V_in / V_out for a simple didoe/op-amp circuit

[Mo]

Homework Statement

Question attatched in image file.

Homework Equations

The Attempt at a Solution

A1.b) Input will be a sine wave 6Vpp . When the input goes positive, D1 conducts and short circuits the ‘first’ resistor. D2 doesn’t conduct (it acts as an open circuit) and so Vout = voltage across R in parallel = 3V.

When the input goes negative D1 doesn’t conduct and D2 does. Therefore, the voltage goes through the ‘first’ resistor but the second resistor in parallel is short circuited. Therefore Vout = 0v since the diode is ideal.A1.C) The output from the op-amp will be = to vin . -1 (negative feedback) .When vin goes positive, D conducts and Vout = -0.7V

When Vin goes negative D acts as an open circuit, so the Vout = 1.5V

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Obviously the answers above somewhat incomplete since a graph of v_in/v_out is also needed, but i just wanted someone to check if i have got the right idea.

regards,
Mo.

 

[berkeman]

Nice job. Your solutions look complete and correct to me.

 

[piercas]

Your explanation is generally correct. The diodes in the circuit act as a voltage limiter, preventing the output from exceeding a certain voltage (usually around 0.7V for a silicon diode).

In the first part, when the input is positive, D1 conducts and the voltage across R1 is shorted, resulting in Vout = 3V. When the input is negative, D2 conducts and the voltage across R2 is shorted, resulting in Vout = 0V.

In the second part, the op-amp acts as an inverting amplifier with a gain of -1. When the input is positive, D1 conducts and the output is limited to -0.7V. When the input is negative, D2 conducts and the output is limited to 1.5V.

It would be helpful to include a graph of V_in/V_out to better visualize how the circuit is behaving. Overall, your understanding of the circuit appears to be correct.