Capacitor Feb. 9

At the beginning of class we reviewed hysteresis which was helpful because I am able to understand the concept a bit better. Here is my attempt to briefly go over the concept. If the input voltage is greater than the out-put voltage (V+ > V-) then the out put voltage is equal to their difference (V+ - V- = 6V- -6V = Vout = 12V in this circuit). Likewise, if the input voltage is less than the out-put voltage (V+ < V-) then Vout = -12V. An op amp with positive feedback would be given by the equation Vout / 2. Thus if the out-put voltage is less than 6 V (V- < 6V) then Vout = 12V but if the output voltage is greater than 6V (V- > 6V) then Vout = -12V. This "relationship" between input (V+) and out-put (V-) forms the pattern of a hysteresis graph.

The big topic of the day was capacitors! A capacitor stores energy as an electric charge (accumulates current and is expressed as voltage). In a circuit the current most keep going so what ever current goes in must come out. As t time approaches "infinity"the current goes to 0 near the voltage boundary line. In other words, the capacitor will charge until it reaches the max voltage. The graph will rapidly increase and then slow down when gets closer to the max voltage.

Here you can see the graph (yellow line) move up and down between the max(12V) and min (-12V) as the voltage shown in blue reaches the limit.

After building the model of the circuit we were able to once again see how Vout switched back and forth between -12V and 12V. Later we added a noise piece which we were able to control with the potentiometer.

Circuit without the noise piece.