EEE133 Electronic Devices and Circuits Lecture 13-15 Week 5 Tues 18th/21th March 2014 DK6 11-1 pm/3-4 pm

In this 3-hr lectures, we will go through the following sections:

1. Finish up the half-wave rectifier (ripple factor) (50 minutes)
2. Terminal characteristics of diodes (35 minutes)
3. Graphical / iterative (15 minutes) 
4. Piecewise linear, constant voltage, small signal analysis (50 minutes)

For the notes, I had prepared supplementary notes to help you understand the lectures that I had delivered on Solid State part. Please see chapter 1 of the following link:

Lecture 1-15 Notes EEE133 with supplementary lectures for Solid State part


The following link is the lecture that I delivered:

Lecture 13-15 EEE133 18th March 2014

In class, when I explained the substitution in calculating the filtered output voltage $V_{OF}(s=j {\omega}_o)$ where $\omega_o$ is the operating radian frequency, one of you were puzzled by my statement and requested for further elaboration.

I did explain that it is due to the transformation from time domain to frequency domain. I also did explain that in general $s=\sigma + j \omega_o$ but in our case we assume $\sigma = 0$ which is under steady state condition in which there is no attenuation in the system or the decaying part in your system. Phasor transform has the same condition as Fourier Transform (Fourier is Laplace special case). Majority of you may not see the line of reasoning but I hope you carry this question in your Control System class, Signal and System class as well as in your Circuit II class in the future.

As  $\sigma = 0$, it definitely implies that we can represent capacitor and inductor as a non-lossy element giving rise to the following relation in the frequency domain:

$Z(j\omega_o)=\frac {1} {j\omega_o C}$ for the capacitor

and
 
$Z(j\omega_o)= {j \omega_o L}$ for the inductor.


I also did explain to you the justification for using the Phasor domain, i.e. you want to analyze your circuit in an algebraic manner rather than solving tedious differential equation.

If you have a power supply connected to series R and L, the following equation holds true:

$L \frac {\delta i} {\delta t} + Ri = V_m cos(\omega t + \phi) $

which will result in the following solution (from Nilsson and Riedel pp 362):

$i = \frac {V_m} {\sqrt{R^2+{\omega}^2}} cos(\phi - \theta) e^{- \frac {R} {L} t} + \frac {V_m} {\sqrt{R^2+{\omega}^2}} cos(\omega t + \phi - \theta)$.

The above tedious calculation leads us to simplify it by using the Phasor domain.