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Given the following ideal diode circuit:

enter image description here

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?

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  • \$\begingroup\$ Provided the 2 V battery is also ideal, then I agree with you. \$\endgroup\$ – Tom Kuschel Jan 11 at 18:28
  • \$\begingroup\$ Yes, it's okay. Your solution would be a bit better if you have at least one of the conditions include equality. \$\endgroup\$ – Spehro Pefhany Jan 11 at 18:48
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Your thought pattern is correct. Here is a spice simulation showing the result.

enter image description here

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  • \$\begingroup\$ Please comment on why this was down voted. OP clearly showed effort/understanding and was looking for confirmation. \$\endgroup\$ – EasyOhm Jan 27 at 7:09

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