit. However, at high frequencies, the Xc value becomes smaller and significant and we will not be in the position to ignore. While dealing with the electronic devices at higher frequencies, two capacitance come into picture. They are

1) Diffusion Capacitance
2) Depletion Capacitance, also called as transition capacitance.

Difference between diffusion capacitance and depletion/transition capacitance

1. Depletion Capacitance

Depletion capacitance will be dominant  in reverse bias region.

The capacitance of a parallel plate capacitor is given by C =εA/d.

Under reverse bias condition, the depletion region acts as  a parallel plate capacitor with the depletion region width as d, and it's effective area as A in the above equation.

Depletion width(d) increases when the reverse bias voltage increases, so the depletion capacitance  decreases with increase in reverse bias voltage.

2. Diffusion Capacitance

Diffusion Capacitance will be dominant in forward bias condition.

CD = τID / ηVT where VT=KT/q

The diffusion capacitance decreases with decreasing current and increasing temperature.


Note: Both this capacitance will come into picture at higher frequencies and one can ignore them at lower frequencies.

Semiconductor theory | Reverse saturation current and forward bias voltage

The behavior of PN junction is different for reverse saturation current and forward bias voltage when the temperature is increased. Let's study the behavior of PN junction with examples.


Q) A Silicon PN junction at a temperature of 20 degree celsius has a reverse saturation current of 10pA. The reverse saturation current at 40 degree celsius for the same bias is approximately

Ans : 40 pA. Reverse saturation current doubles for every 10 degree rise in temperature

At 20 degree celsius-->10pA
At 30 degree celsius  --->20pA
At 40 degree celsius --->40 pA.


Q) A Silicon PN junction biased with a constant at room temperature. When the temperature is increased by 10 degree celsius, the forward bias voltage across the junction decreases by?
Ans : Decreases by 25mV

For every degree rise in temperature, the voltage decreases by 2.5mV.

For 10 degree rise in temperature ---> 2.5mV*10=> Voltage decrease by 25mV 

Current flowing through a capacitor is Displacement current. *PROOF*

Gauss Law detailed explanation

Maxwell Equations and Light

 And Maxwell said

and there was Light.

                                                                

How to draw Root Locus diagram?

1) There are 4 poles and no zeros. P=4 and Z=0. N=4. There will be 4 branches in root locus

2) All 4 branches will start from open loop poles and terminate at infinity.

3) The branches will terminate at infinity along straight line asymptotes whose angles are determined by

where q=0..N-1

Angle of asymptotes =45, 135,225,315

4) The asymptotes meet the real axis at Centroid

Centroid= Sum of real parts of poles-sum of real parts of Zeros/(P-Z)=0-1-2-3/4=-1.5

5) Breakaway point and Break in  point  is calculated by solving dk/ds=0

Solving this gives s=-1.5, -0.381, -2.619. (Note : -1.5 is not valid breakaway point point because it doesnt lies on root locus)

6) The value of k and the point at which the root locus branch crosses the imaginary axis is determined by applying Routh Criterion to the characteristic equation. The roots at the intersection point are imaginary.

 s^4 | 1     11          k

 s^3 | 6       6            0

 s^2 | 10         k          0   

 s^1 | (60-6k)/10   0       

 s^0 |   k                        

60-6k=0 --> kmax=10

Auxiliary  equation : 10s^2+k=0

At k=10, s=+j and s=-j                  

7) Root Locus 

  

This is the root locus diagram for the given transfer function

Summary

The system is absolutely stable for 0<k<10 and at k=10, the system is marginally stable and for k>10, system is unstable.