Calling all brainiacs

[NA4BH]

You electrical engineer types, go take you a free course at MIT.

http://news.yahoo.com/blogs/technolo...225905903.html

[N2RJ]

Listened/watched a lot of the lectures on iTunes. Really good stuff.

[W2NAP]

very interesting.

[W1GUH]

Calling all Brainiacs?

LIke this?

[N1LAF]

[N1LAF]

[NQ6U]

participants are expected to have a grasp of calculus, linear algebra, and advanced high school level physics.

So much for that.

[N1LAF]

Basic Electrical Engineering concepts...

Kirchoff current and voltage laws

1. Sum of voltages in a loop equals 0
2. Sum of currents at a node is 0

1. Voltage Law - Voltage Loop equals 0. Draw a battery with a light bulb across it. Draw the polarities across the battery and light bulb. Starting from the battery positive terminal, in a loop of the battery and light bulb, you have the '+' of the light bulb, then the '-' of the battery. +Vlightbulb - VBattery = 0

Vlightbulb = Vbattery

2. Current Law - Sum of currents at a node is 0. Rewritten, the sum of currents entering a node is equal to currents leaving the node. Example, two resistors tied togther at one end. The tie point is a node. Current flowing through R1 is the same as the current flowing through R2.

[N1LAF]

Impedance of Resistor is R, resistance.

Impedance of an inductor is Ls, where s = jw, and w = 2(pi)(frequency). Higher the frequency, higher the impedance.

Impedance of a capacitor is 1/(Cs), where s = jw, and w = 2(pi)(frequency). Higher the frequency, lower the impedance

[N1LAF]

Vout/Vin = Transfer Equation

Suppose we have a resistor between Vin and Vout, and a capacitor from Vout to ground. Vout load is very high impedance, infinite for practical purposes. Current through the resistor is (Vin-Vout)/R. This same current flows through the capacitor, which is Vout/(1/Cs), which also is VoutCs. Current in equals current out, and the load is infinite, so no current flow through Vout. From this we have the equation of current entering the node is equal to current leaving the node. (Vin-Vout)/R = Vout(Cs)

(Vin-Vout)/R = Vout(Cs)

Vin R - Vout R = Vout Cs

Vin R = Vout(R + Cs)

Vin = Vout (R+Cs)/R = Vout (1 + RCs)

Vout/Vin = 1/(1 + RCs), where s = jw = 2(pi)(f)

Decible = 10Log(Vout/Vin)

Corner frequency: -3dB

10 Log(0.5) = -3dB

Vout/Vin = 0.5 = 1/2

At corner frequency, (1 + RCs) = 2, RCs = 1

For RCs = 1, s = 1/RC, 2(pi)f = 1/RC

The corner frequency, fc, is equal to 1/(2RC(pi))

Here, in the equation, Vout/Vin = 1/(1+RCs), where s = 2(pi)f, as f increases, (1 + RCs) also increases, and 1/(1+RCs) will decrease.

This is an example of a low pass filter.