Given the following ideal diode circuit:
I would like to find the output voltage measured on the wires on the right. The ac source Vs on the left is given as:
$$Vs=8\sin{wt}$$
Here is my attempt at a solution:
Let's suppose that the diode is on, so it represents a short circuit and the voltage measured on the wires is 2Volts. In this case the current of the loop (clockwise direction) is given by:
$$I=\frac{V_s-2}{1000}$$ Knowing that the current can't flow from the cathode to the anode, this equation is valid only if Vs is greater than 2. Otherwise, the output voltage is equal to Vs since the diode represents an open circuit.
In total:
$$V_{out}=2, Vs>2$$ $$V_{out}=Vs, Vs<2$$
Since I haven't touched on these problems for a while, I'm not so sure about the result as well as my approach to the problem. Is my solution correct? Is my approach with the current equation good for such problems?
No comments:
Post a Comment